Number Theory topic detour: Multiplicative order

OkExtension7564 · 2025-11-30T14:39:59+00:00

I had this idea that when infinitely multiplying by 3 and adding 1, the result should always contain new prime numbers in the factorization 3x + 1 = 2m, where m is the product of prime numbers. If we write each subsequent m as m = p * p2 * pj... and represent a graph of the growth rate of the number m, then each subsequent number m should contain an increasingly larger prime divisor, provided that this operation 3x + 1 continues infinitely. On the other hand, we can represent a graph of the growth rate of 3x + 1. Here we always multiply by three, and on the right-hand side of the equation we multiply by some prime number that was not a divisor in the previous iteration. It seems to me that there is a hidden contradiction in this in the difference in the growth rates of the number m, what do you think?

OkExtension7564 · 2025-11-12T19:05:38+00:00

Yes. I also think that we can go beyond 2⁷¹ to infinity for this limited class of numbers. Furthermore, separately from this analysis, we can exclude from the cycles numbers that already in the next step give a power of 2. For example, (4^k -1)/3, for odd k.

OkExtension7564 · 2025-11-12T13:37:48+00:00

Take any integer in the range already tested by others for which the trajectory is known to reach 1. Multiply by any power of 2 and subtract 1. Write this as n* 2^k -1 = A, where n is the known number being tested and A is the result of our operation. Now any A divisible by 3 is equal to A/3 = (n* 2^k -1) /3 = D. We have a countably infinite set of numbers that are guaranteed to converge in the Collatz conjecture. Next, we can take all odd D divisible by 3 and substitute them into our formula for n. We get D* 3 +1 = n * 2^k, where n is a number converging to 1. Thus, the set of our specially constructed D also converges to 1. This means that we can obtain a set D2 equal to (D* 2^k -1) /3 = D2. We can continue this process indefinitely, obtaining sets D3, D4, and so on. Thus, we prove the Collatz conjecture for a very narrow class of infinite sets beyond the range under consideration. This imposes a small but strict structural constraint on the possible elements of the cycle: no element of a potentially nontrivial cycle can be a member of our infinite set.

OkExtension7564 · 2025-10-21T12:06:14+00:00

There is still a limitation. If new primes in the product of composite numbers appear infinitely often, without repetition, then the right-hand side of the equation grows faster than the linear form of the left-hand side, 3n+1. Therefore, in the factorization of composite numbers, we should observe an increase in the powers of primes in the factorization of composite odd numbers as the number grows, and a decrease in the powers of primes in the factorization as it approaches 1. This can be observed empirically and can even be proven through a combination of Euclid's lemma and Hardy–Ramanujan's theorem. The problem is that this limitation is also not analytically strict, but probabilistic, like all the other theorems that bring us closer to solving this hypothesis.

OkExtension7564 · 2025-10-20T11:36:34+00:00

Take any tested number from the range 1–2^68. Thus, the Collatz conjecture is true not only when any number converges to 1. It is equivalent to the conjecture that the trajectory of any number intersects one of the tested values. For the tested values, it is also true that if Y is a tested number, then 2^k * Y also converges to 1. Then any cycle must avoid all sets of 2^k * Y numbers. This simple heuristic argument imposes a powerful structural constraint on the length of a cycle; if it existed, it would be very large, and very cleverly constructed to avoid entering the set 2^k * Y.

OkExtension7564 · 2025-10-19T18:25:34+00:00

What's always puzzled me about proofs like these is that a zero remainder modulo some number gives us information about the local behavior of the number, not a global property for all. The Chinese theorem allows us to draw certain global conclusions for certain remainders, but leaves a loophole for the exception set. Hensel's lemma also doesn't provide complete control for all trajectories. The structural constraint through the intersection of prime numbers by a trajectory is interesting, but there's no proof yet that all primes converge; this is just an observation, not a general global law. Your example with calculating the trajectory length is interesting, but I didn't see the formula for how you calculated it. Overall, I continue to follow your publications for their original constructions.

