I think I know why Collatz does that thing it does

https://news.ycombinator.com/item?id=43658951

by stuartriffle • 1 year ago

3 5 1 year ago

news.ycombinator.com

The normal term for this is "manic delusion of grandeur", so please help me out: 3n walks congruence classes +1 steps each one out of alignment /2 never changes that fact

Accumulating co-primeness is the "memory" that stops n from repeating. This is invariant under 2^k, which allows it to make forward progress through the chaos. Exhausting congruence classes mod 3 forces n to a power of 2; game over.

I asked AI to prove those things, and it did. I assume.

The only question then would be if 1.5n can grow the residue vector quickly enough to outrun the exhaustive walk. I asked AI that too, and got back a one page p-word. I'm not even going to type it.

I can sweet-talk AI into agreeing with damned near anything, so I'm stuck. This is the only forum I know with consistently thoughtful conversation, and I can't think my way out of this one, and I have real work to do.

Is there a mathematician in the house? The p-word is in a comment.

stuartriffle4• 1 year ago

### Key Analytical Components 1. Forbidden Congruence Classes: - For each odd prime $$ p $$, after encountering $$ n \equiv a \mod p $$, all numbers congruent to $$ a \mod p $$ are eliminated from future trajectories. This creates a growing set of forbidden residues $$ \mathcal{F}_p \subseteq \mathbb{Z}/p\mathbb{Z} $$.

2. Rate of Congruence Coverage: - Ergodic Iteration: For a fixed prime $$ p $$, the map $$ n \mapsto 3n + 1 \mod p $$ permutes residues. Since $$ 3 $$ and $$ p $$ are coprime, iterating $$ f(n) $$ cycles through all $$ p $$ residues, exhausting $$ \mathbb{Z}/p\mathbb{Z} $$ in $$ \leq p $$ steps. Thus, residues modulo $$ p $$ are eliminated at a rate proportional to $$ p $$.

3. Growth Bound vs. Prime Density: - Growth Rate: Trajectories grow by at most $$ \frac{3}{2^k} $$ per cycle (where $$ k \geq 1 $$). Worst-case net growth is $$ \log_2 3 \approx 1.58496 $$-exponential. - Prime Density: The number of primes $$ \leq N $$ is $$ \pi(N) \sim \frac{N}{\log N} $$, growing slower than $$ N $$.

4. Synchronization of Elimination: - Overlap Argument: For $$ n $$ increasing to $$ N $$, primes $$ p \leq \log N $$ dominate potential factors. The number of such primes is $$ \sim \frac{\log N}{\log \log N} $$, which grows slower than $$ \log N $$. Since residues modulo each $$ p $$ are exhausted in $$ p \leq \log N $$ steps, the elimination rate ($$ O(\log N) $$) outpaces the introduction of new primes ($$ O\left(\frac{\log N}{\log \log N}\right) $$).

5. Inductive Confinement: - Base Case: For $$ n_0 \leq C $$, finite congruence eliminations force termination (empirically verified). - Inductive Step: Assume all $$ n \leq K $$ terminate. For $$ n > K $$, each step either reduces $$ n $$ directly or accumulates forbidden residues for primes $$ \leq K $$. Since primes $$ \leq K $$ are eliminated in $$ O(K) $$ steps, $$ n $$ must reduce before surpassing $$ K^{1.58496} $$, maintaining $$ n \leq K $$ inductively.

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### Formal Statements Theorem 1 (Finite Congruence Elimination): For any prime $$ p $$, after at most $$ p $$ iterations, $$ \mathcal{F}_p = \mathbb{Z}/p\mathbb{Z} $$. Thus, $$ n $$ cannot retain factors of $$ p $$ indefinitely.

Theorem 2 (Exponential-Prime Decoupling): For $$ n > N $$, the rate of congruence elimination (linear in $$ p $$) exceeds the rate of prime introduction (sub-linear in $$ N $$). Specifically, $$ \sum_{p \leq N} p \sim \frac{N^2}{2 \log N} \quad \text{vs.} \quad \sum_{p \leq N} \frac{N}{\log N} \sim \frac{N^2}{\log^2 N}, $$ showing elimination dominates.

