arxiv: 2605.14213 · v1 · submitted 2026-05-14 · 🧮 math.CO

Recognition: no theorem link

A short proof of Mathar's 2013 recurrence conjecture for the reversible-binary-string sequence A032123

Tong Niu

Authors on Pith no claims yet

Pith reviewed 2026-05-15 02:51 UTC · model grok-4.3

classification 🧮 math.CO

keywords binary stringsreversal orbitsBurnside lemmaP-recursive recurrencebinomial coefficientsOEIS A032123recurrence relations

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The pith

Mathar's conjectured order-5 recurrence holds for the sequence counting binary strings up to reversal.

A machine-rendered reading of the paper's core claim, the machinery that carries it, and where it could break.

The paper proves that the number a(n) of length-2n binary strings with n black beads, considered up to reversal, satisfies the specific order-5 linear recurrence conjectured by Mathar in 2013. It starts from the closed form obtained by applying Burnside's lemma to the two-element reversal group, splits a(n) into a central binomial term plus an even-n half-binomial term, and shows that Mathar's operator applied to each term separately produces a polynomial identity that evaluates to zero. A reader would care because the result converts a conjectural relation into a proven computational tool for the sequence without requiring fresh combinatorial counting at each step.

Core claim

The sequence defined by a(n) = 1/2 (binomial(2n, n) + [n even] binomial(n, n/2)) satisfies the recurrence n(n-1)a(n) - 2(n-1)(3n-4)a(n-1) + 4(2n^2-14n+19)a(n-2) + 8(n^2+5n-19)a(n-3) - 16(n-3)(3n-10)a(n-4) + 32(n-4)(2n-9)a(n-5) = 0 for all n >= 6. The proof proceeds by verifying that the order-5 operator annihilates each summand of the closed form after the binomial coefficients are replaced by their known elementary recurrences and the resulting expressions are reduced to polynomial identities that vanish identically.

What carries the argument

Burnside closed form splitting a(n) into central binomial and even-case half-binomial terms, to which Mathar's order-5 linear operator is applied separately.

If this is right

The recurrence can be used to compute the sequence for all n >= 6 without direct enumeration.
Both the central binomial coefficient and the half-binomial term are annihilated by the same order-5 operator after the substitution step.
The supplementary symbolic verification confirms the polynomial identities hold exactly.
Numerical checks for n up to 5000 match the closed form to the recurrence output.

Where Pith is reading between the lines

These are editorial extensions of the paper, not claims the author makes directly.

The same reduction-to-polynomial-identity technique may shorten proofs for other conjectured recurrences on sequences whose closed forms are linear combinations of binomial coefficients.
It indicates that minimal-order P-recursions for orbit-counting sequences can often be read off once the Burnside expression is known.
Similar direct substitutions could be tried on related reversal or dihedral actions in other combinatorial sequences.

Load-bearing premise

The closed form obtained from Burnside's lemma on the reversal action correctly enumerates the distinct orbits.

What would settle it

An integer n >= 6 at which the closed-form expression for a(n) fails to make the left-hand side of the given recurrence equal zero.

read the original abstract

For the OEIS sequence A032123, the number of length-$2n$ black-and-white strings with $n$ black beads, considered up to reversal, R. J. Mathar contributed in November 2013 the conjectured order-5 P-recursive recurrence \[ \begin{aligned} &n(n-1)\,a(n) - 2(n-1)(3n-4)\,a(n-1) + 4(2n^{2}-14n+19)\,a(n-2) &\qquad + 8(n^{2}+5n-19)\,a(n-3) - 16(n-3)(3n-10)\,a(n-4) &\qquad + 32(n-4)(2n-9)\,a(n-5) \;=\; 0, \qquad n \ge 6. \end{aligned} \] We give a short proof. Burnside's lemma applied to the reversal action gives the closed form $a(n) = \tfrac{1}{2}\bigl(\binom{2n}{n} + [n \text{ even}]\binom{n}{n/2}\bigr)$; the two summands satisfy elementary recurrences of order $1$ and $2$ respectively; and Mathar's order-5 operator, applied to each summand separately, reduces to a polynomial identity that simplifies to zero after a brief calculation. The supplementary archive includes a SymPy script which verifies the polynomial identities symbolically and checks Mathar's recurrence numerically for $n = 6, \ldots, 5000$.

