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Questions tagged [eulerian-numbers]

For questions about the Eulerian numbers $A_{n,k}$, defined as the number of permutations in the symmetric group $S_n$ having $k$ descents. Not to be confused with Euler’s number $e$ or the Euler numbers $E_n$.

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Compute the integral $\int_0^1\int_0^1\ldots\int_0^1 f(x_1 + x_2 + \ldots + x_n)\,dx_1\,dx_2\ldots dx_n $

A mysterious result, probably by Euler himself, goes as follows: If $n$ is a positive integer and $f:\mathbb R \rightarrow \mathbb C$ is integrable on the open interval $(0, n)$, then $$\int_0^1\...
dohmatob's user avatar
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10 votes
1 answer
390 views

Positivity of a certain sum of Stirling numbers

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....
Luis Ferroni's user avatar
9 votes
2 answers
1k views

Polylogarithms of negative integer order

The polylogarithms of order $s$ are defined by $$\mathrm{Li}_s (z) = \sum_{k \geqslant 1} \frac{z^k}{k^s}, \quad |z| < 1.$$ From the above definition, derivatives for the polylogarithms ...
omegadot's user avatar
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8 votes
1 answer
298 views

Bounds on the difference between the polylogarithm with negative base and the gamma function

Trying to understand intuitively the Gamma function I started to think of it as a way to measure how much each factorial power "helps" $x^n$ in the infinite sum of $e^x$, thus trying to ...
Alejandro Quinche's user avatar
8 votes
0 answers
231 views

odd property of Eulerian numbers

One of the curious features of Pascal's triangle is that each row contains a two power number of odd entries. In fact, the precise number is $2^{b(n)}$ where $b(n)$ denotes the sum of the bits of $n$...
user2052's user avatar
  • 2,437
7 votes
1 answer
790 views

A nicer recurrence for the Eulerian polynomials.

I was perusing the subject of Eulerian polynomials. I'm assuming the definition that the Eulerian polynomial is defined by $C_n(t)=\sum_{\pi\in S_n}t^{1+d(\pi)}$, where $d(\pi)$ is the number of ...
Clara's user avatar
  • 1,536
7 votes
0 answers
255 views

Proof for a summation-procedure using the matrix of Eulerian numbers?

I've discussed a procedure for divergent summation using the matrix of Eulerian numbers occasionally in the last years (initially here, and here in MSE and MO but not in that generality and thus(?) ...
Gottfried Helms's user avatar
6 votes
1 answer
214 views

The relation of the Bernoulli numbers to the Catalan numbers

The Bernoulli numbers $B_n$ are the backbone of calculus, and according to B. Mazur, they "act as a unifying force, holding together seemingly disparate fields of mathematics." The Catalan ...
Peter Luschny's user avatar
6 votes
2 answers
223 views

Lerch-$\small \zeta(\varphi,0,-n)$ of integer *n* purely real and imaginary ($\small \zeta_\varphi (-n)^2 $ is real) for $\small n \ge 2$?

Are the Lerch-$\zeta(\varphi,0,-n) $ of integer n (for shortness I use the notation of my earlier question $\small \zeta_\varphi(-n)$) periodically purely real and imaginary: $\zeta_\varphi (-n)^2 $ ...
Gottfried Helms's user avatar
6 votes
2 answers
141 views

An identity related to the second-order Eulerian numbers.

Recently, some of the remarkable properties of second-order Eulerian numbers $ \left\langle\!\!\left\langle n\atop k\right\rangle\!\!\right\rangle$ A340556 have been proved on MSE [ a , b , c ]. ...
Peter Luschny's user avatar
6 votes
1 answer
310 views

Second-order Eulerian numbers, Lambert's W function, and Schröder's fourth problem

Recently, some of the remarkable properties of second-order Eulerian numbers $ \left\langle\!\!\left\langle n\atop k\right\rangle\!\!\right\rangle$ A340556 have been proved on MSE [ a , b , c ]. ...
Peter Luschny's user avatar
6 votes
1 answer
180 views

Series power function over exponential function

A typical exercise from calculus is to show that any exponential function eventually grows faster than any power function, i.e. $$ \lim_{k \to \infty} \frac{k^a}{b^k} = 0 \qquad \text{ for } a,b>1.$...
Alperino's user avatar
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5 votes
2 answers
278 views

Why does $\frac{1}{2}A_n(2)$ count the number of ordered set partitions?

