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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.
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!
As a ring the binds them is a good way to look at getting a formula that works.
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.
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?
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.
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.
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.Quote:
Originally Posted by YesNo
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.Quote:
Originally Posted by YesNo
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.
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.
Have you taken the DeuceHound (mod 2) for a test ride?
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.
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.
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.
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.
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.
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/0B96...ew?usp=sharing
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.
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.
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.
It could be done this way.
F2!{2k+Q}=
[k-11∑ [2n]]-1+ F2!{Q}.
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.Quote:
Originally Posted by 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.
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.Quote:
Originally Posted by 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/0B96...ew?usp=sharing
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.
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.
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/0B96...ew?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.Quote:
Originally Posted by YesNo
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.
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.
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.
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.
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.
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.
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...ibFormula.html
I like to use math.stackexchange. Here is something about that closed form there: http://math.stackexchange.com/questi...onacci-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.
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.
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.
Congratulations, 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?
Here is an update pdf of the notebook: https://drive.google.com/file/d/0B96...ew?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.