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### Finite differences of Stirling numbers

Let s(n,k) and S(n,k) denote the Stirling numbers of the first (with signs) and second kinds, respectively. Next consider the sequence |s(n+2,n)| which begins: (2,11,35,85,175,...) . Using this to ...
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### Stirling number identity via homology?

This is a question about the well-known formula involving both types of Stirling numbers: $\sum_{k=1}^{\infty}(-1)^{k}S(n,k)c(k,m)=0$, where $S(n,k)$ is the number of partitions of an $n$-element set ...
My previous question was solved in a very elegant way, hopefully this (seemingly more complicated) case is also easy for experts. I need the inequality $\Big(\prod^r_{i=1}p_i\Big)\sum^n_{j=0}(-1)^j\... 1answer 695 views ### A bound involving Stirling numbers of the second kind and the asymptotics Let$S_{n,r}$denote the Stirling number of the second kind. Define$A_{n,r}:=\frac{\binom{n+r-1}{n}(n+r)!}{S_{n+r,r}r!}$. I want to prove:$A_{n,1}\ge A_{n,2}\ge..\ge A_{n,r}\ge \lim_{r\to\infty} ...
Stirling numbers of the second kind can be expressed by means of a simple hypergeometric (considering $n$ fixed) sum $$S_2(n,k) = \frac{1}{k!}\sum_{j=0}^{k}(-1)^{k-j}{k \choose j} j^n. \qquad (1)$$ ...