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A short proof of Mathar's 2021 recurrence conjecture for the Lehmer-Comtet diagonal A045406

  • Tong Niu
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Abstract

For OEIS sequence A045406, the column-2 diagonal of the Lehmer-Comtet triangle A008296, R. J. Mathar contributed in September 2021 the conjectured order-2 P-recursive recurrence a(n) + (2n−7) a(n−1) + (n−4)2 a(n−2) = 0, n ≥ 5. We give a short proof. Detlefs's harmonic-number closed form a(n) = (−1)n (2 Hn−3 − 3)(n−3)! for n ≥ 3 collapses the left-hand side, after factoring out (−1)n (n−5)! (n−4), to a polynomial identity in n with coefficient Hn−4. The Hn−4-coefficient simplifies to (n−3) − (2n−7) + (n−4) = 0 (using Hn−3 = Hn−4 + 1/(n−3) and Hn−5 = Hn−4 − 1/(n−4)); the constant remainder is $0$ 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 = 5, …, 5000.

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Cite

BibTeXRIS
BibTeX
@misc{niu2026short,
  title = {{A short proof of Mathar's 2021 recurrence conjecture for the Lehmer-Comtet diagonal A045406}},
  author = {Tong Niu},
  howpublished = {arXiv},
  year = {2026},
  month = sep,
  doi = {10.48550/arxiv.2605.12839},
  eprint = {2605.12839},
  archivePrefix = {arXiv},
  url = {https://arxiv.org/abs/2605.12839},
}
RIS
TY  - UNPB
TI  - A short proof of Mathar's 2021 recurrence conjecture for the Lehmer-Comtet diagonal A045406
AU  - Tong Niu
PY  - 2026
DA  - 2026/09/10
PB  - arXiv
DO  - 10.48550/arxiv.2605.12839
UR  - https://arxiv.org/abs/2605.12839
AB  - For OEIS sequence A045406, the column-2 diagonal of the Lehmer-Comtet triangle A008296, R. J. Mathar contributed in September 2021 the conjectured order-2 P-recursive recurrence a(n) + (2n−7) a(n−1) + (n−4)^2 a(n−2) = 0, n ≥ 5. We give a short proof. Detlefs's harmonic-number closed form a(n) = (−1)^n (2 H_n−3 − 3)(n−3)! for n ≥ 3 collapses the left-hand side, after factoring out (−1)^n (n−5)! (n−4), to a polynomial identity in n with coefficient H_n−4. The H_n−4-coefficient simplifies to (n−3) − (2n−7) + (n−4) = 0 (using H_n−3 = H_n−4 + 1/(n−3) and H_n−5 = H_n−4 − 1/(n−4)); the constant remainder is $0$ 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 = 5, …, 5000.
ER  -