What is the limit of $\frac{a_{2n+1}}{a_{2n-1}}$, where $a_{n}$ satisfies $a_0=0,a_1=1,a_{n}=\frac{\sum_{k=0}^{n-1}a_{k}a_{n-k}}{n}$? The sequence $a_n$ is defined as follows:
$$a_0=0,\ a_1=1,\ a_{n}=\frac{\sum_{k=0}^{n-1}a_{k}a_{n-1-k}}{n}\left(n\geq2\right)$$
What is the limit: $$\displaystyle\lim_{n\rightarrow\infty}\frac{a_{2n+1}}{a_{2n-1}}$$
Numerical computation shows that the limit reaches approximately $2.467401559$, which is actually $\frac{\pi^2}{4}$. How to prove it?
And I also noticed that if we change the value of $a_0$ and $a_1$, the limit will change accordingly. but I cannot find the relationship between the limit and the initial terms.
 A: If you add the "for $n\ge2$", for the generating function $$f(x)=\sum^\infty_{k=0}a_kx^k,$$ you get the differential equation $$f'(x)=a_1-a^2_0+f(x)^2$$ with initial condition $f(0)=a_0$. The solution is elementary, it's
$$f(x)=\frac{a_0+b\tan bx}{1-a_0\frac{\tan bx}b},$$ where $b=\sqrt{a_1-a^2_0}$. With $a_0=0$ and $a_1=1$, that's just $\tan x$, with radius of convergence of its power series $R=\frac{\pi}2$, and your limit is $R^2$, naturally.
Edit: sorry for some confusion in my notation, I'm trying to find a form easier to analyze in the general case. It's not so easy, because the "next singularity to $0$" may come from a pole of $\tan bx$, or from a zero of $1-a_0\frac{\tan bx}b$. If $a_0=a_1=1$, we have the limiting case $b=0$ and $$f(x)=\frac1{1-x},$$ so $R=1$. 
A: Let us put $ f(x)=\sum_{n\geq 0} a_n x^n $, such that
$$f'(x)=\sum_{n\geq 1}na_n x^{n-1},\qquad f(x)^2 = \sum_{n\geq 0}\left(\sum_{k=0}^{n}a_k a_{n-k}\right)x^n $$
lead (by comparing the coefficients of $x^{n-1}$ in both sides) to
$$ f'(x) = 1+f(x)^2 $$
hence to $f(x)=\tan(x)$. This is an analytic function in a neighbouhood of the origin, whose MacLaurin series has a radius of convergence equal to $\rho=\frac{\pi}{2}$ (it is enough to locate the closest singularity). We have $a_{2m}=0$ since $\tan(x)$ is an odd function, and $a_{2m+1}>0$ by strong convexity. 
$$\lim_{m\to +\infty}\frac{a_{2m+1}}{a_{2m-1}}=\rho^2.$$
