Stack Exchange Network

Stack Exchange network consists of 174 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers.

Visit Stack Exchange

The tag has no usage guidance.

1
vote
1answer
84 views

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 ...
6
votes
1answer
337 views

Some strange multinomial averaging

How do I prove : $\sum_{j=2}^{n} (-1)^j {\frac {M(n+j,j;2)}{j!}} = (-1)^n n! + 1$? where $M(n+j,j;2)$ is the multinomial sum $M(n+j,j;2) = \sum_{t_1 + t_2 + \dotsc + t_j = n+j, t_k \geq 2} {n+j \...
3
votes
2answers
406 views

Proof of identity involving Stirling numbers of the second kind

While computing conditional expectations of certain functionals of a Poisson white noise field (details are long and probably irrelevant), I've stumbled upon the need to use the following identity ...
8
votes
1answer
165 views

Singular values of Stirling numbers matrix

Consider the Stirling numbers of the first kind $s(i, j)$, and form a matrix $S_1(n),$ where the $(i, j)$th entry is $s(i, j)$. (IMPORTANT NOTE the indices start at $0,$ so this matrix is $(n+1)\times ...
12
votes
0answers
348 views

What is this sequence?

This is again a question that I asked at Stack Exchange, but got no answer so far, so I am trying here. Let: $$ a_n=\sum_{k\ge0}(k+1) {n+2\brack k+2}(n+2)^kB_k$$ $B_k$ is the Bernoulli number. ${n\...
17
votes
4answers
1k views

Are surjections $[n]\to [k]$ more common than injections $[k]\to [n]$?

Here's an interesting inequality involving binomial coefficient and Stirling numbers of the second kind that I believe holds for all $n,k$: $$ k^n {n \choose k} \leq n^k {n \brace k} $$ On the left-...
4
votes
0answers
683 views

A conjecture about the degrees of special polynomials

Define the congruence "modulo m" on exponential Taylor series as $$ \sum_{n=0}^\infty \frac{a_n}{n!}x^n \equiv \sum_{n=0}^\infty \frac{b_n}{n!} x^n \mod m \iff \forall n: \frac{a_n-b_n}{m}\in \mathbb{...
3
votes
0answers
101 views

Multiple integral evaluation involving Stirling numbers and Riemann zeta function

Hello Mathoverflow community, how are you doing? I just wanted to know if anything is known about the following integral: $$K_n(m) = \overbrace{\int_0^1 \dots \int_0^1}^{n-\mathrm{times}} \left(-\...
7
votes
1answer
228 views

Inequality for Stirling numbers of the second kind

I stumbled upon the following inequality which, I believe, is true. I was able to prove it for small k, but I have no proof for the general case. Any help is welcome. Let $n\geq k\geq 1$ then $$\...
15
votes
1answer
528 views

Positivity of a finite sum involving Stirling numbers

In my research in theoretical physics, I have arrived at some coefficients $a_{n,m}$ depending on two integers, $n\geq 1$ and $0\leq m\leq n$: $$ a_{n,m}=\sum_{j=0}^{n-1} {2j \choose j+m} \left(\frac{...
0
votes
1answer
155 views

Proof of Stirling number symmetric formulas [closed]

I'm looking for a reference to a proof of formulas 6.26 and 6.27 in Concrete Mathematics: $\def\sone#1#2{\left[#1\atop #2\right]} \def\stwo#1#2{\left\{#1\atop #2\right\}} $ $$\stwo{n}{n-m} = \sum_k \...
1
vote
1answer
192 views

Sum of Stirling numbers with exponents

I have a trouble with the following sum $\sum_{i=0}^n\binom{n}{i}S(i,m)3^i$, where $S(i,m)$ is the Stirling number of the second kind (the number of all partitions of $i$ elements into $m$ nonempty ...
2
votes
1answer
342 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 ...
0
votes
2answers
139 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 ...
8
votes
1answer
411 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 $x$, $\...
2
votes
1answer
290 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:...
1
vote
2answers
431 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, ...
4
votes
1answer
254 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 $(\...
0
votes
1answer
162 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 ...
6
votes
1answer
575 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 ...
5
votes
1answer
415 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|...
3
votes
1answer
484 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 $...
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 ...
3
votes
0answers
532 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\...
4
votes
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} ...
11
votes
5answers
2k 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 $$S_2(n,k) = \frac{1}{k!}\sum_{j=0}^{k}(-1)^{k-j}{k \choose j} j^n. \qquad (1)$$ ...