Suppose $A_n(x)$ denotes the Eulerian polynomial. Is there a combinatorial proof that $\frac{1}{2}A_n(2)$ counts the number of ordered set partitions? By this I mean a set partition of a set of $n$ ...
Koji Hamada's user avatar
5 votes
1 answer
303 views

A Stirling number identity representing the second-order Eulerian numbers.

Graham, Knuth, and Patashnik give in CMath a thorough introduction to the Stirling numbers. On table 250 and table 251, they compile two pages of Stirling number identities. Of course, there are many ...
Peter Luschny's user avatar
5 votes
0 answers
71 views

Connection between Eulerian numbers and number of elements in set of uniform variables greater than the mean?

I was recently investigating the following question: Given $n$ independent $\text{Unif(0, 1)}$ variables $U_1,\ldots,U_n$, let $m$ be the number of elements in $[U_1,\ldots,U_n]$ that are greater than ...
Emily Boyajian's user avatar
5 votes
0 answers
223 views

Interpretation of Eulerian numbers using the principle of inclusion-exclusion

Eulerian numbers, denoted $e_{n,k}$, are defined as the number of permutations of $[n]$={1,2,...,n} such that there are k "ascents". (For example, the permutation 23541 of [5] would have 3 ascents, ...
settheorynoob's user avatar
5 votes
0 answers
109 views

$\mathbb{Z}$-Polynomials in an Enumeration Identity

I've conjectured the following identity: For $1 \leqslant k \leqslant l \leqslant n$ and $m \in \mathbb{N}$, \begin{align} \sum_{1 \leqslant i_1 < \cdots < i_l \leqslant n} i_{k}^{m} = \sum_{j = ...
user02138's user avatar
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4 votes
2 answers
847 views

A closed form of Eulerian numbers

The following identity involving Eulerian numbers is well-known: \begin{equation} A(n,m)=\sum_{k=0}^{m}(-1)^k \binom{n+1}{k} (m+1-k)^n. \end{equation} where $A(n,m)$ is the number of permutations $(\...
Kuai's user avatar
  • 1,503
4 votes
2 answers
188 views

Difference of the Stirling cycle numbers and the Stirling set numbers

Denote by $\left\langle\!\! \left\langle k\atop j\right\rangle\!\! \right\rangle$ the second-order Eulerian numbers A340556. Define $$ \left| n\atop k\right| = \sum_{j=0}^k \left( \binom{n + j - 1}{...
Peter Luschny's user avatar
4 votes
1 answer
181 views

References/Proof of the conjectured identity for the Stirling permutation number $\left\{{n\atop n-k}\right\}$

While working with a combinatorics problem, I conjectured that $$ \left\{{n \atop n-k }\right\}=\sum_{p=0}^{k-1}\bigg\langle\!\!\bigg\langle{k\atop k-1-p}\bigg\rangle\!\!\bigg\rangle \binom{n+p}{2k}, ...
Sangchul Lee's user avatar
4 votes
1 answer
144 views

A recurrence of the second-order Eulerian polynomials

Recently, some of the remarkable properties of second-order Eulerian numbers $ \left\langle\!\!\left\langle n\atop k\right\rangle\!\!\right\rangle$ A340556 have been proved on MSE [ a , b , c ] ...
Peter Luschny's user avatar
4 votes
3 answers
363 views

Recurrence formula for the Eulerian and derangement polynomial

For the Eulerian polynomial $$A_n(x)=\sum_{\pi\in S_n}x^{\mathrm{des}(\pi)}$$ it is well known, that we have the nice recurrence formula $$A_n(x)=\sum_{k=0}^{n-1}\binom{n}{k}A_k(x)(x-1)^{n-1-k}.$$ I ...
Richard's user avatar
  • 293
4 votes
1 answer
160 views

Can that two double series representations of the $\eta$/$\zeta$ function be converted into each other?

