Generating Function of Riordan numbers

Question

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).

Answer 1

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

$$\frac{1+w-\sqrt{1-2w-3w^2}}{2w(1+w)}.$$