Cosmology

[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 2^n/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 2^n/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/Montgo...iplication.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/Montgo...iplication.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 2^k 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 2^k)+2, is always two.

(C) Conjecture: When any number of the form 2^k)-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).