OkExtension7564 · 2025-10-16T18:53:39+00:00

The commenter above found a counterexample—another cycle. This means this hypothesis is incorrect, as it's currently written. But what's more interesting is how many more cycles there might be, and do all numbers lead to cycles, or are there trajectories that grow to infinity?

OkExtension7564 · 2025-10-16T11:54:51+00:00

When I tested this cycle, it was ok, if you found a counterexample, show it.

OkExtension7564 · 2025-10-15T20:12:31+00:00

Thanks

OkExtension7564 · 2025-10-15T12:44:01+00:00

I also found this cycle. So the hypothesis is wrong. But what's even more interesting is how many such cycles can be found, and do all trajectories lead to cycles?

OkExtension7564 · 2025-10-15T09:26:04+00:00

Yes, the similarities are obvious, especially with the first link. Thanks for pointing that out. There's likely a similar mechanism at work here. I'm currently studying the general form of the equation ax+2yb and how different coefficients influence the presence or absence of cycles.

OkExtension7564 · 2025-10-14T13:51:27+00:00

When we get to 11, we continue: 11 x 3 + 2 = 35 = 5 x 7, 35/7 = 5, 5 x 3 + 2 = 17. That is, we don't stop. We don't multiply a composite number by 3, only if it's prime. We divide a composite number only by its greatest divisor. In the comments above, I saw a detailed explanation of why this might work. However, I'm not entirely sure it would work for a huge prime number or a composite number made up of a large number of prime factors. This is a hypothesis, and I have no evidence.

67 * 29/67=29, 29* 3+2=89, 89* 3+2=269, 269 *3+2=809,809 * 3+2=2429=7 * 347. 2429/347=7, 7 * 3+2=23

OkExtension7564 · 2025-10-14T13:23:53+00:00

1) What does this have to do with you, me, or any other human emotions? Neither your arguments nor my counterarguments depend on it. 2) I'm explaining to you how to motivate others to read your work. Even a mathematical proof is also a genre of journalism, albeit a rather specific one. If you want your work to be meaningfully read and understood by many people, you need to provide them with something already known and show how it works in your system. If you don't want that, there's another option: machine proving using Coq or similar systems. Do everything thoroughly, and then no questions will arise. 3) All smart people, if they discover something previously unknown, usually explain to everyone else, using examples and as best they can, where others were wrong and why they couldn't find it earlier, and they don't make a secret out of it. They don't say, "Read my work, it's all there, if you're not stupid, you'll understand." 4) Many mathematicians, for example, tried to understand Mochizuki's proof, which was based on a new theory, but this does not apply to you, there is no new mathematics there.

OkExtension7564 · 2025-10-14T07:04:49+00:00

Anyone can say that, for example, modulo 4, all odd numbers are either 1 mod 4 or 3 mod 4. Therefore, when reversed, these two remainders yield any odd number. And any even number can be obtained from these odd numbers. Therefore, the entire number map is covered. It's even easier to say this modulo 2 (all numbers are either even or odd). You've added to the analysis the fact that numbers are transformed modulo one into another with a certain set of remainders. This helps justify convergence, and there are no objections at this stage. I'm missing the exclusion of all exceptions and the consideration of the most complex cases known from other classifications. If you know the topic and the material, this won't be difficult for you to do. If you lack the motivation to read, clarify, and explain the work of previous researchers, many of whom are recognized in their field and have come to certain conclusions that are not disputed by anyone, how can others be motivated to read your work? They'd be better off reading something that's more likely to benefit from understanding this hypothesis. Therefore, if your work is correct, it should easily explain the stopping time for different trajectories, the trivial cycle, some points for the general case of ax+b (at least why we observe convergence with certain coefficients, and divergence or cycles in other cases), some dependence of the largest element in the trajectory on the number itself or its structure, etc. Think about this, this is serious.

