Cosmology

[desiresjab]

By the way, if anyone looks at the Martinson paper be aware that there are two mistakes in it. Specifically, it once lists 2 as a quadratic residue of 19, and it once lists 67 as a 4n+1 number. These mistakes can stall an amateur, as they did me.

[Danik 2016]

I don´t understand anything about number theory, but I´m glad that this scientific thread is alive again.

[Dreamwoven]

I am, too, Danik!

[desiresjab]

I have a dear friend who regularly likes to blast science and technology, and even math. Sometimes I fight back, but more often I let it go. It is useless to argue with anyone about what something is. But I find that most people who rail against science have misconceptions about what it is and what it is supposed to do. In short, one might call it the art of numerical observation. Poets and novelists are keen observers, too, but not normally numeric observers. Scientific observation is tied to numerics because of the world around us--the world and things in it quantify naturally, once humanity taught itself the knack. Objects fall the same speed every time, so through repeated experiment men were able to quantify that speed and finally find a formula for it.

Some people, honestly, expect way too much of mathematics and science, but when asked for suggestions they come up with the same old criticisms. Do they expect all scientists to drop what they are doing and go look for a ghost or proof that aliens built the earth's ancient pyramids?

The legitimate mathematicians plug along, as they always have, noting patterns in innocent numbers. Brains like Fermat, Euler, Gauss, Eisenstein and Reimann and many others, had built up quite a cache of these number patterns in three hundred years. Lo and behold, almost every one of them has a reflection in nature or a direct expression taken from a pine cone, a seashell or a flower stalk, or at least has a very strong application. It is a fact that much of the action in our everyday world of man and nature can be compressed into a simple forumla, a number pattern. These patterns were there, someone had to find them, someone had to eventually realize their applicability to some corner of our universe. It is a cause not for blame but celebration.

Will numbers prove as useful in the study of so called spiritual phenomena, dark matter, time travel, astral travel, dream awareness, consciousness itself? Will it be able to handle what physicists dig up? I suspect it will be useful for the things it is now useful for and some of what science uncovers. We may discover another tool. Math is very fatalistic. Math is a grand tautology.

I feel the biggest laws of the universe are yet undiscovered, even barely suspected. I think this has to be the case when our observations are only impeccable concerning 4% of the stuff in the universe, yet almost totally ignorant of the other 94%. The fruit does not hang so low anymore.

How gigantic was it when Newton discovered the laws of 4% of our stuff, and then later when Einstein replaced the model? An actual scientific breakthrough in any of the fields mentioned above would be huge. I expect something odd when someone finally lays a finger on these mysteries. We may find what is holding together those galactic clusters which are moving too fast is a form of consciousness. Sometime in the future the discovery of a consciousness particle would not shock me. I expect the strange out of the universe.

[YesNo]

I feel the biggest laws of the universe are yet undiscovered, even barely suspected. I think this has to be the case when our observations are only impeccable concerning 4% of the stuff in the universe, yet almost totally ignorant of the other 94%. The fruit does not hang so low anymore.


The problem with that 94% of the supposed missing stuff is that it may not be there. All that we may need to do is reformulate the mathematical gravitation theory and do away with the need to find dark matter. And since we haven't found any, so far, maybe it doesn't exist at all.

Einstein did something like this in the early 20th century. At that time astronomers were looking for a planet they called Vulcan near Mercury that should exist if Newton's laws were correct which would explain the orbit of Mercury. Einstein's modification of Newton's gravitational theory made the search for Vulcan unnecessary.

I'm getting that account of Vulcan from John Moffat's "Reinventing Gravity". He has a new theory of gravity that should make dark matter and black holes unnecessary. Of course, if someone finds dark matter that would shoot down his new theory.

[desiresjab]

The problem with that 94% of the supposed missing stuff is that it may not be there. All that we may need to do is reformulate the mathematical gravitation theory and do away with the need to find dark matter. And since we haven't found any, so far, maybe it doesn't exist at all.

Einstein did something like this in the early 20th century. At that time astronomers were looking for a planet they called Vulcan near Mercury that should exist if Newton's laws were correct which would explain the orbit of Mercury. Einstein's modification of Newton's gravitational theory made the search for Vulcan unnecessary.

I'm getting that account of Vulcan from John Moffat's "Reinventing Gravity". He has a new theory of gravity that should make dark matter and black holes unnecessary. Of course, if someone finds dark matter that would shoot down his new theory.

Reformulate? Hmmm. Not so sure about that. Whatever dark matter & dark energy turn out to be, they represent new phenomnena. I believe theories are reformulated when they are pretty close but off. Any theory of gravity does not even get us close to understanding the phenomena we are observing. But, yes, it could even turn out you are right. I am doubtful we will do away with theories of gravity altogether, and since it would need serious modification to fit today's observations, what we will have around is a modified one. Of course it could also turn out that our threries of gravity are essentially correct, that DM and DE are a new type of phenomena requiring a new structure piled on top of our theories of gravity.

[desiresjab]

Maybe a reformulated theory of gravity would occur if we discovered new features of gravity that could account for the phenomena. There may be types of neighborhoods where gravity behaves differently. There are not supposed to be, the way the theory is formulated, but the universe is full of surprises and I believe it will continue to be. Maybe there is more than one type of gravity--a Higgs Boson stock split of sorts.

[YesNo]

Reformulate? Hmmm. Not so sure about that. Whatever dark matter & dark energy turn out to be, they represent new phenomnena. I believe theories are reformulated when they are pretty close but off. Any theory of gravity does not even get us close to understanding the phenomena we are observing. But, yes, it could even turn out you are right. I am doubtful we will do away with theories of gravity altogether, and since it would need serious modification to fit today's observations, what we will have around is a modified one. Of course it could also turn out that our threries of gravity are essentially correct, that DM and DE are a new type of phenomena requiring a new structure piled on top of our theories of gravity.

I think we need to keep some distance from media reports about what is or is not real in the universe when it comes to dark matter, dark energy, black holes or a singularity at the big bang.

Regarding the current need for dark stuff, the following seems to be true: the current evidence from viewing the rotation of galaxies has falsified Einstein's theory of gravity in a way so big that measurement inaccuracies do not account for the discrepancy in the prediction and the observations.

There are two ways around the problem and, from what I understand from reading Moffat's book, many people are pursuing both approaches:

1) Einstein's theory is correct. That means there exists dark stuff, but we cannot detect it. As people look they eliminate possible candidates for this dark stuff and these negative results are valuable.

2) Einstein's theory is not correct. We need a new theory of gravity. However, that theory of gravity is not easy to come by. Moffat mentioned some of the notable failures. He does think his version is sound and fits the observations.

[desiresjab]

You cannot be a phyicist without a thorough knowledge of calculus. What you have to know if you are a number theorist is modular arithemetic. Many people do not know what that is. If they look it up, they are told it is "clock arithemetic," and so it is, as far as that goes. If I say to you, "It is twenty-five o'clock," you will easily figure out it is one o'clock.

Mathematicians call it conguence theory. That is what Gauss named it. The notation looks like this 16≡4 (mod 12). Translated into English that means 4 is the remainder when 16 is divided by 12. Just as in normal arithemetic, this is equivalent to 16-4≡0 mod(12). However, we could not say (16/4)≡1 (Mod12), as in normal arithemetic, since the remainder when 4 is divided by 12 is 4. Twelve is called the modulus, because Gauss knew Latin.

Usually, the modulus is a prime number, but it does not have to be. There are a few more traps to watch out for and exceptions to know when dealing with composites, the theory a little more extended. We will keep it prime.

To show its usefulness, let us consider an easy problem.

1 Use mod notation find the last digit of 340. (Hint): In other words, the remainder when 340 is divided by 10.

To solve this with mod notation we first have to know a simple law:

If ar≡b (mod m), then ars≡bs. We merely need to factorize the exponent and use this law.

34≡1 (mod 10). Then 34(10)≡110 (mod 10).

The answer is one

****

Let's look at one slightly harder.

2 What is the last digit in 720?

74≡1 (mod 10). Therefore 74(5)≡15 (mod 10).

The answer again is one.

* * * * *

Maybe someone can solve this next one.

3 Find the last digit in 79?

[desiresjab]

The calculational difficulties grow fast with only a little increase in the base. This number is probably too large to find the last digit on your calculator.

What is the last digit in 1920?

Factor the way easiest for calculation.

192≡1 (mod 10), 192(10)≡1 (mod 10)

The answer is one.

[YesNo]

Maybe someone can solve this next one.

3 Find the last digit in 79?

79 (mod 10) = 73(3) (mod 10).

Using Google sheets, 73 = 7*7*7 = 343 and 343 (mod 10) = 3.

Using your example,

73(3) (mod 10) = 33 (mod 10) = 27 (mod 10) = 7

[desiresjab]

79 (mod 10) = 73(3) (mod 10).

Using Google sheets, 73 = 7*7*7 = 343 and 343 (mod 10) = 3.

Using your example,

73(3) (mod 10) = 33 (mod 10) = 27 (mod 10) = 7

Good deal, old boy, you are right there with me. That one may have been trickier than the first several examples. This type of manipulation is fundamental to doing number theory. Many other fundamental ideas and techniques are indispensable. Every one that is learned and studied adds a speck to one's basic understanding of our number system and numbers in general.

Many people either forgot or never quite realized that something as routine as $867 is nothing more than an algebraic equation with the unknowns filled in as 10.

867=8(10)2+6(10)1+7(10)0

[desiresjab]

Banned again for what I am not doing. I don't get it. And I have forgotten how to extract myself from this mess.

[desiresjab]

Hmmmm....Which math notation did I use that it did not like? Could it be as simple as parentheses where it did not understand them?

[desiresjab]

Will it take a letter to a numerical exponent?

a6

[desiresjab]

Will it take aa and (a)a and a(a)?

[desiresjab]

Okay, I am stumped.

[desiresjab]

I will submit this post piece by piece until I find what is preventing it from passing. It will be done when I put QED at the bottom.

I may as well keep going with simple but intriguing ideas found within everday numbers. The next technique is one I would not expect anyone without experience in number theory to find, so I will go ahead and show the techniqe in a solution first, then explain its general workings. I know I can count on Yes/No, I hope others will follow the reasoning, as well, since it seems to be the way God thinks about some things, to mix metaphor and mathematics.

Prove that for every prime number p, there is a multiple of p whose every digit is 9. To me, at least, this law is not intuitively apparent without a studied understanding of numbers first. I suppose a Newton or a Gauss might look at it for the first time and see it immediately, but not most of us. Perhaps I am elevating my own dimness by suggesting it would take a Gauss or Newton. I still think it would require a massively bright person to work this out on their own without former experience in related problems.

A concrete example first, using 7 as our prime. We notice that N, 1/7=.142857... This is part of the technique, in case we are not Newton or Gauss and did not think of this approach. The three dots signify that the numbers after the decimal point comprise the repetend, and keep repeating.

When we multiply by(10)6, we get the digits of the repetend out front of the decimal point, but the repetend behind the decimal is unchaged, like this:

N=.142857.

(.142857...)(10)6=142857.142857...=1000,000N.

We know this is a hundred thousand of this repetend. If we subtract one of these repetend, like so:

(142857.142857...)-.142857...=142857=999,999N.

In other words, 999999/142857=7 and of course, 7·142857=999999, showing that seven has such a multiple.

Okay, everything got through except the general description at the last. I will try to fix that and post it next.

QED

[desiresjab]

In general: N ·10R where R is the length of the repetend of the prime p and N the repetend itself, the above technique must yield a similar result of a string of 9's of length R.

R will always be a divisor of p-1, or it will be p-1 itself.

There was a proof of the general case to go along with this, but since I have screwed up my own home document trying to fix this problem, I will now have to fix that.

That's all. Hopefully, it will be understood without the proof of the general case.

[desiresjab]

Here is the rest of it, not so much a proof as a clarification.

N10R-N=[9(10)R+9(10)R-1...+9(10)0]. This is indeed a string of 9's. Now,


Z=999999N, then Z/999999=N=1/p. Therefore p=999999/Z, and p·Z=999999.

[YesNo]

In other words, 999999/142857=7 and of course, 7·142857=999999, showing that seven has such a multiple.


Nice result.

I've seen the technique used as a way of showing that a repeating decimal can be represented as a ratio of integers. This shows that such numbers are "rational". However, I did not realize that this implied that any prime has a multiple where the product is all nines. Now that you point it out, it makes sense that this should be the case.

[desiresjab]

I will give the forum three days to work on a problem that took me longer than that to solve. If more time is requested, I will give it gladly.

* * * * *

If I summed all the digits of the decimal representation of 44444444, and called that number C, then summed all the digits of that number, what would the resulting number D be, exactly.

It looks like this: AA=B, the sum of the digits of B=C, and the sum of the digits of C=D. Find D.

Can anyone do this problem? Give it a try.

[Dreamwoven]

I will give the forum three days to work on a problem that took me longer than that to solve. If more time is requested, I will give it gladly.

* * * * *

If I summed all the digits of the decimal representation of 44444444, and called that number C, then summed all the digits of that number, what would the resulting number D be, exactly.

It looks like this: AA=B, the sum of the digits of B=C, and the sum of the digits of C=D. Find D.

Can anyone do this problem? Give it a try.
I wouldn't be able to get beyond the neat way you present the problem, desiresjab.

[desiresjab]

I wouldn't be able to get beyond the neat way you present the problem, desiresjab.

Well, this problem is no slouch. I believe I found it in an old math Olympiad exam. It was a bonus question for the brightest teenagers on earth. That is kinda disgusting, isn't it? Although I have had possession of the problem for years, I did not look at it much until I began to collect some tools. Then it took a long time of laying it down for long spells and forgetting all about it. A problem I want to solve but have no way to approach can drive me crazy if it stays in my mind. I just loved the shape of this problem and what it was asking for looked so impossible.

What solving stuff like this comes down to is having the tools. There are always properties, laws and techniques which are related somehow and can be used to excavate the answer. If you have no idea about the existence of some property which will lead you right to the answer you are in for a long haul, probably an impossible one. Only cats like Leibniz could solve such a monster unprepared, independently rediscovering such laws and properties as are needed along the way.

I will show how this one is done soon. Right now I have personal obligations begging to be resolved around the house.

[desiresjab]

Sum the digits of the decimal representation of 44444444, then sum the digits of the new number. The form of the problem looks like this:

AA=B, the sum of B's digits equals C, the sum of C's digits equals D. Find D, exactly.

To find the number of digits in the decimal representation of 44444444, in other words how long the number is, the algebraic technique is to take the log of and add 1, while chopping whatever remains behind the decimal, leaving us with a whole number:

AA, like so:

↓{1+log AA}↓=↓{1+A log A}↓=16211, in the case of

44444444.

The actual calculations are as follows:

4444(log 4444)=16210.707879.... After we chop the decimal, which is irrational and goes on forever without pattern, and add 1 we know the huge number AA has 16211 digits. This number is far larger than the number of atomic particles in the universe, which is only in the neighborhood of 10120 particles, tops.

How can we sum the digits of B when we do not even know the number? We detour creatively, by supposing every one of those 16211 digits to be a 9, so that when we sum them they will equal 9(16211)=145899.

145899, then, is the upper limit for C, it can be no larger. But we are allowed to pretend again. We pretend that all six digits of C are equal to 9.

6X9=54. Ah, do you know what that is? It is an upper limit for D, the number we are after.

Suddenly, we are somewhere. One can almost smell solutions, but how do we get there? The answer lies again in observation and technique, in that order.

When any huge numbers are multipled, it is always easy for us to ascertain what the last digit is. In the case of 4444 times itself, no matter how many times successively we perform the operation, the last digit is always 4 or 6. Even powers of 4444 end in 6, and odd powers end in 4. Of course 4444 as an exponent is an even power, so 44444444 ends in a 6, and when divided by 10 would leave a remainder of 6.

We must detour now for some pretty facts, lest we arrive at our final destination with our route still shrouded in mystery.


* * * * *

Any time you sum the digits of a number X, that sum S will remain congruent to X (mod 9) through successive summing operations. This only happens with 9 because we use base 10. In base 8, 7 would have this same property through successive summing operations. Any base. (The above preservation property is also true for 3 in base 10).

Notice that once we ascertained the number of digits of 44444444, summing the digits successively is the only operation we have performed.

Successive powers under a modulus (any particular divisor ) always bend back and reapeat themselves in a cycle. This is called a power residue cycle, a cyclotomic number, to throw in a fancy term dangerously. They are certainly cyclical, but I do not know if that makes them cyclotomic. The host confesses.

A Modulus does not allow any number in its system to be as large as itself. The modulus is king. Larger numbers are bent back by division until only a remainder is left. Any whole number, when divided by 10, for instance, will leave one of ten remainders: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9. That's it. The modulus can systematically reduce any giant number to one smaller than itself. And as far as the king is concerned, any larger number than itself which it can divide is equal to zero. 30=0 (mod 10).

Let's look at the power residue set of 4444, (mod 9). It is okay to begin with the zero power of 4444, which we know is always 1 for any number, and 1 will always be the first number in the power resudue set of any number.

44440=1 (mod 9)

44441=16=7 (mod 9)

44442=19749136=40=4 (mod 9)

44443=87765160384=55=1 (mod 9)
...
...
...



44444444=1 (mod 3)

9(493)=4437, means 4444=7 (mod 9)

44444444=7 (mod 9)

At its 4444th power, 4444 is congruent to 7 (mod 9) and 1 (mod 3).

In fact, the power residue set for this number (mod 3) is

{1, 1, 1, 1, 1....}, and on any power our huge exponential number only equals 1 (mod 3).

The pattern listed above vertically for (mod 9) power residues has produced 1 again at power 3, so we know the complete cycle, which goes 1, 7, 4, repeating every three powers until the highest power is reached. Also notice that to take the congruency of a number with respect to 9 in our base 10 system, we only need to sum its digits then take the remainder when divided by 9.

The more convenient way for calculation is to begin the pattern on the first power so that the power residue set cycles like this from power 1:

{7, 4, 1, 7, 4, 1...}

* * * * *

Gathering what we know so far:

Our target number is less than or equal to 54 and is congruent to 7 (mod 9).

That becomes a small set.

{7, 16, 25, 34, 43, 52}

We have gone from fifty-four possiblities to six.

Now, as was the case with 9, the value, the magnitude, of a number is the same as the sum of its digits (mod 3), so we are further able to say that the correct answer must be congruent to 1 (mod 3). Only three numbers now qualify.

{7, 16, 34}

Now the judging becomes more demanding, more appropriately the search for a tool or a property to distinguish Miss America out of the three.

We know that B ends in a 6. In symbols B=6 (mod 10). But how does this help us? 10 cannot play the same trick that 3 and 9 did, for the preservation of its congruency does not happen. (It would happen on 10 only for base 11, where the factor 5 would also exhibit the property). Ah, but maybe 6 can play the trick. Actually, it cannot help us that way, either.

The simple observation that works is that any number which leaves a remainder of 1 when when divided by 3 and a remainder of 7 when divided by 9, has to leave a reaminder of 4 when divided by 6. But the English of that is so long and messy. Check this out.

If D=1 (mod 3) and= 7 (mod 9), then D=4 (mod 6).

Only one number from the last set qualifies:

{16} is the answer. D=16

One can only assume several of the interior digital places of C are 0 and 1.

Since 16 is smaller than expected, let me recheck the steps. The mistake, if any, is probably in following that (7, 4, 1) cycle correctly to the last power.

I may be back.

I think I can proclaim it correct, since I validated my stepping on the power residue set of 7, 4, 1. At the 4443rd power the cycle is on 1, which puts it on 7 for the 4444th power.

Since we know C is less than 145899 but still a six digit number, I envision a number built along the following lines:

100456. It makes perfect sense now.

[desiresjab]

I had to scratch the next post out and replace it with this. Everything is all right. I know the answer is correct.

[YesNo]

To see if I understand, you have done the following:

1) Calculated the number of digits in 44444444 using a logarithm. You found there to be 16211 digits.

2) Found an upper bound on the sum of those digits by assuming they are all 9's, the highest value each digit could be. The max sum is 145899 which would be the maximum value of C. We know there are 6 digits in that sum.

3) We want the sum of the digits of C and so can find an upper bound on this as done in (2) by setting all the digits to 9. That max sum is 54.

4) You note that 44444444 (mod 9) = 7 and the sum of the digits of that number is also 7 (mod 9). This reduces the possibilities for D to a number in this set: {7, 16, 25, 34, 43, 52}

5) You note that 44444444 (mod 3) = 1. This reduces the possible value of D to one of {7, 16, 34}.

6) I am still trying to think through the last part, but I will leave this till later.

[desiresjab]

To see if I understand, you have done the following:

1) Calculated the number of digits in 44444444 using a logarithm. You found there to be 16211 digits.

2) Found an upper bound on the sum of those digits by assuming they are all 9's, the highest value each digit could be. The max sum is 145899 which would be the maximum value of C. We know there are 6 digits in that sum.

3) We want the sum of the digits of C and so can find an upper bound on this as done in (2) by setting all the digits to 9. That max sum is 54.

4) You note that 44444444 (mod 9) = 7 and the sum of the digits of that number is also 7 (mod 9). This reduces the possibilities for D to a number in this set: {7, 16, 25, 34, 43, 52}

5) You note that 44444444 (mod 3) = 1. This reduces the possible value of D to one of {7, 16, 34}.

6) I am still trying to think through the last part, but I will leave this till later.


Your confusion at step six may be because indeed I have overlooked something. 16 and 34 both meet all three qualifications, i.e. =7 (mod 9), =1 (mod 3), and =4 (mod 6).

So really, our set of prospects still has two members left:

{16, 34}.

Now I have to think of something else.

[desiresjab]

Ha! ha! The laugh is on me. But I think I have the fix through a different method on step six of Yes/No's listing of my steps.

The idea is to go to base 11 and cast out 10's. We do not actually need to go to base 11, for that would involve adding precisely one new symbol (usually designated simply as a), so let's just say we did and not, because we know 44444444 is the same value whether it is expressed in base 10 or base 11, and so would also yield the same value when 10's were cast out, whether that identical value used identical digits in the same order in the two systems or not. This way we will know how the sum of our number (mod 10) behaves and we are enabled to preserve our value through successive operations of summing the digits. These operations can be imaginary too, since they preserve through. When we cast out 10's, just as we cast out 9's in base 10, we are left with 6, which has identical real value in base 10 or base 11.

By this reasoning, this time I believe infallible, 16 is still the correct answer

[desiresjab]

A Chinese parcel delivery worker just made a significant discovery concerning Carmichael numbers. Those are an important class of composite number that is hard to factor and often fools primality tests for Fermat primes. One of the beauties of number theory is that discoveries like this one are often made without the use of advanced tools.

[YesNo]

The casting out 10's seems promising. I haven't tried it but I expect one should be able to distinguish which of those three numbers is the correct answer.

Do you have a link to the information about the discovery concerning Carmichael numbers? I was in Madison, Wisconsin, yesterday and stopped by some of the many used book stores in that university town and found Carmichael's "The Theory of Numbers" and "Diophantine Analysis". I was planning on going through that book making a Jupyter notebook out of some of the problems I found interesting.

[desiresjab]

The casting out 10's seems promising. I haven't tried it but I expect one should be able to distinguish which of those three numbers is the correct answer.

Do you have a link to the information about the discovery concerning Carmichael numbers? I was in Madison, Wisconsin, yesterday and stopped by some of the many used book stores in that university town and found Carmichael's "The Theory of Numbers" and "Diophantine Analysis". I was planning on going through that book making a Jupyter notebook out of some of the problems I found interesting.

The links I found were distinctly uninteresting. They said less than I have. A Chinese friend told me about it. "Chinese amateur Carnichael numbers", worked as a search phrase.

I believe the casting out of 10's does work. I am very happy with it because I have never seen that used before. It will be embarrassing if my logic is wrong.

The book I am after is called 52! I had a book going on a similar subject when I heard about this one being published. I want to see if he did a better job than I think I could have done. I still may finish mine, if I do not think too much of his effort. I could order it online, but I do not make any financial transactions over the internet. If you find this one score it for me and I will pay cost plus shipping it to me.

[YesNo]

I searched for "52!" but I only came up with links to "52 factorial". Do you know the author or something more about the book?

I'm not familiar with casting out 10's. If I get some time I will see if the idea would work. However, you might want to put in the details of the proof.

It doesn't hurt to publish your own book even if you liked the other one. However, it is good to have a list of references and that book may be one of them.

[desiresjab]

I searched for "52!" but I only came up with links to "52 factorial". Do you know the author or something more about the book?

I'm not familiar with casting out 10's. If I get some time I will see if the idea would work. However, you might want to put in the details of the proof.

It doesn't hurt to publish your own book even if you liked the other one. However, it is good to have a list of references and that book may be one of them.

Michael Wayne Cottle write 52!

I was not familiar with casting out 10's, either. It was just an idea that came to me.

[YesNo]

Michael Wayne Cottle write 52!

I was not familiar with casting out 10's, either. It was just an idea that came to me.

You can find the book on Amazon. The kindle edition is 99 cents.

He sounds like an interesting character.

Tell me more about casting out 10's. I don't really follow it.

[desiresjab]

Is there a certain part of the demonstartion I might explain more clearly, Yes/No?

The biggest leap of faith comes in going to base 11 so we can see what happens to the sums (mod 10), without actually ever going there. I reason that a number expressed in one base versus another does not have more factors of 10. The digit 6 is congruent to this set

(mod 10) {6, 16, 26, 36...} and this one (mod 11) {6, 17, 28, 39...}

It seems to me that for a number A to leave a remainder smaller than itself equal to 6 in (mod 11), that remainder can only be 6, though the original number in base 11 might have ended in a digit other than 6. In fact, the number may end with the digit 5 in base 11, like 105 does. The reasoning for theoretically going to base 11 is to preserve the nature of (mod 10) sums over the operation of summing the digits of the powered number several times, as they are for 9 in base 10. I believe we can only guarantee this with a modulus which is one less than the base.

