How to evaluate
$$\sum_{k=0}^n k*c(n,k)$$
Where $c(n,k)=|s(n,k)|$ is the stirling number of the first kind without a sign.
Evaluate weightned stirling number sum - $\sum_{i=0}^n k*c(n,k)$
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combinatorics
summation
stirling-numbers
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2You can try differentiation of the polynomial $$x^{\underline{n}}=\sum _{k=0}^n c(n,k)x^k,$$ where $x^{\underline{n}} = x(x-1)\cdots (x-n+1)$ – 2017-02-17
1 Answers
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Differentiation of the falling factorial is definitely the way to go, as @Phicar noted. Given that we know $$x^{\underline{n}}=\sum _{k=0}^n S_n^k\,x^k$$ All we have to do is let $x \to -x$, multiply by $(-1)^n$, and differentiate both sides and to get $$(-1)^n(-x)^{\underline{n}}(H_{-n-x}-H_{-x})=\sum _{k=0}^n k\,x^{k-1}S_n^k(-1)^{n-k}=\sum _{k=0}^n k\,x^{k-1}c(n,k)$$ Now just let $x=1$. The LHS will simplify significantly when you do.