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In a calculation of some momenta of random matrices (GOE), I encounter a functional equation, in the form of a second-order recursion, $$L(s+1)=L(s)+2s(2s+1)L(s-1).$$ Is it familiar to someone ? Is there an explicit formula for the solutions ? By explicit, I mean in terms of the Gamma function for instance.

Just in case, I am interested in the solution generated by $L(0)=1$ and $L(1)=-\frac18$ .

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  • $\begingroup$ @ChristianRemling Oups ! I fix it. Actually, it is $L(s-1)$ instead of $L(s+1)$ in the RHS. $\endgroup$
    – Denis Serre
    Commented Feb 17 at 16:59
  • $\begingroup$ This should succumb pretty easily to generating functions; you can get the $2s(2s+1)L(s-1)$ term by doing the usual differentiation (see e.g. section 2.4 of math.cmu.edu/~ploh/docs/math/2011-228/… ). This'll give you a second-order ODE for $\mathcal{L}(x)=\sum_nL(n)x^n$ that you may have more luck finding. $\endgroup$
    – Steven Stadnicki
    Commented Feb 17 at 18:00
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    $\begingroup$ FWIW Mathematica gives $$L(s)=2^{s-3}\left(\left(\frac{17}{\sqrt{2}}-9\right)\Gamma\left(s+1\right)+\frac{17(-1)^s}{\sqrt{\pi}(s+1)}\Gamma\left(s+\frac{3}{2}\right){}_2F_1\left(1,s+\frac{3}{2};s+2;-1\right)\right).$$ $\endgroup$
    – KStar
    Commented Feb 17 at 22:55

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As already pointed out by @KStarGamer in the comment above, Mathematica can sum the recursion using RSolve[]. The result can be rewritten in terms of the regularized incomplete Euler beta function $I_x(a,b)$ and reads, for arbitrary $L_0=L(0)$ and $L_1=L(1)$, \begin{align} \tag{1a}\label{eq:1a} L(s) &= 2^s s!\left[\frac{2 L_0-L_1}{\sqrt{2}} \big(1 - I_{-1}\left(s+1,\tfrac{1}{2}\right)\big) -L_0 + L_1\right] \\ \tag{1b}\label{eq:1b} &= 2^s s!\left[\frac{2 L_0-L_1}{\sqrt{2}} I_{2}\left(\tfrac{1}{2},s+1\right) -L_0 + L_1\right] \end{align} According to Wikipedia,

The regularized incomplete beta function is the cumulative distribution function of the beta distribution, and is related to the cumulative distribution function $F(k;n,p)$ of a random variable $X$ following a binomial distribution with probability of single success $p$ and number of Bernoulli trials $n$: $$F(k;\,n,p)=\Pr \left(X\leq k\right)=I_{1-p}(n-k,k+1)=1-I_{p}(k+1,n-k).$$

This might link to the OP's original problem.

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    $\begingroup$ I apologize that I never learnt softwares such a Mathematica. I should have, but I grew up in the prehistoric times (without personal computers). Thanks a lot ! $\endgroup$
    – Denis Serre
    Commented Feb 20 at 20:01

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