Somer-Krizek uniform-subsequence conjecture

Somer and Krizek, Conjecture 4.2, printed page 8, proposes the following. For every k >= 2, let an integer sequence satisfy

w(n+k) = a1*w(n+k-1) + ... + ak*w(n)

for every n >= 0. Let p be prime, e >= 1, and E > 0. Assume that gcd(ak,p)=1, that the least positive period modulo p^e is exactly p^e E, and that every residue modulo p^e occurs exactly E times in that least period. For every positive g coprime to p and every nonnegative s, put r=E/gcd(g,E). The conjecture asserts that every residue occurs exactly r times among w(s+ng) for 0 <= n < p^e r.

The paper proves the corresponding statement for second-order recurrences and asks whether it extends to all higher orders. Example 4.3 gives a supporting third-order instance. The conjecture is therefore a precise test of whether the second-order mechanism survives at the first new order.

The official journal PDF and DOI record identify the paper as INTEGERS 25 (2025), A1, version 1. The bounded public-source review recorded in issue 8695 checked the official 2025 and 2026 journal listings, the authors' available later work, and exact title, DOI, author, conjecture-number, and topic searches. It found no correction, proof, or refutation of the exact claim in that scope. This is not an exhaustive literature or priority claim.

Take k=3, coefficients (0,0,-1), and initial values (0,1,2), so w(n+3)=-w(n) for every n. The integer period is (0,1,2,0,-1,-2). Modulo three it becomes (0,1,2,0,2,1), has genuine least period six, and contains every residue exactly twice. Thus p=3, e=1, and E=2 satisfy the full hypotheses.

For g=2 and s=0, one has r=2/gcd(2,2)=1. The first three sampled residues are (0,2,2), so residue one occurs zero times rather than once. The candidate was released in the public handoff at issue 7333, comment 5733245266; this dossier credits that handoff and makes no first-discovery claim.

A proof that the displayed third-order sequence fails the recurrence, lacks least modular period six, lacks two copies of each residue in its full orbit, or has one copy of every residue in its first three step-two samples would falsify this route. The explicit six-term word makes each condition finite except the all-index recurrence, which follows from six-periodicity together with w(n+3)=-w(n) at every residue class modulo six.

• Primary source: DOI 10.5281/zenodo.14679256, printed page 8, Conjecture 4.2.
• Source module: D5/S1/Recurrence/Periodic/SomerUniformSubsequenceRefutation.lean.
• The formal claim retains every quantified source parameter and hypothesis.
• The public result negates that complete claim and retains the actual six-code word in its raw type.
• The certificate proves the all-index recurrence, exact least period, full-orbit counts, and the failed sampled count.

theorem. The explicit third-order recurrence refutes the universal higher-order conjecture. It does not challenge the paper's proved second-order theorem and asserts no corrected higher-order criterion.

The literature review is bounded and does not establish exhaustive absence of corrections, unpublished work, paywalled citing papers, or unindexed resolutions. The released public handoff is credited for the candidate; no claim of independent first discovery or worldwide publication priority is made.