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4 votes
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Positivity of a finite sum involving Stirling numbers of the first kind

Past days I've been trying to prove that certain polynomials have positive coefficients. After a lot of thinking, I came up with a formula for each coefficient individually, and they are not that ugly....
Luis Ferroni's user avatar
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2 votes
1 answer
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Show that $\sum_{i=0}^{2k} [ {n\choose i+1} + (-1)^{i+1}{n+i+1\choose i+1} ] \sum_{j=0}^i {i\choose j}(-1)^j (i+1-j)^{2k} =0.$

Let $u(k,j) = 1$ if $j=0$, $0$ if $j > k$, or else it is $j*u(k-1,j-1) +(j+1)*u(k-1,j) $. Prove that $ \sum_{i=0}^{2k} {n \choose i+1} u(2k,i) +\sum_{i=0}^{2k} {-n-1 \choose i+1} u(2k,i)=0. $ ...
Benjamin L. Warren's user avatar