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A short proof of Mathar's 2020 recurrence conjecture for the generalized-Stirling sequence A001711

Combinatorics 2026-05-13 v1

Abstract

For the OEIS sequence A001711, contributed by N. J. A. Sloane long before the on-line era and identified there as the diagonal T(n+4,4)T(n+4, 4) of a generalized-Stirling triangle, R. J. Mathar contributed in February 2020 the conjectured order-2 P-recursive recurrence a(n)(2n+5)a(n1)+(n+2)2a(n2)  =  0,n2. a(n) - (2n+5)\,a(n-1) + (n+2)^{2}\,a(n-2) \;=\; 0,\qquad n \ge 2. We give a one-page proof. Detlefs's harmonic-number closed form a(n)=14(n+3)!(2Hn+33)a(n) = \tfrac{1}{4}(n+3)!\,(2 H_{n+3} - 3) collapses the left-hand side, after dividing through by (n+1)!/4(n+1)!/4, to a polynomial identity of nn with coefficient Hn+2H_{n+2}. The harmonic-number coefficient simplifies to (n+3)(2n+5)+(n+2)=0(n+3) - (2n+5) + (n+2) = 0 (using Hn+3=Hn+2+1n+3H_{n+3} = H_{n+2} + \tfrac{1}{n+3} and Hn+1=Hn+21n+2H_{n+1} = H_{n+2} - \tfrac{1}{n+2}); the constant remainder is 30=0-3 \cdot 0 = 0 for the same reason. The supplementary archive contains a SymPy script verifying both pieces symbolically, the e.g.f.\ expansion against the harmonic closed form, and Mathar's recurrence numerically for n=2,,5000n = 2, \ldots, 5000.

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Cite

@article{arxiv.2605.11351,
  title  = {A short proof of Mathar's 2020 recurrence conjecture for the generalized-Stirling sequence A001711},
  author = {Tong Niu},
  journal= {arXiv preprint arXiv:2605.11351},
  year   = {2026}
}

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8 pages