Questions tagged [stirling-numbers]

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

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Understanding distribution using Stirling number of the second kind

I have adapted this question from this question, I don't fully understand the answers given there/ Understanding distribution using Stirling number of the second kind inclusion exclusion approach ...
Bazman's user avatar
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simplifying a sum containing the binomial coefficient and stirling numbers second kind [duplicate]

I'm working on this sum, I don't have to find out the actual value, I just have to simplify it as much as I can. But the problem is that I don't know how far I can simplify it before reaching the ...
impressive305's user avatar
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Is this proof of the Lah Numbers Polynomial Identity With Induction on $n$ correct?

The Lah numbers satisfy the following rising-falling factorials relation \begin{equation} x^{\overline{n}} = \sum_{k=0}^{n} L(n, k) x^{\underline{k}}, \end{equation} where $x^{\overline{n}}$= $x(x+1)\...
User303131's user avatar
3 votes
1 answer
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Partial sum of Stirling numbers of first kind

After this question, that introduced me to Stirling numbers, I have the following question. For the (signed) Stirling numbers of first kind $s(k,i)$ and $k>M$, I would like to know the value or a ...
MathBug's user avatar
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Cardinality of a certain subset of group $S_k$ defined in connection with cycle decompositions

I am interested in a subset of the permutation group of $k$ elements, $\Sigma_k$. Any element in $\Sigma_k$ can be decomposed into disjoint cycles in a unique way. Conversely, if we take a partition ...
MathBug's user avatar
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Could this closed form for negative integer values of a series involving $\zeta(s)$ be proven?

The following series expression: $$f(s) = \sum_{n=2}^\infty \frac{\zeta(n)-1}{n^s} \tag{1}$$ has well known closed forms for $s=0$ and $s=1$, however also seems to converge for $\Re(s) < 0$. After ...
Agno's user avatar
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4 votes
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Find $\sum^{n}_{k=1} \binom{n}{k} \frac{(-1)^k}{k^a}$

I have been trying to find a general representation of the following finite sum. $$ S(n,a) = -\sum^{n}_{k=1} \binom{n}{k} \frac{(-1)^k}{k^a} $$ The sum seems to be related to Generalized Harmonic ...
Aidan R.S.'s user avatar
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Looking for a proper name for this vector subspaces

Let $K^n$ be the vector space of rows of length $n$ over $K$. A subspace $S$ is called coordinately complete if for every $i\ (1 \leq i \leq n)$ there is an element in $S$ with a nonzero $i$th ...
Nikolay Hodyunya's user avatar
3 votes
2 answers
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Euler numbers in terms of second kind Stirling numbers and in terms of the Touchard polynomials

I recently managed to prove these equalities: $$E_n=\sum_{k=0}^{n}S(n,k)\frac{-k!\sqrt{2}}{\sqrt{2}^{k}}\cos(\frac{3\pi}{4}(k+1))$$ $$E_n=2\int_{0}^{\infty}e^{-t}\cos(t)T_n(-t)dt$$ where $E_n$ are the ...
tomascatuxo's user avatar
3 votes
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Expectation Value of n Bernoulli Variables

Let $X_1,...,X_n$ be i.i.d. Bernoulli Variables with $$E(X_i)=p$$ for $i=1,...,n$ and $$X=\sum_{i=1}^n X_i \, .$$ Furthermore let $k \in \{0,1,...,n\}$ and consider the Expectation Value $$E_\pi=E\...
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Why my proof that $\lim_{x\rightarrow\infty}\frac{x!}{x^x}=1$ is wrong

It is well known that $$\lim_{x\rightarrow\infty}\frac{x!}{x^x}=0$$But according to the definition of the Stirling numbers of the first kind, the falling factorial could be written as $$(x)_n=\sum_{k=...
Kamal Saleh's user avatar
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Combinatorial proof for Stirling number of 1st kind

This question is duplicate. But I have some questions about the already posted answers. And temporarily I don't have enough reputation to comment, so I post one new question. Sorry for that. This one ...
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Gaussian approximation of collision time

In this answer there is a claim that $$\frac{n!}{(n-k)! n^k} \approx e^{-\frac{k^2}{2n}} \tag{1}$$ which is then used to approximate the sum over $k=1,\ldots, n$ via $$\sum_{k=1}^n \frac{n!}{(n-k)! n^...
angryavian's user avatar
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$S(n+1, k) = \sum_{i=k-1}^{n} \binom{n}{i} S(i, k-1)$

