Is there an analytic expression for this recursive sum? $C_n = \sum _{k=0}^{n-1}C_k C_{n-1-k}$ Is there an analytic expression for this recursive sum ? Say ,
$C_n = ?$
\begin{align*}
C_n =& \sum _{k=0}^{n-1}C_k C_{n-1-k}  \\
    =& C_0C_{n-1} + C_1C_{n-2}+\cdots+C_{n-1}C_0
\end{align*}
 A: Hint. One may recall the Cauchy product:
$$ \left(\sum_{n=0}^\infty A_n \right)\!\left(\sum_{n=0}^\infty B_n \right) = \sum_{n=0}^\infty \left(\sum_{k=0}^n A_k B_{n-k}\right), \tag1
$$
then use it with $A_n=B_n=C_nx^n$ to get 
$$ \left(\sum_{n=0}^\infty C_n x^n\right)^2 = \sum_{n=0}^\infty \left(\sum_{k=0}^n C_k C_{n-k}\right)\!x^n =\sum_{n=1}^\infty \left(\sum_{k=0}^{n-1} C_k C_{n-1-k}\right)\!x^{n-1}. \tag2
$$
By hypothesis, the latter series is equal to $\displaystyle  \sum_{n=1}^\infty C_n x^{n-1}$,  thus $(2)$ rewrites
$$ \left(\sum_{n=0}^\infty C_n x^n\right)^2 = \sum_{n=1}^\infty C_n x^{n-1} \tag3
$$$$ x\left(\sum_{n=0}^\infty C_n x^n\right)^2 = \left(\sum_{n=0}^\infty C_n x^n\right)-C_0 \tag4
$$
Setting $\displaystyle f(x):=\sum_{n=0}^\infty C_n x^n $,  the ordinary generating function of the numbers $C_n$, then $(4)$ reads
$$
x\:(f(x))^2=f(x)-C_0 \tag5
$$ giving 
$$
f(x)=\frac{1-\sqrt{1-4 C_0 x}}{2 x} \tag6
$$ for appropriate $C_0$.
Using the binomial theorem, $(6)$ leads to
$$
f(x)=\sum_{n=0}^\infty \frac1{n+1}\binom{2n}{n}\times (C_0)^{n+1}\times x^n \tag7
$$ and we conclude by identification between $\displaystyle f(x)=\sum_{n=0}^\infty C_n x^n$ and $(7)$ to get

$$
C_n=\frac1{n+1}\binom{2n}{n}\times (C_0)^{n+1}. \tag8
$$ 

The case $C_0=1$ gives Catalan's numbers.
A: Yes, this is known as the Catalan sequence:
with $C_0 =1$:
$$
C_n = \frac{(2n)!}{n!(n+1)!}
$$
