I am calculating the square of the Pascal matrix (i.e. the infinite lower triangular matrix with entries $A_{nk}={n\choose k}$) using the theory of Riordan arrays, and have obtained (or so I believe) the generating function for its $k$-th column: $$\frac{1}{1-2t} \left( \frac{t(1-t)}{1-2t} \right)^{k}$$ Now, to calculate the element $A_{nk}$, I need to obtain the explicit formula for the $n$-th coefficient of this generating function, $$[t^{n}]\frac{1}{1-2t} \left( \frac{t(1-t)}{1-2t} \right)^{k}=(*)$$ which looks like a dounting problem, unless I miss some clever way to do that. I've tried using the properties of the coefficient operator $[t^n]$, and have obtained
$$(*)=[t^n] t^k \frac{1}{1-2t} \left( \frac{1-t}{1-2t} \right)^k=[t^{n-k}] \frac{1}{1-2t}\left( \frac{1-t}{1-2t} \right)^k=$$ $$=[t^{n-k}] (1-t)^k \frac{1}{(1-2t)^{k+1}}$$ $$=\sum_{m=1}^{n-k} [t^m](1-t)^k[t^{n-k-m}] \frac{1}{(1-2t)^{k+1}}$$ $$=\sum_{m=1}^{n-k}(-1)^{m}{k \choose m}[t^{n-k-m}] \frac{1}{(1-2t)^{k+1}}$$
but I am not sure what should I do next and if this is the right path at all.
Edit: I have remembered the formula
$$\frac{1}{(1+z)^{\alpha}}=\sum_{k=0}^{\infty} \frac{(\alpha)^{(k)}}{k!}z^k$$ where $(\alpha)^{(k)}$ denotes the rising factorial, hence, $$\frac{1}{(1-2t)^{k+1}}=\sum_{d=0}^{\infty} \frac{(-1)^d2^d(k+1)^{(d)}}{d!}t^k$$ and we have $$=\sum_{m=1}^{n-k}(-1)^m {k \choose m}(-1)^{n-k-m}2^{n-k-m} \frac{(k+1)^{(n-k-m)}}{(n-k-m)!}$$ $$=(-1)^{n-k}\sum_{m=1}^{n-k}{k \choose m}2^{n-k-m} \frac{(k+1)^{(n-k-m)}}{(n-k-m)!}$$
But I still have no idea how to transform this into something understandable. Since the problem itself is taken from the first problem set to an enumerative combinatorics course, given out before the introduction of generating functions, the answer should be more or less nice.
Edit 2: Using the identity ${-n \choose k}={n+k-1 \choose k}(-1)^k$, I managed to further "simplify" the expression to $$(-1)^{n-k}\sum_{m=0}^{n-k}2^{m}{k+m \choose m}{k \choose n-k-m}$$
Still don't understand if this is simplifiable any further, and if the overall strategy was right.
Edit 3: Tried to follow the advice by @Marko_Riedel and calculating the residue of $(*)\frac{1}{t^{n+1}}$ at $1/2$. Unfortunately, this does not seem to look simple:
$$\frac{1}{k!}\sum_{m=0}^{k}{k \choose m}\left(\frac{d}{dt}\right)^{m}\frac{1}{t^{n-k}}\left(\frac{d}{dt}\right)^{k-m}(1-t)^{k}$$ evaluated at $t=\frac{1}{2}$ $$=\frac{1}{k!}\sum_{m=0}^{k}{k \choose m}(-1)^m (n-k)^{(m)}2^{n-k-m}\sum_{d=0}^{k}{k\choose d}(-1)^{d}(d)_{k-m} \frac{1}{2^{d-k-m}}$$
I am too tired to check r.n., but it appears to me that by Vandermonde's convolution this can be shown to be exactly the previous formula up to a sign.