Editorial analysis

A structured set of objections, weighed in public.

Desk editor's note, referee report, simulated authors' rebuttal, and a circularity audit. Tearing a paper down is the easy half of reading it; the pith above is the substance, this is the friction.

Desk Editor's Note private letter to a colleague

This paper gives a short, verifiable proof of Mathar's 2013 recurrence conjecture for OEIS A032123 by reducing the operator to zero on each closed-form summand.

read the letter

The main takeaway is that the authors settle the open recurrence for this one sequence with a direct calculation. They start from the standard Burnside count under reversal, which produces a(n) = 1/2 times the central binomial coefficient plus the middle binomial when n is even. They then apply Mathar's order-5 operator separately to each summand, using the known low-order recurrences satisfied by those binomial terms, and reduce the result to polynomial identities in n that cancel to zero. The supplement supplies a SymPy script that performs the symbolic simplification and confirms the recurrence numerically up to n=5000. This setup is straightforward and leaves little room for hidden errors. The even-odd case split is handled explicitly, which keeps the algebra manageable. The work rests on elementary facts about binomial coefficients rather than any fitted parameters or self-referential steps, so the circularity burden stays low. The central claim therefore holds up. The limitation is the narrow focus. The closed form itself follows from routine Burnside application and was already available, so the new piece is mainly the explicit verification that the specific operator annihilates it. No broader method for producing recurrences or handling other group actions appears. This paper is aimed at people who follow OEIS sequences or study P-recursive relations for small combinatorial counts. A reader who needs the recurrence for A032123 or wants a template for similar verification steps will get direct value from it. The thinking is clear and the evidence is reproducible, so the paper deserves a serious referee even though its scope is limited. I would send it to peer review.

Referee Report

0 major / 1 minor

Summary. The paper gives a short proof of Mathar's 2013 conjectured order-5 P-recursive recurrence for OEIS A032123 (reversible binary strings with n black beads in length 2n). Burnside's lemma on the two-element reversal group yields the closed form a(n) = ½(binomial(2n,n) + [n even] binomial(n,n/2)); the central and middle binomial coefficients satisfy known low-order recurrences; and direct substitution shows that Mathar's order-5 operator annihilates each summand separately by reducing to polynomial identities in n that vanish identically.

Significance. The result supplies an elementary, self-contained verification of the recurrence that rests only on the standard Burnside count and the classical recurrences for binomial coefficients. The supplementary SymPy script performs the exact symbolic reduction of the polynomial identities and confirms the recurrence numerically to n=5000, providing reproducible evidence that strengthens the combinatorial interpretation of the sequence.

minor comments (1)

In the displayed recurrence, the line breaks and alignment of the coefficients could be tightened for readability; a single displayed equation with explicit + signs on each line would improve clarity.

Simulated Author's Rebuttal

0 responses · 0 unresolved

We thank the referee for the positive report, the accurate summary of the proof, and the recommendation to accept. We are pleased that the elementary approach via Burnside's lemma and the classical binomial recurrences was viewed as self-contained and reproducible.

Circularity Check

0 steps flagged

No significant circularity identified

full rationale

The paper derives the closed form a(n) = ½(binomial(2n,n) + [n even] binomial(n,n/2)) directly from Burnside's lemma on the two-element reversal group, a standard and independent combinatorial argument. It then shows that Mathar's order-5 operator annihilates each summand separately by reducing the action to polynomial identities in n that evaluate to zero, with symbolic verification supplied in the supplement. No parameters are fitted to data, no self-citations are used as load-bearing premises, and the recurrence verification does not presuppose the target result. The derivation chain is therefore self-contained against external benchmarks.

Axiom & Free-Parameter Ledger

0 free parameters · 1 axioms · 0 invented entities

The proof rests on Burnside's lemma (standard in group actions on sets) and the elementary recurrences for binomial coefficients; no free parameters are fitted and no new entities are postulated.

axioms (1)

standard math Burnside's lemma counts the number of orbits under a finite group action by averaging the number of fixed points
Applied to the two-element reversal group acting on the set of binary strings with exactly n black beads.

pith-pipeline@v0.9.0 · 5589 in / 1251 out tokens · 38198 ms · 2026-05-15T02:51:24.114460+00:00 · methodology

discussion (0)

Reference graph

Works this paper leans on

11 extracted references · 11 canonical work pages · 1 internal anchor

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