Solving recursion of a complex function I am trying to find a closed form formula for the following recursive function:
$$f_n(h)= \sum_{i=1}^{n-1} \binom{n-2}{i-1} \cdot (0.5)^{n-2} \cdot [ (f_{n-i}(h-1)\cdot \sum_{j=0}^{h-1}f_i(j)) + (f_{i}(h-1)\cdot \sum_{j=0}^{h-2} f_{n-i}(j))] $$
The base cases are the following:
$$ f_1(h)= \begin{cases} 
      1 & h=0 \\
      0 & otherwise
   \end{cases} \\
   f_2(h)= \begin{cases} 
      1 & h=1\\
      0 & otherwise
   \end{cases} 
$$ 
I have been trying to use the generating functions technique, but I have been unsuccessful so far and I was wondering if anyone has suggestions into how to solve this problem. 
Thank you for your help in advance
Edit:
I added the base cases
 A: Define $g_k(m) := \sum_{j=0}^m f_k(j)$. Then the given recurrence becomes
\begin{split}
g_n(h)-g_n(h-1) &= 0.5^{n-2} \sum_{i=1}^{n-1}\binom{n-2}{i-1} [(g_{n-i}(h-1)-g_{n-i}(h-2))g_i(h-1)+(g_{i}(h-1)-g_{i}(h-2))g_{n-i}(h-2)] \\
&=0.5^{n-2} \sum_{i=1}^{n-1}\binom{n-2}{i-1} [(g_{n-i}(h-1)g_i(h-1)-g_{i}(h-2)g_{n-i}(h-2)].
\end{split}
Consider the generating function
$$G_h(x) := \sum_{n\geq 1} g_n(h) \frac{x^{n-1}}{(n-1)!}.$$
The initial conditions imply that $G_1(x)=1+x$ and $G_2(x)=1+x+\frac{x^2}2+\frac{x^3}{12}$.
Then the recurrence takes form:
$$G_h'(x) - G_{h-1}'(x) = G_{h-1}(x/2)^2 - G_{h-2}(x/2)^2$$
or
$$G_h'(x) - G_{h-1}(x/2)^2 = G_{h-1}'(x) - G_{h-2}(x/2)^2.$$
Unrolling the last recurrence, we get that for any $h\geq 2$
$$G_h'(x) - G_{h-1}(x/2)^2 = G_{2}'(x) - G_{1}(x/2)^2=0.$$
That is,
$$G_h'(x) = G_{h-1}(x/2)^2.$$
It seems that there is no simple expression for the solution to this recurrence, although we may notice that $\lim_{h\to\infty} G_h(x)=e^x$.
P.S. For a fixed $h$, the generating function for $f_n(h)$ can be expressed as
$$\sum_{n\geq 1} f_n(h) \frac{x^{n-1}}{(n-1)!} = G_h(x)-G_{h-1}(x).$$