So I need help with this: I need to prove $S(n+1, k) = \sum_{i=k-1}^{n} \binom{n}{i} S(i, k-1)$. I know that if I take an extra element $a_{n+1}$ to a set of n elements then I have 2 options. Either $...
Ari1234567's user avatar
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Stirling Numbers Exponential Generating Function Induction

I was reading the solution to a question written here, and it uses a fact which can be proved by induction. The question is to show that an EGF for Stirling Numbers of the second kind with fixed $k$, ...
Jeremy's user avatar
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Identity connecting Stirling numbers of both kinds with second order Eulerian numbers

Setting. On page 270 of the great book Concrete Mathematics by Graham, Knuth and Patashnik, the second order Eulerian numbers $\langle\!\langle {n\atop k}\rangle\!\rangle$ with $n,k\in\mathbb{Z}$, $n \...
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$S(n,k) \le c(n,k)$ [duplicate]

I am studying "A Walk Through Combinatorics" and I encountered this equation $$$$$S(n,k) \le c(n,k)$ $$$$ The author wants us to prove that. I can verify this equation holds via induction ...
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Bijective proof for Stirling numbers of the first kind [duplicate]

Setting. Let s(n,k) be the Stirling number of the first kind counting the number of permutations on a $n$-set with exactly $k$ cycles. For all non-negative integers $m,n$ there is the identity $$\sum_{...
azimut's user avatar
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Integral as a product of integrals

I have conjectured that for all functions $$ g(t) = \sum_{n=0}^{\infty}\frac{a_n}{n!}t^n, $$ it holds that $$ \int_0^{\infty}\frac{(g(tx)-g(0))^j}{j!}e^{-t} dt = \frac{1}{j!a_0a_1...a_j}\prod_{r=1}^j(\...
Artur Wiadrowski's user avatar
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algebraic or combinatorial proof to an identity of Stirling numbers of the first kind

We denote Stirling numbers of the first kind as $s(n,m)$ counting the ways we can put $n$ people in $m$ cycles. What is a combinatorial/algebraic proof to this identity? $$s(n+1,m+1)=\sum_k{s(n,k). ^...
Mason Rashford's user avatar
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1 answer
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Sum involving Stirling numbers of the second kind and factorial

Is it possible to give a good upper bound for the sum $\sum\limits_{k=0}^n k!\,{n+1\brace k+1}$, where ${n+1\brace k+1}$ is a Stirling number of the second kind. I am aware of the relation $\sum\...
J.Summer's user avatar
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4 votes
2 answers
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Conjecture: If $f(x)=\prod\limits_{k=1}^n(x-k)$ then $\lim\limits_{n\to\infty}\frac1n [\text{largest root of $f(x)=f'(x)$}]=\frac{e}{e-1}$.

I was thinking about the polynomial $f(x)=\prod\limits_{k=1}^n(x-k)$. I noticed that if we draw the graphs of $y=f(x)$ and $y=f'(x)$ together, the $x$-coordinate of the rightmost intersection seems ...
Dan's user avatar
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Is there a combinatorial proof of this binomial-type formula involving Stirling numbers?

Suppose $\partial$ and $t$ are elements of a noncommutative ring which satisfy $[\partial,t]=1.$ Then one can show using an induction argument that \begin{equation} (t\partial)^k=\sum_{i=0}^k{k \brace ...
D. Brogan's user avatar
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Polylogarithm further generalized

Here I proposed a generalized formula for the polylogarithm. However, because of a slight mistake towards the end, visible prior to the edit, I was unaware that it yields just a result of an integral ...
Artur Wiadrowski's user avatar
2 votes
2 answers
101 views

Identity involving binomial coefficients, Stirling numbers of the second kind, and generalized hypergeometric series

I have come across the following identity that involves binomial coefficients $\binom{n}{k}$, Stirling numbers of the second kind $\left\{ m \atop j \right\}$ and generalized hypergeometric series $_p ...
Sherlock9's user avatar
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Stirling numbers of second kind recurrence relation on $k$

In my previous question on the properties of $\sum_{k=0}^nk^a\binom{n}{k}$, a comment directed me to the answers on this related question with formula $$ \newcommand{\stirtwo}[2]{\left\{{#1}\atop{#2}\...
Sherlock9's user avatar
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Riemann zeta function at negative integers as combination of Stirling numbers and Gregory coefficients

