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



Solving recursion of a complex function
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
$$\sum\_{n\geq 1} f\_n(h) \frac{x^{n-1}}{(n-1)!} = G\_h(x)-G\_{h-1}(x).$$


