Cosmology

[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)+..+2^k-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 (2^k)! is 2^k-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{(2^k+Q)!}, Q=F₂{(2^k-Q)!}, because Q=(2^k-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.

(2^k-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{2^k}=(2^k-1), of course.

Yes, friends, the Ruler Sequence not only repeats itself backwards from 2^k, it also repeats itself forward from 1 with exactly the same numbers and forward from 2^k 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 2^k 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{(2^k)!}+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 2^k 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.