RESEARCH / Focused target
Nonvanishing Indices of the Binary Digit Hankel Determinants at t = −2
RELEASED FOUNDATIONS / PROPOSED CONNECTION
Research connections
All source anchors (1)
- D5/S3/Combinatorics/DigitHankel/BinaryDigitHankel.resultAbsent from this graph
Problem
The question
Bartosz Sobolewski and Maciej Ulas, Hankel determinants of weighted binary sums of digits, arXiv:2607.09376v1, Section 5.1, the case d = 2 of Conjecture 5.7, stated in the sentence following it. For u = Σ_j ε_j 2^j with binary digits ε_j put S(u,t) = Σ_j ε_j t^j and H(n,t) = det[S(i+j,t)]_{0≤i,j<n}; let n_k = ⌈2^{k+2}/3⌉, so that n_k = 2, 3, 6, 11, 22, 43, …. The statement is that for every n ≥ 2, H(n,−2) ≠ 0 if and only if n ∈ {n_k − 1, n_k, n_k + 1} for some k ≥ 0.
Motivation
Our foothold
The theorem D5/S3/Combinatorics/DigitHankel/BinaryDigitHankel.result establishes the equivalence for every n ≥ 2.
Gap
Missing bridges
Pre-registration issue 12663 records the literature screen. The paper proves the vanishing on a structured set (Theorem 5.4) and a family of nonzero indices (Theorem 5.10), and states that the difficulty lies in the nonvanishing outside that set. The arXiv record has a single version; web and GitHub searches found no proof. This is a bounded negative finding.
Route
Proposed approach
- Sparse kernel vectors with a binary carry pattern show that H(n,−2) = 0 at every index strictly between the triples.
- A binary-carry conjugation of determinant one reduces the Hankel matrix to a sparse block form; paired elimination gives reflection recurrences that relate H at the triple around n_k to the triple around n_{k−1}, through an auxiliary bordered determinant.
- A simultaneous induction shows that the auxiliary determinants at the two ends vanish, so no cancellation occurs in the recurrences, and evaluates H at all three members of every triple as explicit signed powers of two.
Falsifier
What would falsify this route
The statement would fail if some recurrence coefficient vanished, or if an auxiliary determinant at an endpoint were nonzero for some k so that cancellation became possible.
Evidence
Evidence to collect
An independent referee implementation computed H(n,−2) by exact integer determinants for n ≤ 200 and checked every intermediate lemma of the proof.
Triage
Scope assessment
theorem; the statement is the case d = 2 of Conjecture 5.7 of arXiv:2607.09376v1, quantified over every n ≥ 2.
- Proved (formalized): H(n,−2) ≠ 0 exactly at the indices n_k − 1, n_k, n_k + 1, with the values at these indices given as explicit signed powers of two.
- Open: the general case d ≥ 2 of Conjecture 5.7, that H(n,2ζ) = 0 exactly on A_d for every primitive d-th root of unity ζ.
ASSUMED-UNVERIFIED
Unverified assumptions
The literature screen is limited to the arXiv record, web and GitHub searches and the repository checks.