Skip to main content

Questions tagged [stirling-numbers]

Filter by
Sorted by
Tagged with
0 votes
2 answers
187 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 ...
user3639557's user avatar
9 votes
1 answer
568 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 $x$, $\...
adam W's user avatar
  • 191
2 votes
1 answer
338 views

Simple approximation to a sum involving Stirling numbers?

I have also posted this question at https://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:...
user1748601's user avatar
2 votes
3 answers
529 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, ...
Nadia's user avatar
  • 21
4 votes
1 answer
385 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 $(\...
guaraqe's user avatar
  • 157
0 votes
1 answer
194 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 ...
liaomingxue's user avatar
9 votes
1 answer
825 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 ...
Gottfried Helms's user avatar
5 votes
1 answer
585 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: $$\pm\sum_{k}(-1)^k|...
Adam 's user avatar
  • 1,327
5 votes
3 answers
1k 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 $...
Adam 's user avatar
  • 1,327
29 votes
3 answers
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 ...
Gary Kennedy's user avatar
4 votes
0 answers
578 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 $\Big(\prod^r_{i=1}p_i\Big)\sum^n_{j=0}(-1)^j\...
Dmitry Kerner's user avatar
4 votes
1 answer
886 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} ...
Dmitry Kerner's user avatar
13 votes
4 answers
3k views

Simple/efficient representation of Stirling numbers of the first kind

Stirling numbers of the second kind can be expressed by means of a simple hypergeometric (considering $n$ fixed) sum $$S_2(n,k) = \frac{1}{k!}\sum_{j=0}^{k}(-1)^{k-j}{k \choose j} j^n. \qquad (1)$$ ...
Fredrik Johansson's user avatar

1
2