I need to find $F_{n}$ in : $$ F_{n} = \sum_{i=1}^{n-1} (F_{i}\cdot F_{n-i}) , F_0 = 0 , n>=2 $$
This equation screams convolution , I think , but I find it as a quite long solution sometimes.
Here, I first tried to play with the indexes :
$$ F_{n-1} = \sum_{i=1}^{n-2} (F_{i}\cdot F_{n-1-i}) $$
But this doesn't seem to do anything productive .
So convolution maybe :
Let $$ A(x) = \sum_{n=0}^{∞} F_{i}\cdot x^n $$
Let $$ B(x) = \sum_{n=0}^{∞} F_{n-i}\cdot x^n $$
And $$C(x) = A(x)\cdot B(x) = \left(\sum_{n=0}^{∞} F_{i}\cdot x^n \right)\cdot \left(\sum_{n=0}^{∞} F_{n-i}\cdot x^n\right)$$
But I can't see how this helps me here , any hints and/or ideas?
Thanks
For convenience set $F_0=0$, and let
$$f(x)=\sum_{n\ge 0}F_nx^n=\sum_{n\ge 1}F_nx^n\;.$$
Then
$$f(x)^2=\sum_{n\ge 2}\sum_{k=0}^nF_kF_{n-k}x^n=\sum_{n\ge 2}\sum_{k=1}^{n-1}F_kF_{n-k}x^n=\sum_{n\ge 2}F_nx^n=f(x)-F_1x\;,$$
and we have $f(x)^2-f(x)+F_1x=0$. Solve this as a quadratic in $f(x)$:
$$f(x)=\frac{1\pm\sqrt{1-4F_1x}}2\;,$$
and since $f(0)=F_0=0$, we must choose
$$f(x)=\frac{1-\sqrt{1-4F_1x}}2\;.$$
The generating function for the Catalan numbers $C_n$ is
$$\frac{1-\sqrt{1-4x}}{2x}=\sum_{n\ge 0}C_nx^n\;,$$
so
$$\begin{align*} f(x)&=F_1x\left(\frac{1-\sqrt{1-4F_1x}}{2F_1x}\right)\\\\ &=F_1x\sum_{n\ge 0}C_nF_1^nx^n\\\\ &=\sum_{n\ge 0}\frac{F_1^{n+1}}{n+1}\binom{2n}nx^{n+1}\\\\ &=\sum_{n\ge 1}\frac{F_1^n}n\binom{2n-2}{n-1}x^n\;, \end{align*}$$
and $$F_n=\frac{F_1^n}n\binom{2n-2}{n-1}\;.$$