Calculating the regularized sum $$\sum_{n=1}^{\infty}n^{k}$$ in two different ways gives the following relation: $$\zeta(-k)=(-1)^{k-1}\sum_{\ell=1}^{k}\ell!S_{2}(k,\ell)G_{\ell+1}$$ where $S_{2}(k,\...
Grosseteste's user avatar
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$\dfrac{\mathrm{d}^n}{\mathrm{d}x^n}\dfrac{e^{ax}}{\ln(cx)}$ and summation with Stirling number of the first kind

I would like to calculate the $n$-th derivative of $\dfrac{e^{ax}}{\ln(cx)}$ I tried to calculate it in this way: $$(fg)^{(n)}(x)=\sum_{k=0}^{n}\binom{n}{k}f^{(k)}(x)g^{(n-k)}(x)$$ $$\frac{\mathrm{d}^...
Math Attack's user avatar
2 votes
2 answers
69 views

Generatingfunctionology 1.6 Stirling Number

I am reading Herbert Wilf's Generatingfunctionology these days. And on Page 21, Section 1.6 Another 2-Variable Case, I was confused by the operation he did on formula (1.36), which is: ... The ...
aidenclx's user avatar
1 vote
1 answer
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Why if $\chi (G)=k$, then we have $S(G,k)=\frac{P(G,k)}{k!}$?

Let $G$ be a finite graph of order $n$. Chromatic polynomial of $G$ is defined by $$P(G,\lambda )=\sum_{k=\chi (G)}^{n}S(G,k)(\lambda)_k $$ where $\chi (G)$ is the chromatic number of $G$ and $$(\...
Mahtab's user avatar
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0 answers
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Placing letter in circles using Stirling number of first kind

Place $4$ identical $a's$ and $4$ identical $b's$ in $3$ circular tables such that two of these circular tables have $3$ seats and these two tables are identical, the rest one has $4$ seats. How many ...
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Is there a simple formula for the Stirling numbers of the first kind?

It is well-known that the Stirling numbers $S(m,n)$ of the second kind are given by $$\frac{1}{n!}\sum_{j=0}^n (-1)^{n-j}{n\choose j} j^m$$ Is there such a simple formula also for the Stirling ...
ray's user avatar
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1 vote
1 answer
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Another formulation for Stirling numbers of the second kind

I find another formulation for Stirling numbers of the second kind: Let $n\ge k\ge 1$. Denote by $$\mathbb N_<^n := \{ \alpha = (\alpha_1,\cdots,\alpha_n): 0\le \alpha_1\le\cdots\le\alpha_n, \...
Dreamer's user avatar
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2 votes
2 answers
129 views

Formula involving the Stirling numbers of the second kind

I was studying chapter 4 of the book "Generatingfunctionology" by Herbert S.Wilf when I got stuck in the proof of the following equality, $$ \sum_{k=1}^{n} \Big\{ \begin{matrix} n \\ ...
Matteo Aldovardi's user avatar
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Stirling numbers of the first kind: A combinatorial interpretation problem

One of the most popular interpretations of the unsigned Stirling number of the first kind ${n\brack n-k}$ is the number of permutations of $n$ elements with exactly $n-k$ permutation cycles. I have ...
Poisson's user avatar
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Evaluate $\sum\limits_{n=0}^\infty\operatorname W(e^{e^{an}})x^n$ with Lambert W function

$\def\W{\operatorname W} \def\Li{\operatorname{Li}} $ Interested by $\sum_\limits{n=1}^\infty\frac{\W(n^2)}{n^2}$, here is an example where Lagrange reversion applies to a Lambert W sum: $$\W(x)=\ln(...
Тyma Gaidash's user avatar
1 vote
1 answer
62 views

Euler Numbers in Terms of Stirling Numbers of the Second Kind

It's well known that: (https://en.wikipedia.org/wiki/Euler_numbers#In_terms_of_Stirling_numbers_of_the_second_kind) $$E_{n}=2^{2n-1}\sum_{\ell=1}^{n}\frac{(-1)^{\ell}S(n,\ell)}{\ell+1}\left(3\left(\...
tomascatuxo's user avatar
6 votes
4 answers
197 views

Closed form of $ \sum_{k=0}^{n} (-1)^{n-k}\binom{n}{k}2^k k^m$?

