Let $l$ denote a positive integer and $m$ be an integer $-l \leq m \leq l$. I would like to prove the following identity:
$$\sum_{0 \leq j \leq \left\lfloor\frac{l - m}{2}\right\rfloor}\sum_{0 \leq k \leq \left\lfloor\frac{l - m}{2}\right\rfloor}2^{-2(j + k)}(-1)^k\frac{\Gamma\left(\frac{1}{2} + l - m - j - k\right)}{j!k!(l - m - 2j)!(l - m - 2k)!\left(\frac{1}{2} + l - j\right)} = \frac{\sqrt{\pi}}{2^{l - m - 1}(2l + 1)(l - m)!}$$
For the time being, I have a rather indirect proof of this result but I would like to know whether this could be attained by a straightforward computation.
Let's do some transforms on this sum. First
This makes our sum \begin{multline} \frac{\sqrt{\pi}}{2^{d-1}d!}\sum_{j=0}^{\lfloor d/2\rfloor}\left(\frac{1}{a-2j}\frac{1}{j!\Gamma(\frac{d}{2} + 1-j)\Gamma(\frac{d+1}{2}-j)}\right. \\\left.\times\sum_{k=0}^{\lfloor d/2\rfloor}\frac{\Gamma(\frac{d}{2}+1)\Gamma(\frac{d+1}{2})\Gamma\left(\frac{1}{2} + d- j - k\right)}{\Gamma(\frac{d}{2}+1-k)\Gamma(\frac{d+1}{2}-k)}\frac{(-1)^k}{k!}\right) \end{multline} Note that the $k$ sum is a hypergeometric function, so we have $$ \frac{\sqrt{\pi}}{2^{d-1}d!}\sum_{j=0}^{\lfloor d/2\rfloor}\frac{1}{a-2j}\frac{\Gamma(\frac{1}{2}+d-j)}{j!\Gamma(\frac{d}{2} + 1-j)\Gamma(\frac{d+1}{2}-j)}\,_2F_1\left(-\frac{d}{2},\frac{1-d}{2};j+\frac{1}{2}-d;1\right) $$ Now, as it happens, $$ \,_2F_1\left(-\frac{d}{2},\frac{1-d}{2};d-j-\frac{1}{2};1\right) = \frac{\Gamma(j)\Gamma(j-d+\frac{1}{2})}{\Gamma(j-\frac{d}{2})\Gamma(\frac{1-d}{2}+j)}. $$ We can use this with Gamma reflection, $\Gamma(z)\Gamma(1-z) = \pi/\sin(\pi z)$ to get $$ \frac{\sqrt{\pi}}{2^{d-1}d!}\sum_{j=0}^{\lfloor d/2\rfloor}\frac{(-1)^j}{a-2j}\frac{\sin(2\pi j)}{2\pi j} = \frac{\sqrt{\pi}}{2^{d-1}d!a} = \frac{\sqrt{\pi}}{2^{l-m-1}(2l+1)(l-m)!}. $$ The simple form comes from the fact that $\mathrm{sinc}(2j) = 0$ for all but $j=0$.