gpt 6 astra pro attempt on Math-62

Collatz Conjecture

Definitions and mathematical statement

Define $T:{\mathbb{N}}_{>0}\to{\mathbb{N}}_{>0}$ by \[ T(n)=\begin{cases}n/2,&2\mid n,\\3n+1,&2\nmid n.\end{cases} \] The Collatz assertion is $\forall n\ge1\ \exists k\ge0$ such that $T^k(n)=1$, where $T^0$ is the identity. Reaching one enters the cycle $1,4,2,1$. The iteration here uses the unaccelerated map; no division by an additional power of two is included in an odd step. A negative resolution may be an orbit entering a different cycle or one that never repeats and never reaches one. Verifying any finite initial range does not decide the universal assertion.

Formulation and concept references: [S1].

Short English statement

Does repeatedly halving even integers and replacing odd integers by three times the number plus one always eventually reach one?

Research attempt

Current frontier.

Tao's theorem says that almost all starting values, in logarithmic density, have an orbit reaching a bound tending arbitrarily slowly to infinity. It does not say that every orbit reaches one; the quantifier distinction survives in the revised manuscript [S1]. An exact finite computation cannot remove the exceptional set. We pursue a deterministic descent certificate and test whether adding residue information to a logarithmic potential can make it monotone. The resulting obstruction is proved below and does not depend on statistical modeling of parity bits.

Attack route.

One possible proof would establish that every odd $n>1$ eventually descends below $n$, after which strong induction finishes the conjecture. Another would construct a well-founded potential. A natural computable candidate is $V(n)=c\log n+f(n)$, where $f$ is a bounded correction, possibly depending on finitely many residues. A third route is a finite cover by variable-length descent certificates; its missing requirement is coverage of all exceptional residue cylinders, not just almost every cylinder. We test the potential route first, including fixed-block variants.

Attempt.

For odd positive $n$ define \[ U(n)=\frac{3n+1}{2^{v_2(3n+1)}}. \] The original unaccelerated orbit reaches one if and only if its odd-to-odd orbit under $U$ does. The omitted intermediate integers are exactly the forced successive halvings, so no additional orbit is introduced; $U(1)=1$ represents the original $1,4,2,1$ cycle.

Lemma (arbitrarily long expanding words). For integers $L>R\ge1$, set $n_0=2^L-1$. Then \begin{equation}\label{wsref-0-65-71-31-38-38-3a-72-69-73-65} U^j(n_0)=3^j2^{L-j}-1\quad(0\le j\le R), \end{equation} and all the first $R$ valuations equal one. Indeed, if $n_j=3^j2^{L-j}-1$, then $3n_j+1=2(3^{j+1}2^{L-j-1}-1)$. For $j<R<L$ the parenthesis is odd. This proves the formula by induction. Every step in this segment increases the odd value.

Proposition (bounded-correction obstruction). There do not exist $c>0$ and a bounded function $f$ on the positive odd integers for which \begin{equation}\label{wsref-0-65-71-31-38-38-3a-69-6d-70-6f-73-73-69-62-6c-65} V(U(n))\le V(n)\quad\text{for every odd }n>1, \qquad V(n)=c\log n+f(n). \end{equation} Suppose $|f|\le M$. Telescoping (eq188:impossible) along (eq188:rise) gives \[ c\log\frac{3^R2^{L-R}-1}{2^L-1}\le2M. \] First choose $R$ with $cR\log(3/2)>2M$, and then let $L\to\infty$. The left side tends to $cR\log(3/2)$, a contradiction. This rules out even nonincrease, hence also strict decrease. It applies to every bounded correction, not only periodic corrections.

A fixed block length does not repair the obstruction. If $V(U^r(n))\le V(n)$ holds for all sufficiently large odd $n$, apply it at $n_0,n_r,\ldots,n_{(j-1)r}$ along an expanding word of length $jr$. Letting $j$ grow again contradicts boundedness of $f$. For a residue correction there is an even more direct test: choose \[ n_0=2^{r+1}mt-1\quad(t\ge1). \] The first $r$ iterates all satisfy $n_j\equiv-1\pmod m$, so a function depending only on $n\bmod m$ cancels exactly between the endpoints while the logarithm strictly increases.

We next record the exact affine formula behind a more flexible residue search. Prescribe positive integers $a_1,\ldots,a_r$, and put $A_j=a_1+\cdots+a_j$, $A_0=0$. Whenever the first $r$ valuations equal that word, \begin{equation}\label{wsref-0-65-71-31-38-38-3a-77-6f-72-64} U^r(n)=\frac{3^rn+B_r}{2^{A_r}},\qquad B_r=\sum_{j=0}^{r-1}3^{r-1-j}2^{A_j}. \end{equation} This follows from $2^{a_{j+1}}n_{j+1}=3n_j+1$. The numerator congruence requiring the final quotient to be odd singles out one odd residue modulo $2^{A_r+1}$; reversing the recurrence shows that it gives exactly the prescribed valuations. Thus the word has density $2^{-A_r}$ among odd integers. This is a finite-cylinder statement, not independence along one infinite integer orbit.

For such a word a certified descent occurs whenever \[ 2^{A_r}>3^r,\qquad n>\frac{B_r}{2^{A_r}-3^r}. \] The affine offset matters: a negative logarithmic drift alone does not imply descent for every integer in the cylinder. A proposed proof by residue certificates must cover every odd integer greater than a verified finite base range with some word meeting this inequality. We have not obtained such a cover. Formula (eq188:rise) proves that no fixed maximum word length can give a descent certificate for all large inputs.

Stress test.

The expanding words are finite. They neither produce a divergent orbit nor a nontrivial cycle: after the prescribed word, a much larger power of two may divide the next iterate. Our obstruction therefore does not refute Collatz. It refutes a specific universal monotonicity requirement, even though that requirement initially looks compatible with average negative drift. Conversely, almost-everywhere stopping results do not rule out an exceptional orbit. A minimal counterexample would have to avoid every verified descent cylinder, but no compactness principle says a residual set of $2$-adic cylinders contains, or does not contain, a positive integer with the required infinite trajectory.

The checking script verifies long exact expanding segments, the affine word formula, and the single-residue assertion for bounded word lengths. None of these computations is described as checking the entire conjecture. We make no claim that the obstruction lemma has not appeared elsewhere.

Outcome.

Outcome: Exploratory approach with unresolved gap. The proposed bounded-correction logarithmic Lyapunov strategy fails, and the failure has a precise arbitrary-length construction. Exact variable-word descent certificates remain viable, but a proof must allow unbounded word lengths or a genuinely unbounded correction and must establish universal coverage. That missing deterministic control, rather than a finite-range computation or an average drift calculation, is what would be required to resolve the target.

Sources

• S1 Terence Tao, Forum of Mathematics, Pi 10 $2022$, e12. https://arxiv.org/abs/1909.03562. Catalogue formulation source.