<s>[INST] Sum of all Products on Catalan numbers how can I simplify this?
let:
$$
C_n = {{2n \choose n}\over n+1}
$$
find:
$$
\sum_{P_1 + P_2 + ... + P_k = r} \left(\prod_{j = 1}^k C_{P_j}\right)
$$
thanks!
 [/INST] Following @RobertIsrael we have the generating function of the Catalan
numbers
$$f(z) = \frac{1-\sqrt{1-4z}}{2z}$$
and the quantity in question is
$$[z^r] f(z)^k.$$
This has the integral representation
$$[z^r] f(z)^k
= \frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{r+1}} \left(\frac{1-\sqrt{1-4z}}{2z}\right)^k
\; dz.$$
Now put $1-4z = w^2$ so that $1/4-w^2/4 = z$ and
$dz = -w/2 \; dw$ to get
$$-\frac{1}{2^{k+1}} \frac{1}{2\pi i}
\int_{|w-1|=\gamma}
\frac{4^{r+k+1}}{(1-w^2)^{r+k+1}}
(1-w)^k w
\; dw
\\ = - 2^{2r+k+1} \frac{1}{2\pi i}
\int_{|w-1|=\gamma}
\frac{1}{(1+w)^{r+k+1}} \frac{1}{(1-w)^{r+1}}
w \; dw
\\ = (-1)^r 2^{2r+k+1} \frac{1}{2\pi i}
\int_{|w-1|=\gamma}
\frac{1}{(1+w)^{r+k+1}} \frac{1}{(w-1)^{r+1}}
w \; dw
\\ = (-1)^r 2^{2r+k+1} \frac{1}{2\pi i}
\int_{|w-1|=\gamma}
\frac{1}{(1+w)^{r+k+1}} \frac{1}{(w-1)^{r}}
\; dw
\\ +  (-1)^r 2^{2r+k+1} \frac{1}{2\pi i}
\int_{|w-1|=\gamma}
\frac{1}{(1+w)^{r+k+1}} \frac{1}{(w-1)^{r+1}}
\; dw.$$
Now we require the derivative
$$\left(\frac{1}{(1+w)^{r+k+1}}\right)^{(q)}
= (-1)^q \frac{(r+k+q)!}{(r+k)!}
\frac{1}{(1+w)^{r+k+q+1}}.$$
This yields for the two pieces
$$(-1)^r 2^{2r+k+1}
(-1)^{r-1} \frac{(2r+k-1)!}{(r-1)!(r+k)!} \frac{1}{2^{2r+k}}
+ (-1)^r 2^{2r+k+1}
(-1)^{r} \frac{(2r+k)!}{r!(r+k)!} \frac{1}{2^{2r+k+1}}
\\ = -2 {2r+k-1\choose r-1}
+ {2r+k\choose r}
\\ = {2r+k\choose r}
\left(1-2\frac{r}{2r+k}\right)
\\ = \frac{k}{2r+k} {2r+k\choose r}.$$
Here we have used the fact that $\sqrt{1-4z} = 1 - 2z - 2z^2 - \cdots$
so that  with $z$  rotating once  around the  origin the  variable $w$
rotates once around the value one, with a contour that may be deformed
to a circle.
</s>