Prove that $\sum_{n=1}^{\infty}(\frac{2}{9})^na_n=\frac{1}{3}$ Let $a_n(n\geq1)$ be the sequence of numbers defined by the recurrence relation$\hspace{1cm}a_1=1,\hspace{1cm}a_n=a_{n-1}a_1+a_{n-2}a_2+...+a_2a_{n-2}+a_1a_{n-1}$
Prove that $\sum_{n=1}^{\infty}(\frac{2}{9})^na_n=\frac{1}{3}$

Let $\alpha=\sum_{n=1}^{\infty}a_nt^n$
$\alpha^2=\sum_{n=1}^{\infty}a_nt^n\sum_{n=1}^{\infty}a_nt^n$
 $a_{n-1}a_1+a_{n-2}a_2+...+a_2a_{n-2}+a_1a_{n-1}$ is the coefficient of $t^n$ in the expansionn of $\alpha^2$
I could not further solve this problem.How should i solve it further?

Let $f(x)$ be $\sum_{i=1}^\infty a_ix^i$. Note that $$f(x) \cdot f(x) = \sum_{i=1}^\infty a_ix^i \cdot \sum_{i=1} a_ix^i = \sum_{i=2}^\infty \sum_{j=1}^{n-1} a_ja_{n-j} x^i = \sum_{i=2}^\infty a_ix^i = f(x)-a_1x=f(x)-x$$
Now we have $f(x)^2-f(x)+x=0$, or $f(x)=\frac{1 \pm \sqrt{1-4x}}{2}$.
Since $f(0)=0$, we have $f(x)=\frac{1-\sqrt{1-4x}}{2}$.
This gives $f(\frac{2}{9})=\frac{1-\sqrt{\frac{1}{9}}}{2}=\frac{1}{3}$ as desired.
For similar recursion trick see Catalan Numbers.