So when I sum all these digits in (base 11), (mod 10), I should get the real last digit. I already know what B is (mod 10), but without going to base 11, I could not say anything about the last digit of D, however, because the result of the summing operation would not be preserved. The sum of a number's digit ought to stay congruent to the same set (mod 10).

I have never consulted anyone because I don't like to do that, since I am trying to solve problems myself. I know a math phd who I have played music with. At this point I would not mind giving him a call, to see if my reasoning has been correct. He will quickly see other ways to do it, so I will have to keep him on track. Even if I got the right answer, I want to be sure my reasoning was not faulty somewhere.

Because the problems I choose are hard for me, that makes them fun even when they are torturous. I always learn a lot and correct myself a lot.

[YesNo]

So there are three candidates to consider, {7, 16, 34}, in base 10. We got that set by considering 44444444 = 7 (mod 9). Wouldn't we have to rewrite 44444444 in base 11 and find out what the digit was (probably not 7) modulo ten in base 11?

My suspicion is that converting to a new base might not help with the solution, but I don't know.

[desiresjab]

It is easy to verify that B ends in 6, base 10. Simply by multiplying single 4's together and watching the last digit. It only alternates between 4 on odd powers and 6 on even powers. If one doubted that technique, one could go here http://www.javascripter.net/math/calculators/100digitbigintcalculator.htm and calculate the number directly and observe all 16211 digits--a number I was relieved to see matched my own.

Then the other day I went over to another specialized calculator that was supposed to calculate the sums of digits. I pasted in the digits of our AA=↓B↓=C and it gave me back a measley sum of seventy-six thousand and something. This was only a five digit number, and so I felt it had to be wrong. I wrote the publishers of the calculator and told them so. Now I begin to waffle on my own judgement there.

C represented a maximum value of the sum of B's digits. Maybe it was six digits long in the case of that maximum, but only five digits in reality. Then I would have gotten my maximum for D from these over estimates, which is perfectly fine, since all I was looking for at that point was a maximum.

* * * * *

I was so convinced the sums of digits calculator was wrong that I did not even bother to add up the number it gave as C. Had I done this, I would have noticed that the sum of the digits of 72,601=16, like I did just minutes ago. It was seventy-two thousand and something, not 76 and something. I did not notice the details. All I thought I saw was the wrong answer for C sitting there.

This means I must have arrived at the right answer despite some faulty conclusions. I was protected by the part of my reasoning which was correct--that D was a maximum of 54 and conguent to 7 (mod 9) and 1 (mod 3). It did not matter at all that I thought C was a six digit number instead of a five digit one. It did not interfere with getting the answer.

It seems the answer is settled, then, by brute force and somewhat inadvertantly. It also seems highly unlikely that I would have come up with this particular answer through erroneous reasoning. All my work pointed to this answer.

My base 11 speculations are open for discussion. They twist the brain like a sponge.

* * * * *

I don't know if I mentioned this, but since 4447 is a prime, I thought there might be some means of working backwards through Fermat's little theorem to 4444. It was a fruitless avenue, but I may have only taken a wrong turn.

* * * * *

The key piece of reasoning is in realizing that the congruence class of the sums of digits is preserved mod 9 across multiple operations. The other was understanding power residue sets and their behavior.

I am satisfied with the discussion down to the set {7, 16, 34}. I had already eliminated 7, but I cannot remember how right at this moment, so it can stay in the discussion.

[desiresjab]

My speculation about base 11 was correct!

The letter A in the following has nothing to do with the letter A in AA, but is the letter of the alphabet used to represent 10 in base 11.

The base 11 digits of a number X will always add up (mod 10) to something with the same congruency (mod 10) as the decimal representation of X.

Base
10----11

96----88
196--169
296--24A


Whatever those digits base 11 add up to, we know they will be congruent (mod 10) to whatever the base 10 representation is congruent to. And we further know that the congruence class is preserved across the operation of summing the digits of the base 11 representation. This is how we know with certainty that the last digit of D is 6 in decimal representation, just as it is the last digit of B. I believe that is all of it.

[desiresjab]

The former topic feels so wrapped up that I feel like another problem to untangle how God thinks is due. I believe the following was from a prep test for math Olympiad.

Prove that (2m)! (3n)! is always an integer.
..................(m!)2 (n!)3



(2m)! (3n)! The bottom factors easily, not the top.
m! m! n! n! n!



m! 2mΠm+1 n! 3nΠn+1 after cancellation, this becomes:
m! m! n! n! n!



2mΠm+1 3nΠn+1
m! n! n!



2(m) 2(m-1)...(m+1) 3(n) 3(n-1)....2(n) 2(n-1)...(n+1)
...m...(m-1)...(1)--------(n)....(n-1)......(n)...(n-1)...(1)


All the terms above are factors in both numerator and denominator. I had to use to dots and dashes to make them line up properly for illustrative purposes.

We see that any prime up to M in the denominator will have a double in the numerator, and will so be cancelled. Any prime up to N in the denominator has both a double and a triple in the set of numbers n+1 through 3n, and so both powers of N! are cancelled, leaving a whole number in the numerator and 1 in the denominator for both M and N.

This was the first time I ever figured out how to factor a factorial notatively.

[YesNo]

The factorial problem seems to work as you described.

I also find it hard to trust the results of calculators especially when the numbers get large. Many things can go wrong: implementation, computer hardware, programming. I am currently using Python for other purposes and it looks like this should work well with number theory. You can get Python by installing the anaconda distribution at https://www.continuum.io/downloads. The software is free.

This should do multi-precision arithmetic. I also interface it with jupyter notebooks which comes with the anaconda distribution.

I checked the answer with Python to your previous problem and I also got 16 as you did. Here are copies of the results:

num = 4444**4444
total = sum( [ int(char) for char in str(num) ] )
72601
sum_of_total = sum( [ int(char) for char in str(total) ] )
16

I didn't know how to code the sum of digits and so I searched for a solution and used this one, that is, I didn't come up with it on my own: https://www.codecademy.com/en/forum_questions/502cd6b170453d00020186c8

[desiresjab]

Yes, sir, I believe we have the 4444 problem settled. The best part of it was what was learned along the way. After finishing a problem the next next problem is to find the next problem. One that is solvable is needed, but which will require considerable effort. One could work on the Goldbach conjecture, the twin prime conjecture or Brocard's problem, but would never have a reasonable chance of even making progress. But if one chooses an easier problem, it could lead to discoveries which might be of assistance on those unsolved questions later on.

I may throw in another Olympiad problem to hold us until a truly fascinating problem comes along.

[desiresjab]

The following is a type of problem I find extraordinarily difficult. Other people may see the answer fairly quickly. But I look at this thing and I am baffled where to even start. I have seen problems of this type which are even more brutal. I am sure there are number theoretic techniques to solve them, for I found this problem again in a prep test for math Olympiad. You see, I have a functional problem as a mthematician and a human being--if a technique looks ugly and cumbersome, I avoid it. I seem to be seeking the beautiful in mathematics. I post the following problem because it is so opposite to that, to me. It is quite brutal from my perspective. And since I have no techniques to solve it, it is a head-on brain against problem sort of deal. There is probably also a solution out of formal logic. Here the beast is:

Every man in a village knows instantly when another's wife is unfaithful, but never when his
own is. Each man is completely intelligent and knows that ev-
ery other man is. The law of the village demands that when
a man can PROVE that his wife has been unfaithful, he must
shoot her before sundown the same day. Every man is com-
pletely law-abiding. One day the mayor announces that there
is at least one unfaithful wife in the village. The mayor always tells the truth, and every man believes him. If in fact there are exactly forty unfaithful wives in the village (but that fact is not known to the men,) what will happen after the mayor's announcement?

[desiresjab]

Let this one stay up a for a while. I know some smart people would like to think about it. I may know the answer, but I am not exactly sure, either.

[desiresjab]

I think the question is rather poorly formed. It leaves a certain taste of ambiguity, especially what is in parentheses. That is why I am going to attempt to answer it already, and get it out of here. Then I will move on to a really really difficult one of this type which is well formed.

One of the villagers is probably the local mathematician. He asks each man in the village, including the mayor and himself, to count the number of wives they recall to have been unfaithful. They must only write down their own name and that number on their piece of bark. Then he collects each piece of bark and spreads them out for all to see. All forty men who recall only thirty-nine adulterous wives, must shoot their own before sundown. Even the mayor and the mathematician may end up shooting their wives. This works even if there are only forty-one men in the entire village.

Assuming that to be correct, it was not really that hard, I guess. Let us move on to a real monster, which I do not expect to be able solve at all.

[desiresjab]

Try this one on for size, folks. It is a real baffler. Yet it does have number information which can obviously be used to solve it. I have worked on this one before, and I see from my notes that my work ended in confusion and uncertainty. I will give it another shot, after trying to determine what I was up to before. Sometimes it is quite excruciating to reconstruct your own logic from forgotten work, especially if the logic happened to be wrong! I am on the line now. But to tell you the truth, I have no confidence at all on this one.



Two positive integers are chosen. The sum is revealed to logician A, and the sum of the
squares is revealed to logician B. Both A and B are given this information and the information
contained in this sentence. The conversation between A and B goes as follows: B starts
B: ` I can't tell what they are.'
A: ` I can't tell what they are.'
B: ` I can't tell what they are.'
A: ` I can't tell what they are.'
B: ` I can't tell what they are.'
A: ` I can't tell what they are.'
B: ` Now I can tell what they are.'

(a) What are the two numbers?

[desiresjab]

I know that thing is possible. It just hurts my head like h3ll, though.

[YesNo]

If all the men in the village are completely law-abiding, who is having sex with the unfaithful wives?

I might be missing something about the second problem. We know the sum of the two positive integers, call it S. Consider all the possibilities as A runs from 1 through S - 1 of the pairs A and S - A. Square each of these, A2 and (S - A)2, and see if the sum of those two squares equals the other known value. When it does then one has the two numbers.

[desiresjab]

I would approach this through two possible connections--the generalization of Fermat's theorem on the sums of two squares, in conjunction with the Pythagorean theorem. Only one might be necessary, or maybe neither. It could be wrong altogether, but this is where I would begin to sniff. It is necessary to know exactly what both professors learn from each answer of the other. One could perhaps walk upwards to the correct sum of squares this way. Brutal, but it could work. I am thinking perhaps the Pythagorean theorem can provide a shorcut to the answer, once one understands what is happening with each answer the professors give.

What you have to start with for this investigation is knowledge of just which numbers can be expressed as the sum of two squares. This is not too hard to remember: Those numbers exactly which either have no prime (4n+3) factors, or all prime (4n+3) factors in the prime factorization of the number are to an even power. That is, any number, as long as it prime 4n+3 factors are all to an even power. Either that or it has no such factor at all. Only such numbers can be expressed as the sums of two squares.

[YesNo]

I suppose we could shorten the brute force result with some algebra. Let the two unknown numbers be X and Y. We are given the sum of those numbers, X + Y, and the sum of the squares, X2 + Y2. As an example we can say the sum of the numbers is 6 and the sum of their squares is 26. We can change these parameters later. What are the numbers X and Y?

Since X + Y = 6, we know X = 6 - Y.

We can do the following transformation X2 + Y2 = (6 - Y)2 + Y2 = 36 - 12Y + Y2 + Y2 = 36 - 12Y + 2Y2.

We are given that 36 - 12Y + 2Y2 = 26, so we can subtract both sides of the equation by 26 and get the following quadratic equation: 10 - 12Y + 2Y2 = 2(Y - 1)(Y - 5) = 0. There will be two solutions to this equation. If they are integers then we have the solutions we want. We can see that Y could be either 1 or 5. X would be the opposite.

[desiresjab]

I suppose we could shorten the brute force result with some algebra. Let the two unknown numbers be X and Y. We are given the sum of those numbers, X + Y, and the sum of the squares, X2 + Y2. As an example we can say the sum of the numbers is 6 and the sum of their squares is 26. We can change these parameters later. What are the numbers X and Y?

Since X + Y = 6, we know X = 6 - Y.

We can do the following transformation X2 + Y2 = (6 - Y)2 + Y2 = 36 - 12Y + Y2 + Y2 = 36 - 12Y + 2Y2.

We are given that 36 - 12Y + 2Y2 = 26, so we can subtract both sides of the equation by 26 and get the following quadratic equation: 10 - 12Y + 2Y2 = 2(Y - 1)(Y - 5) = 0. There will be two solutions to this equation. If they are integers then we have the solutions we want. We can see that Y could be either 1 or 5. X would be the opposite.

Your algebra is beautifully done and succinct. It tells us if we plug in the right number we will get back the number we should. It just does not tell us if we have selected the right input number in the first place to make the two professors volley back and forth for six separate "I don't knows" before professor B has enough information to answer the question.

I learned a lot from looking at the following sequences. It is a list of the numbers which can be represented as the sums of two squares. Zero counts In symbols: 62+02 is how one gets 36 as the sum of two squares, etc. The bottom rows are the sums of squares. The numbers above them are the number of representations for that number (how many different ways it can be represented as the sum of two squares.)

1, 1, 1, 1, 1, 1,, 1,, 1,,, 1,,, 1,,, 1,,, 1,,, 2,,, 1,,, 1,,, 1,,, 1,,, 1,,, 1,,, 1,,, 1,,, 1
1, 2, 4, 5, 8, 9, 10, 13, 16, 17, 18, 20, 25, 26, 29, 32, 34, 36, 37, 40, 41, 45,


1,,, 2,,,, 1,,, 1,,,1,,,, 1,,, 1,,, 2,,, 1,,, 1,,, 1,,, 1,,, 1,,, 1,,, 1,,, 2,,, 1,,, 1,,,1,,, 2
49, 50, 52, 53, 58, 61, 64, 65, 68, 72, 73, 74, 80, 81, 82, 85, 89, 90, 97, 100


Something pops out immediatey: all the numbers representable in more than one way are divisible by 5.

If the sum of B's squares were anything less than 25, he would immediately know the decomposition, for there is only one composition of them. Both A and B realize this.

Let us assume the actual numbers are 5 and 0.

B knows A has a 3+4 or 5+0. A knows B has 13, 17 or 25.

As soon as B speaks, A knows B does not have a sum of 13, or 17, so he must have a 25. Since A's sum is 5, he knows the proper sum is 5+0.

* * * * *

Let us assume the actual numbers are 3 and 4.

B knows A has 3+4 or 5+0. A knows B has 49, 37, 29 or 25.

B says he doesn't know. This tells A B has 25.

A says he knows.

* * * * *

The same thing seems to happen at 50. The first number with enough complexity to keep the professors volleying might be 100. At least A does not know the answer when B speaks for the first time. Or perhaps it is the first number representable in three different ways, which I read was either 285 or 385, I think the latter.

The proper sum of squares must force the number of returns in the volley to seven responses total.

[desiresjab]

I wonder if each pair of responses eliminates one root. But that would be a cubic equation, since there are three pairs of call and response, and then B says he knows the answer. So far, as far as I can get is after B speaks, A also says he doesn't know, though I cannot determine what further information that gives B. That is for the sum of squares 100.

[YesNo]

I did not understand what the going back and forth was with A and B each saying they did not know the answer. Apparently they are not allowed to give each other the values they know. Each of them knows a limit on the possible answers without knowing each other's information.

[desiresjab]

My computer keeps freezing up. I lost a long post, but the solution was not in it. As soon as B speaks, A can eliminate all numbers but 50 and 100. But to B the set of possibilities in A's mind could be 50, 100, or 169. For, yes, 169 has two decompositions: 132+02, and 122+52.

And we see a doubly representable sum of squares does not have to be divisible by 5 after all. We have two intersecting sets sets {50, 100} and (100, 169), for 100 can be made from a sum of 10 or 14. Only A knows he is holding a sum X+Y=10, for instance, and not 14.

I will keep the posts relatively short to avoid freeze ups. To be continued as an editorial extension...right here

* * * *

[desiresjab]

The problem might be that the two sets {50, 100} and (100, 169) intersect. I have no idea whether non intersecting sets of doubly representable sums of squares are even possible. Their intersection causes problems for me. If non intersecting sets are possible, at this point I would have to guess the problem could be solved only with them, for I have trouble seeing where either one gains enough additional knowledge through the response when the sets intersect.

[desiresjab]

Oh, boy. All our questions and more are answered in this little paper. The authors give examples of sums of squares with 1, 2, 3, 4, 6 (Can't remember if 5 was there) unique expressions, or factorizations, as these authors more properly call them.

A thorn in the path to watch for is their not counting 02 in their method of calculation. This means the standard formula (which I found elsewhere) will give different values. One only needs to subtract 1 from the total number of expressions.

The standard forumla for the number of expressions of any natural number to any power is:

rk(n)=2[d1(n)-d3(n)].

Now n is the sum of squares itself, k is the power, which may be higher than 2, and d1 and d3 signify the number of actual factors or divisors of n, prime or otherwise, which are respectively congruent to 1 or 3 (mod 4). A congruency of 2 or 0 (mod 4) is not germane, and not included in the calculations.


http://www.rowan.edu/colleges/csm/departments/math/facultystaff/osler/110%20SUM%20OF%20TWO%20SQUARES%20IN%20MORE%20THAN% 20ONE%20WAY%20MACE%20Small%20changes%20Oct%2008%20 %20Submission.pdf

[desiresjab]

Now as far as the two professor problem goes, I have made a terrible, amateurish oversight. I always confess to my oafish ignorance and embarrassing oversights, especially the real whoppers, because they make such good copy.

Essentially, words two and three of the problem nail down a simple constraint I flat overlooked in my zeal to forge ahead. Our ignorance always forces learning upon us when we persist. I cannot be emabarrassed for what I have learned, but only for what I used to not know, like yesterday.

We learned a lot, and the problem will be quite different when we return to it tomorrow--with some elements ejected from the set of prospects.

[YesNo]

http://www.rowan.edu/colleges/csm/departments/math/facultystaff/osler/110%20SUM%20OF%20TWO%20SQUARES%20IN%20MORE%20THAN% 20ONE%20WAY%20MACE%20Small%20changes%20Oct%2008%20 %20Submission.pdf

This article made me realize that whenever one is talking about sums of squares one should think Pythagorean theorem and circles.

[desiresjab]

This article made me realize that whenever one is talking about sums of squares one should think Pythagorean theorem and circles.

Quite true. Remember that is what the Martinson article did.

The two professor problem is too hard for me unless I find another line of attack. Plus, it is not a fun problem for me, but has become a minor obsession anyway. For recreation I have to do beautiful problems. At least they are beautiful to me. I wonder if anyone agrees with me over what "looks" good in mathematics. Here is one.

When is xpq divisible by (xp)q? Prove.

[desiresjab]

The last problem is very easy of course. Let pq=a, let pq=b.

From basic algebra:

xa=xa-b.
xb

Problems like this are good for refreshing yourself on technique.

The answer is for all x except 0.

[desiresjab]

Here is a pretty good one. It is countable, if you figure out how to do it. (Hint): There is also a technical way to do this.

Four hundred people are standing in a circle. You tag one person, then skip k people, then tag another, skip k, and so on, continuing until you tag someone for the second time. For how many positive values of k less than 400 will every person in the circle get tagged at least once?

Have fun.

[YesNo]

It looks there would be many ways in which at least one person would not be tagged and that would depend on k being a divisor of 400. If k is relatively prime to 400 everyone should get tagged eventually.

But what about those values of k that have some factors that also divide 400, but other factors that don't? For example 6? Just checking 6 and 10 that would skip some as well.

So I assume the value would be the number of values of k relatively prime to 400. That would be given by Euler's phi function. I searched for a way to calculate that value: https://www3.nd.edu/~sevens/13150unit10.pdf

However, this isn't a proof that the above is correct, just some comments suggesting that it should be.

[desiresjab]

Yes. Euler's phi function is the right function. It will read off the answer directly. The actual work would look like this.

Φ(400)=400Πp|n(1-1/2)(1-1/5)=400(4/10)=160.

Obviously, divisors of 400 cannot do the job. But every k relatively prime to 400 will circle the table leaving a different remainder (mod Φ(400)) at the end of each revolution.

[YesNo]

I was thinking more about gravity lately.

It occurred to me that finding something in the universe with a dark concentration of gravity larger than three solar masses would trigger the black hole portion of Einstein's gravitational theory. If one finds something like that then the theory says all the matter in that region collapses to a point of no radius. There is no counter source of energy, such as fusion in a star, able to keep the radius larger than zero. It vanishes into a singularity of the theory, a place that effectively looks like the theory is dividing by zero.

What can one conclude if one finds such concentrations of dark gravity? I think the most reasonable thing to conclude is that the theory is wrong. Just because the theory has a singularity doesn't mean that reality has singularities. I also suspect that a theory with singularities has got problems from within. Those singularities imply that the theory can reach a point where it stops serving as an explanation.

Now the theory also has problems explaining the rotation of galaxies. They move too fast to stay together without more matter (which is also dark). But one hasn't, so far, been able to find that dark matter. It looks like the theory fits the universe of the early 20th century well. That universe didn't have galaxies, dark gravitational sources (aka black holes) nor was it believed to be expanding.

[Danik 2016]

I can´t follow your complex calculations but I wonder if this article or the cosmology journal as such might be interesting for this thread:
http://journalofcosmology.com/JOC26/indexVol26CONTENTS.htm

[desiresjab]

I can´t follow your complex calculations but I wonder if this article or the cosmology journal as such might be interesting for this thread:
http://journalofcosmology.com/JOC26/indexVol26CONTENTS.htm

Yes/No and I will send our observations right over to those people.

[YesNo]

I can´t follow your complex calculations but I wonder if this article or the cosmology journal as such might be interesting for this thread:
http://journalofcosmology.com/JOC26/indexVol26CONTENTS.htm

I do find the idea that "black holes" are more like MECOs (magnetospheric eternally collapsing objects) to be interesting. https://en.wikipedia.org/wiki/Magnetospheric_eternally_collapsing_object

MECOs solve one problem with black holes: there is no mass for a zero radius object since the mass has been radiated away throughout eternity. A black hole has a large amount of mass but its radius is zero. It also resolves the problem of a black hole collapsing faster than the speed of light to that point. I liked how such a massive point (black hole) was called a "mathematical myth".

Supposedly there is an effort to build an earth based set of radio telescopes on each side of the globe, called an Event Horizon Telescope, making the resolution power of such telescopes larger that what currently exists (https://en.wikipedia.org/wiki/Event_Horizon_Telescope) They would be able to point to the radio source at the center of our galaxy known as Sagittarius A*. Some believe this to be a black hole, but it failed to "eat" a cloud of matter passing by it a year ago as predicted, if it were the kind of black hole they imagined it to be.

John Moffat's theory does away with black holes entirely, but I don't know how this is done. Besides black holes, there is also the anomaly (based on Einstein's theory) of a too rapid rotation of galaxies as well as the singularity (like a black hole) at the beginning of the universe. Moffat's theory does not merge gravity with the other three forces found at the atomic level (electromagnetic, strong and weak nuclear forces). However, that might be a plus for his theory.

[Danik 2016]

Yes/No and I will send our observations right over to those people.
Seems a good idea!

[Danik 2016]

I do find the idea that "black holes" are more like MECOs (magnetospheric eternally collapsing objects) to be interesting. https://en.wikipedia.org/wiki/Magnetospheric_eternally_collapsing_object

MECOs solve one problem with black holes: there is no mass for a zero radius object since the mass has been radiated away throughout eternity. A black hole has a large amount of mass but its radius is zero. It also resolves the problem of a black hole collapsing faster than the speed of light to that point. I liked how such a massive point (black hole) was called a "mathematical myth".

Supposedly there is an effort to build an earth based set of radio telescopes on each side of the globe, called an Event Horizon Telescope, making the resolution power of such telescopes larger that what currently exists (https://en.wikipedia.org/wiki/Event_Horizon_Telescope) They would be able to point to the radio source at the center of our galaxy known as Sagittarius A*. Some believe this to be a black hole, but it failed to "eat" a cloud of matter passing by it a year ago as predicted, if it were the kind of black hole they imagined it to be.

John Moffat's theory does away with black holes entirely, but I don't know how this is done. Besides black holes, there is also the anomaly (based on Einstein's theory) of a too rapid rotation of galaxies as well as the singularity (like a black hole) at the beginning of the universe. Moffat's theory does not merge gravity with the other three forces found at the atomic level (electromagnetic, strong and weak nuclear forces). However, that might be a plus for his theory.
What fascinates me is that theories today have to change very quickly to keep pace with the high amount of new discoveries.

[desiresjab]

Prove that for n ε N, (n!)! is divisible by n!(n-1)!.

Here is a compact little gem. Divisibility is a fundamental concept in number theory, so it is always good to practice. At the moment, I do not see an answer, but there should be a fairly simple way of showing that the denominator will always be smaller than or equal to the denominator and is a whole number, which is what this problem asks for in plain English. Probably a factorization which allows something to be cancelled on the bottom. It has to be put into such a form first.

[desiresjab]

All the fancy math in the world is only fancy accounting. The speed of a falling object towards earth owes x to mass, t to time spent falling and r to air resistance; current owes so much to voltage and so much to resistance, and they have a relationship which can be stated in symbols:

I=E/R. Current equals voltage divided by resistance. Accounting.

Sometimes events in the universe are so fancy that it takes highly fancy accounting to account for them. Sometimes specific new accounting tools have to be developed to account for something, but it is always just accounting. Look how fancy and specific some of the tools of mathematicians are. They could not be used to do your bookkeeping. They do the bookkeeping for various aspects of the behavior of objects in the universe which you cannot use everyday arithmetic for.

