The sequence A001711 satisfies a(n) - (2n+5)a(n-1) + (n+2)^2 a(n-2) = 0 for n >= 2, shown by substituting the closed form (1/4)(n+3)! (2 H_{n+3} - 3) and verifying that harmonic coefficients and constant terms both cancel to zero.
A short proof of Mathar's 2013 recurrence conjecture for the Meixner sequence A214615
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abstract
For the OEIS sequence A214615, defined by $a(n) = M_{n}(1)$ where $M_{n}$ is the $n$-th Meixner polynomial satisfying $M_{n+1}(x) = x\,M_{n}(x) - n^{2}\,M_{n-1}(x)$, R.~J.~Mathar contributed on 6~March 2013 the conjectured order-2 P-recursive recurrence $a(n) - a(n-1) + (n-1)^{2}\,a(n-2) = 0$ for $n \ge 2$. We give a one-page proof. The exponential generating function $F(t) = \exp\!\bigl(\arctan t\bigr)/\sqrt{1+t^{2}}$ satisfies the first-order linear ODE $(1+t^{2})\,F'(t) = (1-t)\,F(t)$, and Mathar's recurrence then falls out by reading off the coefficient of $t^{n}/n!$. Both steps are short. The supplementary archive includes a SymPy script that checks the ODE identically and the recurrence numerically up to $n = 500$.
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A short proof of Mathar's 2020 recurrence conjecture for the generalized-Stirling sequence A001711
The sequence A001711 satisfies a(n) - (2n+5)a(n-1) + (n+2)^2 a(n-2) = 0 for n >= 2, shown by substituting the closed form (1/4)(n+3)! (2 H_{n+3} - 3) and verifying that harmonic coefficients and constant terms both cancel to zero.