I would be grateful if you checked my solution:
$$ \sum^{\lfloor j/2 \rfloor}_{q=0} \sum^{\lfloor k/2\rfloor}_{s=0} \frac{2^{k-2s}A^{q+s}}{q!s!(k-2s)!}\delta_{j-2q,k-2s} $$
Here $\lfloor \rfloor$ means the floor function. Due to Kronecker delta, I can write:
$$ q=s+\frac{j-k}{2} $$
Then the double sum can be turned into a single sum:
$$ \left( \lfloor \frac{j}{2} \rfloor +1 \right) \sum^{\lfloor k/2\rfloor}_{s=0} \frac{2^{k-2s}A^{2s+(j-k)/2}}{(s+(j-k)/2)!s!(k-2s)!} $$
Is this right? Also, $s+\frac{j-k}{2}$ should always be integer, because $q$ is integer, but no such condition is specified in the original expression, so how is this ensured?
Maybe I need to modify the relation betwee $q$ and $s$ like this:
$$ q=s+\lfloor \frac{j}{2} \rfloor-\lfloor \frac{k}{2} \rfloor $$
I should also warn you that this expression will be used in another double sum with $j$ and $k$ as indices.