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RESEARCH / Focused target

Richman's congruence for the deranged-matching numbers

Literature status: not recheckedRepository record: proved

RELEASED FOUNDATIONS / PROPOSED CONNECTION

Research connections

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Problem

The question

OEIS A053871 is the sequence a(n) = 2(n − 1)(a(n − 1) + a(n − 2)) with a(0) = 1, a(1) = 0. Its FORMULA section states

Conjecture: if m == n (mod q) for q odd, then (-1)^ma(m) == (-1)^na(n) (mod q). - Harry Richman, Aug 29 2023

Issue #10354 fixes the reading: a is the sequence of the recurrence in the entry's name, q is a positive odd integer, and the congruence is in the integers. The statement settled is that for every odd q and all m ≡ n (mod q), (−1)^m a(m) ≡ (−1)^n a(n) (mod q).

Motivation

Our foothold

The entry reads a(n) as the number of deranged matchings of 2n people and as the n-th central moment E[(X² − 1)^n] of the chi-squared distribution with one degree of freedom, X a standard normal variable; its links include Costa, Dobrescu and Fox, Chiral Abelian gauge theories with few fermions (arXiv:2001.11991). The frozen declaration D5/S3/Combinatorics/DerangedMatchingCongruence.result shows that the signed moments are periodic with period q modulo every odd q, which proves the conjecture.

Gap

Missing bridges

Issue #10354 preregisters the statement and its literature check before any formalization. The entry (revision 173, 2026-04-23) still lists the conjecture. The derangement numbers A000166 satisfy an analogous congruence proved in their entry from a first-order recurrence; A053871 has a second-order recurrence with a coefficient depending on n, which that argument does not cover. Johnston, Deranged matchings: proofs and conjectures (arXiv:2209.11319), treats asymptotics in its abstract; its full text was not read. A web search for congruences of deranged matchings found no proof. These readings are not-found-in-searched-scope; they do not establish an exhaustive worldwide literature search or priority.

Route

Proposed approach

  1. b(n) = (−1)^n a(n) satisfies b(n + 2) = 2(n + 1)(b(n) − b(n + 1)), b(0) = 1, b(1) = 0.
  2. Let c(n) = Σ_k (−1)^k C(n, k) (2k − 1)!! and d(n) = Σ_k (−1)^k C(n, k) (2k + 1)!!. Pascal's rule gives c(n + 1) = c(n) − d(n), and (2k + 1)!! = (2k + 1)(2k − 1)!! with k C(n + 1, k) = (n + 1) C(n, k − 1) gives d(n + 1) = c(n + 1) − 2(n + 1) d(n). Hence c satisfies the recurrence of b with the same initial values, and b = c.
  3. For odd q and k ≥ 1, 2^k C(q, k) (2k − 1)!! = q(q − 1)⋯(q − k + 1) · C(2k, k) is divisible by q, and 2 is invertible modulo q, so b(q) = c(q) ≡ 1 = b(0).
  4. b(q + 1) = 2q(b(q − 1) − b(q)) ≡ 0 = b(1).
  5. The coefficient 2(n + 1) has period q modulo q, so by induction b(n + q) ≡ b(n) for every n, and b(m) ≡ b(n) whenever m ≡ n.

Falsifier

What would falsify this route

An odd q and n with b(n + q) ≢ b(n) (mod q) would refute the statement; steps 3–5 exclude it.

Evidence

Evidence to collect

Exact integer computation reproduces the entry's data and confirms b(n + q) ≡ b(n) (mod q) for all odd q < 300 with n + q < 900 (112500 cases) and for all pairs m ≡ n (mod q) with odd q < 60 and m, n < 400. Controls fail as expected: without the sign every odd q from 3 to 39 has a counterexample, and so has every even q from 2 to 18.

The canonical source is D5/S3/Combinatorics/DerangedMatchingCongruence.lean. Its public declarations are a, claim, and result. The frozen module state has statement identity sha256:748b6648d9f5c5f183cce045fa49e0edc8597878aa79885f93746323a7e31043. The result declaration has statement identity sha256:10ebaebf3eb0fddc30c27972480b1f173b46ea99b11029be73c6c138597b392d. The Freeze event is sha256:19520b9cd195fcf3ab685f3864b8bc8116f55d27e30332843b9429885180740f and has no project-level frozen prerequisites. The proof uses only the standard axioms propext, Classical.choice and Quot.sound; no sorry, native_decide, or new axiom.

Triage

Scope assessment

theorem; resolution proved, as read in #10354. The public theorem has proof_shape: content: the alternating-sum form of b, the divisibility of its terms at n = q and the periodicity induction are new propositions on the live path of the proof and are not instances of pinned lemmas. admission_basis: open-problem-resolution under preregistration issue #10354. There is no atom and no digestion coverage edge. The result is a uniform theorem for every odd q, so utility: none applies.

ASSUMED-UNVERIFIED

Unverified assumptions

The bounded literature check does not establish exhaustive worldwide novelty, priority, or the absence of an independent proof; the full text of arXiv:2209.11319 was not read.

Dossier & anchor history

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