Corollary (Termination Guarantee): For $$ n > N $$, the trajectory encounters all primes $$ \leq \log N $$ within $$ O(\log^2 N) $$ steps, forcing $$ n < N $$ before accumulating new primes beyond $$ N $$.

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### Conclusion By inductively eliminating residues modulo primes at a rate exceeding $$ n $$’s growth, every trajectory must eventually descend below an arbitrary $$ N $$, collapsing to 1. While gaps exist in quantifying overlap decay and induction formalization, this framework aligns with observed behavior and modular constraints.

Final Answer \boxed{The systematic elimination of prime congruence classes, coupled with bounded exponential growth, confines trajectories to finite descent, forcing convergence under the Collatz process.}]

fjfaase• 1 year ago

Maybe a math oriented forum, might be a better place to post it.

Do you think you could write up the proof in Lean?

stuartriffle4• 1 year ago

I asked about it on Math StackExchange twice, both posts were deleted within minutes as off-topic.

I formalized the proofs and came back, but I'm apparently banned for a week because I had two posts deleted.

I am fully aware that what I present is impossible, but here we are. It should be easy to contradict the argument, it's very simple. I can write it on a napkin. I'm pulling my hair out here [what's left of it].

stuartriffle4• 1 year ago

Absolutely not, but ChatGPT generated some Lean, I posted it in a comment. I'm not set up to run that, downloading stuff now. I guess we'll see.

[edit] I have deleted the half-ass ChatGPT generated Lean because it doesn't compile yet and that's lame. Working on it.

stuartriffle4• 1 year ago

Here are the basic claims that lead to the p-word, all proven with high-school math, validated by AI. Please, someone, find the error. I need to get some sleep.

# I. Exhaustive Walk Over Prime Congruence Classes

Theorem 1.

Let p be an odd prime with p ≠ 3. Define an affine map f: ℤ/pℤ → ℤ/pℤ by

  f(x) = 3x + 1 (mod p).

Since gcd(3, p) = 1, f is an invertible affine transformation. In particular, f is a permutation of ℤ/pℤ. Moreover, if we write f in the form f(x) = 3(x + c) with the unique constant c = 1/3 (interpreted in ℤ/pℤ) (i.e. by solving 3x + 1 = 3(x + c)), it follows that f is conjugate to the multiplication-by-3 map on ℤ/pℤ and hence its cycle structure is completely determined by the order of 3 modulo p. In particular, for any starting value α ∈ ℤ/pℤ not equal to the unique fixed point, the orbit {α, f(α), f²(α), …} is a (cyclic) subgroup of ℤ/pℤ and hence "exhausts" its cycle.

Proof.

Since 3 is invertible modulo p, the affine map f is invertible and thus a permutation. A standard fact about maps of the form f(x)=ax+b on a finite field is that they are conjugate to the linear map x ↦ a·x. (One may define y=x + b⁄(1−a) so that f becomes y ↦ a·y.) Consequently, the cycle structure of f is the same as the cycle structure of the multiplication-by-3 map, whose nonzero orbits are cyclic of length equal to the multiplicative order of 3 modulo p; the fixed point of f is the unique solution of x=3x+1, viz. x = −(1)/(2) (in ℤ/pℤ).

∎

# II. Removal of Congruence Classes (Modulo 3)

Theorem 2.

For every integer n, we have

  3n + 1 ≡ 1 (mod 3).

Thus, if one defines L(n)=3n+1 (the “linear part” of the Collatz map), then for every n ∈ ℤ we have L(n) ∈ {1} (mod 3); in other words, the mapping L “eliminates” the residue classes 0 and 2 modulo 3. Proof.

For any n ∈ ℤ, note that 3n ≡ 0 (mod 3) and hence 3n+1 ≡ 1 (mod 3).

∎

# III. Invariance Under Division by Powers of Two

Theorem 3.

