It is well-known that the Stirling numbers $S(m,n)$ of the second kind are given by $$\frac{1}{n!}\sum_{j=0}^n (-1)^{n-j}{n\choose j} j^m$$ Is there such a simple formula also for the Stirling numbers $s(m,n)$ of the first kind? The one in Abramowitz-Stegun $$s(m,n)=\sum_{k=0}^{m-n}(-1)^k {m-1+k\choose m-n+k} {2m-n\choose m-n-k} S(m-n-k,k)$$ is a bit cumbersome.Thanks.
2026-03-27 22:12:38.1774649558
Is there a simple formula for the Stirling numbers of the first kind?
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