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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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We prefer questions not to be entirely contained in titles. – Pedro Tamaroff Oct 2 '12 at 20:56

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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Thank you! What is s(n, k)? EDIT: They are the Stirling numbers. – abw333 Oct 2 '12 at 21:28
@abw333 The Stirling number of the first kind. – Alexander Shamov Oct 2 '12 at 21:35

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