Is there a closed form of the above equation or something that simplifies it? Here is the same equation copied: $$\sum_{k=0}^{n} (-1)^{n-k}\binom{n}{k}2^k k^m$$ It looks very similar to the Stirling ...
Baklava Gain's user avatar
1 vote
2 answers
148 views

Trying to understand a proof relating Stirling number of the second kind to its exponential generating function

In Cameron's Combinatorics, he set out to proof the below identity: $$\sum_{n\ge0}\frac{S(n,k)t^n}{n!}=\frac{(exp(t)-1)^k}{k!}$$ He proceeds as follows: $$\begin{align*}\sum_{n\ge0}\frac{S(n,k)t^n}{n!}...
Dave Clifford's user avatar
0 votes
3 answers
103 views

limit to infinite of a factorial number

I have this problem and I solved it like this, but muy teacer told me that it's wrong so I dont know who to do it. Prove that $\lim _{n \to \infty}\frac{(n!)^22^{2n}}{(2n)!\sqrt{n}}$. I use the ...
Mingyu's user avatar
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1 vote
1 answer
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Alternative roots of generalized Laguerre polynomials

$\require{\physics}$ Hi, I am wondering if it is possible to approximate the roots of the generalized Laguerre polynomial $L_n^{(\alpha)}(x)$ not with respect to $x$ but with respect to $n$, i.e. ...
Ivan R.'s user avatar
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2 votes
1 answer
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Simple integer sequence with unknown direct generator

I have a sequence (well a sequence of sequences) that I can't figure out how to generate directly, despite the fact I can generate it recursively Specifically, I can write the elements of a matrix of ...
b3m2a1's user avatar
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1 vote
0 answers
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Number of integer sequences with specific conditions proof

I'm trying to prove that the number of integer sequences of length $n$: $(a_1, a_2, \ldots, a_n)$ such that $0 \leq a_i \leq (n-i)$, $1 \leq i \leq n$ and exactly $k$ entries are equal to $0$ is given ...
user492754's user avatar
2 votes
1 answer
58 views

$10$ students, $16$ lessons, each lesson one student is chosen to be asked, probability that every student will be asked after $16$ lessons

$10$ students, $16$ lessons, each lesson one student is chosen to be asked, probability that every student will be asked after $16$ lessons. First solution: Our sample space is strings of length $16$ ...
romperextremeabuser's user avatar
4 votes
1 answer
96 views

Cleaner form for transformation involving Stirling numbers

I have a specific transformation involving treating the Stirling numbers of the first kind as a matrix, given by, for some positive $b$ $$ C^{(b)} = (S_1^{(b)})^{T} \left(J_b\ \left(\frac{1}{2^{i}}{{b}...
b3m2a1's user avatar
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0 answers
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A problem involving an identity relating Bernoulli numbers and Stirling numbers of the first kind

Problem: Given two positive integers $t,l \in \mathbb{N}$, find a sequence of integers $\{ \gamma_j \}_{j = 1..(t+l)}$ such that for all values $a = 1, \ldots, t+l$, $$(-1)^l \frac{B_{t+l+1-a}}{t+l+1-...
cdkaram's user avatar
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0 answers
46 views

Showing $\sum_{i=0}^n c(n,i) = n!$ with cardinality of Symmetric Group

I've come across an identity for Stirling Numbers of the first kind: $$\sum_{i=0}^n c(n,i) = n!$$ Where $c(n,k)$ is the number of permutations on [n] with exactly k cycles. From a previous study in ...
Oran's user avatar
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0 answers
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Is there a double series expansion of $\ln^n(\ln(a)+2\pi i n_2-\ln(w)),w=2\pi i n_1;n_1,n_2\in\Bbb Z$?

Problem: Via Lagrange reversion with $2$ sets of branches: $$f(z)=ze^{e^z}\implies f^{-1}_{n_1,n_2}(z)=2\pi i n_1+\sum_{n=1}^\infty\frac1{n!}\frac{d^{n-1}}{dz^{n-1}}\ln^n(\ln(a)+2\pi i n_2-\ln(w))\...
Тyma Gaidash's user avatar
1 vote
0 answers
62 views

Given $x$ find minimum positive $N$ for which the following is composite $ 1^x+2^x+3^x+4^x+\cdots+N^x $

Problem: Given positive integer $x$ find minimum value of positive integer $N$ starting which the following is always composite $$ 1^x+2^x+3^x+4^x+\cdots+N^x $$ My Thoughts: This is a followup to this ...
sibillalazzerini's user avatar
5 votes
0 answers
156 views

Inequality for triple product of Stirling numbers

Let ${n\brace k}$ be the Strirling number of the second kind, such that ${n+1\brace k} =k{n\brace k}+{n\brace k-1}$ with ${0\brace 0}=1$. Let $j,p,n$ integers such that $1 \le j \le p \le n$. I ...
René Gy's user avatar
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