Questions tagged [stirling-numbers]
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52
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How to find the coefficient of $x^k$ in the expression $\prod_{p=2}^n (1+xp) $
I got this general formula for $ n\in N$ (I showed it here)
$$\int_0^1 \left(\frac{x}{1-x} \ln x \right)^n dx=n \sum_{p=0}^{n-1}a(n,p+1) (-1)^{n-p} \zeta(p+2)+n! $$
where $a(n,k)$ is the coefficient ...
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0
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107
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Inequality for 2-associated Stirling numbers of the second kind
Let $S_2(n,k)$ denote the 2-associated Stirling number of the second kind for $n$ objects and $k$ blocks, with $n$ being at least two. That is, we partition $n$ labeled objects into $k$ unlabeled ...
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3
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Evaluating a sinusoidal series
Define the sequence of functions
$$f_n(x)=\sum_{m=n}^\infty(-1)^m\frac{x^{2m}}{(2m+1)!} {m \choose n} $$
Is there a closed form expression for arbitrary $n$? It is clear that the result should assume ...
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Is there any literature on $\sum_{i=1}^{k} \left[ {k \atop i} \right] \zeta(i+1) $?
As per these questions, I'm trying to evaluate $$\sum_{n=2}^{\infty} \big{(} \zeta(n)^{2}-1 \big{)} = 1+ \sum_{m=2}^{\infty} \frac{H_{-\frac{1}{m}}}{m}. $$
Here, $H_{x}$ is a generalized Harmonic ...
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1
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Prime divisibility of Stirling numbers of first kind
Prove that $p\mid\genfrac[]0{}{p^w}k$ where $p$ is an odd prime, $w \in \mathbb{N}$, $1<k<p^w$ and $k \neq p^v$ for some positive integer $v<w$. This has to be already done I just can't find ...
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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
$$\left(...
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1
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305
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Identity involving Stirling number of the second kind
I'm looking for a citable reference for the following identity involving the Stirling numbers of the second kind $S(n, k)$ stated in Equation (27): For $n \geq 2$,
$$
\sum_{m=1}^n S(n, m) (-1)^m (m-1)!...
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2
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Divisibility of Stirling numbers
It is well known that if $p$ is prime, Stirling numbers of the first and second kind, $s_1(p,k)$ and $s_2(p,k)$, are divisible by $p$ if $1<k\le p-1$ (Lagrange ; easiest is working in $\mathbb F_p$ ...
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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, ...
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2
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773
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Determinant of matrix with Stirling numbers as elements
After noticing that the determinant of an $n \times n$ matrix $A_n$ with elements $a_{i,j}=i^j$, $1 \le i \le n$, $1 \le j \le n$, is the superfactorial (product of the first $n$ factorials), I wanted ...
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Bell polynomial with variables 1 and 0
Let $B_{n,k}(x_1,\cdots,x_{n-k+1})$ be the Bell polynomial.
If $x_1=\cdots=x_{n-k+1}=1$, we know that $B_{n,k}(x_1,\cdots,x_{n-k+1})=S(n,k)$, where $S(n,k)$ is the Stirling number of second kind.
...
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Could you please confirm or deny two identities involving weighted Stirling numbers of the second kind?
In the paper [1] below, among other things, Carlitz introduced weighted Stirling numbers of the second kind $R(n,k,r)$. He also proved that the numbers $R(n,k,r)$ can be generated by
\begin{equation*}%...
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Ask for a reference or a proof of a combinatorial identity $\sum_{k=0}^n\binom{2n+1}{2k}\binom {k}{m} =2^{2(n-m)}\frac{2n+1}{2(n-m)+1}\binom{2n-m}{m}$
Could you please recommend a reference to or supply a proof of the following identity \eqref{combin-ID-Maclaurin}, or \eqref{first-equiv-form}, or \eqref{combin-ID-Mac-Equiv}, or \eqref{combin-ID-Mac-...
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Counting permutations with a fixed number of descents and an extra condition
I am computing the volumes of certain polytopes and it turns out that knowing a "closed formula" for the following number would help a lot.
Determine the number of permutations $\sigma\in \...
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3
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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 $...