The awesome thing is, sometimes the accounting tools are discovered and developed before the phenomena themselves are observed!

As for equations, the same old usual suspects apply across every field. An extraordinary number of phenomena around us are described by ordinary first degree equations. The equations that account for other phenomena are more complex and of a higher degree. Quadratic equations bit off another big hunk of what is understandable and can be accounted for.

Even the eight-miles-to-a-side matrices used in the attempts by physicists to simulate or reproduce consciousness, are but attempts to account for consciousness the way -16t2 accounts for the position of a falling object.

[desiresjab]

When Newton fully generalized the binomial theorem for the first time, what was he up to? This provides one of the best examples of the accounting tool being developed before the phenomenon it accounts for is even suspected, in this case phenomena many times over. The bionomial theorem is one of the most universally applicable from biology to backgammon to sociology. To Newton it was an exercise in arithmetic. He could have had no idea of the landslide of applications to come, even if he felt certain there would be some applications somewhere or th'other.

I am always comforted that at least a few of our brightest are doing pure mathematics. We are not the first. We could ask ourselves a big question. How did the ancients do it?

How did east Injun mathematicians before the time of Christ come up with at least a way to calculate individual binomial coefficients? We cannot be sure they did not already have the general method Newton developed. An even greater question is: Why? Why would they do it? It is almost certain they had no application. In fact, it looks like humans had to develop the ideas behind the binomial theorem at least three times in our history, forgetting it twice, apparently because it had no use.

The same thing that drove Newton must have driven those second century B.C. east Injuns and twelfth century Moslems who were onto the theorem before. It was about the behavior of numbers. I think they were simply playing at a form of play that requires a lot of concentrated mental focus.

Now, if IQ measures anything it measures the ability to do this kind of abstract accounting. It seems unlikely to me that ancient east Injuns would have measured out at an average (over the entire population) higher than modern Injuns on IQ tests, an average we know is lower than the west, lower than China and lower than Ashkenazi Jews, yet they did get this job done, i.e. the binomial theroem, the coefficients of its expansion.

This tells me human populations with an average as low as 85 will get the job done. They will occasionally produce geniuses enough standard deviations to the right to produce mathematical acheivement. This lower average was perhaps unable produce enough geniuses to sustain mathematics through startup technologies it could be the accounting firm for.

However, we should not kid ourselves that the average European slopping through mud on his way to mass in the fourteenth century would have an IQ of 100 by modern standards, if we had a time machine and could test them. In point of fact, this person is likely right on par with a real dummy these days.

But this European had one mighty advantage that none before him had possessed--Gutenberg. Ideas could be disseminated as never before. Without Gutenberg mankind might have forgotten the binomial theorem a third time.

[YesNo]

Prove that for n ε N, (n!)! is divisible by n!(n-1)!.


I was thinking of using a mathematical induction argument on this, but I only got as far as the base cases for n = 1, 2 and 3. However, I don't know how to get the inductive step to work algebraically. That is given that the statement is true for n, how can I algebraically manipulate the factorials so that it has to be true for n + 1?

So, I looked up the problem. Here is a combinatorial solution that I am not totally convinced of, but I assume is correct: http://math.stackexchange.com/questions/1602292/prove-that-n-divisible-by-nn-1

What I liked about that solution was the way the person who answered it rewrote the problem as a fraction and then reinterpreted it as a combinatorial problem that must have an integer as a solution. Hence the numerator is divisible by the denominator.

One might also be able to use the gamma function which would allow one to rewrite the problem as integrals. But I didn't see how that would make the inductive step any easier to calculate.

[desiresjab]

I was thinking of using a mathematical induction argument on this, but I only got as far as the base cases for n = 1, 2 and 3. However, I don't know how to get the inductive step to work algebraically. That is given that the statement is true for n, how can I algebraically manipulate the factorials so that it has to be true for n + 1?

So, I looked up the problem. Here is a combinatorial solution that I am not totally convinced of, but I assume is correct: http://math.stackexchange.com/questions/1602292/prove-that-n-divisible-by-nn-1

What I liked about that solution was the way the person who answered it rewrote the problem as a fraction and then reinterpreted it as a combinatorial problem that must have an integer as a solution. Hence the numerator is divisible by the denominator.

One might also be able to use the gamma function which would allow one to rewrite the problem as integrals. But I didn't see how that would make the inductive step any easier to calculate.

I cannot ever remember using induction successfully with a problem other than homework assignments long ago. It is not one of the techniques. The notation is difficult and requires brain-wracking precision. It is not one of my techniques, which I will eventually regret.

Some of these problems come from textbooks, and the student is expected to use the same techniques studied in each section to solve the problems. You might have noticed I seldom do this. I use my own bag of preferred tricks on almost everything, picking up additional information as I go.

One thing I find torturous is long, algebraic manipulations looking for derivations. This was Euler's style. I avoid that when I can. Induction would be a good example. Several times in my solutions, however, I do point to induction as the final step of a process without actually doing it, if I am pretty sure of myself.

* * * * *

Both are merely factorials, it should suffice to show the denominator is smaller than or equal to the numerator, since any smaller factorial divides a larger one.

n!(n-1)!=1n! 2n! 3n!...(n-1)n!

(n!)!=n!(n!)!Π(n+1) i)

n!((n!)!Π(n+1)i)
1n! 2n! 3n!...(n-1)n!

Obviously, a factor of n! could be canceled in this fraction. That does not yet show the top is at least equal to the bottom.

The concept is the one we used before with this notation. The numerator consists of factors which are all higher multiples of all the factors of the denominator.

Look at the fraction again. The numerator is making itself (n!)!Π(n+1) i) copies of n!, leaving at least one copy after cancellation of every factor in the denominator, since .

Note: The subscript and superscript on the multiplication operator capitol Pi are not powers and only show where to begin and end the multiplication sequence. I am unable to put both of them on the same side of the operator, or I would.

[YesNo]

Even the eight-miles-to-a-side matrices used in the attempts by physicists to simulate or reproduce consciousness, are but attempts to account for consciousness the way -16t2 accounts for the position of a falling object.

I was thinking about this while we were walking around Oak Park looking at early Frank Lloyd Wright houses. I don't think it is possible to reproduce consciousness with a mathematical structure.

One might be able to simulate some aspects of consciousness or find correlates of consciousness in the brain, but at some point this fails to reproduce consciousness. The reason is because the mathematical structure, based on determinism and randomness, cannot make a choice. However, consciousness could be characterized as having an ability to make a choice, no matter how limited. That implies that the property of making a choice cannot be mapped to a deterministic-random structure.

This property of making a choice is not the "hard" problem of consciousness. That would have to do with subjective experience and "qualia". That so-called hard problem limits consciousness to sentient animals (and perhaps plants) who can be expected to have subjective experiences.

This is a harder problem of consciousness and allows anything, including indeterministic quantum reality, to be conscious in its own way with or without qualia.

Of course the counter argument is that nothing is able to make a choice which is why the indeterminism of quantum reality is shocking. If one defines choice as being outside a deterministic/uniformly random structure, then that quantum indeterminism can be interpreted as making a choice. But we don't have to deal with reality directly at that level. We see it as larger clumps where it behaves more predictably. We can make tables, chairs and computers out of it and think it is all mathematically predictable because those tables, chairs and computers as tables, chairs and computers don't make choices. They are not conscious as such.

[desiresjab]

I was thinking about this while we were walking around Oak Park looking at early Frank Lloyd Wright houses. I don't think it is possible to reproduce consciousness with a mathematical structure.

One might be able to simulate some aspects of consciousness or find correlates of consciousness in the brain, but at some point this fails to reproduce consciousness. The reason is because the mathematical structure, based on determinism and randomness, cannot make a choice. However, consciousness could be characterized as having an ability to make a choice, no matter how limited. That implies that the property of making a choice cannot be mapped to a deterministic-random structure.

This property of making a choice is not the "hard" problem of consciousness. That would have to do with subjective experience and "qualia". That so-called hard problem limits consciousness to sentient animals (and perhaps plants) who can be expected to have subjective experiences.

This is a harder problem of consciousness and allows anything, including indeterministic quantum reality, to be conscious in its own way with or without qualia.

Of course the counter argument is that nothing is able to make a choice which is why the indeterminism of quantum reality is shocking. If one defines choice as being outside a deterministic/uniformly random structure, then that quantum indeterminism can be interpreted as making a choice. But we don't have to deal with reality directly at that level. We see it as larger clumps where it behaves more predictably. We can make tables, chairs and computers out of it and think it is all mathematically predictable because those tables, chairs and computers as tables, chairs and computers don't make choices. They are not conscious as such.

Good post. But then, I did not say I thought it was possible or impossible, just that that is what the boys and girls with the eight-milies-on-a-side matrices were trying to do, some very fancy accounting.

Because we are unable to define consciousness adequately it would be harder to reproduce and know if we had. Movies are full of rebellious machines.

Personally, I would not be surprised if some connection between consciousness and prime numbers exists. That is why, of all the unsolved problems in elementary mathematics, I consider the Goldbach conjecture the most important, for it is unlikely that it would be proven without shedding much light on the additive properties of primes, something almost nothing is known of.

I will post a problem now my bag of tricks seems insufficient for. I will have to pick up some new tricks, unless I notice a way I have missed. The problem looks simple enough.

Let n=n1+n2...+nk where n are nonnegative integers. Prove that the quantity

n! ............................... Is an integer.
n1! +n2!...nk!

[desiresjab]

Something is wrong! I must have stated the problem wrong. Maybe it asks for under what conditions, because 5+2=7, but 5!+2! does not divide 7!

I will have to recheck the wording of this problem, since a counter example is so available.

[YesNo]

I checked it doesn't work with 5, 2 and 7 using a Jupyter notebook as you noted. However, 3, 4 and 7 works.

I think the following is true: The sum of the factorials of two consecutive positive integers divides the factorial of their sum.

It doesn't work for 0, 1 and the sum 1 so the integers have to be positive.

A more general statement might work: The sum of the factorials of consecutive positive integers divides the factorial of their sum. I don't have a proof for this. It is just a conjecture. It looks like a nice pattern.

I wonder if the following works? Does the sum of the factorials of consecutive primes divides the factorial of their sum? That would be another nice pattern. If it is true.

[desiresjab]

I have to find the question and read it again. I think it is related to De Polignac's formula, because that is the section I found it under. It was also a solved problem.

Those other patterns are interesting. The answer I have to give to them is: I don't know. It is the kind of stuff you might find in the sources I am consulting.

In the meantime, here is a problem that is a killer. It might not look like it, but it is.

Prove that 7 divides (22225555+55552222).

It cannot be done what seems the intuitive way of reducing everything in sight (mod 7).

That is the scary part. I always thought you could reduce with impunity any time you felt like it or it was handy. Apparently that is not the case, otherwise 34+43 would be divisible by 7, but it is not. This simple looking problem shows the traps involved with congruence theory. I wish I knew why 34+43 does not work, but the truth has not sprung on my brain yet.

I did not solve this problem, I looked at the answer. It took a while before their answer made sense, because I have a stubborn brain. I see why their answer is correct, but I do not see why the simpler version is incorrect. It is another problem in itself. I will not be able to let go of this until I understand it. I foresee torture.

I will let you look at this one for a spell while I go look up the other one.

[desiresjab]

I made a horrible oversight. No wonder the proposition was not true as stated. All the separate nk's sum to n all right. but then he asks if the product of those numbers (not the sum as I printed) divides n! and is an integer. It now looks easy, but the Polignac Formula and method they are using is complicated.

Now that I have righted the mistake in the question itself, I am wondering if some of my old tricks might solve this one easier than all of De Polignac's torturous Eulerian algebra. Will get back.

[desiresjab]

Prove that if n1+n2.... +nk=n, then

......n!.........
n1! n2!... nk!

is an integer.

That is the problem stated correctly. I does not seem like a monster now. Maybe it is, though. Later.

[YesNo]

Not having the plus sign in the denominator makes this look easier. The first n1 easily cancels.

For my last conjecture about consecutive primes, a counterexample would be 5 and 7 showing the conjecture is false.

Edit: What you are asking is if the coefficients of the terms in the "multinomial theorem" are integers. Here is some discussion of it: https://en.wikipedia.org/wiki/Multinomial_theorem

They would be expected to be integers since they are involved with counting elements in sets. That would mean that the fraction above is an integer.

[desiresjab]

I have been away for a few days.

What you say about the coefficients is quite likely true. While I was away I only thought about the problem in my head, no paper or computers. My goal on many such problems is to reduce the solution to a simple diagram and a little algebra. Just how elementary can this grade of a problem be made to look? Look how far Eisenstein reduced QR with his simple diagram. This is a much simpler problem. The diagram will be a lot simpler. I already know the diagram. What I have left to do is the little bit of algebra. I am rather tired after my drive. It might be tomorrow.

[desiresjab]

x+y=n...n-y......n-x

*
*
*
*............*
*............*..........*
*............*..........*
*............*..........*

n............x..........y


x+y has as many units left over at the top as n-x has total units. Those three large numbers as a product will be greater than what the three smaller numbers in y! produce. This proves the numerator is always greater than or equal to the denominator, a condition for an integer.

n will always have y units left over at the top after it matches every unit of x. In the case above, n has three units left over, i.e. three consecutive numbers. Being of equal length as y, that stretch of consecutive integers is forced to contain every factor of puny y factorial, always getting the first one of "1" for free. Getting that first factor of "1" for free guarantees that it will not miss any factors of y factorial.

Now, if we imagine placing y directly on top of x, we see that 5 is a multiple of 1, 6 is a multiple of 2 and 3, and 7 is a prime. 7 had to be a prime, because there were no more factors, which had just been used and could of course not be used again the very next number.

This proves they will always divide evenly.

This "proof" is intuitively strong, I believe, but of course it is algebraically weak. I could use fewer words, but I prefer to be as clear as possible for anyone making an effort to follow. A modern algebraist would not use many words at all. Wordy proofs are sort of antique in nature. But the fact is, if you can show it to yourself verbally, you will truly see it.

Suppose I had chosen 6 and 1 for the additives, instead of 4 and 3. That is the easiest case of all, because 1! affects nothing, and we can plainly see that 6! is a factor of 7!.

I am experimentally certain one could go a step further and show that (4!)(3!) is a factor of (5!)(2!), (5!)(2!) is a factor of 6!... right on up the line as far as it goes. I can see that but I have not demonstrated it.

[YesNo]

I trust algebraic proofs as much as I trust the results of a computer program. There could be something wrong in either that has little to do with the original problem. Being convinced is the intuitive part of the proof.

One way that I think about factorials is to view them as the number of ways to order n objects. In the first position one could have n choices. After that choice there are n - 1 choices for the next position and so on all the way down to the last piece and there is only one way to order it. Multiply all those together and you get n! For the multinomial formula coefficient the numerator is the way to order n objects. The denominator is the way to order subsets of those objects separately.

I was looking at Carmichael's "Theory of Numbers" and he approaches the problem by considering the highest power of a prime p in n! which I am not familiar with but sounds interesting. Then if the highest power of a prime in the denominator is less than or equal to the highest power of the prime in the numerator the fraction would be an integer. This might be more useful than the multinomial coefficient approach since it could answer more questions.

I agree that it is good to find multiple ways to solve something. The easier, the better.

[desiresjab]

I trust algebraic proofs as much as I trust the results of a computer program. There could be something wrong in either that has little to do with the original problem. Being convinced is the intuitive part of the proof.

One way that I think about factorials is to view them as the number of ways to order n objects. In the first position one could have n choices. After that choice there are n - 1 choices for the next position and so on all the way down to the last piece and there is only one way to order it. Multiply all those together and you get n! For the multinomial formula coefficient the numerator is the way to order n objects. The denominator is the way to order subsets of those objects separately.

I was looking at Carmichael's "Theory of Numbers" and he approaches the problem by considering the highest power of a prime p in n! which I am not familiar with but sounds interesting. Then if the highest power of a prime in the denominator is less than or equal to the highest power of the prime in the numerator the fraction would be an integer. This might be more useful than the multinomial coefficient approach since it could answer more questions.

I agree that it is good to find multiple ways to solve something. The easier, the better.

Could not agree more. I have looked at the link now and what we have here is the same recipe taken out of context. Really, every time one is working with factorial formulae it would be wise to check for this connection.

Carmichael may be using Polignac's formula. That is what it does, I believe.

Factorials are beasts. It is easy enough to see what they are and what they designate for counting (very good description of their role in the multinomial, by the way), but when you want to compare them to something like a power it becomes sticky. One reason I appreciate Wilson's theorem so much is that it relates factorials to powers. That is cool.

It is time to move on. My own urge is toward Borcard or Goldbach. I have learned a lot from Brocard's problem. The Goldbach conjecture, however, is difficult to make even a quarter inch of progress on. I suppose a place to start would be with Ramanujan's amazing formula for summing additive partitions. No one believed such a formula was even possible.

When looking at problems which minds like his have already considered, one can only hope they were in a great hurry that day.

[desiresjab]

If all the mysteries of a simple number line were solved, the human race might become more advanced than the alien civilizations we envision in our fiction. Complete power over a number line would be god-like. The solving of some problems would contribute mightily to acheiving this power. The Reimann conjecture is high on the list. I think of Goldbach's conjecture as very important.

The twin prime conjecture is one that fascinates people. We know something about twin primes from studying QR on here for so long. Since twin primes are a mixed couple in terms of 4n+1 and 4n+3, their combined Legendre symbols will be (1)(-1)=-1.

Working on unsolved problems there is generally little or nothing to report. Many a doctoral dissertation in math has explored some tiny area of these problems. Any progress is counted a success.

[YesNo]

I didn't see a reference to Polignac, but it looks like the same argument after checking Wikipedia.

We should probably just assume the Riemann hypothesis is true. Then derive some consequences from it and check if they are true or not. Inadvertently someone might discover that it is false by finding a consequence of it that is false.

I am still working on the Sierpinski problem. Or rather, I think about it off and on. There isn't a lot of work getting done.

[desiresjab]

The Reimann conjecture may be overrated in terms of what the impact of its solution would be. I do not know for sure. Researchers can already assume it is true and take it from there. That is done with a lot of propositions. I believe it is usually called the "weak form," if a proof in the chain is missing. Can anyone say what else would be immediately true if the Reimann hypothesis were true? Perhaps it is a hugely significant problem on its own.

It seems to me that a proof of Goldbach's conjecture would almost necessarily lead to a fuller understanding of the additive nature of primes. Is there even such a theory to be had? What the mathematical world needs right now is Ramanujan. If Goldbach is solvable, I think Ramanujan had the best chance. Had he lived beyond his twenties it might be solved now.

There are two biographies of mathematicians I can recommend. The Man Who Loved Only Numbers and The man Who Loved Only Infinity, about respectively Paul Erdos and Ramanujan. They are both fantastric reads in my opinion.

I turned back to Brocard's problem and immediately made progress on an area of it I was not equipped to several years ago. I am going to hang with it again for a while and see if anything else pops out for me. Originally, I said all I wanted to do was see if I could get to the same vista where Ramanujan had stood before he gave up on this problem. Perhaps I have done that, for the present stage seems impassable. But of course no one knows where Ramanujan stood, so I must trudge onward.

[YesNo]

According to Wikipedia, Brocard's problem (https://en.wikipedia.org/wiki/Brocard%27s_problem) is related to the "abc conjecture": https://en.wikipedia.org/wiki/Abc_conjecture. However, I don't see the connection at the moment.

This article searching for solutions was also cited: http://www.math.uiuc.edu/~berndt/articles/galway.pdf

There is also some discussion on stack exchange: https://math.stackexchange.com/search?q=brocard%27s+problem

[desiresjab]

I can see where it might relate. Almost everything in number theory seems to relate anyway, and it is still Diophantine equations. They also use Legendre symbols from QR in the research. Since so much relates, that is what the smart guys have been doing interrupted for a couple of hundred years, which is exactly the reason and the only reason we know the importance of particular unsolved problems in immediately solving other unsloved problems. It is amazing how the brilliant boys and girls keep knocking chips off these problems until someone gets a finished sculpture. They do it with brilliance in various areas of high math. The Japanese who claims a proof of the abc conjecture invented entirely new methods, from the reports, which went far outside number theory, perhaps any theory. I still have to look at it to see if I can make monkey of anything he says.

In the meantime, I trudge along. I am within sight of a new perch from which to see the problem.

[desiresjab]

Most of the progress I thought I made on Brocard was illusory, further thought showed, though there was a bit of an increase in understanding. Still, I am essentially where I was last time I left on working on it after all. I feel a real shortage of tools. The problem with known tools is that everyone who is greater than me has tried them.

[YesNo]

There is a proof of the abc conjecture that people are trying to verify although the proof is very long: http://phys.org/news/2016-08-abc-proof.html

If that proof convinces others then understanding it may be more important than Brocard's problem. However, I am still trying to understand why the abc conjecture is relevant to Brocard's problem.

[desiresjab]

There is a proof of the abc conjecture that people are trying to verify although the proof is very long: http://phys.org/news/2016-08-abc-proof.html

If that proof convinces others then understanding it may be more important than Brocard's problem. However, I am still trying to understand why the abc conjecture is relevant to Brocard's problem.

I am aware of that proof. It uses properties of number classes that are outside conventional mathematics. Elliptic curves are still generating solutions to unsolved problems. These guys are using group theory, abstract algebra, moduli and Elliptic curves and topology. His proof uses almost no calculus, so is elementary, which means anything but simple, as I have been trying to convince people for a long time.

Anyone who is not an expert in the above fields may as well forget actually understanding his proof. I know that leaves me out. Of course I will poke around with it anyway.

[desiresjab]

Along with all sorts of other strange mathematical objects and operations, such as rings, ideals, kernels and cosets, he uses something called theatres. This involves treating certain groups as if they were abstract topological fields without the labeling. Mochizuki provides this new labeling. There is a distortion during the operations of multiplication and addition in the ring that he measures and accounts for outside the group.

It does not pop out at me why the abc is linked to Brocard. But everytime I read about this stuff in depth I learn something new. For instance, I finally understand clearly what an algebraic integer is. Ideal rings are much clearer now too. Immediately I see a connection of theirs to Brocard. Sorry I have to be so cryptic. You see, I am slightly luny and still intend to solve this thing before the abc conjecture does it sweepingly!

It is truly amazing to me how far number investigations have been taken. Now Mochizuki has added some new beasts to the zoo.

[YesNo]

Based on the likelihood that the abc conjecture is true, then the number of solutions to Brocard's problem is finite. The only question remaining is to either show that all the solutions have already been found or to find another one which should have more than 20 digits. Another contribution would be to find an algorithm faster than computing quadratic residues to check that a solution does not exist.

[desiresjab]

Based on the likelihood that the abc conjecture is true, then the number of solutions to Brocard's problem is finite. The only question remaining is to either show that all the solutions have already been found or to find another one which should have more than 20 digits. Another contribution would be to find an algorithm faster than computing quadratic residues to check that a solution does not exist.

I have an approach to Brocard I have not seen per se. An instinct tells me it is solveable and that I am on a good course. I chose this problem long ago because its shape was pleasing to me. I like factorials. Right now I am at an impasse, looking for a way around. One is always at an impasse on unsolved problems, then suddenly a little progress is made. Often these seeming advances are illusory, the result of a mistaken notion one realizes later. So progress really is slow, but that makes any advance exciting.

[YesNo]

It is always good to come up with a simpler solution than the ones known.

Regarding the abc conjecture, I can see how the form relates to Brocard's problem with is n! + 1 = m2. Here a = n!, b = 1 and c = m2. The product of all distinct primes in the product abc is close to c, that is the product of the distinct primes in this product (n!)(1)(m2) is about m2. I can see that m would be larger than n, but why wouldn't it have some primes in common with n!?

[desiresjab]

It is always good to come up with a simpler solution than the ones known.

Regarding the abc conjecture, I can see how the form relates to Brocard's problem with is n! + 1 = m2. Here a = n!, b = 1 and c = m2. The product of all distinct primes in the product abc is close to c, that is the product of the distinct primes in this product (n!)(1)(m2) is about m2. I can see that m would be larger than n, but why wouldn't it have some primes in common with n!?

Because m2 being one greater than n! can share no factors with it. Euclid used the same idea in his proof of the infinitude of primes.

You set Brocard into the form correctly. I still cannot see why it would have an impact on Brocard's problem, because the task there is to show if the difference between a square and a factorial can ever be exactly 1, other than the three known cases. Researchers are trying to bound the function from above.

There is a relationship between factorials and triangular numbers I find fascinating.

(2n)!=2n k=1∏n T2k-1.

Stunning!! What this really says is: factor out n powers of 2 from (2n)!, then multiply it by all the oddly labled triangular numbers which have been multiplied together all the way up to one less than twice the value of n, and you will be back at 2n factorial.

[YesNo]

Right. I can see now that m2 must be relatively prime to n! because any factor of n! that divides m2 must also divide 1.

The relationship with triangular numbers is interesting.

[desiresjab]

Right. I can see now that m2 must be relatively prime to n! because any factor of n! that divides m2 must also divide 1.

The relationship with triangular numbers is interesting.

It seems a queer relationship offhand and I have to investigate it.

[desiresjab]

It seems a queer relationship offhand and I have to investigate it.

I have some of it already. It is tying in with what I am doing and what I was almost able to see last time I quit the problem.

But still, why on earth is there a connection between factorials and the product of these oddly labled triangular numbers? I have not seen that part yet.

You have been so attentive and perspicuous I have to give you something. The hypothetical very large factorial in Brocard's problem has to be equal to 8 times some single triangular number. That fact should not have eluded me for so long, for I have long had everything I needed to realize it, but I am rather slow at this business. I miss things a math teacher would see routinely.

Can you figure out why the factorial must be equal to 8 times some single triangular number for there to be another pair of Brown numbers? I think you can. I believe it is within your range, from what you have shown me.

