Convolution of binomial coefficients As part of a (SE) problem I've been working on, I came up with this expression:
$$
\sum_{i=0}^M\binom{M-1+i}{i}\binom{M+i}{i}
$$
I'd like to get a closed form for this, but after a considerable amount of time searching my references and online sources (not to mention the time I've spent bashing this into other equally opaque equivalences), I've come up empty. Does anyone have a clue? I'll be happy to link to the original if asked, but the expression more or less tells the story.

Here is an alternate representation of the sum.
Suppose we seek to evaluate
$$S_M =
\sum_{q=0}^M {q+M-1\choose M-1} {q+M\choose M}.$$
Introduce
$${q+M\choose M} = \frac{1}{2\pi i}
\int_{|w|=\epsilon}
\frac{(1+w)^{q+M}}{w^{M+1}} \; dw.$$

and furthermore introduce
$$[[0\le q \le M]]
= \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1+z+z^2+\cdots+z^M}{z^{q+1}} \; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{z^{M+1}-1}{(z-1)z^{q+1}} \; dz$$
which controls  the range so we may  let $q$ go to  infinity to obtain
for the sum
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{z^{M+1}-1}{(z-1)z} 
\frac{1}{2\pi i}
\int_{|w|=\epsilon}
\frac{(1+w)^{M}}{w^{M+1}} 
\sum_{q\ge 0} {q+M-1\choose M-1} \frac{(1+w)^q}{z^q}
\; dw\; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{z^{M+1}-1}{(z-1)z} 
\frac{1}{2\pi i}
\int_{|w|=\epsilon}
\frac{(1+w)^{M}}{w^{M+1}} 
\frac{1}{(1-(1+w)/z)^M}
\; dw\; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{z^{2M+1}-z^M}{(z-1)z} 
\frac{1}{2\pi i}
\int_{|w|=\epsilon}
\frac{(1+w)^{M}}{w^{M+1}} 
\frac{1}{(z-(1+w))^M}
\; dw\; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{z^{2M}-z^{M-1}}{(z-1)^{M+1}} 
\frac{1}{2\pi i}
\int_{|w|=\epsilon}
\frac{(1+w)^{M}}{w^{M+1}} 
\frac{1}{(1-w/(z-1))^M}
\; dw\; dz.$$
Extracting the residue at $w=0$ we obtain
$$\sum_{q=0}^M {M\choose M-q} {q+M-1\choose M-1} \frac{1}{(z-1)^q}.$$
There are two contributions to the outer integral here, the first is
$$\mathrm{Res}_{z=1}
\sum_{q=0}^M {M\choose q} {q+M-1\choose M-1} 
\frac{1}{(z-1)^{q+M+1}} \sum_{p=0}^{2M} {2M\choose p} (z-1)^p
\\ = \sum_{q=0}^M {M\choose q} {q+M-1\choose M-1} 
{2M\choose M+q}.$$
The second is
$$\mathrm{Res}_{z=1}
\sum_{q=0}^M {M\choose q} {q+M-1\choose M-1} 
\frac{1}{(z-1)^{q+M+1}} \sum_{p=0}^{M-1} {M-1\choose p} (z-1)^p$$
and this is easily seen to be zero.
The product of the three binomials is
$$\frac{M!}{q!\times (M-q)!}
\frac{(q+M-1)!}{(M-1)! \times q!}
\frac{(2M)!}{(M+q)!\times (M-q)!}.$$
which yields
$$\frac{M}{M+q}
\frac{(2M)!}{q! \times (M-q)! \times q! \times (M-q)!}.$$
which finally gives for the sum
$$M {2M\choose M}
\sum_{q=0}^M \frac{1}{M+q} {M\choose q}^2.$$
