A short proof of Mathar's 2020 recurrence conjecture for the generalized-Stirling sequence A001711
Fuente:
arXiv
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| Formato: | Preprint |
| Publicado: |
2026
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| _version_ | 1866913115432550400 |
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| author | Niu, Tong |
| author_facet | Niu, Tong |
| contents | For the OEIS sequence A001711, contributed by N. J. A. Sloane long before the on-line era and identified there as the diagonal $T(n+4, 4)$ of a generalized-Stirling triangle, R. J. Mathar contributed in February 2020 the conjectured order-2 P-recursive recurrence \[ a(n) - (2n+5)\,a(n-1) + (n+2)^{2}\,a(n-2) \;=\; 0,\qquad n \ge 2. \] We give a one-page proof. Detlefs's harmonic-number closed form $a(n) = \tfrac{1}{4}(n+3)!\,(2 H_{n+3} - 3)$ collapses the left-hand side, after dividing through by $(n+1)!/4$, to a polynomial identity of $n$ with coefficient $H_{n+2}$. The harmonic-number coefficient simplifies to $(n+3) - (2n+5) + (n+2) = 0$ (using $H_{n+3} = H_{n+2} + \tfrac{1}{n+3}$ and $H_{n+1} = H_{n+2} - \tfrac{1}{n+2}$); the constant remainder is $-3 \cdot 0 = 0$ 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 $n = 2, \ldots, 5000$. |
| format | Preprint |
| id |
arxiv_https___arxiv_org_abs_2605_11351 |
| institution | arXiv |
| publishDate | 2026 |
| record_format | arxiv |
| spellingShingle | A short proof of Mathar's 2020 recurrence conjecture for the generalized-Stirling sequence A001711 Niu, Tong Combinatorics 05A15, 05A19, 11B37, 11B73, 11B83 For the OEIS sequence A001711, contributed by N. J. A. Sloane long before the on-line era and identified there as the diagonal $T(n+4, 4)$ of a generalized-Stirling triangle, R. J. Mathar contributed in February 2020 the conjectured order-2 P-recursive recurrence \[ a(n) - (2n+5)\,a(n-1) + (n+2)^{2}\,a(n-2) \;=\; 0,\qquad n \ge 2. \] We give a one-page proof. Detlefs's harmonic-number closed form $a(n) = \tfrac{1}{4}(n+3)!\,(2 H_{n+3} - 3)$ collapses the left-hand side, after dividing through by $(n+1)!/4$, to a polynomial identity of $n$ with coefficient $H_{n+2}$. The harmonic-number coefficient simplifies to $(n+3) - (2n+5) + (n+2) = 0$ (using $H_{n+3} = H_{n+2} + \tfrac{1}{n+3}$ and $H_{n+1} = H_{n+2} - \tfrac{1}{n+2}$); the constant remainder is $-3 \cdot 0 = 0$ 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 $n = 2, \ldots, 5000$. |
| title | A short proof of Mathar's 2020 recurrence conjecture for the generalized-Stirling sequence A001711 |
| topic | Combinatorics 05A15, 05A19, 11B37, 11B73, 11B83 |
| url | https://arxiv.org/abs/2605.11351 |