Let $L_n^{(\alpha)}(x)$ denote a generalized Laguerre polynomial. I would like to have a closed form for the expression $$\frac{L_n^{(\alpha)}(1)}{L_{n-1}^{(\alpha)}(1)};$$ where we set $x=1$ and both $n$ and $\alpha$ are positive integers.

I'm aware, as explained in this answer on Math.SE, that the quotient can be expressed as a continued fraction. But this holds for any $x$, and I was hoping to get a closed form in my case, given its specificity. Mathematica doesn't seem to be of any help.

I would be grateful for any comment or idea.

  • 1
    $\begingroup$ Even when $\alpha=1$, the ratio you seek has no closed formula. In fact, $L_n(1)$ can be given as a binomial sum but lacks a closed form expression. I'm not sure what you were hoping for. $\endgroup$ – T. Amdeberhan Nov 18 '16 at 1:28
  • $\begingroup$ Thanks for your comment! I wasn't aware that already $L_n(1)$ lacks a closed form expression. Given $L_0^{(\alpha)}(1) = 0$, $L_1^{(\alpha)}(1) = \alpha$ and the recurrence relation $$\frac{L_{k+1}^{(\alpha)}(1)}{L_{k}^{(\alpha)}(1)} = \frac{2k+\alpha}{k+1} -\frac{k+\alpha}{k+1}\frac{L_{k-1}^{(\alpha)}(1)}{L_{k}^{(\alpha)}(1)} ,$$ I was hoping that an explicit solution could be given. (I mean, it is clear that this recurrence relation is solved by Laguerre polynomials, but the hope is that something easier could come up!) $\endgroup$ – Giovanni De Gaetano Nov 18 '16 at 8:25

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