English

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

Combinatorics 2026-05-15 v1

Abstract

For the OEIS sequence A032123, the number of length-2n2n black-and-white strings with nn black beads, considered up to reversal, R. J. Mathar contributed in November 2013 the conjectured order-5 P-recursive recurrence n(n1)a(n)2(n1)(3n4)a(n1)+4(2n214n+19)a(n2)+8(n2+5n19)a(n3)16(n3)(3n10)a(n4)+32(n4)(2n9)a(n5)  =  0,n6. \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)=12((2nn)+[n even](nn/2))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 11 and 22 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,,5000n = 6, \ldots, 5000.

Keywords

Cite

@article{arxiv.2605.14213,
  title  = {A short proof of Mathar's 2013 recurrence conjecture for the reversible-binary-string sequence A032123},
  author = {Tong Niu},
  journal= {arXiv preprint arXiv:2605.14213},
  year   = {2026}
}

Comments

10 pages