Does $\sum_{k = 1}^{n-1} \frac{\binom{2k}{k}\binom{2n-2k}{n-k}}{k(n-k)}$ have a simpler expression? 
*

*The following constants have appeared in my recent research and I was wondering if they have a simpler expression $?$.

*I have calculated several values but I don't see any obvious combinatorial answer.

$$
\sum_{k = 1}^{n-1}
\frac{\displaystyle\binom{2k}{k}\binom{2n - 2k}
{n - k}}{k\left(n-k\right)} =\ ?
$$
and more general
$$
\sum_{\substack{k_{1}\ +\ k_{2}\ +\ \cdots\ +\ k_{r}\ =\ n
\\[1mm] k_{1}\,,\ k_{2}\,,\ \ldots\ k_{r}\ \ge\ 1}}
\frac{\displaystyle\binom{2k_{1}}{k_{1}}
\binom{2k_{2}}{k_{2}}\cdots\binom{2k_{r}}{k_{r}}}{k_{1}\,k_{2}\ldots k_{r}} =\ ?
$$
 A: This sum has the form of a convolution so can be done by taking a product of generating functions. We can get the generating function we need by starting from
$$\sum_{k \ge 0} {2k \choose k} x^k = \frac{1}{\sqrt{1 - 4x}}$$
(which can be proven using e.g. the generalized binomial theorem) and subtracting $1$, dividing by $x$, and then integrating to give
$$F(x) = \sum_{k \ge 1} {2k \choose k} \frac{x^k}{k} = \int_0^x \frac{1 - \sqrt{1 - 4t}}{t \sqrt{1 - 4t}} \, dt.$$
The substitution $u = \sqrt{1 - 4t} \Rightarrow u^2 = 1 - 4t, 2u \, du = -4 \, dt$ gives
$$\int_1^{\sqrt{1 - 4x}} \frac{1 - u}{ \left( \frac{1 - u^2}{4} \right) u} \left( - \frac{u}{2} \right) \, du = -2 \int_1^{\sqrt{1-4x}} \frac{1}{u+1} \, du = 2 \ln 2 -2 \ln (1 + \sqrt{1 - 4x}).$$
This can be rewritten
$$2 \ln \frac{2}{1 + \sqrt{1 - 4x}} = 2 \ln \frac{1 - \sqrt{1 - 4x}}{2x}$$
where the expression inside the logarithm is, curiously, the generating function of the Catalan numbers. Writing $f_r(n)$ for the sum you're interested in, we then have
$$\boxed{ \sum_{n \ge 0} f_r(n) x^n = F(x)^r = \left( 2 \ln \frac{1 - \sqrt{1 - 4x}}{2x} \right)^r }.$$
If we consider a two-variable generating function summing over all $r$ this gives
$$\boxed{ \sum_{r, n \ge 0} f_r(n) x^n \frac{t^r}{r!} = \exp(F(x) t) = \left( \frac{1 - \sqrt{1 - 4x}}{2x} \right)^{2t} }.$$
These are both funny generating functions and I can't say that I've seen them before. I don't think there will be nice closed forms but there should be asymptotics at least. This second generating function might have a nice combinatorial interpretation but I'd have to think for a bit about what it is.
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\on}[1]{\operatorname{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
&\bbox[5px,#ffd]{%
\sum_{\substack{k_{1}\ +\ k_{2}\ +\ \cdots\ +\ k_{r}\ =\ n
\\[1mm] k_{1}\,,\ k_{2}\,,\ \ldots\ k_{r}\ \ge\ 1}}
{\ds{{2k_{1} \choose k_{1}}
{2k_{2} \choose k_{2}}\cdots{2k_{r} \choose k_{r}}} \over k_{1}\,k_{2}\ldots k_{r}}}
\\[5mm] = &
\sum_{k_{1}\,,\ k_{2}\,,\ \ldots\ k_{r}\ \ge\ 1}
{\ds{{2k_{1} \choose k_{1}}
{2k_{2} \choose k_{2}}\cdots{2k_{r} \choose k_{r}}} \over k_{1}\,k_{2}\ldots k_{r}}\,\bracks{z^{n}}
z^{\sum\limits_{j = 1}^{r}\,\, k_{j}}
\\[5mm] = &
\bracks{z^{n}}\bracks{\sum_{k = 1}^{\infty}
{\ds{2k \choose k} \over k}\,z^{k}}^{\, r}
\\[5mm] = &\
\bracks{z^{n}}\bracks{\sum_{k = 1}^{\infty}
\ds{-1/2 \choose k}\pars{-4}^{k}\, z^{k}\int_{0}^{1}t^{k - 1}
\,\,\dd t}^{\, r}
\\[5mm] = &\
\bracks{z^{n}}\bracks{\int_{0}^{1}\sum_{k = 1}^{\infty}
\ds{-1/2 \choose k}\pars{-4zt}^{k}
\,\,{\dd t \over t}}^{\, r}
\\[5mm] = &\
\bracks{z^{n}}\bracks{\int_{0}^{1}
{\pars{1 - 4zt}^{-1/2} - 1 \over t}
\,\dd t}^{\, r}
\\[5mm] = &\
\bbx{2^{r}\bracks{z^{n}}\bracks{\vphantom{\LARGE A}%
\ln\pars{2} - \ln\pars{1 + \root{1 - 4z}}}^{\, r}} \\ &
\end{align}
I guess it would be a quite difficult task to extract the coefficient of $\ds{z^{n}}$ from the last expression. I leave it here while we find some sort to handle the above mentioned task.