By an analysis of the matrix of Eulerian numbers(see pg 8) I came across the representation for the alternating Dirichlet series $\eta$: $$ \eta(s) = 2^{s-1} \sum_{c=0}^\infty \left( \sum_{k=0}^c(-1)^...
Gottfried Helms's user avatar
4 votes
0 answers
270 views

How to prove this equality about Eulerian numbers? [duplicate]

I want to prove the following equality where $A(k,m)$ is the Eulerian number : $$\forall k\ge0,\sum_{k=0}^{\infty}n^k x^k = \frac{\sum_{m=0}^{k-1}A(k,m)x^{m+1}}{(1-x)^{k+1}}$$ I previously proved ...
Bérénice's user avatar
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4 votes
0 answers
104 views

About $t$-analogue of the Eulerian polynomials.

A certain way to define the $t$-analogue of the Eulerian polynomials $C_n(x)$ is by $$ C_n(x,t)=\sum_{\pi\in S_n}x^{\text{des}(\pi)+1}t^{\text{maj}(\pi)} $$ where $des(\pi)$ is the descents in $\pi$, ...
Clara's user avatar
  • 1,536
3 votes
3 answers
273 views

Simplify binomial sum

It is well-known that for any integer $j$ $$ \sum _{i=0}^k (-1)^{i} \binom{k}{i} (j-i)^k = k! $$ But I wonder if there's a closed form for the following sum $$ \sum _{i=j}^k (-1)^{i} \binom{k}{i} (j-i)...
LeafGlowPath's user avatar
  • 7,243
3 votes
2 answers
166 views

Sum of the series $\sum_{k=0}^\infty k^ne^{-k}$ for a positive integer $n$

How can we calculate the sum of the series $\sum_{k=0}^\infty k^ne^{-k}$ for a positive integer $n$? I tried: $$\sum_{k=0}^\infty k^ne^{-k} =(-1)^n \sum_{k=0}^\infty \frac{d^n}{dy^n}\bigg|_{y=1}e^{-ky}...
blancket's user avatar
  • 1,770
3 votes
2 answers
220 views

How to prove Eulerian Identity?

This is a question in the book Concrete Mathematics of Graham and Knuth about Eulerian numbers, Stirling Numbers of the second kind and binomial coefficients. I have been thinking about it for a lot ...
Sean's user avatar
  • 41
3 votes
1 answer
79 views

The value of the second-order Eulerian polynomials at $x = \frac{-1}{2}$.

Recently, the second-order Eulerian polynomials $ \left\langle\!\left\langle x \right\rangle\!\right\rangle_n $ have been discussed on MSE [ a , b ]. $$ \left\langle\!\left\langle x \right\rangle\!\...
Peter Luschny's user avatar
3 votes
1 answer
1k views

Half of the binomial theorem

The binomial theorem states that the generating function $\sum_{k=0}^n {n \choose k} x^k$ is equal to $(1+x)^n$ for any $n$. For a given $n$, let $$B(x)=\sum_{k=0}^n {2n+1\choose k} x^k.$$ That is, ...
Sam Spiro's user avatar
  • 415
3 votes
1 answer
187 views

An associated Stirling number identity related to the second-order Eulerian numbers.

A similar Stirling number identity representing the second-order Eulerian numbers can be found at this question. We denote the associated Stirling cycle numbers as $\left[\!\left[ n\atop k\right] \! \...
Peter Luschny's user avatar
3 votes
1 answer
154 views

Does $x\left(\frac{d}{dx}\left(\cdots x \left(\frac{d}{dx} \left( \frac{x}{1-x}\right)\right)\cdots\right)\right)$ have a closed form expression?