[YesNo]

This might work as a way to show that a solution of Brocard's problem is 8 times a triangular number.

Let n and m be integers such that n! + 1 = m2. Then n! = m2 - 1 = (m - 1)(m + 1).

Brocard's problem does not work for n = 1 and so n > 1. Notice that m must be odd because otherwise for n > 1, n! and m have a common factor. That means m - 1 is even. Let 2r = m - 1. Then 2r + 2 = m + 1.

Then (m - 1)(m + 1) = 2r (2r + 2) = 4(r (r + 1)). Multiply this by 2/2 = 1 to get 8(r(r + 1)/2).

A triangular number, T, is the sum of all positive integers less than or equal to that number. That sum can be written algebraically as T(T + 1)/2. This shows that r is the desired triangular number since it has the algebraic form of a triangular number.

[desiresjab]

You took your own way, which is to be expected from different minds. I arrived at this idea from another factorization of the factorial.

Observe: Incidentally (or perhaps not), both 120 and 720 are factorials and triangular numbers. Triangular numbers can be square numbers sometimes, as well. But they can never occur together because a factorial can never be a square, which has an easy proof but is intuitively clear to anyone who thinks about it hard enough. I have been around these exact issues for a while now and I have to an extent humanized them into normal language as far as I have understood them. Some algebra is necessary, as this is not the 12th century.

Now to the heart of it.

8[(n)(n+1)]/2= 4[n(n+1)]=

4n2+4n=2n(2n+2).

Consider: Every whole number is a square root. Every odd integral square root lies between two numbers of the form 2n and (2n+2) for any n.

When two numbers on either side of a number x are multipled together the product is always one less than the square of x. (x-1)(x+1)=x2-1.


The Secret: It does not have to be proven that such and such a square can or cannot exist. The kernel of the problem is whether a very large factorial can ever be factored precisely into the form 2n(2n+2). If it cannot be, then such a breed of square can never repeat itself beyond the three known examples.

The simple power of this factoring approach may be missed. Our factorial is huge because n itself is very large, so factorialized n is really large.

Because they are only two apart on the number line, this makes 2n and 2n+2 next door neighbors in the ordered set of even numbers. It also makes one highly even and the other barely even. These two factors that are expected to produce a factorial, can share only a single factor, i.e. one factor of 2. They are not far enough apart to share anything else.

Can two large next door neighbors in the world of evenness ever contain between them precisely and only all the factors of a factorial? Can two such neighbors exist? If a factorial cannot be factored this way it cannot meet the qualifications of Brocard's problem.

There is quite a bit more. We cannot determine which of the factors is highly even and which is barely even. I call them super factors and SF for short.

* * * *

The possibilities for the last two digits of the SFs are listed below vertically. The middle number might be paired with either factor in its group. Remember, the SFs are very large numbers themselves and these are only the last two digits of the possible SFs. Observe that the digits of the long factors in each group are identical copies of one another except for their last two digits, the one's and ten's positions. The same is true for the SFs in group B.

A
xpnuh000...02
xpnuh000...00
xpnuh000...98

B
jbvfg...52
jbvfg...50
jbvfg...48

In the case of B all the factors of 5 are with the middle factor, but it has only one factor of 2 to make zeros with, and both its possible mates 52 and 48 are highly even without any 5.

In the case of A all the factors of 5 are with the abundance of 2's in the middle SF, and the traditional tail of zeros is observed. The highly even number is in the middle this time and both its possible mates 98 and 02 are barely even.

We can also observe now that our hypothetical integral square root of (n!+1) which lies exactly between and contiguous to the SFs, must have last digits of 49, 51, 99 or 01.

It is hard to say which of the numerous collected facts and observations might next help to further understanding. The proof does not have to be about squares at all. It could be about triangles.

[YesNo]

That is an interesting way of looking at the problem, considering one factor being highly even and the other barely even. Together those two factors should equal n! which has zeros in the units and tens positions after n is greater than or equal to 10 so a factor of 100 is in n!. I can see how one of those two numbers should have many zeros in them. I suspect there should be n/5 number of zeros in n!.

There are trivial cases where a factorial could be a square, such as 0! and 1!, but as soon as one gets larger than 1 there are primes involved. The largest prime less than or equal to n would not have another prime like it in n! to pair up with to make a square out of that number.

You mentioned the factorials that are also triangular numbers like 120 and 720. I wonder when would there be an r such that for some n, n! = r(r+1)/2. That would be like saying the product of positive integers less than n is equal to the sum of positive integers less than r.

[YesNo]

While bicycling in the neighborhood, it occurred to me that you could do something similar to the highly even and barely even with the Brown numbers. As you mentioned, assuming n! = (m - 1)(m + 1) for n > 1, then both m - 1 and m + 1 are even, but since they have a difference of 2, one of them has only one factor of 2 and the other has the rest of the factors of 2 that are in n!. Now take any other prime p > 2 in n!. All powers of that prime will be in either m - 1 or m + 1. That prime cannot be shared across those two factors since they have a difference of 2.

For the three known solutions,

4! = (4) (6) = (22) ((2)(3))
5! = (10) (12) = ((2)(5)) ((22)(3)
7! = (70) (72) = ((2)(5)(7)) ( (23)(32))

One way to show that the solutions are finite is to try to see if this additional constraint forces there to be no solutions after a certain point.

[desiresjab]

That is an interesting way of looking at the problem, considering one factor being highly even and the other barely even. Together those two factors should equal n! which has zeros in the units and tens positions after n is greater than or equal to 10 so a factor of 100 is in n!. I can see how one of those two numbers should have many zeros in them. I suspect there should be n/5 number of zeros in n!.

On the right track, but you have to remember to include those factors in the numbers less than or equal to 100 which can be divided by 5 more than once, 25, 50, 75, 100. There is a formula for this using sigma and the floor function.


There are trivial cases where a factorial could be a square, such as 0! and 1!, but as soon as one gets larger than 1 there are primes involved. The largest prime less than or equal to n would not have another prime like it in n! to pair up with to make a square out of that number.

This is precisely the reason


You mentioned the factorials that are also triangular numbers like 120 and 720. I wonder when would there be an r such that for some n, n! = r(r+1)/2. That would be like saying the product of positive integers less than n is equal to the sum of positive integers less than r.

I made a computational mistake regarding 720. It is not a triangular number. In fact, it is conjectured there are no more triangualr factorials after 1, 6 and 120, but this conjecture remains unproven and in about the same state as Brocard, generating lots of phd dissertations in mathematics as brains gnaw away on various corners of it.

Apparently, the question of are there anymore triangular factorials is related to a more general and deeper problem which, if solved, would suddenly lead to the solution of this problem and perhaps Brocard's too. That is, a general solution to this:

n!=a!b!c!...

Some examples are known, and brains are gnawing away.

* * * * *

Triangulars were likely a dead end for Brocard's problem, unless one could show Brocard's factorial has to be triangular to fulfill the problem. Then there would be a connection.

What is still of major interest is the connection between triangulars and factorials in this identity:

(2n)!=2n k=1∏n T2k-1.

Major interest. What does it say? I think it says that every even factorial is the product of some powers of 2 and some odd labled triangular numbers, which is just a jewel.

(2n)! can be manipulated as follows.

Let me get back. I have to recheck my notes for errors...

[desiresjab]

(A.) (2n)!=2n k=1∏n T2k-1.

(2n)!=2(n)· (2n-1)· 2(n-1)·(2n-3)·2(n-2)...·(2n-2n+1)

Note that 2 can be factored out of every other term. Since there are 2n terms there are n terms from which 2 can be factored. We now withdraw n factors of 2 and place them in front of cap pi, just as in the identity (A.) above, leaving:

(2n)!= 2n[n·(2n-1)·(n-1)·(2n-3)·(n-2)...·(2n-2n+1)]=

Looking now exclusively at the factors above in brackets, all we have to do is rearrange the terms to see what is going on.

[(2n-1)·(2n-3)...·(2n-2n+1)...·(n)·(n-1)·(n-2)...·(n-n+1)]

Simply amazing. What we now have is this:

(2n)!=2n[(n!)·(2n-1)!!].

Caution: Double factorials are not the same as nested factorials. The double factorial means to multiply together all the odd numbers from a certain point down to 1.

I am not yet seeing in the above bold forumla the connection to triangular numbers I was hoping would jump out at me, but the double factorial coming into play is a bit fascinating, as now the factorial and the double factorial multiplied together by definition must have the same prime factorization as a string of odd labled triangular numbers multiplied together, since the two products are the same. This fact needs to be drawn out of the algebra, if I can find the manipulations. I need the manipulations that break what is inside the brackets down to a visible (recognizable) triangular connection that we know is there, again, by definition, from the identity itself.

Note: It is always possible my manipulations, though correct, went the wrong way to uncover the triangular connection, leaving us with a curiosity that is merely interesting.

[YesNo]

This is the first I've heard of double factorials, but they make sense: https://en.wikipedia.org/wiki/Double_factorial

[desiresjab]

This is the first I've heard of double factorials, but they make sense: https://en.wikipedia.org/wiki/Double_factorial

I knew about double factorials, but this is the first time I ever ran into one out in the wild. I had been wondering if they were mere toys or relevant to anything. I should have known. In math everything is always relevant on some level because everything is connected. The road that leads ever on is sometimes hard to find.

All these new connections tell me something else--I am nowhere near the last view of Ramnujan on Brocard's problem. Wherever he stopped and gave up, I have not arrived to yet, if ever I will. To do so was a stated goal when I started the problem.

* * * * *

A friend of mine wonders if the world did not in the long run lose out by Ramanujan's journey to England. I had always assumed that he picked up enough mathematical formality from Hardy and associates at Cambridge to untangle his genius from unnecessary quests and his few incorrect notions to make it a great blessing that he undertook his voyage. There was great value to us living now in Ramanujan's being shepherded toward some of the most important mathematcial problems of his time and of all time and brought right up to specs on the frontiers of research by several of the world's preeminent number theorists who recognized the Injun's awesome abstract power.

The greatest mathematicians always best everyone of their time, doing here and there what was impossible for other great talents. They come up with formulas long thought to be impossible, solve a problem from antiquity or invent new tools. Most of Ramanujan's tools were a personal thing he could not understand himself, so his forte was producing amazing formulas that also looked amazing. When you see them, you know you were not made to go there. You will consider yourself fortunate if years of study garners a half decent understanding of just a few of the identities he pulled from nowhere, directed, he said, by a household goddess.

Ramanujan's formulas are too long and difficult to try to set up on the house word processor. People will have to take a look for themselves. There was a shorter one that was amazing, but I cannot find it.

I guess it is still an open question whether we got lucky or unlucky by Ramanujan's trip to England.

[desiresjab]

I have a small improvement to announce. Improvement sounds much better than correction. In the formula

(2n)!=2n[(n!)·(2n-1)!!d, n+(1 or 2)],

We had to add that subscript of "d" at the end because the double factorial is not a complete one, I realized after something kept nagging me. Going downwards, it begins on (2n-1) and descends to the next odd number after n. So it begins on n+1 or n+2, depending on whether n is even or odd. A full written expression of the function would have to account for this with something like the piecemeal notation of the Legendre symbol using the oversized bracket. That is hard on the house word processor. So is defining the range of cap pi since the superscript and subscript cannot be put up properly at the same time. One or the other, but not both.

On the function above I could have used a "u" subscript for upward. Either way you write it down, the double factorial will start and end in the same place, just the notation encodes it differently.

What can be done with the beast above, I do not know. I cannot see that triangular relationship the way we have it expressed so far.

[YesNo]

That goddess, Namajiri, was a form of Lakshmi. https://en.wikipedia.org/wiki/Namagiri_Thayar

In the good old days, poets would credit muses with what they produced and I think they meant it. Ramanujan still meant it.

[YesNo]

Here is something on triangular factorials that gives an argument that only 1 and 3 such: http://math.stackexchange.com/questions/238462/when-is-a-factorial-of-a-number-equal-to-its-triangular-number

[desiresjab]

In their example the triangle has to equal the factorial, which is very easy to prove. The actual problem we are looking at and the more interesting question is whether any factorials equal any triangle, beyond the three known solutions.

It may be just eerie coincidence, but there are only three known solutions of Brocard too. With all the connections between mathematical objects, it makes one wonder. Brocard's three solutions lead back to three different numbers, though, than the three solutions to the factorial/triangular question. Whew!

That gets me to wondering further if there are numerous such unsolved problems in which there are exactly three known easy solutions and the existence of more can neither be proved or disproved plus computer computations have shown there are no more solutions of the function out to a vast input? In some famous problems does it stop at four solutions? How about five solutions? How are small numbers (or any numbers) distributed over a large number of unsolved problems of this type?

Wouldn't it be nice if all you had to do was plug in the number of known solutions, then plug in the extent to which you had searched for more solutions, and as output you would receive back an answer as to whether any more solutions existed? Someday I believe there might actually be a formula that works on similar principles that can decide which unsolved problems should not be worked on any longer. I know this goes against Godel. But Godel himself went against what were considered unassailable notions.

[YesNo]

I see what you mean by the factorial/triangular problem. I was assuming the n used in the factorial had to be the n used in the triangular number. But they could be different.

If Brocard's problem and the factorial/triangular problem are related there should be some underlying explanation for the relationship. However, the existence of only three solutions suggests that maybe there is such a relationship yet to be discovered.

There were three features about mathematics that Hilbert wanted to show: (1) completeness, (2) consistency and (3) decidabliity. Godel showed that (1) and (2) cannot be achieved, but Church and Turing showed that (3) was not attainable either. https://en.wikipedia.org/wiki/Entscheidungsproblem However, the idea of giving a computer examples of input and correct output and then asking it to create a model based on that training data and then use that to predict what the correct output would be for arbitrary input underlies "machine learning" or "artificial intelligence".

[desiresjab]

1, 3, 5...

Add this string up and you get a square. You always get a square no matter how long or short you make the consecutive string of odd numbers.

Also, all squares are the sum of two consecutive triangular numbers.

Also, cube each consecutive integer and add them up. Then add each consecutive integer again and square the result. In other words:

13+23+33...=(1+2+3+...)2

Also, the difference between the squares of two consecutive triangular numbers is a cube.

Also, since every square number is the sum of consecutive odd integers, so is the square of a triangular number.

This last one could be very important to us, since we have the sum of some odd integers in our derived equation.

[YesNo]

1, 3, 5...

Add this string up and you get a square. You always get a square no matter how long or short you make the consecutive string of odd numbers.

This one makes geometric sense. Start with 1 dot and then to get the next square add a dot to the left and the bottom plus one in the corner. In general if one has a square of side n dots, then one needs n dots on the left side and n dots on the bottom and then one dot in the corner to get the next larger square. That would be 2n + 1 extra dots added to the n2 dots already there. The previous square would have needed n - 1 dots on the side and the bottom and one on the edge or 2n - 1 dots. So the sequence of squares is the sum of the odd integers.



Also, all squares are the sum of two consecutive triangular numbers.

For this one use the closed form of the triangular number, T, as T(T+1)/2. Then algebraically add the closed form of T + 1 to it. One should get a square form.



Also, cube each consecutive integer and add them up. Then add each consecutive integer again and square the result. In other words:

13+23+33...=(1+2+3+...)2

I don't see an explanation for this one. I didn't look for counterexamples.



Also, the difference between the squares of two consecutive triangular numbers is a cube.

I don't see an explanation for this one either, but I did check it for n < 16.



Also, since every square number is the sum of consecutive odd integers, so is the square of a triangular number.

This last one could be very important to us, since we have the sum of some odd integers in our derived equation.

This last one doesn't seem to be correct. So, I will look for a counterexample. Try 4 = 1 + 3. That one works. Try 9 = 3 + 5? That doesn't work, so 9 = 32 is a counterexample.

Edit: I see it now. It is not the sum of two consecutive odd integers but the sum of the odd integers starting with 1 up to some point.

[desiresjab]

This last one doesn't seem to be correct. So, I will look for a counterexample. Try 4 = 1 + 3. That one works. Try 9 = 3 + 5? That doesn't work, so 9 = 32 is a counterexample.

Edit: I see it now. It is not the sum of two consecutive odd integers but the sum of the odd integers starting with 1 up to some point.

Sorry, I did not word that one very well.

The part that you add to get the next and the next and the next figurate number, the Greeks called the gnomon. It is actually a useful word.

With an unending proliferation of relationships, it is no wonder one runs into sudden connections when dealing with figurates.

Every other triangular number is a hexagonal number.

Every pentagonal number is 1/3 of a triangular number.

All even perfect numbers are triangular Tp with prime p.

666 is the largest repdigit triangular number (Bellew and Weger, 1975).

The latter would have been a great problem to work on, if it were not already solved. I envy the people who got to solve it. That is my kind of problem.

The sum of all the triangular numbers up to the nth triangular number is the nth terahedral number.

The sixth heptagonal number minus the sixth hexgonal number is the fifth triangular number.

It seems triangular numbers are generators for any figurate number where gnomons are applicable. If you know the right triangular numbers you can calculate all the other figurates, it does seem.

[desiresjab]

Well, I guess I am not banned.

[desiresjab]

And finally:

1=13, 3+5=23, 7+9+11=33, 13+15+17+19=43,...

This about takes the cake, or is the frosting on the cake.

The additive properties of numbers and their multiplicative properties being friendly but not related by family is part of what keeps numbers so mysterious. There is still a lot of work left to do in the additive properties. Unfortunately, none of it will be accesible to civilians the way the properties of triangles and squares are. I doubt if elliptic equations will become common to people. That is about as likely as eighth graders of the future comfortably reading Finnegan's Wake.

Most of the properties of figurate numbers seem to be additive. There is the exception that the product of sums of squares is also a sum of squares.

In the factorial problem one of the factors of (2n)! is a product of triangular numbers. This just hit me in the head that I have set that problem up wrong. We do not have an upward double factorial. That is the labeling! What we have is an upward sequence of triangualr numbers multiplied together. Excuse me again.

[desiresjab]

And finally:

1=13, 3+5=23, 7+9+11=33, 13+15+17+19=43,...

This about takes the cake, or is the frosting on the cake.

The additive properties of numbers and their multiplicative properties being friendly but not related by family is part of what keeps numbers so mysterious. There is still a lot of work left to do in the additive properties. Unfortunately, none of it will be accesible to civilians the way the properties of triangles and squares are. I doubt if elliptic equations will become common to people. That is about as likely as eighth graders of the future comfortably reading Finnegan's Wake.

Most of the properties of figurate numbers seem to be additive. There is the exception that the product of sums of squares is also a sum of squares.

In the factorial problem one of the factors of (2n)! is a product of triangular numbers. This just hit me in the head that I have set that problem up wrong. We do not have an upward double factorial. That is the labeling! What we have is an upward sequence of triangualr numbers multiplied together. Excuse me again.

I have not made a mistake at all, except to think I made one in the first place. Double excuse me, and the double factorial is still on!! Yes, I derived it that way. I am happy again, but too tired to think math tonight, perhaps..

[desiresjab]

I have not made a mistake at all, except to think I made one in the first place. Double excuse me, and the double factorial is still on!! Yes, I derived it that way. I am happy again, but too tired to think math tonight, perhaps..

What can I say? I am making mistakes left and right. Mistakes can eventually lead to the truth. Careless algebra on a word processor not meant for it, instead of doing the algebra on paper first, is the main reason for the mistakes.

Calculation instead of algebra has shown that within the brackets of the formula

(2n)!=2n[(n!)·(2n-1)!!]

we have a full double factorial instead of a partial one. I do not know if this will make a difference, but it is certainly more neat with the notation and more pure of form. I do not like mistakes, but I like this result. This time it is byond dispute, for indeed

10!=25(5!9!!).

Awfully neat and suggestive, but I still have no suggestions.

[desiresjab]

10!=25(5!9!!) and

10!=25(T1·T3·T5·T7·T9)

Really cool.

[YesNo]

I doubt if elliptic equations will become common to people. That is about as likely as eighth graders of the future comfortably reading Finnegan's Wake.

Not only eighth graders. When it comes to something like Finnegans Wake I ask myself would I rather spend my time trying to understand that or trying to understand quantum physics or gravitational theory or elliptic curves or Brocard's problem or Sierpinski's problem? If one wants an impossible task one might as well choose an interesting one. It might turn out not to be so impossible after all.

[YesNo]

10!=25(5!9!!) and

10!=25(T1·T3·T5·T7·T9)

Really cool.

I checked it in a Google sheet. It worked.

[desiresjab]

I keep asking myself, "How did they know how many factors of 2 to draw out front of the cap pi? We can reverse engineer the forumla pretty easily for 10 simply by taking a factor of 2 from each even number 10 or below. Then we notice two factors of 2 remain above 5 and between 10 , and tqo factors are missing below five. Easy replacement. But how did they know to do it? They would have no reason to do that unless they had a clue from somewhere else.

They derived their formula from something.. "Our" formula came from somewhere too. I want to get to their formula for the above one.

* * * * *

Trying to solve unsolved problems, one lives for such moments-- side trips through wonderland. One understands one will not solve the problem. In trying anyway one runs into questions asked by one's self which open up new vistas and allow learning to go on in the approximate area of the problem when progress has stalled. It happens every time. Now I have this new question to answer, and I will not stop until it is answered. Where is the connection between triangles, factorials and double factorials? Even when I know, I may not be any closer to solving Brocard, but I will have better connections among numbers.

[desiresjab]

I learned a little more. Apparently, double factorials were not even around as a function until roughly mid 20th century. Bringing those two out front of cap pi must have been someone's idea when defining the function. The function is pretty natural, if you look at

n!!=2kk! for even numbers, and

n!!=(2n)!..=..(n!)....
.......2kk!......(n-1)!!

for the odds.

Very reminiscent of what I derived from (2n)! These are attractive to me. I think they look good.

Even and odd double factorials have to be defined separately. Odd ones can weirdly be extended to negative values.

[YesNo]

It doesn't seem obvious to me either why someone would factor out those 2s. However, the exponent makes sense. There will be at least n/2 factors of 2 in n!.

However, if one is multiplying triangular numbers together to see what relationships pop up, someone might have seen that the difference between them and the n! is a curious factor of 2 to the n/2 power and guessed the relationship. The Greeks would have been thinking along geometric lines rather than algebraic ones.

[YesNo]

n!!=2kk! for even numbers


This formula makes intuitive sense. The n!! is n(n-2)(n-4)...2. Factor a 2 out of each of these n/2 factors and you get the 2n/2(n/2)!. Let k = n/2.

[desiresjab]

This formula makes intuitive sense. The n!! is n(n-2)(n-4)...2. Factor a 2 out of each of these n/2 factors and you get the 2n/2(n/2)!. Let k = n/2.

It does make intuitive sense. It is the way we ended up doing it by simply removing a factor of 2 from each even number. But we knew what we were after. We were simply trying to manipulate the formula for (2n)! by manipulating (2n)! to see where we would arrive, i.e. if we could arrive back at the forumla given for it including triangular numbers. Of course we didn't, we ended up at our factorial times double factorial thing, which obviously is that product of triangular numbers in disguise.

Here is the difficulty right now. For odd numbers, when I remove n factors of 2 from the numbers above and below n, I need a proof that the number of such factors remaining above n is exactly the amount needed to replace those taken from below n. This part is not intuitively clear to me and I believe a proof must have been provided at some point.

The key to making it intuitively clear may lie simply in studying the function on even numbers, which I have not done yet. I need a little more time with all these formulas. I am glad there is something to sort out--we cannot grow unless there is. Do I believe this digression will be helpful with Brocard? Probably not, but it is increasing our understanding of numbers, and that will.

[desiresjab]

I have to be away for a few days again. I see the way to settle the question perhaps. If the total 2's in 2n minus the trotal 2's in(2n-1) equaled n and was proven algebraically, I suppose that might do it. Go ahead and prove that while I am gone if you have a mind to.

[desiresjab]

Here is what I know for sure. The higher powers of 2 between n and 2n have to equal the lowest power of 2 between 1 and n. In other words, for n=5, after removing that first layer of 2's there is still 22 left where 8 was. 22 is exactly what was skimmed off between 1 and n. I am asking how they know this to be true in general. I am probably missing something quite basic. Maybe I will find it before I leave for a few days off.

[desiresjab]

Specifically, it needs to be shown that


i=2Σ2n↓2n/2i↓=↓n/2↓

where the arrows indicate the floor function, which is always rounded to the lowest whole value. This may prove to be a lousy way to set the problem up to find the answer. I just know it is correct. The value on the left (for 2n) is without its lowest and most numerous power of two, the value on the right (for n) is calculated only for the value of its lowest power of 2. The two should be equal.

It also says that the factors of 2 in 2n! is equal to 2n, doesn't it? In other words there are n such factors. Which is what we are trying to show. It is a nifty equality. One has to internalize that.

Well, one should prove it first, even though it is known to be true. I cannot see it intuitively, but I believe it is seeable that way.

[desiresjab]

I guess I do not see that intuitively--yet, at least. It is what I said was intuitively clear a few posts ago. But I do not see that the remaining number of 2's between n and 2n after the first layer is peeled off should always be equal to what was peeled off between 1 and n. But maybe that fact is just a consequence of the law. No, it is the law in slightly different expression.

[desiresjab]

A consequence of this would be that whenever n is a power of 2, (n)! will have (n-1) factors of 2. A power of 2 factorialized always has a value of one less than 2k. 4! is guaranteed to have three factors of 2, 8! is guaranteed to have 7, etc. People who have worked in binary know this.

[YesNo]

For even n, n!! skips every other factor one would normally see in n!. That is, it looks like this product: 2*4*6*...*n. There are n/2 factors. Now remove 2 from each of those factors. You get 2n/2(1*2*3*...*n/2) = 2n/2(n/2)!. Let k = n/2 to simplify the notation and you get 2kk!.