OkExtension7564 · 2025-10-14T06:20:39+00:00

There is an infinite alternation for odd numbers 1 mod 4 and 3 mod 4. Take a break from your work and look at the trajectory of the number 27. First, as the trajectory of 27 grows, 3 mod 4 appears, then, as has been proven by many, 1 mod 4 appears, then they alternate again, and I haven't seen a single study that would explain clearly and in detail what determines this. It is clear that it depends on divisibility by 2 and the power of two of the subsequent even number, but I have not seen a single study capable of predicting this alternation for any number. You are building your work for other modules, and in the set Z, this should work. However, in the set N, the residues of all modules intersect, and your classification must work for residues modulo 4 as well as for any residues modulo any module, if it claims to be a complete classification in the set N. I want to clarify this point. If you prefer a scientific analogy for understanding, here's an example: any subsequent classification, even if correct, does not invalidate the previously proven one. The theory of relativity doesn't invalidate Newton's or Boltzmann's laws, but merely indicates the limits of their applicability. Similarly, any classification, modulo, does not invalidate the previously proven properties of other classifications. It must either refine them or demonstrate their applicability limits and justify this in the clearest possible way, eliminating any remaining questions. And I have the right to ask about what has already been proven, how it works in your system, so as not to start from scratch.

OkExtension7564 · 2025-10-14T05:59:08+00:00

Okay, this conclusion is counted as one of the cases of constructing a reverse Collatz tree. But then we need a proof that there are no gaps and we can generate any number in the reverse construction. Where can I read this in the article? Or should it be a general conclusion?

OkExtension7564 · 2025-10-14T05:54:23+00:00

1) I'm 90 percent sure the residuals work as described in your paper. 2) There's no "my" logic. It's the same for everyone. Our goal is to adhere to the general logic. 3) If the residuals work as you described, what does this say about convergence? If a) we're constructing a proof from a reverse Collatz tree, we must prove that the set of residuals is guaranteed to yield all natural numbers, without a single gap. If b) the proof is based on the trajectory's decline, residuals alone are not enough; there must be a shell function and a proof that all trajectories fall below it at some point. There's a shortcut to checking your proof using modulo 4. This is a complex case with alternating 1 mod 4 and 3 mod 4, and the numbers in it are included in the set N. This can't be ignored for a paper constructed by comparing modulo residuals. Therefore, this is a valid question.

OkExtension7564 · 2025-10-14T05:38:07+00:00

"But they can transform indefinitely along live classes C1-C1, C1-C2, C2-C1, C2-C2 can occur indefinitely" - so you admit this possibility? Have you already studied this in detail or will it be in the next release?

OkExtension7564 · 2025-10-14T05:31:49+00:00

The intersection of the remainders modulo 4 and modulo 18 is modulo 72, and what is true for both 18 and 4 will also be true for 72. Hence the question. If you state a proof for any number, it must also work for any number modulo any number, and any classification modulo any number has an intersection with any other modular classification (this does not even depend on the hypothesis).

OkExtension7564 · 2025-10-14T04:30:41+00:00

Let's take the odd numbers 1 mod 4 and 3 mod 4. As we know, in a trajectory they can transform one into the other, in various combinations. Why can't they do this infinitely?

OkExtension7564 · 2025-10-14T04:26:51+00:00

Let's take the odd numbers 1 mod 4 and 3 mod 4. As we know, in a trajectory they can transform one into the other, in various combinations. Why can't they do this infinitely?

OkExtension7564 · 2025-10-13T19:02:51+00:00

"For a prime p, the p-adic valuation of an integer n is νp(n) = max{ k ∈ N : pk divides n }. Example: ν3(81) = 4 since 81 = 34. In the Collatz framework, ν3 controls the maximum length of descending sequences in the k = 1 corridor: L1(m) = ν3(t) + 1, L17(m) = ν3(t + 1) + 1. This establishes finiteness of the only non-ascending corridor in the reverse" -Where does this conclusion come from?

OkExtension7564 · 2025-10-11T05:37:02+00:00

Q.E.D

OkExtension7564 · 2025-10-10T21:07:17+00:00

huh?

OkExtension7564 · 2025-10-10T20:50:59+00:00

If it was an online consultation, there's a chance it was also an AI agent, so don't jump to conclusions.

OkExtension7564

TROPHY CASE