Does $\displaystyle \underbrace{x\left(\dfrac{d}{dx}\left(\cdots x \left(\dfrac{d}{dx} \left( \dfrac{x}{1-x}\right)\right)\cdots\right)\right)}_{\text{$x \frac{d}{dx}$ $m$ times}}$ have a closed form?...
Puzzled417's user avatar
  • 6,956
3 votes
1 answer
368 views

Eulerian Polynomial Generating Function Proof

The generating function for the Eulerian polynomials is $$\frac{t-1}{t-e^{(t-1)x}} = \sum_{n=0}^\infty A_n(x) \frac{t^n}{n!}$$ where $A_n(x)$ is the $n^{th}$ Euler polynomial and $$A_n(x) = \sum_{k=...
PC_greg's user avatar
  • 81
3 votes
1 answer
60 views

Is my use of derivatives correct in this infinite sum?

I was inspired by a question that seeked to find a generating function for the sum of powers. We set $s(n,p)=\sum_{k=0}^n k^p$, now we are looking for $G(x,p)= \sum_{n=0}^{\infty}s(n,p)x^n$. ...
GuPe's user avatar
  • 7,338
3 votes
1 answer
133 views

Coefficients of a generating function

I need a bit of help. I was solving the form of the coefficients of the generating function $\sum_{n}n^m r^n$. Then I started building the indefinite sum $\sum n^m r^n \delta n$ through recursive ...
Masacroso's user avatar
  • 30.7k
3 votes
1 answer
195 views

Do these ratios of the Eulerian number triangle converge to the logarithm of x?

Consider the matrix $A_3$ with the definition if $n=k$ then $A_3(n,k)=\binom{n-1}{k-1}=1$, else if $n\ge k$ then $A_3(n,k)=\frac{\binom{n-1}{k-1}}{1-x}$ else $A_3(n,k)=0$. $\binom{n-1}{k-1}$ means the ...
Mats Granvik's user avatar
  • 7,426
3 votes
0 answers
61 views

The second-order Eulerian numbers meet the Clausen numbers

The (generalized) Clausen numbers A160014 are defined as $$\operatorname{C}_{n, k} = \prod_{ p\, -\, k\, |\, n} p \quad (p \in \mathbb{P})$$ where $\mathbb{P}$ denotes the primes. The classical ...
Peter Luschny's user avatar
3 votes
0 answers
166 views

Permutation statistics in multiple rows

Usually we study the statistics of a permutation written in one row. Is there any result for the statistics of a permutation written in multiple rows? Let me give an example in order to be more clear: ...
WunderNatur's user avatar
2 votes
1 answer
98 views

What is the connection between Eulerian numbers and this identity?

It is stated in this pdf that for $X_j \text{ iid},$ uniformly distributed on $[0,1],$ $$\dfrac{1}{n!} \left\langle n \atop k \right\rangle = P\left(\sum_{j=1}^{n}X_j\in[k,k+1]\right)$$ without ...
martin's user avatar
  • 9,008
2 votes
2 answers
375 views

Identities for sums of Eulerian numbers

Let $[n] := \{1,2,\dots,n\}$. An Eulerian number, $A(n,k)$, denotes the number of permutations $\sigma \in S_n$ of $[n]$ such that exactly $k$ numbers have the property that $\sigma(i) < \sigma(i+...
Celal Bey's user avatar
  • 516
2 votes
1 answer
186 views

A combinatoric identity involving Eulerian and Stirling numbers

Let $$ A_{n}(x)=\sum_{k=0}^{n}\left\langle\begin{array}{l} n \\ k \end{array}\right\rangle x^{n-k} $$ Eulerian Polynomials and $$ \left\langle\begin{array}{l} n \\ k \end{array}\right\rangle $$ are ...
user1062's user avatar
  • 421
2 votes
1 answer
87 views

