A Binomial Identity simplify I want to symplify
$$ \sum_{\ell=1}^{k} \frac{1}{\ell}\sum_{m=1}^{\min\{\ell,k-\ell\}}\binom{\ell}{m}\binom{k-\ell-1}{m-1}.
$$
 A: Too long for a comment. It is advantageous to write your quantity as:
$$
c_{ij}^k  =k\sum_{\ell=1}^{k-1}\sum_{m=0}^{\ell}\sum_{c=0}^m\frac{1}{\ell} \left(-1\right)^{\ell-i} \binom{\ell}{m}\binom{k-\ell-1}{m-1}\binom{m}{c} \binom{k-2m}{i + j -2c -\ell}\binom{m}{\ell+c-i},
$$
where $k$ is assumed to be larger than $1$.
According to numerical experiments the quantity can be expressed by the following closed form:
$$
c_{ij}^k=\begin{cases}
\hphantom{-}\binom{k}{i},& j=0\text{ or } j=k,\ 1\le i\le k-1;\\
-\binom{k}{j},& i=0\text{ or } i=k,\ 1\le j\le k-1;\\
\hphantom{-}\hphantom{-}0,& \text{in all other cases},
\end{cases}\tag1
$$
which can be written in one line as:
$$
c_{ij}^k=(\delta_{j0}+\delta_{jk})\binom ki-(\delta_{i0}+\delta_{ik})\binom kj.\tag2
$$
Hope, this can help.
A: By way of an extended comment in response to a pers. comm. / request. A conjectured alternate representation of the sum (here $i=p$ and $j=q$) is given by
$$k (-1)^p \mathrm{Res}_{z=0} z^{p-1}
[w^{p+q}] (1+w)^{k}
[v^{k}] (1+v)^{k-1} \\ \times
\sum_{\ell\ge 1} \frac{(-1)^\ell}{\ell}
z^{-2\ell} w^\ell (1+w)^{-2\ell}
v^\ell (1+v)^{-\ell}
(z(1+w)^2+v(1+z)(z+w^2))^\ell.$$
At this point the term in $1/\ell$ introduces a logarithm and the power term in $\ell$ does not factorize / collect easily in $z$ or $w.$
