I know this is very specific, but is there a way to represent the expression
$$\frac{3}{5} \sum_{n = 0}^\infty \left(\frac{2}{5}\right)^n \sum_{j = 0}^n {n \choose j} \delta_{2j - n, k}$$
in terms of the variable $k \in \mathbb{Z}$? I think I'm just after some way to find out which values of $j,n$ and $k$ the delta function is nonzero, and then I'll be able to write the expression in terms of $k$ since the series converges. Thanks!
Note that I'm only answering your easier question of how to get rid of the Kronecker delta. In the inner sum, the only nonzero term occurs when $2j - n = k$ and hence $j = (k+n)/2$. (There is no nonzero term if the latter quantity isn't an integer.) So the only nonzero terms in the outer sum occur when $n \equiv k \pmod{2}$. For instance, if $k$ is even, say $k = 2m$, then the sum becomes after reindexing
$$ \frac{3}{5}\sum_{n=0}^\infty \left(\frac{2}{5}\right)^{2n}\binom{2n}{m+n} $$
(or, if you'd prefer, change the $m$ to a $k/2$ to express it in terms of $n$). The series looks similar if $k = 2m+1$:
$$ \frac{3}{5}\sum_{n=0}^\infty \left(\frac{2}{5}\right)^{2n+1}\binom{2n+1}{m+n+1}. $$