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Sum of the Stirling numbers of the second kind multiplied by $k$ and falling factorials
I am looking for closed forms, or at least a good approximation for
$$f(n) = \sum_{k=1}^{k=n} \genfrac\{\}{0pt}{}{n}{k}(n)_kk$$
I know that
$$\sum_{k=1}^{k=n} \genfrac\{\}{0pt}{}{n}{k}(n)_k = n^n$$
I ...
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Sum with Stirling numbers of the second kind
Let $wt(n)$ be A000120, number of $1$'s in binary expansion of $n$ (or the binary weight of $n$)
and
$$n=2^{t_1}(1+2^{t_2+1}(1+\dots(1+2^{t_{wt(n)}+1}))\dots)$$
Then we have an integer sequence given ...
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Prove that $ \sum_{i=0}^{2k}( {n+R-1\choose R+i} + (-1)^{i+1}{ n+R+i\choose R+i } )\sum_{j=0}^i {i\choose j}(-1)^j(i+1-j)^{2k}=0 $
For all $k,R \in \mathbb{N}$ fixed, prove that $ \sum_{i=0}^{2k}( {n+R-1\choose R+i} + (-1)^{i+1}{ n+R+i\choose R+i } )\sum_{j=0}^i {i\choose j}(-1)^j(i+1-j)^{2k}=0 $. I'm quite sure this is true but ...
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Show that $\sum_{i=0}^{2k} [ {n\choose i+1} + (-1)^{i+1}{n+i+1\choose i+1} ] \sum_{j=0}^i {i\choose j}(-1)^j (i+1-j)^{2k} =0.$
Let $u(k,j) = 1$ if $j=0$, $0$ if $j > k$, or else it is $j*u(k-1,j-1) +(j+1)*u(k-1,j) $. Prove that $ \sum_{i=0}^{2k} {n \choose i+1} u(2k,i) +\sum_{i=0}^{2k} {-n-1 \choose i+1} u(2k,i)=0. $ ...
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What is the inverse Laplace transform of $\frac{(1/s)_{n}}{s} $?
Introduction
So far, I have found (p. 5) the following generating functions of the unsigned Stirling numbers of the first kind:
\begin{equation} \tag{1} \label{1} \sum_{l=1}^{n} |S_{1}(n,l)|z^{l} = (z)...
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Stirling number bounds and polynomials and the Lambert $W$ function
Let $s(n,k)$ be the (signed) Stirling numbers of the first kind. The polynomials
$$L_n(x)=\sum_{j=1}^ns(n,n+1-j)\dfrac{x^j}{j!}$$
enter in the asymptotic expansion of the Lambert $W$ function, see for ...
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1
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Estimation of a sum involving Stirling's number of second kind and binomial coefficient
Let $S(n, j)$ be Stirling's number of second kind. Let $p\in [0,1]$ and $m \in N$.
Bound from above the following sum:
$$
\sum_{j=0}^m S(n,j) {m \choose j}\, j! \, p^j
$$
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On the equation involving Stirling numbers of the second kind ${n\brace a}{m\brace b}={k\brace c}$, and its solutions satisfying certain requirements
In this post we denote the Stirling numbers of the second kind as ${r\brace s}$ and we consider the proposal to ask if the equation of the title has infinitely many solutions $${n\brace a}{m\brace b}={...
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Sum of divisors of Stirling numbers of the second kind
In this post we denote the Stirling number of the second kind as ${n\brace k}$, I add as reference the article Stirling numbers of the second kind from the encyclopedia Wikipedia. And we denote the ...
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What is this sequence?
This is again a question that I asked at Stack Exchange, but got no answer so far, so I am trying here.
Let:
$$ a_n=\sum_{k\ge0}(k+1) {n+2\brack k+2}(n+2)^kB_k$$
$B_k$ is the Bernoulli number. ${n\...
- 30.6k
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Showing this formula counts these things
I'm writing an article, and I got stuck trying to prove that some numbers are positive. I have a relatively good intuition for guessing what an expression is counting, but in this case I'm not being ...
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Positivity of a finite sum involving Stirling numbers of the first kind
Past days I've been trying to prove that certain polynomials have positive coefficients. After a lot of thinking, I came up with a formula for each coefficient individually, and they are not that ugly....
