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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 ...

**8**

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**1**answer

358 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**

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**1**answer

289 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**

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**1**answer

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### 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**

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**1**answer

595 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

**1**answer

175 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**

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**1**answer

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

**0**answers

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

**1**answer

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**

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**1**answer

236 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 ...

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vote

**2**answers

381 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**

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**2**answers

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### 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 ...

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**1**answer

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**

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**0**answers

139 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 ...