For odd n, n!! skips just like for the even n. It looks like this product: 1*3*5*...*n. This is the same thing as multiplying all the numbers together less than or equal to n and then dividing out the even ones: (1*2*3*...*n)/(2*4*6*...*(n-1)). But that is n!/(n-1)!!. Note that n-1 is even since n is odd. We already have a way to write n!! when it is even and so we get the following for n odd: n!! = n!/(n-1)!! = n!/(2(n-1)/2((n-1)/2)!). Let k = (n-1)/2 and this simplifies to n!/(2kk!).

[YesNo]

I was thinking about those double factorials. They are a way to split a factorial into the even factors and the odd factors in this manner:

n! = n!! (n-1)!!

For the even n, we can factor out the n/2 powers of 2 and continue the process. For example, if n is even then n!! = 2n/2(n/2)! Now take the (n/2)! and split it into even and odd factors. Repeat the process until all the factors of 2 are removed from n! leaving 2 to some power and the odd factors.

I suppose one could also do this for other primes but this notation is specific to 2. What one gets is the prime factorization of n!.

[desiresjab]

On my trip it became crystal clear where the oversight lay that was causing my disagreement with the formula. Math requires lots of solitude and good marijuana. I had overlooked the detail in my reckoning that the lower half of the double factorial was being filled in with odd small numbers. Here is the formula again:

(2n)!=2n[n!(2n-1)!!].

Only n factors of 2 are needed to produce the full double factorial with its complete lower end because that is exactly how many factors of 2 are contained in the upper interval. I have not seen it proved anywhere that the upper interval will always contain exactly n factors of 2, but having now examined the mechanics of the details, I fully accept that it does and has been proven elsewhere, probably by induction. It is one of those facts of life of numbers I did not know. There are plenty of them I did not know before but came to understand. I like this formula and I am glad we stopped to consider it. Whether it can help in the quest for Brocard awaits more solitude and marijuana. I believe there is a connection and I believe I see a hint of it through the smoke, sir. It is really a matter of connecting notations with the problem.

Maybe more vision will enable me to see why the part of the formula in brackets ( the factorial times the double factorial) is equivalent to multiplying the oddly labled traingular numbers together. I look forward to seeing that, and I suppose it is what I must stick with for the moment.

[YesNo]

I was trying to make sense out of the abc conjecture. I think I have a basic understanding of what it is trying to say primarily from this wikipedia article: https://en.wikipedia.org/wiki/Abc_conjecture

What I don't see at the moment is why it implies that Brocard's problem has only finitely many solutions, but this reference is supposed to provide the key: http://www.mat.univ.szczecin.pl/files/brocardramanujan.pdf

[desiresjab]

I was trying to make sense out of the abc conjecture. I think I have a basic understanding of what it is trying to say primarily from this wikipedia article: https://en.wikipedia.org/wiki/Abc_conjecture

What I don't see at the moment is why it implies that Brocard's problem has only finitely many solutions, but this reference is supposed to provide the key: http://www.mat.univ.szczecin.pl/files/brocardramanujan.pdf

I do not see it clearly yet, either. One first needs to study Diophantine equations to get the basics of their solution and become familiarized with the usual methods in the field.

I am still trying to figure out the triangular connection to the other problem, but I am looking at this problem too. I would like to understand the abc conjecture better, so I suppose I will. You never know when the insight will come, except that it will be when you are concentrating your best.

[desiresjab]

As an intersting extension of the (2n)! problem, I have realized it is possible to determine the exact number of factors of 2 in the interval (1, 2n) without recourse to the floor function which is the the usual manner. I have derived by observation the piecemeal formula that works for all values of x≥1, and I would have no need of the floor function, then, to determine the number of 2's in virtually any factorial. If I do not lay it out in a list it will be harder to conceptualize. Near the end of the list I realize I need only include even values in the range (1, 2x) to get a formula.

(2·4)!=8·7·6·5║·4·3·2·1
...........3....1......2....1


(2·5)!=10·9·8·7·6·║5·4·3·2·1
............1.....3....1......2....1


(2·6)!=12·11·10·9·8·7║·6·5·4·3·2·1
............2.........1....3......1.....2...1


(2·7)!=14·13·12·11·10·9·8·║7·6·5·4·3·2·1
............1........2.........1....3......1....2. ...1


(2·8)!=16·15·14·13·12·11·10·9║·8·7·6·5·4·3·2·1
............4........1........2........1.......3.. ..1....2....1

(2·9)!=18·16·14·12·10║·8·6·4·2·1
............1...4....1...2...1...3.1.2.1


(2·10)!=20·18·16·14·12·║10·8·6·4·2·1
..............2...1...4....1...2....1..3.1.2.1


Certain patterns now appear.

1 for even x, the two partitions have the same number of elements

2 For odd x the upper partition gets a bonus factor of 2 among its elements.

3 Every number divisible by 2 in x≥(x/2) has a double in the upper partition with exactly one more factor of 2.

4 Where x is a power of 2, the lower partition has one less factor of 2 than the upper, so the count in this special case is easy. The general method below, however, still applies.

5 To determine y in the equation below, merely count how many even elements in x are≥(x/2), and add one to the value of y for each case. If x is odd, we add one additional bonus factor, and we are done.

2x+y+k, where y is the number of even elements in the interval (1, x) that are ≥(x/2), and k=0 when x is even, k=1 when x is odd:

[desiresjab]

We already knew we needed no floor function to determine the number of factors of 5 in a factorial (for we simply count the number of 0's in the tail), but now now we have a way to avoid the floor function to determine the total number of factors of 2, as well. The fact that it is simpler and works for all factorials is impressive. It is probably an easy consequence of things I already knew, but I never put this together until now.

[YesNo]

As an intersting extension of the (2n)! problem, I have realized it is possible to determine the exact number of factors of 2 in the interval (1, 2n) without recourse to the floor function which is the the usual manner. I have derived by observation the piecemeal formula that works for all values of x≥1, and I would have no need of the floor function, then, to determine the number of 2's in virtually any factorial. If I do not lay it out in a list it will be harder to conceptualize. Near the end of the list I realize I need only include even values in the range (1, 2x) to get a formula.

(2·4)!=8·7·6·5║·4·3·2·1
...........3....1......2....1


(2·5)!=10·9·8·7·6·║5·4·3·2·1
............1.....3....1......2....1


(2·6)!=12·11·10·9·8·7║·6·5·4·3·2·1
............2.........1....3......1.....2...1


(2·7)!=14·13·12·11·10·9·8·║7·6·5·4·3·2·1
............1........2.........1....3......1....2. ...1


(2·8)!=16·15·14·13·12·11·10·9║·8·7·6·5·4·3·2·1
............4........1........2........1.......3.. ..1....2....1

(2·9)!=18·16·14·12·10║·8·6·4·2·1
............1...4....1...2...1...3.1.2.1


(2·10)!=20·18·16·14·12·║10·8·6·4·2·1
..............2...1...4....1...2....1..3.1.2.1


Certain patterns now appear.

1 for even x, the two partitions have the same number of elements

2 For odd x the upper partition gets a bonus factor of 2 among its elements.

3 Every number divisible by 2 in x≥(x/2) has a double in the upper partition with exactly one more factor of 2.

4 Where x is a power of 2, the lower partition has one less factor of 2 than the upper, so the count in this special case is easy. The general method below, however, still applies.

5 To determine y in the equation below, merely count how many even elements in x are≥(x/2), and add one to the value of y for each case. If x is odd, we add one additional bonus factor, and we are done.

2x+y+k, where y is the number of even elements in the interval (1, x) that are ≥(x/2), and k=0 when x is even, k=1 when x is odd:

I checked this with Python. My program may be wrong.

I don't know if it generalizes as you are suggesting or not.

Here is the code:

from math import factorial
from sympy.ntheory import factorint

for i in range(1,11):
b = factorint(math.factorial(2*i))
print("2 times",i,"factorial has",b[2],"factors of 2.")

Here are the results:

2 times 1 factorial has 1 factors of 2.
2 times 2 factorial has 3 factors of 2.
2 times 3 factorial has 4 factors of 2.
2 times 4 factorial has 7 factors of 2.
2 times 5 factorial has 8 factors of 2.
2 times 6 factorial has 10 factors of 2.
2 times 7 factorial has 11 factors of 2.
2 times 8 factorial has 15 factors of 2.
2 times 9 factorial has 16 factors of 2.
2 times 10 factorial has 18 factors of 2.

Here are some further values:

2 times 11 factorial has 19 factors of 2.
2 times 12 factorial has 22 factors of 2.
2 times 13 factorial has 23 factors of 2.
2 times 14 factorial has 25 factors of 2.
2 times 15 factorial has 26 factors of 2.
2 times 16 factorial has 31 factors of 2.
2 times 17 factorial has 32 factors of 2.
2 times 18 factorial has 34 factors of 2.
2 times 19 factorial has 35 factors of 2.
2 times 20 factorial has 38 factors of 2.

[desiresjab]

Looking again at the bracketed part of the formula

(2n)!=2n[n!(2n-1)!!]

for the equivalence between that and triangular numbers, let us make a set of the oddly labled triangular numbers

(1, 1+2+3, 1+2+3+4+5, 1+2+3+4+5+6+7,...)=

(1, 6, 15, 28,...)

Each new element adds the next two successive integers, which means we are adding an odd number to the total already there, so the parity of the set strictly alternates. We only need to show that the above set multipled together has identical prime factorization as the part of the formula in brackets.

1, 2 and 3 of the lower factorial are got from 1 and 6. The 4 comes from the 28, leaving a 7. This 7, along with the 5 and the 3 in the prime factorization of 15, completes the double factorial starting from (2x-1), as the 1 at the end is superfluous as a multiplier and does not change the product.

We have shown it, now we need to prove it for every case.

We note that the above ordered set shares alternating parity as a feature with the factors of a factorial. We note further that factors of prime n are introduced on the nth element of the set of oddly labled triangular numbers. We note that any remaining numbers after these factors are factored out of the triangular numbers are odd, and along with factors still remaining in the triangular set, form a double factorial beginning on 2x-1.

That is as close as I can get to a proof right now.

[YesNo]

Here is a paper discussing the relationship between factorials and triangular numbers. I have only skimmed the introductory part, but it looks like it is discussing a similar problem to the one you are addressing: http://www.integers-ejcnt.org/l50/l50.pdf

He called the factorials the "additive analogs of factorials" which makes sense, but I did not think of it that way before.

[desiresjab]

I checked this with Python. My program may be wrong.

I don't know if it generalizes as you are suggesting or not.

Here is the code:

from math import factorial
from sympy.ntheory import factorint

for i in range(1,11):
b = factorint(math.factorial(2*i))
print("2 times",i,"factorial has",b[2],"factors of 2.")

Here are the results:

2 times 1 factorial has 1 factors of 2.
2 times 2 factorial has 3 factors of 2.
2 times 3 factorial has 4 factors of 2.
2 times 4 factorial has 7 factors of 2.
2 times 5 factorial has 8 factors of 2.
2 times 6 factorial has 10 factors of 2.
2 times 7 factorial has 11 factors of 2.
2 times 8 factorial has 15 factors of 2.
2 times 9 factorial has 16 factors of 2.
2 times 10 factorial has 18 factors of 2.

Here are some further values:

2 times 11 factorial has 19 factors of 2.
2 times 12 factorial has 22 factors of 2.
2 times 13 factorial has 23 factors of 2.
2 times 14 factorial has 25 factors of 2.
2 times 15 factorial has 26 factors of 2.
2 times 16 factorial has 31 factors of 2.
2 times 17 factorial has 32 factors of 2.
2 times 18 factorial has 34 factors of 2.
2 times 19 factorial has 35 factors of 2.
2 times 20 factorial has 38 factors of 2.

What I meant to say was that the second interval has an extra factor over the first interval for every even element from 1 to x for which x≥(x/2), plus one more if x is odd. Subtract that amount from the second interval, and you have the amount in the first interval. Then you add the first and second intervals together for the total. The algorithm has to work but I am usually messy or plain incorrect with such notation out of the blocks. The verbal correction is right. I will get to the notation later.

[desiresjab]

Here is a paper discussing the relationship between factorials and triangular numbers. I have only skimmed the introductory part, but it looks like it is discussing a similar problem to the one you are addressing: http://www.integers-ejcnt.org/l50/l50.pdf

He called the factorials the "additive analogs of factorials" which makes sense, but I did not think of it that way before.

That looks like a good paper. I don't think I have read it before.

But right now let us peek at the first part of the abc conjecture in your link. I was able to figure out the first part for A as a non-square so far.

As long as the prime p≤n, p will divide n!

This means n!≡0 (mod p). So in the expression

n!+A≡m2 (mod p), if n!≡0, then A≡m2. That is why n!+A≡m2 (mod p), as he says, reduces to A≡m2 (mod p).

Bounded from above by a constant means bounded from above by a horizontal line, so that only a finite number of solutions in whole numbers could lie between that line and the input values themselves.

I hope that helps if there was any confusion on that part. I will try to get to the far longer and more difficult looking part later. I have to leave for another few days soon.

[YesNo]

This means n!≡0 (mod p). So in the expression

n!+A≡m2 (mod p), if n!≡0, then A≡m2. That is why n!+A≡m2 (mod p), as he says, reduces to A≡m2 (mod p).


That would be one way to approach searching for more solutions. The term n! + A would have to be a quadratic residue modulo primes larger than that value for it to be equal to a square.

[desiresjab]

My interpretation of that first proof was that it eliminated non-quadratic A's from consideration by showing they contradicted the earlier assumption. But perhaps they assumed there was a solution to show there were finitely many more, at best. Some of the attendant conjectures to the problem will have to be understood in order to understand abc. For instance, I see the algebra perfectly, but I do not see at all yet how their proof for the non-quadratic variety of A bounded it from the top by a constant. Seems to me the constant would be zero, because there would be no solutions of that variety, is how I take it.

This means we are riders in the same boat when it comes to grasping the connection between Brocard's problem and the finite solution set implied by the abc. What I showed was the extent of my reading of that article so far. I will do more, but I have to find the time. It appears that to get a good grasp of the abc I will need to pick up a lot of additional information and skills. Fine. That is how I do it anyway. Problems within the problems I select always force me to the books to learn more. I understand there are problems completely beyond my brain power to even get a good purchase on. I wonder if this is one of those, or if we can get close to the understanding the big boys and girls have?

[desiresjab]

As I mentioned, there are many ancillary propositions and ideas to understand before one can have a solid amateur's grasp of the abc conjecture. Szpiro's equation looks like a horror to grasp. I don't even know what e represents in it. Are they talking about the famous e of calculus, or have they assigned something else to e? Every single element of the proof must be grasped to have the understanding we want, but this relies on many other propositions that support this one. And of course the radical is not the radical from high school algebra.

This problem is a bit different from the style of number theory I have worked on. It brings in many basics, like the GCD. They have to be at your command. To be perfectly honest, I had forgotten that any integer not of the form 4t+2 could be expressed as the difference of two squares. The article reminded me of that. I believe it is necessary to have everything one knows about squares, GCD's, LCM's, divisor functions, Euler's phi function, the Euclidian algorithm, the extended Euclidian algorithm et al, at one's fingertips during such an investigation. Otherwise, one can spend a great deal of time and frustration on aspects of a problem that would be obvious if one had only remembered some basic function or proposition from earlier in one's studies. I like to avoid painful algebra when I can. It cannot always be avoided. I am delving downward on this one. Inch by inch, progress will show--I hope, at least.

In the meantime, I think I will go over and peek at your link on factorials and triangular numbers. That subject is almost like recreation now, compared to the foreign difficulty of the abc. I have a reasonable command of triangles and factorials as cohorts now. I expect to understand the article with much greater ease than I would have before our own investigations, if I could have at all, that is. I am hoping to see a proof out of him that I can follow.

[YesNo]

The link on factorials and triangular numbers gives a proof of what you were trying to show for even factorials. Basically, think of a triangular number written in closed form. That is, the sum of the first n numbers can be represented by n(n+1)/2. Since we skip the even triangular numbers and get rid of the 2 in the denominator with the 2n/2 factor, this looks like it should prove the result.

I am beginning to understand the abc conjecture. Take relatively prime integers a + b = c. Define rad(abc) = rad(a)rad(b)rad(c) to be a product of the unique primes in the product abc. This product is typically larger than c, but sometimes it is not. Sometimes rad(abc) < c and there are infinitely many triples (a,b,c) such that rad(abc) < c. However, there are (so I hear) no known example where (rad(abc))2 < c. That is if we raise rad(abc) to some power larger than 1, no matter how small that "larger than one is", then we get only finitely many triples that would work, that is, (rad(abc))1+epsilon < c has only finitely many triples (a, b, c) for which that relationship holds.

That is the abc conjecture. For this to apply to Brocard's problem, we would need to establish that for all n, n! + 1 = m2 is such that (rad(n!)rad(1)rad(m2))1+epsilon is always less than m2 for some epsilon. Of course, I might be misunderstanding all of this.

[desiresjab]

The link on factorials and triangular numbers gives a proof of what you were trying to show for even factorials. Basically, think of a triangular number written in closed form. That is, the sum of the first n numbers can be represented by n(n+1)/2. Since we skip the even triangular numbers and get rid of the 2 in the denominator with the 2n/2 factor, this looks like it should prove the result.

I am beginning to understand the abc conjecture. Take relatively prime integers a + b = c. Define rad(abc) = rad(a)rad(b)rad(c) to be a product of the unique primes in the product abc. This product is typically larger than c, but sometimes it is not. Sometimes rad(abc) < c and there are infinitely many triples (a,b,c) such that rad(abc) < c. However, there are (so I hear) no known example where (rad(abc))2 < c. That is if we raise rad(abc) to some power larger than 1, no matter how small that "larger than one is", then we get only finitely many triples that would work, that is, (rad(abc))1+epsilon < c has only finitely many triples (a, b, c) for which that relationship holds.

That is the abc conjecture. For this to apply to Brocard's problem, we would need to establish that for all n, n! + 1 = m2 is such that (rad(n!)rad(1)rad(m2))1+epsilon is always less than m2 for some epsilon. Of course, I might be misunderstanding all of this.

Continued misunderstandings to overcome are what I depend on to inch along. Which reminds me to retract all my glorious declarations of a new way to find the 2's in the interval (1, n). It does not work. I am very curious about that, and for a while may be delayed there. Is it even a solved problem? That is, is there an explicit, dependable formula relating the number of 2's in each partition to each other with perhaps an exception or two for some particular classes of number.?That is what I was trying to do. Working late into the night after traveling, I did not even see the exceptions popping up in my list. I plead fatigue blindness.

I have to decide what interests me most right now. That is the other way I inch along--by taking on only what interests me. The abc is new and different and a huge bite, but I do not mind, because I was only avoiding it because previous cursory inspections of it had left a definite impression of a problem which itself was very difficult to understand. What was the problem, and what was it trying to acheive? That is where we are spending our time right now--trying clearly to figure out what the problem is, and after that to figure out precisely how these bounding ideas which play so large a part in their discussion suggest there are at most finite solutions.

I have to wonder if they are using calculus to squeeze out these bounds but giving an overview strictly in terms of algebra. I did not see an integral sign or a differential in the entire discussion. Usually I do not see them, but I know they are being used in much of the work. For reasons not clear to me top researchers often use the complex number system combined with calculus, simply called complex analysis, to research many of the major problems in number theory. When you see explanations of the Reimann hypothesis you seldom see anything to indicate the complex number system, but for a fact the Reimann hypothesis is a statement in the complex number system.

[YesNo]

I haven't finished reading the various papers on the multitude of topics we have discussed. Usually I need to read them multiple times over a period of days before some dream clarifies what is going on even if I do finish one of them. Most I haven't even read once.

I think the Riemann hypothesis uses complex numbers because the sum of reciprocals of the primes converges while the sum of the reciprocal of integers does not and that converging sum of prime reciprocals can be extended to the zeta function which is defined over the complex numbers. The only advantage of the Riemann hypothesis (that I see at the moment) is getting a bound on the estimate of the number of primes in a range. That provides another way to disprove the hypothesis: find a range with more primes than the hypothesis says should be there.

The abc conjecture probably does not involve calculus to my knowledge at the moment. However, the epsilon is taken from calculus as a standard symbol for some small value that is determined by the choice of another small quantity called delta. If epsilon is 1 in the abc hypothesis, then the exponent for rad(abc) is 1 + epsilon = 1 + 1 = 2. The abc hypothesis would say that given epsilon = 1 there are only finitely many triples of integers, a + b = c, such that rad(abc)1 + 1 < c. At the moment that finite number is unknown but it could well be 0, because no examples are known. If I remember right, the triple requiring the largest epsilon would have the exponent about 1.66 or so. That is epsilon would be 0.66.

This leads to a computational problem similar to finding larger and larger primes: find the triple with the largest epsilon. It must exist since there are only finitely many triples (a, b, c) that make the inequality (rad(abc))1+epsilon < c. In order to measure which is the top triple the idea of "quality" is defined. One could just say the quality is c/rad(abc), which directly compares the two values of interest, but that leads to rather large numbers and so logs are put around the numerator and denominator and so the quality becomes q(a,b,c) = log(c)/log(rad(abc)). At least that is what I think motivates that definition.

The hardest part of the computation is that one needs to be able to factor a, b and c to find the distinct primes so one can compute the rad of those numbers which is the product of the distinct primes in those co-prime integers. Computations are limited by the numbers we can factor in a reasonable amount of time.

[desiresjab]

I have not finished everything either. I loaded too much on the plate and now I feel myself getting tired. Absolutely it takes multiple times reading through these articles to tease out understanding. I get a scrap here and a scrap there. What I realize because I know myself is that I will keep after various aspects of these problems like a fanatic, pushing myself, because they have challenged me now. These problems taunt me, saying I am not even intelligent enough to understand them correctly, and I accept the challenge on tentative sea legs.

A person feels there must be an upper bound on their own intelligence, too. Yet one has had a lifetime to reason out that people are bad judges of their limits and should dismiss all limits concerning their own potential from active duty for a better life.

So, where shall this mathematical butterfly settle for a while, then, if I insist? The factorial/double factorial problem is still haunting me--not for its triangular connection anymore but searching for those 2' in the interval (1, n).

[YesNo]

While looking at how to search for additional examples of Brocard's problem, I found reference to Montgomery modular multiplication that looked interesting: http://www.hackersdelight.org/MontgomeryMultiplication.pdf

Even if the abc conjecture is true, and I assume it is, the next step would be to find all the finite solutions to n! + 1 = m2. One way to get that result is to find an upper bound on the possible solutions and then do a comprehensive search.

[desiresjab]

While looking at how to search for additional examples of Brocard's problem, I found reference to Montgomery modular multiplication that looked interesting: http://www.hackersdelight.org/MontgomeryMultiplication.pdf

Even if the abc conjecture is true, and I assume it is, the next step would be to find all the finite solutions to n! + 1 = m2. One way to get that result is to find an upper bound on the possible solutions and then do a comprehensive search.

400,000,000,000!

If there are any more solutions, they are beyond that number, I believe I read.

In the meantime, I have made progress. I am still working. My DeuceHound Model-2, nicknamed DH (mod 2), is nearly ready for mass production. I hope to have it on the shelves by mid Christmas season.

I must rule the ruler function.

[YesNo]

Yes, I think various people have checked it that far. But one could argue that that's not very far. There are only 11 digits in that number. Of course, putting a factorial on that number makes it rather large.

So, what does this DeuceHound Model-2 do?

[desiresjab]

Yes, I think various people have checked it that far. But one could argue that that's not very far. There are only 11 digits in that number. Of course, putting a factorial on that number makes it rather large.

So, what does this DeuceHound Model-2 do?

Four hundred billion factorial might be a competitor for one of the largest numbers we ever had practical use of any kind for.

The DH (mod 2), a deluxe line out of the DeuceMaster series, delivers the most comprehensive factorial unwinder on the market today right to your virtual doorstep.

[desiresjab]

In the relationship between the number of factors of 2 in the interval (n+1, 2n) versus the interval (1, n), there is a chain of beautiful islands of stability within the integers that grow increasingly farther one from the next as increasing powers of 2 recede from our vision into our imagination. These orderly places are centered, in fact, around pure powers of 2.

(A) Conjecture: Whenever N in N! is a power of 2, i.e. some 2k in the expression (2k)!, the difference between the number of factors of 2 in the intervals (2k, k+1) and (k, 1), is one.

(B) Conjecture: The difference in the number of factors of 2 between the same two intervals when the number being factorialized is of the form 2k)+2, is always two.

(C) Conjecture: When any number of the form 2k)-2 is factorialized, the difference between the number of factors in the upper and lower intervals is k-1. The difference in the number of factors between the middle factorial and the lower factorial, is k.

Proof of (A): Observe the ruler function sequence. It represents the number of factors of 2 of each successive even number.

1 2 1 3 1 2 1 4 1 2 1 3 1 2 1 5 1 2 1 3 1 2 1 4 1 2 1 3 1 2 1 6...

The value of any number with an address of the forms 4n+1, 4n+2 or 4n+3 is fully known simply from the address. The only change occurs in numbers with an address of the form 4n. If we were considering all numbers instead of just the set of even ones, we would call them 8n numbers instead of 4n, and so forth, but we need only consider the evens, hence the reduced addressing.

But any address with the form of 4+8n, has a value of 3. Further, any address of the form 8+16n, has a value of 4; any address in the form of 16+32n, has a value of 5, and so forth for each successive positive integer. Only the first appearance of a number in the sequence represents a pure power of 2. The first appearance of 5 in the sequence represents the number 32 itself. All later appearances of 5 represent 32 wrapped up within a larger composite with factors other than 2. If we arrange the above sequence in columns, it will highlight the major properties.