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Evaluation of a complete homogeneous symmetric polynomial related to Stirling number of 2nd kind
It is well known that the complete homogeneous symmetric polynomial $h_{n-k}(1,\,2,\,3, ...,\,k-1,\,k)$ equals $S(n,\,k)$ the Stirling number of the second kind. [Wikipedia]
During a research project ...
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Simplify a double summation involving binomial coeficient
$$T(N,K)=\sum_{i=2}^{K}\sum_{j=2}^{i}(-1)^{i-j}\binom{i}{j}\frac{j^{N+1}-1}{j-1}$$
Is it possible to evaluate the sum for $K=10^7$ efficiently. If we manage to remove one of the sums, it will be ...
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Proof of Stirling number symmetric formulas [closed]
I'm looking for a reference to a proof of formulas 6.26 and 6.27 in Concrete Mathematics:
$\def\sone#1#2{\left[#1\atop #2\right]}
\def\stwo#1#2{\left\{#1\atop #2\right\}}
$
$$ \stwo{n}{n-m} = \sum_k \...
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Closed form for product of Stirling numbers of the second kind
What does the following expression evaluate to:
\begin{equation}
\sum\limits_{k=1}^n \dbinom{n}{k} \cdot k! \begin{Bmatrix} n \\ k \end{Bmatrix} \cdot k! \begin{Bmatrix} n \\ k \end{Bmatrix}
\end{...
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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 ...
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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 \...
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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 $0\leq m\leq n$:
$$
a_{n,m}=\sum_{j=0}^{n-1} {2j \choose j+m} \left(\frac{...
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Singular values of Stirling numbers matrix
Consider the Stirling numbers of the first kind $s(i, j)$, and form a matrix $S_1(n),$ where the $(i, j)$th entry is $s(i, j)$. (IMPORTANT NOTE the indices start at $0,$ so this matrix is $(n+1)\times ...
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Simple/efficient representation of Stirling numbers of the first kind
Stirling numbers of the second kind can be expressed by means of a simple hypergeometric (considering $n$ fixed) sum
$$S_2(n,k) = \frac{1}{k!}\sum_{j=0}^{k}(-1)^{k-j}{k \choose j} j^n. \qquad (1)$$
...
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Proof of identity involving Stirling numbers of the second kind
While computing conditional expectations of certain functionals of a Poisson white noise field (details are long and probably irrelevant), I've stumbled upon the need to use the following identity ...
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Are surjections $[n]\to [k]$ more common than injections $[k]\to [n]$?
Here's an interesting inequality involving binomial coefficient and Stirling numbers of the second kind that I believe holds for all $n,k$:
$$ k^n {n \choose k} \leq n^k {n \brace k} $$
On the left-...
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A conjecture about the degrees of special polynomials
Define the congruence "modulo m" on exponential Taylor series as
$$
\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{...
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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(-\...
- 1,459
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1
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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 ...
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2
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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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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 ...
CommunityBot
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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 ...
CommunityBot
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3
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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 ...
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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}$$
Where $x$, $n$ and $d$ are non-negative integers, $\Delta^d$ is the $d$-th difference with respect to $x$, $\...
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Simple approximation to a sum involving Stirling numbers?
I have also posted this question at https://math.stackexchange.com/questions/486917/simple-approximation-to-a-sum-involving-stirling-numbers. I have an exact answer to a problem, which is the function:...
CommunityBot
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1
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Relations involving Stirling numbers of second kind
While inverting a Laplace transform using Post's inversion formula I found the following expression:
$$
\sum_{k=1}^n S^n_k \ x^k(\alpha)_k
$$
where $S^n_k$ is a Stirling number of second kind and $(\...
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1
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Asymptotic formula for an expression in terms of the second kind of stirling numbers
We have proved that
the limit of $\sum_{k=0}^n r^2k^m / (1+r)^{k+1}$ when n approaches infinity is $\sum_{k=1}^m S(m,k)k!/r^{k-1}$
where S(m,k) is the second kind of stirling number.
Is there a ...
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Alternating sums of alternate Stirling numbers
Does anybody know of any identities or combinatorial interpretations for alternating sums of alternate Stirling numbers?
I am particularly interested in expressions of the form:
$$\pm\sum_{k}(-1)^k|...
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