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OP-6984Status unassessedCombinatorics

Wichmann's Conjecture on Optimal Sparse Rulers

TheoremDB P8790 · Candidate awaiting source review

Inferred from the community field; finer classification has not been reviewed.

Source candidate · not reviewedAs of 2026-09-08Source & evidence ↓

Problem statement

A sparse ruler is described by a finite sequence g=(g1,…,gn)g = (g_1, \ldots, g_n) of positive integers called segment lengths or gaps. Its marks are the partial sums m0=0<m1<⋯<mn=Lm_0 = 0 < m_1 < \cdots < m_n = L where mi=g1+⋯+gim_i = g_1 + \cdots + g_i, and L=g1+⋯+gnL = g_1 + \cdots + g_n is the length of the ruler. A distance k∈{0,1,…,L}k \in \{0, 1, \ldots, L\} is measurable if k=mj−mik = m_j - m_i for some marks mi,mjm_i, m_j.

A sparse ruler is complete if every distance k∈{0,1,…,L}k \in \{0, 1, \ldots, L\} is measurable. A complete ruler is minimal if no complete ruler of the same length has fewer segments. A complete ruler is maximal if no complete ruler with the same number of segments has greater length. A ruler is optimal if it is both minimal and maximal.

For natural numbers rr and ss, the Wichmann ruler W(r,s)W(r,s) is the sparse ruler with segment-length sequence

1r,  (r+1),  (2r+1)r,  (4r+3)s,  (2r+2)r+1,  1r,1^r,\; (r+1),\; (2r+1)^r,\; (4r+3)^s,\; (2r+2)^{r+1},\; 1^r,

where aba^b denotes bb consecutive segments of length aa. This ruler has 4r+s+24r + s + 2 segments and length 4r(r+s+2)+3(s+1)4r(r + s + 2) + 3(s + 1).

Prove that if gg is an optimal sparse ruler with more than 1313 segments, then there exist natural numbers rr and ss such that gg equals W(r,s)W(r,s) or the reverse of W(r,s)W(r,s).

Lean 4

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Formal statement

Lean 4
theorem wichmann_conjecture {g : List ℕ} (hopt : IsOptimal g) (hseg : 13 < g.length) :
    ∃ r s : ℕ, g = wichmannGaps r s ∨ g = (wichmannGaps r s).reverse

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A source attachment or link is not a verified proof. The statement’s correspondence with the original question needs human review.

Proofs & verification

No Lean has been submitted for this problem yet. A Lean file can be attached to a submission for review, though with no target here it cannot be machine-checked.

Progress & source assessment

Source-reported assessment, retrieved 8 September 2026. Frontier Science has not independently reviewed this status.

Open. No signed Lean verification is attached to this problem or its selected resolution.

This record is in the source backlog; its novelty and unresolved status have not passed source review.

References
  1. [1]TheoremDB P8790 link
  2. [2]Source formal statement (not locally verified) link

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