Questions tagged [stirling-numbers]

Question 1

EDIT: Just to be clear ${n+k \brace n}$ denotes the (n+k,n)-th stirling number of the second kind. I have been trying to find a decent form for the generating function $\sum_{n=0}^\infty{n+k \brace n}...

Question 2

this question is intended as a standard place to gather combinatorial interpretations, properties/relations and literature references to be linked to from other questions, in a similar vein to The sum ...

Question 3

I was wondering if it was possible to take an infinite falling power series and create an equivalent power series. For example: $$ \sum_{n=0}^{\infty} a_nx^\underline{n}=\sum_{n=0}^{\infty} b_nx^{n} $$...

Question 4

Context: Mircea Dan Rus's 2025 paper Yet another note on notation (a spiritual sequel to Knuth's 1991 paper Two notes on notation) introduces the syntax $x^{\{n\}}=x!{n\brace x}$ to denote the number ...

Question 5

Let $$ u(x)=e^{1/\ln x}-1 \qquad (x>1), $$ and consider $$ H_5(x)=x\Big(u+\tfrac12 u^2+\tfrac{4}{3}u^3+\tfrac{11}{3}u^4+\tfrac{223}{15}u^5\Big). $$ This comes from the more general family $$ H_M(x)=...

Question 6

It follows that $$\sum_{j = 0}^m\sum_{i = 0}^j(-1)^{i+j}\binom{n}{j}\binom{j}{i}\binom{n-j}{k}((n-i)/k)^m = \binom{n}{k}$$ if $n,m,k \in \mathbb{N}$. I first tried to rewrite the left side to get $$\...

Question 7

Let $f(n)$ be an arbitrary function with integer values. $a(n)$ be an integer sequence such that $$ a(n) = \sum\limits_{k=1}^{n} f(k) {n \brace k} a(k-1), \\ a(0) = 1. $$ $T(n,k)$ be an integer ...

Question 8

I consider a general term for a partition of size $k$, $r-k$. The number of ways is: Choose $k$ objects for the first circle: $\binom{r}{k}$ Arrange the $k$ objects in a circle: $(k-1)!$ Arrange the ...

Question 9

Cross-posted on MathOverflow (here) a long time ago. Let ${n \brack k}$ be the (unsigned) Stirling numbers of the first kind. Let ${n \brace k}$ be the Stirling numbers of the second kind. I ...

Question 10

Let $a(n)$ be A000670, i.e., an integer sequence known as Fubini numbers: number of preferential arrangements of $n$ labeled elements; or number of weak orders on $n$ labeled elements; or number of ...

Question 11

Regarding following polynomials that combine Stirling numbers of the first kind and binomial numbers: $$ F_{l,r} (x) = \sum_{k=0}^{\infty}|S_1(l,k)|\binom{k}{r}(-x)^k$$ which naturally appear in ...

Question 12

Let $T(n,k)$ be A354794, i.e., an integer coefficients known as Bell transform of the sequence $\{m^m \mid m \geqslant 0\}$. Here $$ T(n,k) = \frac{1}{k!} \sum\limits_{j=0}^{k} \binom{k}{j} (n-j-1)^{...

Question 13

I had to solve a combinatorics problem. It can be basically represented as you have m different boxes and n distinctly colored balls. How many combinations of $n$ balls put in $m$ boxes are there, if ...

Question 14

I'm looking for the most efficient way to compute ${n+10^{10} \brace 10^{10}}$ for $1 \leqslant n \leqslant 1000$. Obviosly, the standard formula, namely $$ {n \brace k} = \frac{1}{k!} \sum\limits_{j=...

Question 15

Problem: Let $S(r,n)$ be the Stirling number of the second kind, representing the number of ways to distribute $r$ distinct objects into $n$ identical boxes such that no box is empty. We aim to show ...

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Questions tagged [stirling-numbers]

Ordinary generating functions for the sequence $a_n={n+k \brace n}$ where $k$ is fixed [closed]

the sum $\sum_{j=k}^n{n\brack j}{j\brace k}x^j$

Converting an infinite falling power series into a traditional power series

bijective proof of identity coefficient-extracted from negative-exponent Vandermonde identity, and the upper-triangular Stirling transforms

Approximating $\mathrm{Li}(x)$ (logarithmic integral) with $H_M(x)$

$\sum_{j = 0}^m\sum_{i = 0}^j(-1)^{i+j}\binom{n}{j}\binom{j}{i}\binom{n-j}{k}((n-i)/k)^m = \binom{n}{k}$

Integer coefficients for recursively defined sequence that uses Stirling numbers of the second kind

Number of ways to arrange r distinct objects into 2 identical circles, stirling number of first kind

Symmetric (under the swapping) recursions for Stirling numbers of both kinds

Simple closed form for signed partial sums of Fubini numbers

Stirling Binomial polynomials roots

Is there a way to simplify this closed form for Stirling numbers of the second kind?

Is this equivalent to the $n!S(n,m)$ formula? (using the Stirling numbers of the 2nd kind)

Most efficient way to compute ${n+10^{10} \brace 10^{10}}$ for $1 \leqslant n \leqslant 1000$

Flaw in my approach for $S(r,2)$ (stirling number 2nd kind) [duplicate]

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