1 2 1 3
1 2 1 4
1 2 1 3
1 2 1 5
1 2 1 3
1 2 1 4
1 2 1 3
1 2 1 6...

In fact, this will be the method of our proof. We can slide any length of symmetric grouping from later in the sequence over the top of an earlier one, and see for ourselves what is going on. From one pure power to the next, the entries are identical, except for the last.

1214
1213, the difference is one. Even if we break the entries 1 2 in half and slide the later portion over as in a subtraction, we get,

1 2

2
1, the difference is one.

It always works:

1 2 1 3 1 2 1 5
1 2 1 3 1 2 1 4, the difference is one. This proves directly that the

difference is 1, in the number of factors of 2 between a number k and its double both factorialized, when k is a power of 2, and hence in the intervals (2k, k+1) and (k, 1).

This is all to state the obvious fact that if you multiply any number k times 2, the new number j has exactly one more factor of 2 than k. That would work as a proof, too.

[desiresjab]

Proof of (B): In the shift from 2(16)! to 2(17!), the bottom intervals stay the same, but one more factor of 2 is introduced with the doubling of 17, which was not a member of the set up to 16. Therefore the difference between the number of factors of 2 in the inverval (2k, k+1) and (k, 1), is two, when the factorialized number is of the form 2(2n)+1. See below.


(2·16)!=32·30·28·26·24·22·20·18·║16·14·12·10·8·6·4 ·2·1
..............5...1...2...1...3...1...2....1....4. ..1...2...1..3.1.2.1

(2·17)!=34·32·30·28·26·24·22·20·18·║16·14·12·10·8· 6·4·2·1
..............1...5...1...2...1...3...1...2....1.. ...4...1...2...1..3.1.2.1


The proof is complete. Conjecture (B) is now Theorem (B).

[desiresjab]

Proof of (C):

See the operations below. Further observe that (2·8)! is of the form (2·2k) because it equals (2·23), with k equaling 3 in this case. Therefore (2·7)! is of the form (2·2k-1)!, and we have postulated there should be a difference of k between its upper and lower intervals in their factors of 2.

Note that when the center bar moves backward one position to the right in going from 16 to 14 factorial, the lower interval loses k factors of 2 from a total we already know. The upper interval, by law, will lose only one factor of 2 to its upper interval counterpart in the move backward, also from a total we know. The factors of 2 in the lower interval, then, would be 2k-1-k; the number of factors of 2 in the upper interval would be 2k-1. Subtracting one from the other we get:

(2k-1)-(2k-1-k)=k



(2·7)!=14·13·12·11·10·9·8·║7·6·5·4·3·2·1
.............1........2.........1....3......1....2 ....1


(2·8)!=16·15·14·13·12·11·10·9║·8·7·6·5·4·3·2·1
............4........1........2........1........3. ..1....2....1


The proof is complete. Conjecture (C) is now Theorem (C).

[YesNo]

1 2 1 3
1 2 1 4
1 2 1 3
1 2 1 5
1 2 1 3
1 2 1 4
1 2 1 3
1 2 1 6...


I think I can see the pattern you are trying to explain with these examples. This might result in a recursive algorithm that is simpler than whatever is used at the moment to find the number of factors of 2 there are in n!. But I don't know what people use at the moment.

[desiresjab]

Now we are able to look at any

(2·2k)! or

(2·2k±2)!

and report by inspection alone the power of 2 in the prime factoriaztion of any of the three, as well as the number of 2's in each of their respective upper and lower intervals.

This is a big improvement over using the repeated floor function. Of course it is only for islands of predictable stability that grow ever more sparse out the number line. The factorial islands of stability for 2 are:

(3, 4, 5)!,
(7, 8, 9)!,
(15, 16, 17)!,
(31, 32, 33)!,
(63, 64, 65)!
(127, 128, 129)!...etc., etc., etc.

The Master Factorial Unwinder (model ∞), after its launch, will show that these islands exist for any pure prime power and a neighbor, not necessarily next door. Its larger task is naming all the powers of prime factors in the prime facrorization of a factorial by inspection, as the larger task of the DeuceMaster (mod 2) is to perform this for factors of 2.

[YesNo]

What values do you expect for various intervals? I'll test it with Python.

[desiresjab]

The Ruler Sequence is a pattern I have studied a fair bit. The key to any value in the sequence is simply its address. In the case of 4n+1. 4n+2 and 4n+3 numbers, we only need the street without the street number, so to speak. Determining the fourth column is trickier, but not really.

3's will be found only at addresses that can be put, at lowest, in the form of 4+8n. (In other words, not some 2+4n.)

4's will be found only at addresses of the form 8+16n.

5's will be found only at addresses of the form 16+32n,

and so forth.

I used to believe with you that the sequence could only be mastered in an algorithm. I hope to show that this is false, and to give a general forumla for finding the power of all primes in any factorial. An analog of the Ruler Function for each prime will be of great use, and can perhaps be put into one explicit forumla. That formula will be none other than the

Master Factorial Unwinder (model ∞).

* * * * *

By the way, here is how they calculate the factors of a prime in a number, shown for the prime 2 and the number 69. I calculate how many times each power of 2 that divides 69 (not necessarily evenly) goes into it, and add them all up.

64 goes in once, 32 goes in twice, 16 goes in 4 times, 8 goes in 8 times, 4 goes in 17 times, and 2 goes in 34 times. Add them up.

1+2+4+8+17+34=66.

F2(69)=66

This is the procedure they use now, I mean to say. I hope for a great improvement. If not, finding the island chain of stability for 2's, proving its existence and then conjecturing these islands exist for any prime in a factorial, has been satisfying in itself.

[desiresjab]

Ask yourself: "Why should neighbors farther removed than next door from a pure power of 2 show any less stability and predictability than a next door neighbor does?" The answer is they shouldn't, it is just the next step in a pattern not yet fully identified but totally contingent upon the Ruler Sequence.

[desiresjab]

What values do you expect for various intervals? I'll test it with Python.

I don't understand your question.

[YesNo]

I don't understand your question.

I am going to try to implement your algorithm to make sure it is correct. However, I don't completely understand it at the moment, but some examples would be helpful. I will compare what I get from your algorithm with what I would get from constructing the intervals of the factorial and then factoring them.

[desiresjab]

Give me a while to think. When I rush is when I am excited and make all these mistakes. I may have to go back and edit some posts to amend the conjectures. If the conjectures are wrong, I will actually go back to those posts and make them right with an edit.

I will be able to extend the neighborhood, which is good news.

[desiresjab]

I have made some mistakes but I have straightened them out mentally. It is still valid and even better. I overlooked like a simpleton again that my values were for even numbers factorialized which were on either side of a pure power of 2, which makes them even, as well. Bad but good. This allows me to get the values of the next door neighbors. I thought I was calculating for the next door neighbors before, but I wasn't, I was calculating for the next to the next door neighbors instead. This makes the in-betweens easy, which is the good news.

I will be back once I get every detail straight, not before!

[desiresjab]

Okay, I amended mistakes in previous posts (of which there were very few, it turns out) and am ready to present the beautiful chain of receding, stable islands, which is two factorials wider than we realized before.

Let's use some simple examples. Because 8 is a power of 2, we know 8!=(2·22)!, with the exponent equaling 2 in this case, we know that it will have 22 powers of 2 in its upper interval, and 22-1 powers of 2 in its lower half, for a total of 23-1 factors of 2.

(2·3)!=6·5·4·║3·2·1
...........1....2......1

(2·4)!=(23)!=8·7·6·5║·4·3·2·1
........................................3....1.... ..2....1

(2·5)!=10·9·8·7·6·║5·4·3·2·1
............1.....3....1......2....1

* * * * *

Let (2·2p)=(2k)

Not only can we read directly the values for 10! and 6! by knowing the values for 8! (8 being a power of 2), but we can easily read between the lines now and get the values for 7! and 9!. The relationship of the difference between the number of factors of 2 in the middle even factorial and the lower even factorial (such as 8! and 6!) is always equal to k, when the k is from the pure power of 2, so that difference will not change, either, in going from 6! to 7!.

7! will look identical to the schematic above for 6! in terms of its eveness values, we can see. The factorial between 6! and 8! has the same disparity between its upper and lower halves as 6! does, and the same amount of factors of 2 overall, since when 7 is multiplied times 6! no new factors of 2 appear. Therefore 7! will have the same difference between its upper and lower halves, k-1 in the expression (2·2p)=(2k), as 6! has. The difference between factors in the middle even factorial and the lower even factorial is also equal to k.

By my formula,10!, otherwise known as (2·5)!, should have a difference of two between its top and bottom intervals, and contains one more factor of 2 overall than 8! does. The calculations above, which are really only summations of increasing lengths of the Ruler Sequence starting from its beginning, allow us to fill in visually where 9! factorial would be and verify that its upper and lower intervals are identical to those for 8! for powers of evenness. The difference between the number of factors of 2 in the two succesive factorials 9! and 8! is 0.

This proof is correct. Below, let me try to put the results briefly and clearly.

* * * * *

(2k-2)!=k fewer factors of 2 than 2k, and Ui-Li=k-1.

(2k-1)!=k fewer factors of 2 than (2k)!, and Ui-Li=k-1.

(2k)!=2k-1 total factors of 2, and Ui-Li=1.

(2k+1)!= 2k-1 total factors of 2, and Ui-Li=1.

(2k+2)!=2k total factors of 2, and Ui-Li=2.

The Demonstration is complete.

[desiresjab]

You have to admit, using my method as opposed to using the traditional one, would make calculating these values for large factorials a snap, as long as the factorial fell within our five-wide swath of islands. Even for a factorial as small as 16!, the traditional method would spend considerable effort over mine. Try it. Even if you knew the shortcut secret of the pure power's 2-count, calculating all those values for the surrounding four factorials without my tricks would require a near-Herculean effort. Remember, the number below is a tiny one.

Around every power of 2 factorialized, such a cluster of predictability exists.

16!=20922789888000.

This one simple function of the DeuceMaster (mod 2) enables you to calculate the two different 2-counts (Overall and Ui) for any factorial in the swath from a few basic rules, while others wrestle with serpentine iterations of the Floor Function.

[desiresjab]

What if the formulas and procedures are similar or identical for the factorial swath surrounding the factorialzed powers of any prime, such as 3? This is what I am foreseeing. I still have to finish the DeuceMaster (mod 2), but I am eager to look for the island pattern in higher primes and see how it codes out.

Without precise algebra, it will never work out easily for higher primes. The factorial powers grow large so fast there is no time to look for a stable pattern. Rather, no way to verufy the results by a calculator with 32-bit decimal precision.

[desiresjab]

If we know any number, or any number factorialized, we can figure out its address and then its value. If the number is a power of 2 factorialized, we have theorems which enable us to be quite complete with all calculations concerning factors of 2 for this power and its four neighboring factorials--the two on either side of it.

Give me a power of 2 factorialized, and I can give you back its address, the addresses of its neighbors and all relevant values concerning the factor 2 in the positions of these factorials in the Ruler Sequence. That is childishly easy. The factorial itself will have an address of 4n. The factorial just under it will be a 4n+3 number and will have a positional value (number of factors of 2) of 1 in the Ruler Sequence. The factorial just under that will be a 4n-2 number with a positional value of 2. Just above the power of 2 factorialized, there will be a 4n+1 number factorialized, with a positional value of 1. Just above that will be a 4n+2 number factorialized, with a positional value of 2.

Gve me 10536209805943621, and I can tell you that its double factorialized (2x)!, has x factors of 2 in its upper interval, and how many in total. We may not know any tricks yet to get all the 2-count information from (2x)! and its neighbors, but we can do this much at least, with any number. We see, then, how it might be useful to obtain these values merely from both the magnitude of the address and the form it can be put in, for any number whatsoever.

[YesNo]

The Montgomery modular multiplication uses the idea that a factors of two in a binary computer system can be easily performed through bit movements. It is a way to speed things up.

If you can do this with 2 you might be able to do something similar for any prime.

I am trying to think of some way to use Python to explore this more.

[desiresjab]

I just about have the entire formula. Until then, I can predict with minor calculations all the 2 values for a swath twelve factorials wide surrounding the pure power of 2 factorialized, but surrounding it asymmetrically so far. I can predict 4 values on the lower side and 7 values on the upper side of the pure power, from knowing the pure power, which itself makes 12. I may be able to extend this farther, to at least 16. If I can find one formula that covers any number, I say that is more compact, since the formulas I have seen are more like algorithms, like the Fiobinacci sequence--easy to write down the instructions to the next number, hard to write down an equation.

However, proceeding in the same vein as before, I now easily have a swath 29 factorials wide, man! Fourteen on either side of the pure power. This is just what falls out naturally, and I have a thinking it may not be the limit of what falls out easily.

I could write out rules for these 28 neighbors of the pure power (now in symmetrical neighborhoods again) as simply and easily as I did previously for two neighbors on each side. But really, what I would do is plug them into a computer program which did the work, which I have another thinking would require less computer labor than the Floor Function method, though on the summation algebra for the Ruler Sequence, one is certainly reminded of The Floor Function, so I am hoping mine is not that function still wearing a mathematical disguise, and if it is, that at least it represents another method of finding the F2 for the numbers surrounding pure powers of 2 factorialized.

This is probably nothing new, I just had to figure it out for myself rather than look it up. Come to think of it, the Floor Function is really succinct. Gauss invented the function, so you know it is down to its most terse expression. Perhaps what I have in mind is at least different. I should be able to see that far ahead, but I can't quite. I need to figure a bit more for the whole thing.

Currently, I have extended the swath to a width of 93! consecutive factorials I can calculate F2 for, centered around the power of 2 factorialized position in the Ruler Function, if the factorial is large enough, like 64! is. Once you look at enough of the Ruler Function, you see the symmetry radiate outward from the virgin power of 2.

That is, the first group surrounding the group of four the virgin power is in, ends in 3, the next group farther out on either side, ends in 4, the next group ends in 3 again, the next group ends in 5, therefore the next group must end in 3 again, since every other group does, etc., etc.

I am tired. I will write this all in later. I am close to my own formula, which may or may not be my own. LIkely not. I think the symmetry on top extends all the way up to the K-2 power, but I am not sure of this yet.

[YesNo]

I was thinking of calculating the factorization of n! by storing the factorization of (n-1)! and then adding the factors of n to that. It would take a lot of storage. Or store the individual factors of the numbers up to n and then summing the exponents for each prime less than n. Of course, people usually want algorithms that are fast and start from scratch without looking anything up.

[desiresjab]

I am looking for an explicit function to factor factorials. That is why I want something besideds the Floor Function.

1 2 3 4 5 6 7 8
...2....4....6....8
...1....2....1....3
...2...22 ........23.......................24............... .............................25.

Above are four different metrics I am using.

(20·1k)+ 20(k-1)+ 21(k-2)+ 22(k-3)+..+2k-2(k-k+1)]= F2{N!}

(1·6)+(1·5)+(2·4)+(4·3)+(8·2)+16·1= F2{64!)=F2{26!}=

6+5+8+12+16+16=63.

[desiresjab]

What I am really doing is searching for an expression of factorials which might be more helpful than using a mere N! in Brocard's problem. This could allow me to see why adding one to a large factorial cannot (or can) produce a square.

[desiresjab]

The next task may be an impossible one--to do the same for the prime 3 as I have done for 2 above, and from there on to any prime. At the end I propose to add one and see how it looks. The specialness of 2 as the only even prime may mean it is the only prime that such a feat is possible with. Who would have thunk it, God even hid deep structures within something so seemingly innocent and innocuous as the set of even integers?

Well, now we need to know how odd primes are packed into a factorial. We cannot use their oddness to help. Or can we? We know all these primes will be 4n+1 or 4n+3. Another expression I might use is 4n-1 for 4n+3. Same thing, since we are speaking (mod 4).

I thought of doing 3's out of a similar form (3n)!, but that does not help me in totally factoring (2n)! I need to extract the 3's from (2n)! Though indeed I might be able to get something from (3n)!, it would only do me any good when it came to factoring something like (6n)!, which is a composite of both.

No need to defer the painful algebra, it has to come some time. Here goes not a swan dive but a belly flop, sir.

[YesNo]

For 3, you could split the numbers into two sets, those that 3 divides and those that it doesn't. Those numbers that 3 does not divide can be ignored. Those that 3 does divide, remove one factor of 3 from all of them. Take that set reduced by one factor of 3 and do the same to it: split it into two sets, those that 3 divides and those that it doesn't. Continue until there are no more numbers left to consider.

[desiresjab]

Can't do the algebra until you see the patterns. Here is a Ruler Sequence for 3's.

1 1 2 1 1 2 1 1 3 1 1 2 1 1 2 1 1 3 1 1 2 1 1 2 1 1 4
3 6 9........................27..............54....... ................81


The house word processor makes it difficult for me to put more than one space between letters, or I could highlight the action better by putting in all the numbers without the danged periods. If anyone knows how to do that, I would appreciate the tip.

Anyway, it occured to me that every odd prime would be modeled after this in its own metric, so to speak. The sequence for 5 would be:

11112..11112..11112..11112..11113...

For 7 it would be:

1111112..1111112..1111112..1111112..1111112..11111 12..1111113

Det it? Dot it. Dood.

I have a thinking I can model this pattern (which works for any odd number not just primes) into one formula, simultaneously including mastership over the only even prime 2. One formula to breaK any factorial of all its primes. One Ring to rule them all! One Ring to bind them!

[YesNo]

As a ring the binds them is a good way to look at getting a formula that works.

[desiresjab]

Without further adieu, I can unveil the DeuceHound (mod 2). The formulas I showed earlier are all compacted within the DeuceHound, and could be reproduced for further research, if needed. Yes, all those earlier formulas are elegantly folded--(ahem!)--into the DeuceHound formula.

Call a Measure four units in the Ruler Sequence. That means eight units in natural numbers. The symmetry of the Ruler Sequence on both sides of a virgin power will be very important to us. We know the number of factors of 2 in (2k)! is 2k-1. We could figure them out by our earlier formula, but why do that when we know the shortcut? We only wanted to demonstrate that the earlier formula worked. Suppose we are given:

(26+31)!

We know the numbers on either side of a virgin power in the Ruler Sequence are perfectly symmetrical, as far as they can be. From each highest power in a measure to the highest in the next, moves 4 units in the Ruler Sequence, or one Measure, but that is eight units in natural numbers. Remember, some units of the natural numbers are invisible--not represented in the Ruler Sequence because they are not divisble by 2, but they are there invivibly--so three measures moves us up 24 units in natural numbers. We know those will be 1 2 1 3..1 2 1 4..1 2 1 3..1 2 1 and represent only the even numbers with the invisible odd numbers in between. Moving up one visible unit in the sequence moves us up two units in real numbers. We add three full measures, then the first three quarters of another one. The number we seek, then, is 95, an invisible number. So, we simply need to add

(26-1)+7+8+7+4=63+26=89. F2{95!}=89

to get our answer, because we know F2{95!}=F2{94!}.

In this special case F2{(2k+Q)!}, Q=F2{(2k-Q)!}, because Q=(2k-1-1).

Now the general method. If Q is odd, subtract 1 from it, since it will have the same F2 as its lower neighbor.

(2k-1)+F2{Q-Q (mod 8)}+F2{Q(mod 8)}.

63+F2{25}+F2{6}

63+22+4=89

The last term, if not zero, will now be an even number less than 8. This remainder tells us how much of a partial measure we have to add at the end, if any. We can use the middle term because, after all, the Ruler Sequence merely repeats itself from the beginning when you stop on a virgin power, up to any Q after that virgin power. The first term of the Hound is just F2{2k}=(2k-1), of course.

Yes, friends, the Ruler Sequence not only repeats itself backwards from 2k, it also repeats itself forward from 1 with exactly the same numbers and forward from 2k with exactly the same numbers. Is that amazing, or what?????

Well, that is it, the DeuceHound (mod 2). It may look complicated, but I can explain any detail, because it is simpler than it looks. I may have made a mistake somewhere which you can catch.

[YesNo]

I suppose we could take the 31 and write it as 24+15 and then continue doing that for 15.

How did you get the 7+8+7+4? I assume that was by looking up the results.

It does seem like a simplification. Rather than looking at numbers larger that 2k we can look at the numbers less than that. I don't know how this fits in with current factorization algorithms. Have you looked at any papers on this topic?

[desiresjab]

Now for the surprise I did not see myself until now. The DeuceHound (mod 2) simplifies further in an amazing way to this:

F2{(2k)!}+F2{Q!}.

Wow! How's That for understandable?

In case it is a new expression, you witness it here with a time and date stamp.

[desiresjab]

I am glad I discovered the formula in the last post naturally and last. That way I learned all the steps that go into it--how to expand it backwards into more detailed forms.

[desiresjab]

I suppose we could take the 31 and write it as 24+15 and then continue doing that for 15.

How did you get the 7+8+7+4? I assume that was by looking up the results.

Not at all. I simply plugged in real numbers for k and Q in the general formula, before the formula evolved even further. Then it evolved beyond that to what I present in the next post after the one you are responding to. That is really simple and easy to calculate. You don't even see all the past formulas that are folded into it, except for the General Form Of DeuceHound you are responding to. That reduction is visible. That is what made me see it.


It does seem like a simplification. Rather than looking at numbers larger that 2k we can look at the numbers less than that. I don't know how this fits in with current factorization algorithms. Have you looked at any papers on this topic?

Exactly. And now it is even easier to use. It all comes down to the palindromic nature of the Ruler Sequence. It has symmetries going in so many directions they confused me for a while. Then I realized what the sequence has done up to any perfect power of 2 it will repeat exactly on the way to the next virgin power, where it clicks over on the last number to the new virgin power instead of repeating the last virgin.

[desiresjab]

I have not looked at any or many factorization algorithms. I am too busy. I would like to look at everything, and if I were as I expect some of my descendants to be with a computer implanted in me and integrated with my consciousness, I could. I could not only read it all, I could digest and collate it all.

Math papers are hard to understand. Often those algorithms are for computer and half in computer language I do not understand. Soon I will be outfitted with a new computer with better capabilities. At that time I may look seriously into learning an OOP language good for math. My experience programming is ancienct and on a PO platform, procedural-oriented. I would have to learn object-oriented. I don't know how hard that is for someone with my procedural background.

[YesNo]

Try Python. You can get what you need to get started by downloading the Anaconda distribution. You will also be able to use Jupyter notebooks which contains mathjax. It allows you to format the mathematics more easily and is the way that I suspect is standard today for publication purposes. This mathjax is the same as what is being used on math.stackexchange.

[desiresjab]

Have you taken the DeuceHound (mod 2) for a test ride?

[desiresjab]

I believe I can do the same thing for 3 that I did for 2. Getting all primes together under one roof for the DeuceHound (mod ∞) will be a monster job. I do not know if it is possible. I suspect if it were, it would already have been done.

Keep your eye peeled for unsolved number theory problems for my collection where only a finite number of solutions are known. For Fermat's Last theorem there were two known solutions, 1 and 2. For Brocard's problem there are three known solutions, (4!, √25) (5!, √121) (7!, √5041). There must be quite a few unsolved problems with this setup where a few solutions are known and it is not known if there are any more.

[desiresjab]

Then there is the Beal conjecture where infinite solutions are known, but all of one variety--where A, B and C share a common factor. He offers $1,000,000 for a proff that only this type of solution is possible. He is a billionaire banker to the big cat oil men, and worth, they say) about $8 billion, who jusy happens to be a quality mathematician.

I don't think he would pay anything for my DeuceHound. I would likely be laughed out of the king's cour for bringing something common. But for a solution to Brocard's problem, especially one he could help tidy up with his technical knowledge, he might shell out a million, as well, or some amount of conch anyway. Personally, I do not have any way at all to approach his conjecture. He loves number theory. I do have a way to approach Brocard's Problem, and the approach is improving. It seems like all the work done on Brocard is of the bounding type, or monsterous computer calculations that are able to promise no more soultions up to a certain limit. I will be trying to eliminate particular forms.

[YesNo]

If one could eliminate particular forms that would be something. I haven't tried implementing the DeuceHound 2 yet. I will try to build a jupyter notebook and send you the link to it. It looks like I would need to know the largest power of 2 that divides n! and then I can work with the remainder.

[desiresjab]

The DeuceHound (mod 2) is extraordinarily simple to implement. In plain English, you take the F2 of the perfect power of 2, which we know is always 2k-1, then add to it the F2 of Q, with Q being the amount you have factorialized beyond the perfect power. For instance

F2{(26+31)!}=

F2{26!}+F2{31!}=

F2{64!}+F2{31!}=

63+26=89.

[desiresjab]

The DeuceHound (mod 2) works for any two numbers one puts in for k and Q.

The formula again is (and I would like to dispense with writing so many factorial signs, so I will introduce the new notation of F2! to mean all the factors of 2 in a factorialized number):

F2!{2k+Q)}=F2!{2k}+F2!{Q}.

If anyone has not seen this perfectly, they only need to gaze at the Ruler Sequence for a while. I do not have to look at it to write it down. Neither will you, if you gaze at it carefully after what I have to say about it. From every virgin power, it simply starts all over again. There are all kinds of clever ways the DeuceHound could be used, but just stick with the basics for now, until you see it really does start over from any perfect power of 2. I will make it really easy, and show the Ruler Sequence up to 95 (26+31), which means it will end on an invisible number 95. That's okay. 94 is just fine for our purposes.

1 2 1 3..1 2 1 4..1 2 1 3..1 2 1 5..1 2 1 3..1 2 1 4..1 2 1 3..1 2 1 6..1 2 1 3..1 2 1 4.. 1 2 1 3..1 2 1

The Ruler Sequence up to 94. From 64 to 95, the sequence is exactly the same as it is from 1 to 31. This is its nature. The Ruler Sequence is an awesome Mandelbrotian Object.

