I am trying to solve the following recurrence relation: $$a_n=\sum_{i=0}^{n}\binom{n}{i}^2a_ia_{n-i}$$ Where: $$a_0=0,a_1=1$$
I started by defining a generating function for $a_n$: $$g(x)=\sum_{n=0}^{\infty}a_nx^n$$ $$=x+\sum_{n=2}^{\infty}\left(\sum_{i=0}^{n}\binom{n}{i}^2a_ia_{n-i}\right)x^n$$
I then noticed that:
$$g(x)^2=\sum_{n=2}^{\infty}\left(\sum_{i=0}^{n}a_{i}a_{n-i}\right)x^n$$
But I cannot seem to solve for $g(x)$ by writing it in terms of itself, due to the $\binom{n}{i}^2$ coefficient. Is there a way to do this, or would another approach be more suitable?
Let $b_n:=\frac{a_n}{n!^2}$, so $$\tag1b_n=\frac{a_n}{n!^2} =\sum_{i=0}^n\frac{a_i}{i!^2}\frac{a_{n-i}}{(n-i)!^2} =\sum_{i=0}^nb_ib_{n-i}.$$ So for $h(x):=\sum_{i=0}^\infty b_ix^i$, we formally have the simple equation $$\tag?h(x)=h(x)^2.$$ Oops - we forgot that $(1)$ does not hold for $n=1$ (where it would state $1=b_1=2b_0b_1=0$). So the correct equation needs a fix in the linear term: $$\tag2 h(x)=h(x)^2+x.$$ Writing $h(x)=xk(x)$ (to codify the fact that $b_0=0$), we arrive at $$\tag3 k(x)=1+xk(x)^2,$$ or $$k(x)=\frac{1\pm\sqrt{1-4x}}{2x}.$$ Only the minus-solution converges as $x\to 0$, so $$ h(x)=xk(x)=\frac{1-\sqrt{1-4x}}{2}.$$