Cosmology

Proof of (C):

See the operations below. Further observe that (2·8)! is of the form (2·2^k) because it equals (2·2³), with k equaling 3 in this case. Therefore (2·7)! is of the form (2·2^k-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 2^k-1-k; the number of factors of 2 in the upper interval would be 2^k-1. Subtracting one from the other we get:

(2^k-1)-(2^k-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).

Quote:

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Originally Posted by desiresjab

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...

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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.

Now we are able to look at any

(2·2^k)! or

(2·2^k±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.

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

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.

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.

Quote:

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Originally Posted by YesNo

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

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I don't understand your question.

Quote:

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Originally Posted by desiresjab

I don't understand your question.

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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.

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.

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!

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 (2·2^p)=(2^k)

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 (2·2^p)=(2^k), 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.

* * * * *

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

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

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

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

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

The Demonstration is complete.

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.

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.

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.

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.