[YesNo]

Here's a pdf export of a jupyter notebook that confirms your result. I would like to make it more general. That is given n = 2343542, compute the number of factors of 2 in n! This just checks your example: https://drive.google.com/file/d/0B96SHXzNSomRZ2ZOWHhmMkdiTmc/view?usp=sharing

[desiresjab]

All you have to do is plug your own numbers into the DeuceHound (mod 2). You have to figure a way to code that formula into Python, so you can plug values of your choosing into the DeuceHound for k and Q. I can also provide a formal proof, which is only to bring the formulas and reasoning from former posts together. However, the need for a lot of additions became unnecessary when I noticed that The Ruler Sequence repeated itself from the beginning every time a new power of 2 showed up. But as a throwback, remember when I said the difference between F2{(2k)!} and F2{(2k-2)!} would be k itself. That seems obvious enough, doesn't it? Through such reverse extrapolation we proved the entire premise before. Once my mind has moved on, it is difficult to recall, exactly. But I do know this, writing a formula was going to prove horrendously difficult the way every other measure is 1 2 1 3 and the others keep changing, until I noticed that other symmetry in the Ruler Function--that it repeats itself exactly from the beginning every time it reaches a virgin power, i.e. a power for the first time.

I will show you how it might be done for a fairly difficult number, how one one might produce a Result through Descent.


Suppose we are asked to find F2! for (210+150)!

The first term is always easy. We know that contains 210-1 factors of 2.

For the value 150, we will need to break it up again. We could break it up several times if we wanted to, but that is our choice. We could just take the value of the 150 through the familiar Floor Function, and add it to what we already have, or we could continue with the DeuceHound method, which of course we will do.

F2!{150}=F2!{128}+F2!{22}.

Again. it is easy for us to add up the 11 Ruler Sequence values for 22, but we prefer to decend further:

F2!{150}=F2!{128}+F2!{16}+F2!{4}+F2!{2}.

Now we have descended all the way to the bottom. What it suggests is that we could build the whole thing forward instead of backwards, in as tiny of increments as we felt like handling. We sum

1023+127+15+3+1=1169

Notice that these summands are all powers of 2 minus the quantity one.

Okay, that is it.

[desiresjab]

Now let us look at some more wonderments within the Ruler Sequence. Let us look at just the pure powers of 2 in the sequence. What do you suppose we would get if we added up all those powers to, say, 26, since we have been working with that as an example.

2+4+8+16+32=64.

64-1=F2!{26}.

The above is why the DeuceHound works.We could use this method, too, but we would still need the DeuceHound (mod 2) to caculate for values of Q.

[YesNo]

I think one way to do this is to take the number n represented in binary and then apply the rule to each bit that is set. I may get this ready tomorrow.

[desiresjab]

It could be done this way.

F2!{2k+Q}=

[k-11∑ [2n]]-1+ F2!{Q}.

[desiresjab]

I think one way to do this is to take the number n represented in binary and then apply the rule to each bit that is set. I may get this ready tomorrow.

Yes, it is intimately related to binary. However, I do not know if that is the best way to go about it. Maybe though. I do not know Python or OOP. I would simply do it like my last post, if I were programming a computer to work it out, which may be precisely what you meant. I would use the Descent Technique showed earlier to calculate F2! for Q.

[YesNo]

I think it would be the same as what you are suggesting to do. The binary representation of the number picks out the powers of 2 in that number.

Python is an OOP language. There are classes and inheritance. You don't need to know much about it to make it work. I started using it when my wife was working on machine learning concepts. We built clusters and then created models. I know (next to) jack about how these things are done, but with the Python packages, such as, pandas, sympy, scikit-learn, and so on, one can get results very quickly with very little actually programming. The programming work has all been done in the packages and it has all been optimized by others. You don't have to program these things, only use what others have packaged to explore your ideas. You are giving me ideas on how to use other packages I have not explored before. It is more a matter of looking up what is available and using that to implement an idea. Also you get mathjax which allows you to format text.

I don't know when I will get a general implementation done. I need to look for the right packages that deal with bit maps or representations of numbers to different bases so as to get factors other than 2. I don't want to have to implement that myself because I will probably implement it wrong and it will not be as efficient.

[desiresjab]

I think it would be the same as what you are suggesting to do. The binary representation of the number picks out the powers of 2 in that number.

Python is an OOP language. There are classes and inheritance. You don't need to know much about it to make it work. I started using it when my wife was working on machine learning concepts. We built clusters and then created models. I know (next to) jack about how these things are done, but with the Python packages, such as, pandas, sympy, scikit-learn, and so on, one can get results very quickly with very little actually programming. The programming work has all been done in the packages and it has all been optimized by others. You don't have to program these things, only use what others have packaged to explore your ideas. You are giving me ideas on how to use other packages I have not explored before. It is more a matter of looking up what is available and using that to implement an idea. Also you get mathjax which allows you to format text.

I don't know when I will get a general implementation done. I need to look for the right packages that deal with bit maps or representations of numbers to different bases so as to get factors other than 2. I don't want to have to implement that myself because I will probably implement it wrong and it will not be as efficient.

Your idea of filling bytes would work. You would actually be adding each member of the Ruler Sequence individually that way. One shorter way is to figure the nearest power of 2, then get Q with the method you suggest. That would cut down on a lot of the work. Perhaps you are just looking to empirically verify through a number of different examples. I am so confident I am resting now, watching movies. In a day or two I will be coming back to see if I can do anything about 3's. I am trying to get a good sense of all the figurate connections too.

[YesNo]

One way to speed up the process is to "vectorize" the number. That is represent each bit as an element in a vector and then perform component-wise replacement of the 1 with the desired number. I am still working on that, however, Python's numpy or pandas packages should provide a way to do that. For the moment I have updated the notebook with a function that will do arbitrary numbers using a for loop and numpy's binary_repr to split out the individual bits. It not not an efficient way to do that.

Next steps: (1) vectorize the solution and (2) include other bases besides binary.

https://drive.google.com/file/d/0B96SHXzNSomRMHdzWjJ6ZEJTb2M/view?usp=sharing

[desiresjab]

I have no familiarity with vectorizing components. Once January comes and I get a new system I will be doing some programming. All my programming will involve number functions, as I am not the least interested in programming my heater to go on and off et al. I hate learning a new language, because it is only a tool to get at what I want, but I will dive right in because I do want what I want.

Today I have to try and replace the side mirror on my car. An elk knocked it off as her head smashed into my windshield. If it had been a bull a horn would have killed me. It doesn't do much good to slow down because they will run right to your headlights anyway. You have to lay on the horn. Who can do that when they are swerving with their foot hard on the brake? These are the largest elk in the word--Roosevelt. I hope it did not become bear food. I was only going about 5mph when it hit me. I have lived in deer and elk country all my life and never hit one before now, though I have seen scores along the roads that others killed.

Well, it is not like you simply screw the mirror on from the outside. You have to take the door panel off. I could pay someone, but I refuse to do that on a job I should be able to do myself.

Back to math. I am hesitant to start all the work on 3's, because I do not believe there is much future there. I need one formula that shakes every prime out of a number. Actually, I think there is a way to do that by making some adaptations to the DeuceHound. I think that is hard. I know my brain is going to suffer.

When I look at the Ruler Sequences for 2 and 3, the opening cycle on 2's is so short it makes it hard to write the common pattern. I know it is there and I can see it. Writing it down successfully will be like a Chinese water torture. But when that is done, it should be fairly routine to write code for the other odd primes. That is what I am hoping, anyway.

Knowing myself, It will likely be another few days before I start. I have to rev myself up to attack the hard ones. A fine treasure waits at the end of this rainbow, if I can just find the end. First, I have to make myself look for it.

[desiresjab]

I forsee another difficulty ahead. The DeuceHound requires a number it can convert to decimal. The only way it knows to handle an exponential number is to convert it to decimal first. This would make it a wonderful machine for what today are considered fairly small numbers like billions, trillions or quadrillions, and it is especially designed to handle factorials. But what about numbers that are so large there is no possibility of the whole thing sitting inside a computer at once? I think this is the magnitude of number the calculational number theory guys are working on these days. I believe they have a way of breaking these numbers up, just the way tasks are broken into many modules on some of the large, shared research projects such as searching for the latest champ of primes, or even the Serpenski project you are contributing to (is that shared?), and delivering these modules to home computers across the world to work on while their owners surf the web, or something like that.

Theoretically, the DeuceHound (mod ∞) would still be a useful accomplishement, I guess, because it can handle powers factorialized, of a restricted kind (powers of primes). Otherwise, it can handle large decimal numbers factorializrd. 400,000,000,000!, the current ceiling for Brocard solutions, would be do-able, I believe, on the DeuceHound. A large power of 2 might not be nearby, but a large power of some prime might lie extremely close to 400000000000!, once the machine is (mod ∞). Finding that would be an extra problem. I expect to devise a better way eventually, i.e. to attack any number regardless of the form it is found in. That is a ways ahead.

The problem I forsee with the attempt to add bits is that every measure is not the same. If you just had to add 7 or 8 for each measure, that would be easy, but the values keep growing. Of course, we know what the values will be for even longer stretches than single measures. Fancy formulas or no, it may come down to adding up these stretches systematically until one reaches the desired last summand, anyway, as you have proposed. I simply do not know yet. I am still amazed that I got the DeuceHound (mod 2) to work correctly. I am not used to success.

[YesNo]

I do have one old computer running a Sierpinski sieve for PrimeGrid a distributive computing platform.

That you have broken this up into sums should allow it to be done piecemeal. I added another implementation of the DeuceHound. It only works for 2. Unfortunately it isn't vectorized. I also haven't done any performance tests on it, but I do have a few "old_ways" procedures coded for comparison and testing. https://drive.google.com/file/d/0B96SHXzNSomRQXNqRVdSMUFPV2M/view?usp=sharing

[desiresjab]

I do have one old computer running a Sierpinski sieve for PrimeGrid a distributive computing platform.

That you have broken this up into sums should allow it to be done piecemeal. I added another implementation of the DeuceHound. It only works for 2. Unfortunately it isn't vectorized. I also haven't done any performance tests on it, but I do have a few "old_ways" procedures coded for comparison and testing. https://drive.google.com/file/d/0B96SHXzNSomRQXNqRVdSMUFPV2M/view?usp=sharing

I am quite impressed with your DeuceHound implementation. You understand. I believe what you are calling piecemeal I called the Descent Method--where you got the triangular shape. Remainders are treated the same way as any number. That method never says, "Okay, thirty-one is small enough, I will just add up the Ruler Sequence from there." No sir, that is what a human would do. The DeuceHound finds the smallest power under a remainder and goes to work again, a perfect slave to its method, and perfectly accurate.

Sometimes figuring out an explicit formula for something is harder than devising an algorithm. I have not tried the explicit formula yet, but I will next. Right now I want to show you the rough schematic for the DeuceHound (mod ∞). It is what the formula will say and the algorithm do. It is not exactly a flowchart either. Well. See what you think. Where I am saying "print", in the outline, I just mean add. You get the idea. That will be described in the formula. Here I want the flow of the logic. Did I get it?

* * * * *

We can use the technique of adding bits to conquer 2, because it happens to be 2, and computer bits are binary. Only a general formula for all primes might conquer primes other than 2, however.

So we need a technique that will write (add) each bit for any prime. If a power of x is factorialized:

1 Write (x-1)...1's, then the next integer. This loop will be used on every print command throughout the algorithm.

2 Do that entire process x-1 times, then on the xth time write 1...x-1 times then jump to the next integer, which for the prime 2 (our example) simply means to now write down 1 3, since we had already done our process x-1 times the first time we did it. With the prime 3, we would have had to have written 1 1 2..1 1 2..1 1 3. We would write that x-1 times before we came to a 1 1 4 in the Ruler Sequence for the prime 3. But here 2 is our example, because from it we must get the pattern. For 3 we wrote down 1 1 2 ....(x-1) times (twice), before jumping to the next integer 3. Like this 1 1 2..1 1 2..1 1 3..1 1 2..1 1 2..1 1 4....Before we get to 5, we have to do exactly what we just did, except for the last number, which increases by one.

3 We are back with 2 as our example. Write down everything you have again, except jump to the next integer on the last entry, which means write down the 1 2 1 3 you already have, except for the last entry, where 4 appears for the first time, like so: 1 2 1 3..1 2 1 4

4 Can you guess what to do next? Once more write down everything you have written down, but jump to the next integer on the last entry.

5 When you reach the correct power, add all your entries up.

The above would take care of values which are pure powers of any prime. We are using 2 as our prime in this exegesis, just as a familiar example. For values of Q, if Q were small, we would choose a simple algorithm that merely walked up to that value without any recourse to powers, perhaps. If Q were still uncomfortably large, we would instead use the method of descent with that prime, which I showed before for 2's.

6 Add the value for Q to the sum you already have, and that sum is the power of x for that factorial.

So the infinity model of the DeuceHound would then pull all factors of x exclusively from numbers of the type xk!+Q, with the goal of later enlarging its scope to track down the factors of any number regardless of the form it is presented in, whether this involves finding the nearest power of x to the value one is dealing with, or some other technique later to be discovered.

* * * * *

In a nutshell: it adds 1...x-1 times, then adds the next integer. Now it must loop back and add exactly what it just added to the total except for the last number, before advancing to add the next integer, as in a virgin power. You know how Q is handled. So that's it. In the old Procedural platform I could have done it. There would have been a lot of ghastly plumbing and loops, for sure. I understand the OOP is much cleaner, easier, faster.

The DeuceHound may need a trademark and patent or copyright before long.


* * * * *

I do not think anyone else reads this thread. I am amazed you have followed the reasoning in detail. If it is not too personal a question, may I ask your profession before you became an old man like mysef? I am officially retired, but I have spent a total of at least thirty thousand hours in a chair or standing, doing each of the following to get by: playing poker, playing and teaching guitar; and doing the following for fun or ambition probably an equal ampunt of time: reading prose and poetry, writing prose and poetry, and doing amateur mathematics. My intersts continually pass the torch around to each other. Right now math is dominating again, sometime in the future it will be one of the other interests that dominates for a good spell. Since my interests dominate my life, they are about all I do besides kiss my family a lot, eat and sleep and watch movies on my computer. When a subject dominates me it really does, and I let it gladly. There is no resistance or regret.

* * * * *

What I want to gaze at right now is a number line with powers of numbers highlighted, the first power will not be highlighted, because that would mean highlighting every number. I know what I want, but it is hard to describe. I want to get a sense of how close any power comes to another. I do not know of a rule or a law for that. All numbers are on the line, and first powers are merely black as usual, 2nd powers and beyond would be red. I need the ability to look at long stretches far out the number line to see how powers interact. I did compare powers on 2 and 3 for a ways, and after a while they are not particularly neighborly, sometimes with thousands between a power on 2 and the nearest value that was a power on 3. I have a thinking there must have been powers of other numbers that were closer than that, which fell between those big gaps. I am trying to get a sense of how large we can expect Q to be when powers of all numbers are involved way, way out the nuumber line.

[YesNo]

The piecemeal method is like the descent method, a kind of looping or recursive or fractal or mathematical induction process. An explicit formula would be something like n(n+1)/2 for the nth triangular number. The piecemeal formula just starts adding from 1 to n. However, all the iterating can take time after a while.

With OOP one could create a class and build a set of methods supporting that class inheriting standard methods from the parent class. I didn't have to create a class and methods. I just used what Python methods were available in classes already built. However, I was thinking of building a class for a Sierpinski covering, if I ever get to that.

For the other primes there must be more involved than the base used to represent the number n, but maybe it is not as complicated as it looks at the moment.

As far as my profession, what I do could be called software engineering, computer science, data science, or database development. I have degrees in mathematics, but I do not work as an academic and I don't publish papers.

[desiresjab]

Degrees in math? No wonder you can follow this stuff. I was beginning to get suspicious. Only a technically trained person could or even would follow this thread in such detail. Vectorizing components was a pretty good clue. Maybe you should be writing the thread and I should be the one responding.

Anyway, I am still after the formula for DeuceHound (mod ∞). It is in my brain and will not straighten itself out yet. Perhaps a sigma with double indices might take care of the main routine. As for double sigmas, I truthfully must look up how they are used again. I used to know and forgot if it is simple nesting or something else.

[desiresjab]

Looking at the Ruler Sequence for 3, I worked out the explicit formula for the Tre-Tracker mode of the DeuceHound (mod ∞):

F3(3K!)=3k-1.
................2

Sorry for all the dots. They were the only way the 2 would line up where it needs to be.

Will the formula for every odd prime be as simple as this and furthermore be its mimic, establishing a general formula? It seems intuitive that this should be, yet there may be a difference for powers of 4n+1 and 4n+3 primes factorialized, for instance. There may be two separate formulas.

[YesNo]

So, it is (3k-1)/2? I will check that later today.

When you get your new computer we could exchange jupyter notebooks. We can format using mathjax.

[desiresjab]

I do not know anything about Jupyter and Python right now. For a long time I have desired Mathematica. It seems like it might be the ultimate in math software for home users and is not as expensive anymore. There are other packages available from different vendors. I have not had time to study them either. What do you know about all these math packages? I suppose some of them are good for a ways and then show glaring limitations.

We will definitely exchange some notes via internet channels once we can.

I believe I am close to a formula for all odd primes. The formula for 3 is so compact it still might be useful. Beautiful mathematical objects usually fit somewhere. The general formula may be an ugly girl you avoid when the pretty one is at hand.

When one inputs his prime p and power k, the computer checks whether p is 2 or is odd. If 2, the computer performs the by now familiar DeuceHound (mod 2) algorithm. If odd, it makes sure the input p is a prime. Then it does its thing with its alternative algorithm for odd primes.

I think I can get there within a day or two, but I might be fooling myself. First I may have to figure out the rule for what looks to be quite a cranky sequence I have come across.

Math man, my experience has been it is pretty easy to see and apply the rule of certain kinds of sequences, yet often a heavy challenge to find the formula that exactly describes that easy rule. The Fiobinacci sequence is easy to apply, but finding the formula that gives you any element by address in the sequence did not look particularly easy to me when my eyes hastily traveled over it, with √5 mysteriously appearing and all. I believe it is the same thing with the General Ruler Sequence--applying it is easy, but finding and correctly formulating the general rule requires the brain to do heavy reps. Maybe it is just my brain. More experienced mathematicians would probably struggle far less with these same concepts--which might even be described elsewhere. But investigating them seems like a personal job. I feel it is important to know the number mechanics of what is going on at all levels as thoroughly as you can, the way we investigated Quadratic Reciprocity, when one researches mathematical objects. I do not mind getting there without using a life line if I am able. The challenge is what I am into. Doing what it takes to construct such a function will enrich me along the way.

The 2 machine of the DeuceHound works. I also think the name works. Now we need its odd prime machine to work as handily. Once the formula is cracked, programming it might be easier than cracking the formula. I really do not know. I know in programming you get to work with algorithms. Instead of talking so much I should be trancing, or I have a thinking I will never get the plans for the DeuceHound's main engine drawn.

[YesNo]

Regarding the closed form of Fibonnacci numbers, I don't know how to derive it, but then I would just look it up and try to figure it out from there. Here is a source you might already be familiar with: http://www.maths.surrey.ac.uk/hosted-sites/R.Knott/Fibonacci/fibFormula.html

I like to use math.stackexchange. Here is something about that closed form there: http://math.stackexchange.com/questions/90821/how-to-find-the-closed-form-to-the-fibonacci-numbers It is a good place to get practice asking questions and using mathjax which is rather simple to use once you do it a few times.

One of the problems with Mathematica is that it is proprietary software. Not only does that mean that it costs money but I don't think the code implementing whatever you are using can be seen by you. I prefer open source software for mathematics. Python is free and you can download enough modules to get started through Anaconda's distribution. You can also add your own modules. Also, there are thousands of packages available which means a lot of people are using the language.

[desiresjab]

What I mean to say is, the pattern for both 3 and 5 are easy to define recursively, since each value depends on the last, and that one on the last before it. In the case of 3, an explicit formula was easy to find. 5 seems more stubborn, or different, or maybe I am having number blindness. At 7 I have not looked yet. It may be that 4n+1 types and 4n+3 types are different in this context, which makes sense as it is about powers, and that is why 5 will not be nice. Unfortunately, the next 4n+1 prime to test this hypothesis on is 13. The actual values for these tests grow huge fast and therefore harder for a human to spot a pattern or constant by eye. The DeuceHound may end up with three distinct main engines.

[desiresjab]

Jimeney Christmas! I got it!

When you trod the long way around, reasoning everything out for yourself, it takes a while. Which is the long way of saying I know I have seen this formula before. Whether it was used specifically for this or for some other purpose, I do not know (beautiful mathematical objects often have diverse applicability), but one does not forget such beauty even when there is no understanding of it.

Fp(pk!)=

pk-1
p-1

And this beauty even works for 2. I would have seen it earlier had I listened to my own instincts about the explicit formula for 3 being a template for all the other odd primes (little did I know it would work for the even prime, as well), but as usual I had to bull around in the china shop for a while before seeing the truth.

Well, the DeuceHound (mod ∞) is about ready to go. I suppose it has been for a couple of centuries. Sometimes it is so nice being ignorant, because playing discoverer is a great way to learn. It is infinitely better than finding the formula in a book and trying to figure out why it is so. I know why this one is so.

[Dreamwoven]

Congratulations, desiresjab!

[desiresjab]

Thanks, Dreamwoven. I only managed to labor my way into what has to be a well known fact after much sweat. That is the kind of nifty minor object Fermat was knocking out regularly 350 years ago. I am amazed I was able to find it. If the general case for 3 had not been pretty easy to spot, I am likely to have missed it altogether. Now I am really curious what else that formula might be used for. Just because I saw it before does not mean it was used for this. Something like that can be hard to find. Is there a formula hound on the internet?

[YesNo]

Here is an update pdf of the notebook: https://drive.google.com/file/d/0B96SHXzNSomRZ2ZOWHhmMkdiTmc/view?usp=sharing

This passes the initial test for 2 and 3, but not for 5 and 7. It might be a problem with the way I coded it.

[desiresjab]

Here is an update pdf of the notebook: https://drive.google.com/file/d/0B96SHXzNSomRZ2ZOWHhmMkdiTmc/view?usp=sharing

This passes the initial test for 2 and 3, but not for 5 and 7. It might be a problem with the way I coded it.

I don't know what could be wrong. The programming language is so compact I cannot see anything. The explicit formula should give the value of the sequence up to a certain value for n for one prime. Your program appears to want the sum of all primes up to a certain n. It will do that individually to the primes. Adding the numbers of all the primes together will not give you anything, if that is what I am seeing. Am I looking at the program incorrectly? I just got up. I need time to wake up and examine it more closely.

* * * * *

Now we have the ability to sum the Ruler Sequence for any prime. That is with each prime having the separate metric of itself. I guess what we need for Brocard is one metric to judge all primes against. That is what we do not have. We need the abilty to see how the ordered primes are stuffed into a factorial, how many together we need to meet the conditions of 2n(2n+2), if any amount together can.

If we had a way to judge all primes just from the value of the formula for one prime,like 2, then we would have somethibng that might assist with Brocard. We would not have to change the value of p each time we wanted the facts on another prime.

Meanwhile, I have not been able to even find the expression 2p-1/p-1 anywhere else again. It will turn up. It will not be as handy as some equation like d/r=t, used for everything from baking cookies to electronics, but will have quite a few disparate uses, I am wagering.

[YesNo]

I think I got the description of your algorithm wrong. It isn't the sum over the primes but the sum over the powers of a particular prime dividing n. I'll try to get a correction tomorrow.

Edit: In looking it over it seems I implemented the wrong algorithm.

Edit: I updated the pdf file with what I think is the correct algorithm, but I am still getting discrepancies when I try n = 95 for primes 5 and 7. I printed out intermediate results. https://drive.google.com/file/d/0B96SHXzNSomRZ2ZOWHhmMkdiTmc/view?usp=sharing

[desiresjab]

95 is not a power of 5 or 7, which means you will have some value of Q tagging along. The nearest power of 5 is 125, giving us the opportunity to use subtraction on Q for the first time rather than addition and illustrate another symmetry of the Ruler Sequence. The measures on either side of a virgin power are symmetrical, meaning that measure for meausre the sequence will look the same on either die of the virgin power until, of course, the measure containing the next virgin. This means that even when we are subtracting Q instead of adding it, all we have to do is go to the beginning of the Ruler Sequence and add those values up to and including Q, then subtract them from Fp(5k!). No nasty backwards subtractions in the sequence. We can calculate -Q precisely as if we are calculating Q. An example below with p as 5 and k as 3, goes as follows when I simply add up Q by sight rather than instituting the repetitive breakdown on Q as a computer would do and a complete formula designate (which we call the Descent or Reduction method):

F5(95!)=F5(125!)-F5(30!)=

[(53-1)/4]-7=24

If that is not right, cancel the party hats while there is time, lads!

* * * * *

By a common metric for all primes I mean this: plug in the value 2, for instance, for p, and something else for k, and the DeuceHound spits back the Fp for any prime of your choosing up to the value 2k+Q. 2 might be the right metric, too, because it is unique among primes and has the same relationship to every prime (mod 2). Now, that kind of formula is something we are not even close to yet, do not know is frankly possible, and doubt our toolkit in the search for such a master key, none of which shall stop us but only give us pause to whine a little before we proceed. We can acquire or invent new tools and perspectives where they are missing. Our toolbox is light for those hikes far out the number line.

[YesNo]

If we allow subtraction, then we will need a rule to tell when to use subtraction. I think the number of factors of 5 in 95! is 22. It was part of the result of running old_way(95) in the jupyter notebook. Of course, I might have programmed that wrong.

I get a remainder of 45299530029296875 when I run 95! mod 524. That should be 0 if there are 24 factors of 5 in 95!

