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.
What is the coefficient of $z^k$ in ${z+n-1 \choose n}$ for $1 \leq k \leq n$?
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$\begingroup$
combinatorics
binomial-coefficients
stirling-numbers
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1We prefer questions not to be entirely contained in titles. – 2012-10-02
1 Answers
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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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0@abw333 The Stirling number of the first kind. – 2012-10-02