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29
votes
4answers
1k 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 ...
8
votes
1answer
359 views

Is this a new formula? $\Delta^d x^n/d! = \sum_k \left[ x \atop k\right]{ k+n \brace x + d}(-1)^{x+k}$

$$\frac{\Delta^d x^n}{d!} = \sum_k \left[ x \atop k\right]{ k+n \brace x + d}(-1)^{x+k}$$ Where $x$, $n$ and $d$ are non-negative integers, $\Delta^d$ is the $d$-th difference with respect to ...
5
votes
1answer
293 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: ...
5
votes
1answer
422 views

An infinite set of identities using Stirling numbers 1st kind - are they all zero?

I have the following set of series involving the Stirling numbers 1'st kind and binomials, which can be understood as a set of dot-products of row- and column-vectors of two infinite matrices (where R ...
4
votes
1answer
600 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} ...
4
votes
1answer
180 views

Relations involving Stirling numbers of second kind

While inverting a Laplace transform using Post's inversion formula I found the following expression: $$ \sum_{k=1}^n S^n_k \ x^k(\alpha)_k $$ where $S^n_k$ is a Stirling number of second kind and ...
3
votes
1answer
417 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 ...
3
votes
0answers
498 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 ...
2
votes
1answer
245 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: ...
2
votes
1answer
240 views

Trying to prove a congruence for Stirling numbers of the second kind

This a repost of a question I asked at Stack Exchange, but I got no answer so far, so I am trying here, even though it may not suit the "research level" requirement. Proposition: When $n$ and $m$ are ...
1
vote
2answers
384 views

How this expression leads to the given sequence

Here given is a sequence from OEIS. The sequence is triangle of coefficients from fractional iteration of $e^x - 1$. Few terms are: 1, 1, 3, 1, 13, 18, 1, 50, 205, 180, 1, 201, 1865, 4245, 2700, 1, ...
0
votes
2answers
105 views

Combinatorial Interpretation of Generalized Stirling numbers

I know the combinatorial interpretation of first, and second order Stirling numbers (#of k cycles of n items, and #of partitions n items into k subsets). Is there an interpretation for the generalized ...
0
votes
1answer
145 views

Asymptotic formula for an expression in terms of the second kind of stirling numbers

We have proved that the limit of $\sum_{k=0}^n r^2k^m / (1+r)^{k+1}$ when n approaches infinity is $\sum_{k=1}^m S(m,k)k!/r^{k-1}$ where S(m,k) is the second kind of stirling number. Is there a ...
0
votes
0answers
143 views

Generating function for reciprocals of Stirling numbers?

Is there an (ordinary or exponential) generating function for the reciprocals $\frac{1}{s(n,k)}$ of the Stirling numbers of the first kind? Also, is there some general way to find generating ...