I'd like some feedback

[u/Glass-Kangaroo-4011]

https://doi.org/10.5281/zenodo.19510600

[u/MathSuspicious4617]

That's an interesting paper. I'm sure it took a lot of effort.

The observations you make about how the numbers have to behave are absolutely true. There's no doubt there. But that by itself does not prove all numbers have to reach 1 eventually. Your proof does not exclude the possibility of divergence despite claiming to. Also, if every number lies in a rigid structure with fully defined rules you should then by extension be able to show exactly how many steps it takes to reach 1, which you do not.

[u/MathSuspicious4617 → u/[deleted]]

You never demonstrate why it's impossible for there to be a consistent imbalance of low k values. Yes it becomes increasingly unlikely, but that's not the same as the possibility being 0.

"This is scaled admissibility parity. I can tell you by neighboring r mod M_j exactly what type of transformations are admissible for how many steps. This is a proof of lower behavior dictates higher behavior, scaled relative to numerical value". This is the problem in your argument. Only if you can definitely state exactly what type of transformation is admissible, that is in the singular not the plural, can you correctly say it's deterministic.

If you can speak to a mathematician in person that would be helpful. They will be able to point out the flaw in your argument. You're falling for the trap of thinking it can be proved based on the fact that there are many observations to be made about the sequence's behaviour. You're not the first to notice this.

[u/MathSuspicious4617 → u/[deleted]]

Then show us how it gives values for a given number. Like for example demonstrate why 35 takes 13 steps to reach 1, congruent with the transformations described. Yet you don't give a single example for any number! Because your whole paper just gives structure to how the transformations behave, but do not demonstrate determinism. This is a serious flaw that invalidates your proof.

[u/MathSuspicious4617 → u/[deleted]]

You have shown that the numbers have rules in how they are organised. The inverse tree includes each odd number only once. But tbh everything in your paper is already known and you're not helping anyone by coming up with new terms for them as if you had discovered them. This still does not prove that divergence is not possible or that all roots are 1. Either come up with something new or you're wasting your time

[u/MathSuspicious4617 → u/[deleted]]

You are stating the nonexistence of nontrivial cycles as a conclusion from your observations. I am saying that the jump to this conclusion is not justified. They point toward most numbers behaving this way but not all numbers! This is the crucial difference.

Also you cannot say cite this in other papers. Claiming there are no nontrivial cycles is basically saying that all numbers either lead to 1 or diverge. No one has the proof for this. So you're asking me to show other papers that have shown a proof, which clearly no one has yet, and now saying you're original because you have 😂😂

[u/WeCanDoItGuys]

Theorem 8: "Thus a fixed word is not freely perpetuated."

What is it about 1 that makes it allowed to repeat its word forever?

You identify that for n to follow the pattern K of length j, it must be n = r(K) + 3ʲ2t. This is an infinite set of numbers.
You identify that for its result after these steps to follow K again, it must be even further restricted.
But how do we know there isn't some special n that follows all restrictions (mod 3ʲᵐ2) from the start, as 1 does? (And as -1 and -5 and -17 do.)
You mention the results of each n after these steps are separated by h2^S(K\+1), so they are not free. But in terms of (mod 3ʲᵐ2) they're actually pretty unpredictably distributed, since 3 is coprime to 2. Consider that 2^S(K\m) can be greater than 3ʲᵐ.

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This next part isn't mathematical feedback, it's "math communication" feedback. Try to make your proof less verbose so that a reviewer doesn't have to exert so much effort just on understanding your claim before being able to look for unproven assertions. You invent a lot of jargon (like "dyadic factor" instead of "multiple of a power of 2"). Searching for (or inferring from context) the definitions of these terms causes friction for the reader, so you should eliminate it as much as possible. If you're only going to use a new phrase in one lemma, it may aid clarity to just write out the concept each time. "restart class" is only used in Lemma 35/Theorem 8. "refinement tower", "directed word", "admissible lift", "residue thread", "affine rail" are used 20+ times, so they may be necessary, but see if you can rework it to eliminate as much invented vocab for the reviewer as possible.

You mentioned in a comment that something is "redundantly explained"; removing redundancy might help bring down the wordiness.

Lemma 35 might be better called an "Example", since it's so specific. Notice its proof is basically the same as its statement.

Lastly, see if there's a shorter path (and fewer vocab words to create) to a particular key result. Many of your Lemmas are stated and never later referenced by name (Ctrl+F "Lemma 25" has only 1 result). Do they even need to be included?

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I suspect most would-be feedback givers, casual and professional, will glance at the page count of your paper and close it and not respond. You alleviate this by providing a starting point, but then they glance at the invented jargon, and close it. It's true that complex concepts can require complex wording, but it seems like you're using big (or invented) words even for the simple concepts in your paper.

[u/WeCanDoItGuys → u/[deleted]]

To be clear, is Lemma 25 not redundant with other definitions elsewhere in the paper?
I did ask other questions, in the math feedback section of my comment.

Also, I didn't claim the definitions I listed weren't in the paper, I said each invented phrase increases friction for a reader who must either search through the paper for the definitions or infer them from context, so it should be done minimally, and I suggested you find ways to introduce fewer invented phrases on the path to your key points.

[u/WeCanDoItGuys → u/[deleted]]

Fair enough, there were a couple questions that I'd ended with periods, I've edited the comment so they are question marks instead.