Proof that the Eulerian Numbers $A(n,m)$ are asymptotic to $(m+1)^n$ when $m$ is fixed and $n\rightarrow\infty$

This is stated in many places but seldom proved. The goal is to prove that, for fixed $m\geq 0$, $$\lim_{n\rightarrow\infty}A(n,m)\sim (m+1)^n$$ where $A(n,m)$ are the Eulerian Numbers, namely, the ...
user559748's user avatar
2 votes
1 answer
93 views

The ratio between Central Eulerian Numbers and the sum of Eulerian Numbers at a fixed level converges to zero

The Central Eulerian Numbers are given by the formula $$C(n) = \sum_{j=0}^n(-1)^j(n-j)^{2n-1}\binom{2n}{j}$$ This represent the Eulerian Number $E(2n-1, n)$ where $$E(n, k) = \sum_{j=0}^k(-1)^j(k-j+...
user avatar
2 votes
0 answers
97 views

Understanding the recurrence relation $f(n,k) = kf(n-1, k) + (n-k+1)f(n-1,k-1)$ for the Euler numbers $f(n,k)$

The Euler numbers $f(n,k)$ are given in Generatingfunctionology as the number of permutations of $[n]$ with exactly $k$ increasing runs (exercise 1.18.c). Its recurrence relation is $$f(n,k) = kf(n-1, ...
Xin Yuan Li's user avatar
2 votes
0 answers
51 views

Does the sum $\sum_{k\ge 0}A(n,k)k^t$ have a closed form?

My friend and I are trying to find a closed form for the sum $\sum_{k\ge 0}A(n,k)k^t,$ in which the notation $A(n,k)$ is for the Eulerian number. (An alternating notation is $\left<n\atop k\right&...
Renko Usami's user avatar
2 votes
0 answers
194 views

Simon Newcomb's problem

I am looking for an answer to the following problem. Let $S$ be the multiset $\{1^{d_1},2^{d_2},\dots,m^{d_m}\}$ $A_{S,k}$ is the number of permutations of $S$ with $k-1$ descents and no descent at ...
bronko's user avatar
  • 265
1 vote
2 answers
89 views

Does $\sum_{k=0}^{\infty} \sum_{n=1}^{\infty} x^nn^k =cx-x\ln(x)+x\ln(1-x) $ for $0 < x < 1$ for some real $c$? If so, what is $c$?

Let $g_k(x) =\sum_{n=1}^{\infty} x^nn^k $ and $G(x) =\sum_{k=0}^{\infty} g_k(x) =\sum_{k=0}^{\infty} \sum_{n=1}^{\infty} x^nn^k $. For $0 < x < 1$ is $G(x) =cx-x\ln(x)+x\ln(1-x) $ for some real $...
marty cohen's user avatar
1 vote
1 answer
137 views

Number of Permutation with Pro = k equals the Eulerian Number A(n,k+1)

The problem phrases as follows: Let $\sigma = (\sigma_1, . . . , \sigma_n)$ be a permutation. We say that element $i$ is progressive if $\sigma_i > i$. We write $pro(\sigma)$ for the number of ...
ChubbyRuby's user avatar
1 vote
3 answers
128 views

Power series of $x/(1-ae^{-x})$.

I am looking for a power series expansion for the function $x(1-ae^{-x})^{-1}$ (perhaps for $0<a<1$). Using the Bernoulli numbers, I can write \begin{align*} \frac{x}{1-ae^{-x}} &= \frac{x}{...
Kenneth Ng's user avatar
1 vote
1 answer
160 views

Recurrence $C_n(t)=t(1-t)C'_{n-1}(t)+ntC_{n-1}(t)$ for Eulerian Polynomials?

I was reading about Eulerian polynomials on OEIS, and there is a recurrence given for them, namely: $$ C_0(t)=1 $$ and $$ C_n(t)=t(1-t)C'_{n-1}(t)+ntC_{n-1}(t)\qquad (n\geq 1). $$ How can ...
Clara's user avatar
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