<s>[INST] How to get closed form from generating function? I have this generating function:
$$\frac{1}{2}\, \left( {\frac {1}{\sqrt {1-4\,z}}}-1 \right)  \left( \,{
\frac {1-\sqrt {1-4\,z}}{2z}}-1 \right)$$
and I know that $\frac {1}{\sqrt {1-4\,z}}$ is the generating function for the sequence $\binom {2n} {n}$, and $\frac {1-\sqrt {1-4\,z}}{2z}$ is the generating function for the sequence $\frac {1}{n+1}\binom{2n} {n}$.
Now, I thought that I could substitute those in there, and where they multiply I'll use a summation like this:
$$\frac{1}{2}\left( 1-\frac{1}{n+1}\binom{2n} {n}-\binom{2n} {n} + 
\sum_{k=0}^n \frac{1}{k+1} \binom{2k}{k}\binom{2(n-k)}{n-k}
\right)$$
Could this be right? It doesn't seem to work when I try in Maple. What else could I do?
I already know that the end sequence will be $\binom{2n-1}{n-2}$ if this can help...
 [/INST] Expanding the product, and after some algebra:
$$
   g(z) = \frac{1}{2} \left(\frac{\sqrt{1-4 z}-1}{z}+\frac{1}{\sqrt{1-4 z}}+1\right)
$$
Using generalized binomial theorem:
$$
   [z]^n g(z) = \frac{1}{2} \left(  \delta_{n,0} + \binom{1/2}{n+1} (-4)^{n+1} + \binom{-1/2}{n} (-4)^n \right)
$$
Using 
$$ \begin{eqnarray}
  \binom{-1/2}{n}  - 4 \binom{1/2}{n+1} &=& \frac{\Gamma(1/2)}{n! \Gamma(1/2-n)} - \frac{\Gamma(3/2)}{(n+1)! \Gamma(1/2-n)} \\ 
 &=& (-1)^n (n-1) \frac{\Gamma(1/2+n) \Gamma(n+1)}{\Gamma(1/2) (n+1)!n!} \\ 
 &=& (-1)^n (n-1) \frac{(2n)!}{4^n (n+1)!n!} \\ 
  &=& (-1)^n \frac{(2n-1)!}{2 \cdot 4^{n-1} (n+1)!(n-2)!} = 2 \cdot \frac{(-1)^n}{4^n} \cdot \binom{2n-1}{n+1}
\end{eqnarray}
$$
Hence, using Iverson's bracket:
$$
   [z]^n g(z) = \binom{2n-1}{n+1}  \left[ n \ge 2 \right]
$$
</s>