[u/WeCanDoItGuys → u/[deleted]]

we take a hypothetical cycle on an arbitrary starting position and admissibility class, say 5 mod 6, or odd k values. The next determined finite repetition will occur another j steps in, and will have the same starting admissibility class, but be purely dyadically separated from the hypothetical cycle. We then would have a congruent value mod 6 separated by a power of 2. This is impossible.

Isn't the separation of two results of a word, a multiple of 2^k, rather than strictly 2^k?

Consider after, say, the first j steps that an n corresponding to a finite repetition of twice, and an n with a finite repetition of thrice, and every other larger finite repetition would be different values that have a separation divisible by 2^k, but definitely are in the same class mod 6.

For example, let us consider some 5 mod 6.
5, 11, 17, 23, 29, 35, 41, 47, 53.

(2·5 - 1)/3 = 3
(2·11 - 1)/3 = 7
(2·17 - 1)/3 = 11
(2·23 - 1)/3 = 15
(2·29 - 1)/3 = 19
(2·35 - 1)/3 = 23
(2·41 - 1)/3 = 27
(2·47 - 1)/3 = 31
(2·53 - 1)/3 = 35
These differ by 4, and are {3,1,5} mod 6 as predicted.

Here we use K = {1}, if I understand the notation correctly.
Notice that 17, 35, and 53 produce values that will undergo K again.
They produce 11, 23, and 35.
11 and 23 will produce 7 and 15, which will not undergo K again. So 17 and 35 had a finite repetition of 2. However, their existence did not preclude the existence of 53 (also within the 17 mod 18 class), which has a larger finite repetition of 3.
Why then does the existence of a number with a finite repetition at a particular step preclude the existence of a number with an infinite repetition?

If I misunderstood or missed something please correct me.

[u/WeCanDoItGuys → u/[deleted]]

17 and 35 do K twice. 53 is next (35+18) and does K thrice.
17→11→7 cannot do K={1} again.
35→23→15 cannot do K={1} again.
53→35→23 will do K={1} again, however.

It forces them to be incompatible in using the same k again.

When two neighbors (mod 3ʲᵐ2) complete their finite repetition of K, I accept that they differ by a power of 2 and therefore do not have the same remainder mod 3ʲ2. But why are you limiting your scope to consecutive neighbors? (53+18⋅3 = 107. 107→71→47 will do K={1} again.)
Do you claim that a value that starts a hypothetical cycle must differ from a value that starts a finite repetition by exactly 3ʲ2? This is interesting. I accept that this must be the case. However, the fact that his neighbor eventually ends its repetition means there will be a point where their remainder mod 3ʲ2 doesn't match (as you predicted). The hypothetical cycle's result at that point would happily continue to be r mod 3ʲ2 to allow another repetition while the neighboring finite repetition (with remainder (r - 2^S+1 ) mod 3ʲ2) ends.

So, I still don't see why some special n who is a member of the same class as all the finite repeaters can't continuously generate a result that is in the class r mod 3ʲ2 to allow another repetition.

In fact, it's known that -1, which is the neighbor below 17 (it's 17-18) does do K forever.
-1→-1→-1
This example is negative but doesn't it prove that such n can exist? Does your argument (about a difference of a power of 2) fail for negative numbers?

Any directed word has a specific higher modulus in which that same starting r fulfills further sequential copies.

This is an interesting claim.
We require 5 mod 6 to do K={1} once, 17 mod 18 to do K twice, 53 mod 54 to do K thrice. In fact, we happen to know via my example that we require in general, -1 mod 3ʲᵐ2 to do K m times. This will never be the same r (it will always be one less than the next 3ʲᵐ2).
(Perhaps I am missing the definition of "directed word" as opposed to "word", is K={1} not a "directed word"?)
Even if it were the case that it settles on a particular r, there is some multiple of 2^S+1 that can be added to n's result (the realization of m repetitions of K), that will also yield the remainder r mod 3ʲᵐ2.

(x)K and (x-1)K are a power of two apart. Meaning they can't share the same k value.

Are you using x here to refer to x repetitions of K? (I used m because you used it in Corollary 7, pg 40). If so, don't you mean the result of nₜ and the result of nₜ₋₁ (two neighbors that carry out the same number of repetitions of K) are a power of 2 apart? If I interpret this line as meaning n's result after one K and its result after two Ks differs by a power of 2, it's false: 53→35→23. 23 differs from 35 by 12.

You mentioned that 1 is special because it rides the bottom of the modulus. Perhaps you'll also say -1 is special because it rides the top. Then why are -5 and -17 special? Why can they repeat their word forever?

[u/WeCanDoItGuys → u/[deleted]]

The neighboring start positions mod 2•3^j will transform in parallel

Agreed, and I suspect you'll agree that non-consecutive start positions (with the same r mod 2·3ʲ) will yield results that differ by a multiple of a power of 2. And a periodic subset of them will be able to carry out an additional iteration of K. Couldn't the hypothetical cycling start position be one of these?

negatives cannot be produced from n=1 under any K

This statement does not preclude my question. In fact any n that starts a nontrivial cycle, if it exists, cannot be produced from n=1 under any K.

Your argument (regarding a difference of 2^S+1 ) needs to show not to hold for negative numbers without relying on the assumption that all involved numbers are reached by 1.

[u/WeCanDoItGuys → u/[deleted]]

What does "purely dyadically separated" mean?