An Exact All-Width Plateau for the Three-Summand Tu-Deng Count modulo $2^k-1$

Put $N=2^k-1$ and write $\mathrm{wt}$ for the binary Hamming weight. Let $F_k(t)$ count the ordered triples $(a,b,c)\in\{0,\dots,N-1\}^3$ with $a+b+c\equiv t \pmod N$ and $\mathrm{wt}(a)+\mathrm{wt}(b)+\mathrm{wt}(c)<k$ , the three-summand analogue of the two-summand count of the Tu-Deng conjecture at the same modulus and weight budget. Evaluate $F_k(t)$ exactly on the targets $t$ whose $k$ -bit cyclic word has exactly two zero digits, no two adjacent: is the value the same for every such $t$ at a given $k$ , and what is it?

Result: Proved(see note)
Status: Resolved
AI contribution: AI-discovered
Method: Argument
Field: Combinatorial number theory; binary digit sums and cyclic carries
Posed by: —
Year posed: —
Years open: —
Solved: 2026-08-26
Model: GPT-5.6 Sol (high reasoning), Claude Opus 5 (high reasoning)
Vendor: OpenAI, Anthropic
Collaborators: —
Verification: Site-confirmed
Publication: Announced
Significance: 4 / 100
Disclosed cost: —
Wikipedia: No dedicated article

What was actually shown

Answered in full: for every $k\ge4$ , a target with exactly two nonadjacent zero digits has $F_k(t)=(k+23)3^{k-4}$ , independently of the distance between the zeros. Exact at every width, no error term, no hypothesis on $k$ (Theorem 1.1). This is an evaluation, not an extremal result, and the paper is explicit about the difference: the plateau value is not maximal. At $k=12$ it reads $35\cdot3^8=229{,}635$ while $F_{12}(110101101010)=293{,}499$ at five zero digits, so no global maximizer of $F_k$ is classified. The paper's other results are finite-layer and do not settle the extremal question: balancing monotonicity of $[x^{\le C}]H_t$ holds only for $C\le5$ (Theorem 1.3), and the all-mass statement is Conjecture 8.1, which the paper states outright does not follow from Theorem 1.3. The chamber where zero digits are adjacent is not addressed.

What the AI did

Disclosed in the artifact, on the author line and in a dedicated Section 10. The footnote reads "The mathematics in this manuscript was produced principally by AI systems, and the human author contributed no mathematical content", and Section 10 attributes the work: GPT-5.6 Sol at high reasoning effort "did the majority of the mathematics" - the three-state cyclic-carry transfer matrix, the carry-mass regrading, the exact carry-layer coefficients through mass five, the primary and secondary balancing exchange arguments, the bounded-correlation principle, the receiver-boundary compensation and receiver-opening kernel theorems, the certificate programs, and the Lean 4 development. Claude Opus 5 at high reasoning "supplied direction rather than derivations": framing the problem as the first multisummand case after Tu-Deng, selecting which sub-questions to attack, enforcing the line between proved and conjectured, and running the prior-art search.

The unusual part is what is left over. The human author is anonymous and, on the manuscript's own account, "framed no argument, supplied no proof step, and contributed no mathematical content; the role was to run the systems, collect the output, and publish it." The directing role that would ordinarily be a person's was played by a second model. Section 10 also states that no step has been checked by hand by a human mathematician.

Verification

Site-confirmed: this site reproduced it, not just the authors. Two things were run here, 27-28 August 2026.

Their certificate suite, from a clean checkout, exact integer arithmetic throughout: 33a passes; 33c at --kmax 12 passes in 1 min 14 s; 33d passes its 44,250 frontier cells in 8 min 10 s; 33e passes at 151,200 sign checks. 33f (1,377,000 signs) was not carried to completion - nine insertion types cleared with no failure before it was stopped - as the submission volunteered.

And an independent re-derivation, from the problem statement rather than their code: $F_k(t)=(k+23)3^{k-4}$ is exact for every $k$ from 4 to 11, on two distinct nonadjacent-two-zero targets each. The revision's two new numbers check out too: $F_{12}(110101101010)=293{,}499$ against the plateau's $229{,}635$ , and the Section 3 example where balancing the zero gaps $(1,2,3)$ to $(2,2,2)$ at $k=6$ lowers the count from 231 to 216.

What this does not establish. The theorem is stated for all $k$ and instances $k=4,\dots,11$ were checked, so what is confirmed is the certificates plus a finite range of the claim, not the all-width statement, whose proof is the paper's own short argument. The Lean was not built here, and is uncompiled by the submitter's account too. It is also narrower than its file names suggest, as Section 9.1 now says: the bridge from $F_k$ to the formal objects is assumed, not formalized, CarryConfig taking the digit equation as a hypothesis with next an arbitrary permutation.

Sources

Related entries

Builds onTu-Deng conjecture

Submitted by SilentIbis765 on 27 Aug 2026

An Exact All-Width Plateau for the Three-Summand Tu-Deng Count modulo 2k−12^k-12k−1