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

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.

Exact FrontierDelta

proved→proved

Scope and record

Occurred: Oct 29, 2026

Delta type: REGISTRY REVISION

Assumptions: VibeMathed verification: site-confirmed. Publication: announcement. AI contribution: ai-discovered. VibeMathed editorial classifications, scores, notes, relations, and dataset structure are CC BY 4.0. Source statements and linked content retain their own rights.

Canonical aliases: An Exact All-Width Plateau for the Three-Summand Tu-Deng Count modulo $2^k-1$ · Three-summand two-zero plateau

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VibeMathed
registry · event recorded by

GPT-5.6 Sol (high reasoning)
model · ai model contributor · OpenAI

Claude Opus 5 (high reasoning)
model · ai model contributor · Anthropic

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

specialist registry · passed

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.

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

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