Questions tagged [stirling-numbers]

For questions about the two kinds of Stirling numbers and related topics, such as Lah numbers.

There are two kinds of Stirling numbers:

Each sequence satisfies a recurrence:

\begin{align*} {n+1 \brack k} &= n {n \brack k} + {n \brack k-1} \\ {n+1 \brace k} &= k {n \brace k} + {n \brace k-1} \end{align*}

Lah numbers are closely related, being $$L(n,k) = \sum_{j=0}^n {n \brack j}{j \brace k}$$

739 questions
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Calculating the total number of surjective functions

It is quite easy to calculate the total number of functions from a set $X$ with $m$ elements to a set $Y$ with $n$ elements ($n^{m}$), and also the total number of injective functions ($n^{\underline{m}}$, denoting the falling factorial). But I am…
user50229
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Combinatorial proof of a Stirling number identity.

Consider the identity $$\sum_{k=0}^n (-1)^kk!{n \brace k} = (-1)^n$$ where ${n\brace k}$ is a Stirling number of the second kind. This is slightly reminiscent of the binomial identity $$\sum_{k=0}^n(-1)^k\binom{n}{k} = 0$$ which essentially states…
23
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Generating function with Stirling's numbers of the second kind

It's very easy to prove that: $$\sum_k \left\{k\atop n\right\}z^k=\frac{z^n}{(1-z)(1-2z)...(1-nz)}$$ But this generating function looks very pretty, so my question is: does this identity have some simple combinatorial interpretation?
xan
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Compute $S_n=\sum\limits_{a_1,a_2,\cdots,a_n=1}^\infty \frac{a_1a_2\cdots a_n}{(a_1+a_2+\cdots+a_n)!}$

It is tagged as an open problem in the book Fractional parts,series and integrals. If this proof is valid , I don't have any idea how to get it published so I posted it here . $\displaystyle \sum_{a_1,a_2,\cdots,a_n=1}^\infty \frac{a_1a_2\cdots…
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Stirling Numbers and inverse matrices

Let $s(m,n)$ be the Stirling Numbers of the first kind, $S(m,n)$ be the Stirling Numbers of the second kind. The matrices $$\mathcal{S}_N := (S(m,n))_{N \geq m,n \geq 0} \text { and } \mathcal{s}_N := (s(m,n))_{N \geq m,n \geq 0}$$ are inverse to…
muffel
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Splitting $r^k (r+n)!$ as a sum of factorials

I wanted to split the expression $r^k (r+n)!$ as a sum of factorials, where $k ,n \ \in \ \mathbb{Z} \ ;\ k>0$. For example, $r(r+n)! = (r+n+1)! - (n+1)(r+n)!$ $r^2(r+n)! = (r+n+2)! - (2n+3)(r+n+1)! + (n+1)^2 (r+n)!$ And in general, If $$ r^k…
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Stirling numbers and harmonic number identities

Permit me a brief introduction before I state the question, three questions in fact. Inspired by this MSE link I computed the following harmonic sum identities: $$1/6\, \left( {H_{{n}}}^{(1)} \right)…
Marko Riedel
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17
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Stirling numbers of second type

How can I do a combinatoric proof that for Stirling number of second type the equality if true: $${n\brace k} = \frac{1}{k!}\sum_{i=0}^{k}{k \choose i}i^n(-1)^{k-i}$$
SugerBoy
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16
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A Vandermonde Identity for Stirling Numbers?

I'm facing the problem of trying to express a quantity in the simplest possible way (it is, using the least possible number of sum symbols). $$ \sum_{j=0}^{n} \sum_{\ell=0}^m \frac{1}{j!}\binom{b+j}{j} {j+1 \brack {\ell+1}} {b+2 \brack…
13
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5 answers

Stirling numbers of the second kind on Multiset

Stirling numbers of the second kind $S(n, k)$ count the number of ways to partition a set of $n$ elements into $k$ nonempty subsets. What if there were duplicate elements in the set? That is, the set is a multiset?
13
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Challenging identity regarding Bell polynomials

Note: [2015-03-08] A proof of the identity below was aimed to close the gap of a rather extensive elaboration of this answer of mine. The identity (1) below is part of a more complex one, which is stated in Part 3, (39)-(42) in this follow-up…
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Is there a closed form expression for $\sum_{1\le i_1 \le i_2 \le \dots \le i_k \le n} i_1 i_2 \dots i_k$?

These sums showed up in a probability problem I was working on. They're not quite the Stirling numbers of the first kind since it's possible to have e.g. $i_1 = i_2$. Denoting the sum by $(k\mid n)$ we have the recurrence relation $(k\mid n) =…
Rivfader
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Stirling number

I am trying to evaluate the following finite sum: $$ \sum_{k=1}^{n}(-1)^{k}(k-1)!S(n-1, k-1)(\sum_{i=0}^{k-1}H_{i}), $$ where $S(n, k)$ are the Stirling's numbers of the second kind and $H_{i}$ denotes the $i$ harmonic number. Could you please shed…
user91485
11
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3 answers

Stirling number of the first kind: Proof of Recursion formula

I want to prove this recursion formula for Stirling numbers of the first kind: $$s_{n+1,k+1} = \sum_{i=k}^{n} \binom{i}{k} s_{n,i}$$ But I lack a useful idea. Perhaps someone could inspire me? Kind regards.
10
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Stirling numbers combinatorial proof: $S(m,n)=\frac 1{n!} \sum_{k=0}^{n} (-1)^k\binom nk (n-k)^m$

This is a Homework Question. I am required to give a Combinatorial proof for the following. $$S(m,n)=\frac 1{n!} \sum_{k=0}^{n} (-1)^k\binom nk (n-k)^m$$ Hint given is : Show that $n!S(m,n)$ equals the number of onto functions $f\colon A …
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