2
votes
1answer
228 views

Simple approximation to a sum involving Stirling numbers?

I have also posted this question at http://math.stackexchange.com/questions/486917/simple-approximation-to-a-sum-involving-stirling-numbers. I have an exact answer to a problem, which is the function: ...
4
votes
1answer
245 views

Alternating sums of alternate Stirling numbers

Does anybody know of any identities or combinatorial interpretations for alternating sums of alternate Stirling numbers? I am particularly interested in expressions of the form: ...
3
votes
1answer
391 views

Acyclic orientations of complete graphs in terms of Stirling numbers?

It is well-known that the number of acyclic orientations of $K_n$ is $n!$. Does anybody know of a combinatorial argument for this fact which uses the identity: $$n!=\sum_{k=1}^ns(n,k),$$ where the ...
27
votes
4answers
992 views

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 ...
3
votes
0answers
484 views

A combinatorial bound involving Stirling numbers of the second type

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 ...
4
votes
1answer
551 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} ...