What is a closed form for
$c_m = \sum_{n=|m|}^{\infty} \left(\dfrac{1}{2}\right)^{2n} (-1)^{m+n}{2n \choose m+n}$
with $m$ an integer? (N.B. the starting index is an absolute value.)
I already know a form in terms of a hypergeometric function, but that's really just rewriting the sum as another sum.
This series arose near the end of this answer: https://math.stackexchange.com/a/2984718/441161
For any particular $m$, this sum is scaling and summing the terms down a vertical column of the Pascal's triangle that has alternating signs, and is only operating on columns that have elements in the even rows of Pascal's triangle -- rows with coefficients for $(a-b)^{2n}$.
$m = 0$ corresponds to the center column of Pascal's triangle. $m = 1$ corresponds to the first relevant column to the right of the center column of Pascal's triangle. $m = -1$ corresponds to the first relevant column to the left of the center column of Pascal's triangle. And so on...
If $m \ge 0$ I get $$ \frac{(3-2\sqrt{2})^m}{\sqrt{2}}$$ and if $m < 0$, $$ \frac{(3+2\sqrt{2})^m}{\sqrt{2}}$$