<s>[INST] Generating Function of Riordan numbers I would like to find generating function of $f(n)$, where $f(n)$ is defined as following:
$$f(n)=\sum_k^n \binom{n}{k}(-1)^{n-k}C_k\text{.}$$
With $C_k=\frac{1}{k+1}\binom{2k}{k}$($C_k$ is the $k^{th}$ Catalan's number).
Thanks in advantage for Your help
Good Evening
 [/INST] We have for the sum
$$\sum_{k=0}^n {n\choose k} (-1)^{n-k} C_k
= \sum_{k=0}^n {n\choose k} (-1)^k C_{n-k}
\\ = [z^n] \frac{1-\sqrt{1-4z}}{2z}
\sum_{k=0}^n {n\choose k} (-1)^k z^k
= [z^n] \frac{1-\sqrt{1-4z}}{2z}
(1-z)^n
\\ = \; \underset{z}{\mathrm{res}} \;
\frac{1}{z^{n+1}} (1-z)^n
\frac{1-\sqrt{1-4z}}{2z}.$$
Now put $z/(1-z) = w$ so that $z=w/(1+w)$ and $dz = 1/(1+w)^2 \; dw$
to find
$$\; \underset{w}{\mathrm{res}} \;
\frac{1}{w^{n+1}} (1+w)
\frac{1-\sqrt{1-4w/(1+w)}}{2w/(1+w)}
\frac{1}{(1+w)^2}
\\ = \; \underset{w}{\mathrm{res}} \;
\frac{1}{w^{n+1}} 
\frac{1+w-\sqrt{(1+w)^2-4w(1+w)}}{2w(1+w)}
\\ = \; \underset{w}{\mathrm{res}} \;
\frac{1}{w^{n+1}} 
\frac{1+w-\sqrt{1-2w-3w^2}}{2w(1+w)}.$$
It follows that the desired OGF is
$$\frac{1+w-\sqrt{1-2w-3w^2}}{2w(1+w)}.$$
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