RESEARCH / Focused target
Mathar's recurrence for the surviving walks A026023
RELEASED FOUNDATIONS / PROPOSED CONNECTION
Research connections
All source anchors (1)
- D5/S3/StatisticalMechanics/RandomWalks/SurvivingWalkRecurrence.resultAbsent from this graph
Problem
The question
OEIS A026023 counts the sequences s(0), …, s(n) of nonnegative integers with
|s(i) − s(i − 1)| = 1 and s(0) = 3; by a comment of R. M. Ziff, a(n)/2^n
is the probability that a random walker started at x = 4 has not been
adsorbed at x = 0 by time n. Its formula field records:
Conjecture: (n+4)(n-1)a(n) +(n-1)(n+1)a(n-1) -2(n+1)(2n+1)a(n-2) -4(n-1)(n+1)*a(n-3)=0. - R. J. Mathar, Sep 29 2012
Issue #10761 fixes the reading: a is the entry's name read literally, the
recurrence is asserted for every n ≥ 3 and read in ℤ.
Motivation
Our foothold
The entry still marks the recurrence as a conjecture (revision 47,
2025-10-13). D5/S3/StatisticalMechanics/RandomWalks/SurvivingWalkRecurrence.result
proves it for every n ≥ 3 from the literal walk definition.
Gap
Missing bridges
Issue #10761 preregisters the proof route and the literature check; the
entry's history records no proof, and the papers proving batches of Mathar's
conjectures do not treat A026023. not-found-in-searched-scope.
Route
Proposed approach
- Splitting on the first step,
W(n + 1, x) = W(n, x + 1) + [x ≥ 1] W(n, x − 1)for the numberW(n, x)of walks of lengthnfromx. - The reflection count
Σ_{n < 2d + x + 2, 2d ≤ n + x} C(n, d)satisfies the same recursion by Pascal's rule, so it equalsW(n, x)(Theorem 2.1 of Jianu and Dăuş 2025, entering as a local step of the proof). - At
x = 3:a(2m) = C(2m + 2, m)anda(2m + 1) = 2 C(2m + 2, m). - With
c(m) = C(2m + 2, m),(m + 1)(m + 3) c(m + 1) = 2(m + 2)(2m + 3) c(m); the recurrence expression is12times this relation for oddn, andj + 2times it is a combination of the relations atj + 1andjfor evenn = 2j + 4.
Falsifier
What would falsify this route
The proof would fail if the walk count differed from the reflection count, or
if the combinations of step 4 did not reproduce the recurrence expression for
some n ≥ 3.
Evidence
Evidence to collect
Exact integer computation: the literal definition by brute force matches the
recursion for n ≤ 14, the walk count reproduces the 34 data terms, the
recurrence holds for n = 3, …, 400, and replacing the leading factor n + 4
by n + 5 fails for every such n (issue #10761).
The canonical source is
D5/S3/StatisticalMechanics/RandomWalks/SurvivingWalkRecurrence.lean. Its
public declarations are walks, a, claim, and result. The frozen module
state has statement identity
sha256:623ab7879bf31a0a1aa4630d573480d1ca1de9ff570ac9ceae6e95312d1284cc.
The result declaration has statement identity
sha256:6edcfcebe2f131c3bfce97e661f25c580c694d3a21e0e4f4650fad7e76612054.
The Freeze event is
sha256:7cf33fc8bf2901d835ef1093573a5837afaa23e07771a8369bc7b0c0a4cb1b50.
It has no project-level frozen prerequisite. The proof uses only the standard
axioms propext, Classical.choice and Quot.sound; no sorry,
native_decide, or new axiom.
Triage
Scope assessment
Tier 1 external named conjecture (OEIS formula field), preregistered in issue
#10761 before the probe. theorem; resolution proved. The public theorem
has proof_shape: content: the first-step bijection and the identification
of the walk count with the reflection count (Jianu and Dăuş 2025, Theorem 2.1)
are new propositions on its live path. Admission basis
open-problem-resolution; utility none.
ASSUMED-UNVERIFIED
Unverified assumptions
The bounded literature check does not establish exhaustive worldwide novelty, priority, or the absence of an independent proof.