[desiresjab]

Subtraction is no different from addition. To subtract, you just add. One could, in fact, change the negative sign in front of Q and obtain the same results, which is the whole idea. I am not sure what might be wrong with the program. Let me physically illustrate that the number of factors of 5 in and below 53 is 31, and in and below 30 it is 7 factors of 5.

5.10.15 2025 30
1..1..1..1..2....1..1..1..1..2....1..1..1..1..2... .1..1..1..1..2....1..1..1..1..3

Sorry the multiples of 5 are kind of crowded above the sequence. As you can see, counting backwards from 3 remains the same as counting forward from 1 all the way to just before 3. That is why we can start at the beginning of the Ruler Sequence when subtracting instead of counting backwards from 3. Same thing.

The first 3 in the sequence takes us up to 53. The sixth element in the sequence takes us up to 30. The value of the sequence (added together ) up to 30 is 7. Subtract that from 31 and we have 24, the answer. Notice we could have counted backwards from 53 for the same answer. This means we could also have counted forward six elements from 3 to obtain an identical answer, as well. Hope that helps.

[desiresjab]

One of my beliefs is that mathematics will continue to play a significant role in our comprehension as we go about unfolding the universe for ourselves and exposing deeper and deeper levels of its reality. No one can predict what future mathematics will look like any more than Newton or Fermat could have told you squat about vectors, matrices, topology or complex analysis.

Will this future mathematics consist to a large degree in making headway in classifying and analyzing the set of transcendental numbers? And I have to ask myself, why not? We know almost nothing about them, yet the few we do have cognizance of are of extraordinary importance in the grappling contest with nature. Pi and e are primary to our understanding of nature. Besides those two, which were both finally proven to be transcendental after much effort from great mathematicians, there are a host of other important constants waiting in the wings to be proven transcendental or not. Almost surely every one of them is, but each proof for any individual number suspected to be transcendental is titantically difficult.

How many more important constants await our discovery which are the lynchpins of phenomena we have barely or yet begun to study, like consciousness and quantum reality?

There are more transcendentals in the interval between (1, 2) than there are rational numbers in the universe. In fact, there are more transcendentals in the interval between (1, 1000001/1000000) than there are natural numbers. Make the interval as small as you want, there will always be more transcendental numbers in that interval than there are fractions in the universe.

Why would one not wonder if many great secrets lie unsuspected within this set? It is more numerous than all other sets combined, and all other sets have yielded up great truths about reality.

The unapproachability of this set in general is where the difficulty lies. Liouville finally managed to very cleverly construct some artificial transcendentals. As far as I know, this particular class of transcendental has not proved useful outside of mathematics, though it still could I suppose. No one ever said every transcendental might be useful. In fact, there is no such thing as every transcendental. Their numerosity cannot be counted, or even called numerosity, or even classified beyond Cantor's description of them as having the suspected power of the continuum. Cantor does not know if any set lies between the "countable," infinities and the uncountable continuum or not.

This why I cannot help but feel great paradigm shifts lie ahead which will directly correspond to our increasing understanding of this set of numbers, after the appropriate human lag in time to understand what mathematicians have discovered. This lag is always part of our reality. Leibniz was three centuries ahead of digital computers, but understood already the basic concept of such computing machines. Boole was a century and a half ahead of his time; Archimedes played around with or very near to the fundamental ideas of calculus. Then there was Liouville who managed to artificially construct a transcendental number never seen before. How many centuries will it be until someone goes through the door Liouville could only open a tiny crack?

[YesNo]

I wonder how Liouville constructed that artificial transcendental number. He would have to make sure it could not be the root of a polynomial with rational coefficients. Here is one paper that looks promising but I haven't read it: http://deanlm.com/transcendental/construction_of_a_transcendental_number.pdf

[desiresjab]

I wonder how Liouville constructed that artificial transcendental number. He would have to make sure it could not be the root of a polynomial with rational coefficients. Here is one paper that looks promising but I haven't read it: http://deanlm.com/transcendental/construction_of_a_transcendental_number.pdf

I have not read the link yet either. Here is how I have read Liouville constructed his number. In every digital place of his number he put a 0, unless that digital place was the value of a factorial. He put 1's in the 1rst, 2nd, 6th and 120th digital places, etc. His number then looked like this:

11000100....1000000000...00000001... ...

....................↑ 24th digital place .......↑... 120th digital place.

He had to undertake to prove that this was transcendental. This may be intuitively clear, but I cannot quite make it out. More likely it is a very involved monster to prove his number is a tranny. But wait, he only constructed this number in the first place because he knew beforehand it would be a tranny. He had a concept and carried it out, then. If that was intuitively clear to him, then it must be possible for the same concept to be intuitively clear to us, I would think. We need to reverse engineer it. I believe that is what he must have done, once he realized the concept mentally.

Those occasional 1's mean a remainder of 1 when the Liouville number is divided by powers of 10. Any partial representation of the Liouville number ending in a 1 and divided by powers of 10 smaller than itself would also leave a remainder of 1. Don't ask me what this means. I only see it, I don't know how it fits into his proof. I would say he was a clever man.

[YesNo]

That construction would guarantee it is not a rational number, since the digits do not repeat, but how to show it is not algebraic?

One can come up with infinitely many transcendental numbers by us of the Gelfond-Schneider theorem. If a and b are algebraic numbers with a not equal to 0 or 1 and b not a rational number, then ab is transcendental.: http://sprott.physics.wisc.edu/pickoveR/trans.html I don't know how that theorem was proven either.

[desiresjab]

I wonder how Liouville constructed that artificial transcendental number. He would have to make sure it could not be the root of a polynomial with rational coefficients. Here is one paper that looks promising but I haven't read it: http://deanlm.com/transcendental/construction_of_a_transcendental_number.pdf

After having read the link, I see they point out what intutively makes this number irrational--its repetend never repeats. Repetends and the pattern of digits recurring after a decimal point are essentially the same thing. That makes it at least an irrational. The rest of the proof proves that it is of the transcendental variety of irrational number. Besides a little set theory notation, the proof mainly involves algebraic integers and several bounding theorems from calculus.

[desiresjab]

That construction would guarantee it is not a rational number, since the digits do not repeat, but how to show it is not algebraic?

One can come up with infinitely many transcendental numbers by us of the Gelfond-Schneider theorem. If a and b are algebraic numbers with a not equal to 0 or 1 and b not a rational number, then ab is transcendental.: http://sprott.physics.wisc.edu/pickoveR/trans.html I don't know how that theorem was proven either.

I have not read this link yet. But I have seen this result before, and I must say it always surprised me that the class jump could be made so easily. Somewhere down in the the mechanical process of multiplication which a power is, the numbers are doing exactly what they must when a strictly algebraic irrational is used as an exponent on another algebraic number whether it be rational or not (is the way I took it). These number mechanics I believe we can only see from a higher level of abstraction, we cannot see the clicking of individual numbers in the process and what turns the result into a transcendental, the way such a vision of number mechanics might enable a complete mechanical understanding QR (notice I say might).

That might mean learning the theory of algebraic integers better. In number theory this is where polynomial equations are used to perform the operations of arithmetic upon each other instead of regular numbers doing it to regular numbers. If they are algebraic integers, then I believe (but don't know) they obey all the laws of integers in some form. We should remember this result and stay aware of it whenever we encounter the term.

I believe I remember reading that algebraic integers even have their own version of prime numbers. These may be more than something trivial like √(x+1)=7. I do not know a great deal about the theory of algebraic integers--big ol' polynomials you can treat just like integers in your calculations, is how I think of them.

[desiresjab]

Things which are likely shoo-ins still must be formally proven in mathematics before they can be accepted. One smiles to see that we are free to assume 2π and 2π are transcendental, but we cannot assume ππ is.

[desiresjab]

That weird symbol that did not turn out well in the last post is pi.

[YesNo]

I was looking more at Liouville numbers. This Wikipedia link seems to contain the basic information along with a proof that these numbers are transcendental: https://en.wikipedia.org/wiki/Liouville_number

The proof depends on the concept of the "irrationality measure" of a number which is a measure of how close the number can be approximated by rational numbers. Rational numbers have an irrationality measure of 1. Basically, they are not irrational. Algebraic numbers that are not rational have irrationality measure of 2. Transcendental numbers have an irrationality measure of 2 or greater. Liouville transcendental numbers have infinite irrationality measure. They are kind of extreme as far as this measure goes. It looks like they were constructed to make sure they were as far away from being algebraic as possible which allowed them to be more easily proven to be transcendental.

The algebraic numbers are roots of polynomials with integer coefficients. They include the rationals which are roots of linear polynomials f(x) = rx + s where r and s are integers. Algebraic "integers", a subset of algebraic numbers are roots of such polynomials where the coefficient of the highest power of x is 1, that is, they are roots of "monic" polynomials, such as, x - 7 would have the root 7, an integer.

[desiresjab]

I was looking more at Liouville numbers. This Wikipedia link seems to contain the basic information along with a proof that these numbers are transcendental: https://en.wikipedia.org/wiki/Liouville_number

The proof depends on the concept of the "irrationality measure" of a number which is a measure of how close the number can be approximated by rational numbers. Rational numbers have an irrationality measure of 1. Basically, they are not irrational. Algebraic numbers that are not rational have irrationality measure of 2. Transcendental numbers have an irrationality measure of 2 or greater. Liouville transcendental numbers have infinite irrationality measure. They are kind of extreme as far as this measure goes. It looks like they were constructed to make sure they were as far away from being algebraic as possible which allowed them to be more easily proven to be transcendental.

The algebraic numbers are roots of polynomials with integer coefficients. They include the rationals which are roots of linear polynomials f(x) = rx + s where r and s are integers. Algebraic "integers", a subset of algebraic numbers are roots of such polynomials where the coefficient of the highest power of x is 1, that is, they are roots of "monic" polynomials, such as, x - 7 would have the root 7, an integer.

I thought Liouville numbers could be more closely approximated by rational rumbers than pi and e. It looks like I could get pretty close with Liouville numbers. Just those occasional pesky 1's are in the way of the exact value.

[desiresjab]

I do not know what to look into next. I hate being stuck between projects worse than being stuck on a project. When this happens it is best to go study until a relevant project suggests itself. It helps to have a Brocard or suspected Brocard connection.

[desiresjab]

Time to go to school, folks, for me, that is. I write my own lessons, gleaned from various sources for ideas. I hope you will audit the class. My knowledge of number theory is a patchwork rather than a logical procession up the heirarchy of concepts. I continually find it necessary to backtrack and learn things I missed in my initial excitement to forge ahead. I usually also find it necessary to revisit sites of conflict more than once before concepts sink in. Often, I am satisfied for the moment to glean an important idea of the concept that was easier than I thought it would be, whereupon I will once again set out for amateur waters where actual work (as in absorbing concepts fully) is done in a leisurely fashion more in accordance with sloth. The current topic is such an area of Congruence theory. I have a mediocre understanding of it, now it is time for a complete understanding.

The main idea of the Chinese Remainder Theorem is not hard to understand. It has an analogy in normal algebra—solving a system of equations. Here we solve a system of congruences. This means you will have different modulii. The task is find one modulus that works for everything and solve for n. This turns out to be a delightfully easy concept—you just multiply the various modulii together for the common modulus. The restriction on this is the modulii have to be relatively prime pairwise. This means any two you choose will be relatively prime.

Now, solving a setup problem does not mean one can see everything a theorem implies. Far from it. Learning to see the relvance of basic number theoretic functions in live situations is the more important part of learning them, and comes with experience. After we see how to solve a basic setup problem in this field, we will take a look at something the theorem implies which would have been quite, quite hard to forsee. One would really have to have an instinct for numbers to see this out of the blocks. First, a typical setup type problem.

In a system of linear congruences I could choose my numbers congruent in each modulus at random, and the theorem still guarantees solutions in integers.

* * * * *

Problem:

Find a number n such that when divided by 3 leaves a remainder of 2, when by 5 leaves a remainder of 1, and when divided by 7 leaves a remainder of 1.

This implies that

35n≡70 (mod 140)
28n≡28 (mod 140)
20n≡20 (mod 140)

n would equal 21n-20n, right? It so happens we can set this up. We have

3(35n-28n)=21n

3(70-28)-20=106 (mod 140)=

210-84-20≡106 (mod 140)=

Any number in the same residue class as 106 (mod 140) is a valid solution.

* * * * *

Okay, very cool. Now, what does such a theorem imply that we might not necessarily see right away? What animal would think it means we can solve the following?

Problem:

Can one find one million consecutive integers that are not square free?

[YesNo]

I thought Liouville numbers could be more closely approximated by rational rumbers than pi and e. It looks like I could get pretty close with Liouville numbers. Just those occasional pesky 1's are in the way of the exact value.

Yes, that is how I see it as well, but the idea of "close" is different. If we pick a real number, x, and an arbitrarily small but greater than zero value, epsilon, then there are infinitely many rational numbers (p/q where p and q are integers) close to that x no matter what x we pick. That is |x - p/q| < epsilon for infinitely many values p/q.

So that idea of closeness isn't going to help differentiate rational from algebraic or transcendental numbers since there are infinitely many rationals close to any real number.

One way out of the problem is to let epsilon vary depending on the rational number, p/q. If we replace the constant epsilon with a function of the rational number we could get something like |x - p/q| < f(p/q) = 1/q. This would allow epsilon to vary, but it is still not adequate. There are still infinitely many rational numbers close to any real number, x.

One way to tighten the function is to raise q to some power. If we replace f(p/q) = 1/q1 with f(p/q,u) = 1/qu then if u > 1, according to Wikipedia, I don't quite see it yet, only finitely many rationals could approach any given rational number. If u > 2 then we can say that only finitely many rationals approach even irrational algebraic numbers. If that is the case, then we could use this to distinguish between rationals and irrationals and between irrational algebraic and transcendental numbers. We could define a function of x that gives the precise u value back for which the change occurs between having infinitely many rationals approximate x using this new idea of closeness to having only finitely many rationals approximate it.

[desiresjab]

Yes, that is how I see it as well, but the idea of "close" is different. If we pick a real number, x, and an arbitrarily small but greater than zero value, epsilon, then there are infinitely many rational numbers (p/q where p and q are integers) close to that x no matter what x we pick. That is |x - p/q| < epsilon for infinitely many values p/q.

So that idea of closeness isn't going to help differentiate rational from algebraic or transcendental numbers since there are infinitely many rationals close to any real number.

One way out of the problem is to let epsilon vary depending on the rational number, p/q. If we replace the constant epsilon with a function of the rational number we could get something like |x - p/q| < f(p/q) = 1/q. This would allow epsilon to vary, but it is still not adequate. There are still infinitely many rational numbers close to any real number, x.

One way to tighten the function is to raise q to some power. If we replace f(p/q) = 1/q1 with f(p/q,u) = 1/qu then if u > 1, according to Wikipedia, I don't quite see it yet, only finitely many rationals could approach any given rational number. If u > 2 then we can say that only finitely many rationals approach even irrational algebraic numbers. If that is the case, then we could use this to distinguish between rationals and irrationals and between irrational algebraic and transcendental numbers. We could define a function of x that gives the precise u value back for which the change occurs between having infinitely many rationals approximate x using this new idea of closeness to having only finitely many rationals approximate it.

That will take a lot of thought to reason out. I read your post only once because I am tired. I only partially understood. The idea of a number only being approached by finitely many rationals is foreign to me, which is not bad but hard.

[YesNo]

All numbers have infinitely many rationals within any interval around them. What makes the set finite is that there is an extra constraint on the rational numbers, p/q. Not only must they be close, they must also pass the condition that |x - p/q| < 1/qu where u gets larger than 1. Come to think of it if the Liouville numbers have u arbitrarily large then they will always have infinitely many rational numbers approximating them and fulfilling this new condition. They are an extreme form of transcendental number.

[desiresjab]

I have to be gone for a few days again. I will be thinking about the Chinese Remainder theorem. There is a less painful way to do it. I am trying to understand the precise logic behind that method. Once you get more than three modulii to work with it is real hard to enact the method I showed earlier, because it requires brain-twisting logic. The new method is more straightforward though a little longer. I will understand it before I present it. Understanding of this method would bring us well along on our goal to a complete comprehension of the CRT. It does not enable us to see how far the influence of the theorem spreads or how many situations that look diverse can be handled by it, but it will be a good start.

[YesNo]

Regarding those Liouville numbers, it occurred to me that a way to describe these numbers is to say that they are numbers that can be approximated by a sequence of rational numbers whose denominators are positive, but very small.

The Wikipedia article says: A Liouville number can thus be approximated "quite closely" by a sequence of rational numbers. https://en.wikipedia.org/w/index.php?title=Liouville_number

The metaphor "quite closely" is misleading. One should always be able to find a sequence of rational numbers that approximates the Liouville number even closer than the sequence used to verify that the number is a Liouville number. The only problem with that closer sequence is the denominators of those rational numbers used in that closer sequence would likely be larger than those in the sequence of rational numbers used to show that the number was a Liouville number.

[desiresjab]

I have learned the secret of the Chinese Remainder theorem. The key involves mod inverses, which are hardly ever used in the proof, I believe. Do not have time now to explain it to our throngs of readers, but will do so when I return from my travels in about three days.

Learning these basic number theoretic functions inside out is another key to number theory. One cannot know them just so-so. Inside out, so that when one is applicable, you are sure to see it instead of putting in a lot of wasred effort. There is no other way.

[YesNo]

Problem:

Can one find one million consecutive integers that are not square free?

So each of these million consecutive integers must have at least one prime to the second power?

[desiresjab]

So each of these million consecutive integers must have at least one prime to the second power?

Yes, no repeated factors in a square free prime factorization.

I am back, but too beat to do a good exegesis on the Chinese Remainder theorem, or even think of it now. To only half see it would be utter failure, and I think I half see it, so I will be at rest here until I go on. I have allowed myself to be confused. Of course I cannot allow that. Ahem!

On the matter of transcendentals, my mind wanders far into the future to wonder if any theory can exist to locate important ones. So far we have just run into numbers like pi and e, or discovered them in a sense out of data. In calculus ex is the number that has itself for derivative, and pi the ratio of the circumference of a circle to its diameter. The Feigenbaum constant was discovered after iterations in chaos theory were noticed to converge. Might there be a seive invented by advanced minds of the future to strain out useful transcendentals, rather than having to discover each one in action? Maybe the question is crazy. I have no idea what the system of constraints would be, but the constraints of the Liouville number might be an echo of that theory. Not a scientific observation, I know, just futuristic musing.

[YesNo]

If no repeated factors are allowed, then we could have only a sequence of three consecutive integers since 4 = 22 divides one of every four consecutive integers.

I was thinking about the DeuceHound. I think the general idea of using only n to find all the prime factors of n! is solvable. That is, one does not have to construct n! and then factor it to get the prime factorization. One can get that from working with n itself. Here is a video on the topic that explains one technique: https://www.youtube.com/watch?v=HkAKM2lfvAA The problem with this technique is that it uses an iterative approach by looping through all the powers of a prime in n rather than a closed form to get the number of factors of a prime in n!.

[desiresjab]

If no repeated factors are allowed, then we could have only a sequence of three consecutive integers since 4 = 22 divides one of every four consecutive integers.

I was thinking about the DeuceHound. I think the general idea of using only n to find all the prime factors of n! is solvable. That is, one does not have to construct n! and then factor it to get the prime factorization. One can get that from working with n itself. Here is a video on the topic that explains one technique: https://www.youtube.com/watch?v=HkAKM2lfvAA The problem with this technique is that it uses an iterative approach by looping through all the powers of a prime in n rather than a closed form to get the number of factors of a prime in n!.

I have not read the link yet. I am excited about doing so. Lying in bed a few moments ago I was thinking about the DeuceHound, too. It can already take its place among valid and useful number theoretic functions, claiming its own identity because it is based directly on the Ruler Function. The Achilles heel of the DeuceHound and of Legendre's floor function method is their inability to find primes. They are able to manipulate what they are told are primes, they do not find these primes themselves. Any machine for finding primes and manipulating them would necessarily be vast, since finding and testing for primes is the really difficult part. As long as it is told which numbers are prime, the DeuceHound will do fine.

Storing a list of primes in the computer seems quite crude to me. But that is where the human race is on this job.

[desiresjab]

I looked at the video in the link and the three that followed it. No surprises. He is using the Floor Function algorithm. It is fast and general. The Deucehound is even faster and easier where pure powers of a prime are factorialized, because it is an explicit formula in those cases and Q is equal to zero. Powers factorialized are something that might be found in problems involving the natural sciences. In case Q does not equal zero, we know how to append the value of Q to our total.

With a non-zero value for Q, is the DeuceHound as fast as the Floor method? It is quite close, I think, and has the advantage of an explicit formula for cases where pure powers are factorialized. The DeuceHound has definite similarities to the Floor Function, but is not exactly the same thing, since the Floor Function is not based on the Ruler Function. One is certainly reminded of it as one figures the value of Q, especially.

[desiresjab]

I am ready to wrap up the series on the Chinese Remainder theorem. We would be stunned if such a theorem did not exist, it is such a natural consequence. If x divided by p leaveas a remainder of a, and x divided by q leaves a remainder of b, it is hardly surprising that x divided by ab will also leave some unique remainder as well, is it now? That is the simplicity of the theorem in plain English. As ususal, the situation in mathematical notation is more difficult but more precise. But anyone should keep in mind the plain English interpretation of the Chinese Remainder theorem above when studying its mathematical details.

We already looked at a method that sometimes works easily for a system of exactly three modular equations. Later in this post or the next post we will look at the most general method for solving systems of congruences.

The method below is extremely simple for two modulii, which we now digress a moment to cover.

x≡2 (mod 3)
x≡4 (mod 5)

5x≡10 (mod 15)
3x≡12 (mod 15)

5x-10=3x-12. Now simply solve for x.

5x+2=3x

2x=-2

x=-1=14

Indeed 14≡2 (mod 3) and 14≡4 (mod 5).

This method is so easy, I recommend it whenever there are only two equations in the system.

* * * * *

For those cases when we have three or more modulii, we want to show and explain the method that is at the heart of matters and will always work.

x≡a1b1M+.....arbrM(mod M)
...........m1...........mr

Is the general description.

M is all the modulii multipled together. M divided by any mi is M without that mi. The b's are the inverses (mod mi) of

M
mi

The big question is why do we want these inverses, how did they get involved?

You might say they are involved because mathematicians wanted them involved. In order to isolate the various ai's so they can be added (mod M), it is necessary get rid of the terms around them. This is accomplished by multiplying

M
mi by its inverse (mod mi)

* * * * *

Let's do a classic example where I will flat make up the numbers. It involves an old lady from the village riding her bicycle to town with a large bag of eggs to sell who is knocked over. All but 160 of her eggs are broken. The culprit, a clumsy but honest mathematician, offers to reimburse her on the spot for damages. The old lady however, being an odd sort, does not remember how many eggs she had, but she does remember a few other details. When she counted them by 3's, there was one left over; when she counted them by 4's, there were three left over, when she counted them by 5's, there were two left over; when she counted them by 7's, there were five left over. At ten cents per egg, how much did the poor mathematician have to pay the old lady?

Step1

x≡1 (mod 3)
x≡3 (mod 4)
x≡2 (mod 5)
x≡5 (mod 7)


Step 2

140x≡140 (mod 420)
105x≡315 (mod 420)
84x≡168 (mod 420)
60x≡300 (mod 420).

In the first column we have divided 420 by each modulus, and in the second column multipled that times the value of x its original modulus. We must now find the inverses of 140, 315, 168 and 300 in their original modulii.

Step 3

(140)-1 (mod 3)=2,

(315)-1 (mod 4)=3

(168)-1 (mod 5)=2

(300)-1 (mod 7)=6


Step 4

Multiply each Mi by both its original value (ai) under the old modulus, and its mod inverse under the old modulus. This indeed must isolate the ai's so they can be added (mod M).

(140)(1)(2)+(315)(3)(3)+(168)(2)(2)+(300)(5)(6)=

280+2835+672+9000=12787

12787≡187 (mod 420).

160 of her original 187 eggs are unbroken. That means the clumsy professor broke only 27 eggs. At a dime apiece, he owes the old gal a mere $2.70.

* * * * *

This method will always work. We may not be quite through with the theorem, though, for I have found some other articles that even explain it graphically, which I intend to look at.

The important thing, of course, is that we make ourselves able to recognize when particular number theoretic functions are relevant and useful in situations encountered in the wild. Without this ability we are only able to answer prepared questions that are carefully worded to let us know which function we should be thinking about.

[desiresjab]

Speaking of those little important consequences of theorems which conceal themselves in so many places, I just learned one concerning Fermat's Little Theorem, which, had I known it earlier would have greatly aided my efforts at an original proof of the theorem could I have found a clever, non-circuitous way of proving the egg first without the chicken. Anyway, I think it is quite important and a good illustration of why one must be ever watchful for consequences of the theorems one learns, if one ever hopes to become a master of numbers in the wild. This simple overlooked or at least under appreciated fact is that the theorem offers an alternative way to compute mod inverses when a Є *Zp, for a-1 can be computed as ap-2, since we have a·ap-2=1 by the theorem. Yes, circuituous, but an important fact nonetheless, whether or not it can be used constructively in a valid, original proof of Fermat's Little Theorem.

Now I feel compelled to go back and look at my own work with this new restatement in hand. It is possible I even noticed this before but felt I could not use it precisely because I felt the reasoning would be circuitous--like using a word in its own definition. I do not particularly need to, it is more like a drive to complete something I started and was unable to satisfactorily finish. Practice at proving is one aspect of math I could always use work on. Any function one already knows and can make the connection between is fair game for use in a proof of any theorem, including this old one. My way was a visual demonstration followed by an attempt at algebraic proof. If anything interesting comes of my revisit, I will report.