What is the coefficient of $z^k$ in ${z+n-1 \choose n}$ for $1 \leq k \leq n$? Thanks. I'm currently looking into Stirling numbers of the first kind, as it seems there is a connection.
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Yes, they are just Stirling numbers, up to a sign. $\displaystyle {z+n-1 \choose n} = \frac{1}{n!} z^{n\uparrow}$, where I use the notation $z^{n \uparrow}$ for the rising factorial. On the other hand, $\displaystyle z^{n \uparrow} = (-1)^n (-z)^{n\downarrow} = \sum_k (-1)^{n-k} s(n,k) z^k$, so the coefficient is $\displaystyle \frac{1}{n!} (-1)^{n-k} s(n,k) = \frac{1}{n!} |s(n,k)|$. |
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