Let be $C_k$, $k^{th}$ Catalam's number and $$f(n)=\sum_{k=0}^n\binom{n}{k}(-1)^{n-k}C_k\text{.}$$
I want to prove the following recurrence:
$$f(n+1)+(-1)^{n}=f(0)f(n)+f(1)f(n-1)+\cdots+f(n)f(0)\text.{}$$
I know that there is a similar recurrence involving $C_{n+1}$, that is $\displaystyle C_{n+1}=\sum_{k=0}^nC_{n-k}C_k$.
I wonder if there is a method to prove recurrence, by using Moebius's Inversion Theorem or if there is another way to get recurrence.
Any hints would be really precious, so Thank You very much in advance.
At the following MSE link the generating function of
$$f(n) = \sum_{k=0}^n {n\choose k} (-1)^{n-k} C_k$$
was found to be
$$F(w) = \frac{1+w-\sqrt{1-2w-3w^2}}{2w(1+w)}.$$
The generating function of the RHS of the proposed recurrence is by the Cauchy product the square $F(w)^2$ of this OGF:
$$\frac{(1+w)^2 - 2(1+w)\sqrt{1-2w-3w^2} + 1-2w-3w^2} {4w^2(1+w)^2} \\ = \frac{2-2w^2 -2(1+w)\sqrt{1-2w-3w^2}}{4w^2(1+w)^2} = \frac{1-w-\sqrt{1-2w-3w^2}}{2w^2(1+w)} \\ = \frac{1+w-\sqrt{1-2w-3w^2}}{2w^2(1+w)} - \frac{1}{w(1+w)}.$$
Extracting the coefficient with $n\ge 0$ we get
$$\sum_{q=0}^n f(q)f(n-q) = [w^n] \frac{1+w-\sqrt{1-2w-3w^2}}{2w^2(1+w)} - [w^n] \frac{1}{w(1+w)} \\ = [w^{n+1}] \frac{1+w-\sqrt{1-2w-3w^2}}{2w(1+w)} - [w^{n+1}] \frac{1}{1+w} \\ = f(n+1) + (-1)^n.$$
Note that $$\sqrt{1-2w-3w^2} = 1-w +\cdots$$ as can be seen by evaluating at zero and evaluating the derivative
$$\frac{1}{2} \frac{-2-6w}{\sqrt{1-2w-3w^2}}$$
at zero which means the OGF is analytic at zero and does not have a singularity there. This also follows from it being the square of a function $F(w)$ that is analytic in a neighborhood of zero.