Let p be any odd prime. Suppose A ∈ ℤ is given and p ∤ A. For any integer k ≥ 0, define

  A/2^k

to be the unique element of ℤ/pℤ equal to A · (2^k)⁻¹, where (2^k)⁻¹ is the multiplicative inverse of 2^k modulo p. Then the operation “division by 2^k” is an automorphism of ℤ/pℤ. In particular, if A ≡ c (mod p) then A/2^k ≡ c · (2^k)⁻¹ (mod p), so any congruence property of A modulo p is preserved (up to multiplication by the invertible constant (2^k)⁻¹). Proof.

Because p is odd, 2 is invertible modulo p. Hence, for any k the number 2^k is invertible mod p and division by it is well defined as multiplication by (2^k)⁻¹. Therefore, if A ≡ c (mod p), then A/2^k ≡ c · (2^k)⁻¹ (mod p).

∎

# IV. Monotonic Contraction of Allowed Congruence Classes

Theorem 4.

Let S be a finite set of odd primes, and suppose that for each p ∈ S the iterative process (applying L(n)=3n+1, possibly followed by division by the maximal power of 2) “eliminates” one or more congruence classes modulo p from the set of permitted residues in the long‐term evolution of the sequence. That is, assume that after one iteration the residue of n modulo p lies in a subset R_p ⊂ ℤ/pℤ with |R_p| < p. Then any n is forced to lie in the set of residue vectors

  R = ∏{p∈S} R_p

which is a proper subset of X = ∏{p∈S} ℤ/pℤ. In particular, by the Chinese remainder theorem the total number of residue combinations the future value of n may assume is at most ∏{p∈S} |R_p|, strictly less than ∏{p∈S} p. Thus, under successive iterations the set of “admissible” residue vectors (that is, the overall modular configuration of n) is contracted. Proof.

For each odd prime p in S, the hypothesis is that after an iteration n is forced to lie in a subset R_p ⊂ ℤ/pℤ with |R_p| < p. Then by the Chinese remainder theorem, any integer n is determined modulo M = ∏{p∈S} p by its vector of residues (n_p){p∈S} ∈ X. Since the process restricts n_p to lie in R_p for each p, the number of possible residue vectors is at most ∏_{p∈S} |R_p|, which is strictly less than M.

∎

# V. Forcing an Integer to be a Power of Two

Theorem 5.

Suppose that, under an iterative process derived from the Collatz rule, every odd prime p (or all p in an infinite set) forces the residue of n to be a fixed value—say, 1 modulo p; i.e., n ≡ 1 (mod p) for all such p. Then for every finite set S of odd primes we have n ≡ 1 (mod M), where M = ∏_{p∈S} p. Since this holds for every finite S, it follows that if n > 1 had any odd prime divisor, then for some p dividing n we would have n ≡ 0 (mod p), a contradiction. Hence n is not divisible by any odd prime and must be a power of 2.

Proof.

Assume that for every odd prime p (or for every p in an infinite set P) the iterative process forces n ≡ 1 (mod p). Then for any finite set S ⊆ P, by the Chinese remainder theorem n ≡ 1 (mod ∏_{p∈S} p). If n were divisible by an odd prime q, then taking S such that q ∈ S would force n ≡ 1 (mod q), contradicting n ≡ 0 (mod q). Thus n is not divisible by any odd prime, which implies that n is a power of 2. ∎

# Conclusion

The above theorems rigorously formalize the following modular observations about the linear part of the Collatz iteration (and its interaction with division by powers of two):

- For every odd prime p ≠ 3, the map L(n)=3n+1 acts as a permutation of ℤ/pℤ that exhaustively cycles through the residues (up to its natural cycle structure). - In particular, modulo 3 we have L(n) ≡ 1 for all n, so the non-1 residue classes are eliminated. - Division by any power of two (applied after the linear step) preserves congruence information modulo any odd prime. - If an iterative process restricts the allowable residues for n modulo a collection S of odd primes, then the Chinese remainder theorem implies that the set of possible values for n (modulo ∏_{p∈S} p) is strictly reduced. - Finally, if this process eliminates all “nontrivial” residue classes for all odd primes, then n must be congruent to 1 modulo every odd prime, forcing n to be a power of two.

I think I know why Collatz does that thing it does

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