RESEARCH / Focused target
Richman's congruence for the deranged-matching numbers
RELEASED FOUNDATIONS / PROPOSED CONNECTION
Research connections
All source anchors (1)
- D5/S3/Combinatorics/DerangedMatchingCongruence.resultAbsent from this graph
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
b(n) = (−1)^n a(n)satisfiesb(n + 2) = 2(n + 1)(b(n) − b(n + 1)),b(0) = 1,b(1) = 0.- Let
c(n) = Σ_k (−1)^k C(n, k) (2k − 1)!!andd(n) = Σ_k (−1)^k C(n, k) (2k + 1)!!. Pascal's rule givesc(n + 1) = c(n) − d(n), and(2k + 1)!! = (2k + 1)(2k − 1)!!withk C(n + 1, k) = (n + 1) C(n, k − 1)givesd(n + 1) = c(n + 1) − 2(n + 1) d(n). Hencecsatisfies the recurrence ofbwith the same initial values, andb = c. - For odd
qandk ≥ 1,2^k C(q, k) (2k − 1)!! = q(q − 1)⋯(q − k + 1) · C(2k, k)is divisible byq, and2is invertible moduloq, sob(q) = c(q) ≡ 1 = b(0). b(q + 1) = 2q(b(q − 1) − b(q)) ≡ 0 = b(1).- The coefficient
2(n + 1)has periodqmoduloq, so by inductionb(n + q) ≡ b(n)for everyn, andb(m) ≡ b(n)wheneverm ≡ 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.