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I have $\sum_{j=0}^{k} {n+k-j-1 \choose k-j}{m+j-1 \choose j}$ I have identified that this is a convolution where $a_n = {2n-1 \choose n}$ and $b_n = {m+n-1 \choose n}$. I have written the product of their generating functions $\sum_{n=0}^{\infty} {2n-1 \choose n}x^n * \sum_{n=0}^{\infty} {m+n-1 \choose n}x^n = \sum_{n=0}^{\infty}\sum_{k=0}^n {2k-1 \choose k}{m+n-k-1 \choose n-k}x^n$ but I can't come up with a meaningful identity to solve for the coefficient of $x^n$ in this ordinary generating function.

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Your sum does come from a convolution of series, but not the ones you mentioned. (You seem to have mistakenly identified $k$, $n$ and $m$.) The convolution really comes from the series $f(x)=\sum_k \binom{n-1+k}kx^k$ and $g(x)=\sum_k\binom{m-1+k}kx^k$. These generating functions have well-known forms: $f(x)=1/(1-x)^{n}$ and $g(x)=1/(1-x)^{m}$. So, your sum is $[x^k]1/(1-x)^{n+m}=\binom{n+m-1+k}k$.

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