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\title{On some Euler-Mahonian distributions} 

\author{Angela Carnevale\thanks{Supported
  by the German-Israeli Foundation for Scientific Research and
  Development, grant no.~1246.}\\
\small Fakult\"at f\"ur Mathematik\\[-0.8ex]
\small Universit\"at Bielefeld\\[-0.8ex] 
\small Bielefeld, Germany\\
\small\tt acarneva1@math.uni-bielefeld.de\\}

\date{\dateline{Dec 22, 2016}{Jul 26, 2017}{Aug 11, 2017}}

 \begin{document}
 \maketitle

\begin{abstract} 
 We prove that the pair of statistics $(\des,\maj)$ on multiset permutations is equi\-dis\-tri\-buted with the pair $(\stc,\invv)$ on certain quotients of the symmetric groups. We define an analogue of the statistic $\stc$ on multiset permutations whose joint distribution with the inversions equals that of $(\des,\maj)$. We extend the definition of the statistic $\stc$ to hyperoctahedral and  even hyperoctahedral groups. These functions, together with the Coxeter length, are equidistributed with $(\ndes,\nmaj)$ and $(\ddes,\dmaj)$, respectively.
\end{abstract}

\section{Introduction}
The first result about the enumeration of multiset permutations with respect to the  statistics now called \emph{descent number} and \emph{major index} is due to MacMahon. Let $\rho=(\rho_1,\ldots,\rho_m)$ be a composition of $N \in \N$. We denote by $\Sr$ the set of all permutations of the multiset $\{{1}^{\rho_1},\ldots,{m}^{\rho_m}\}$. The
  \emph{descent set} $\Des(w)$ of the multiset permutation $w =  w_1\cdots w_N\in \Sr$ is
  $\Des(w)=\{i \in [N-1] \mid w_i>w_{i+1}\}$. The descent and major index
  statistics on $S_\rho$ are given by
$$\des(w)=|\Des(w)| \quad\textup{ and }\quad\maj(w)=\sum_{i\in\Des(w)} i.$$  Then (\cite[\textsection 462, Vol.~2, Ch.~IV,
  Sect.~IX]{MacMahon/16})
  \begin{equation}\label{eq:mmpart}
    \sum_{k\geq 0}\left(\prod_{j=1}^m\binom{\rho_j +k}{k}_q\right) x^k=\frac{\sum_{w \in \Sr}{x^{\des(w)}q^{\maj(w)}}}{\prod_{i=0}^{N}(1-xq^i)} \in \Z[q]\llbracket x\rrbracket,
  \end{equation}
  where, for $n, k \in \N$, we set
  \[\binom{n}{k}_p =\frac{[n]_p!}{[n-k]_p![k]_p !}, \qquad [n]_p!=\prod_{i=1}^n [i]_p, \qquad [n]_p=\sum_{i=0}^{n-1} p^i.\] 
 The well-known result about the equidistribution on multiset permutations of the \emph{inversion number} with the major index also goes back to MacMahon. In  \cite{FoataSch/78} Foata and  Sch{\"u}tzenberger  proved that this equidistribution refines, in the case of the symme\-tric group, to inverse descent classes.
A pair of statistics that is equidistributed with $(\des,\maj)$ is called \emph{Euler-Mahonian}.
In \cite{Skandera/01} Skandera introduced an Eulerian statistic, which he called $\stc$, on the symmetric group and proved that the pair $(\stc,\invv)$ is Euler-Mahonian. He pointed out that the word statistic obtained by generalising $\stc$  in the natural way is not Eulerian. He also left open the question of finding a suitable generalisation with this property.

In this note we prove that the joint distribution of $(\stc,\invv)$ on certain quotients of the symmetric group is indeed the same as the distribution of $(\des,\maj)$ on multiset permutations; we use this result to define a statistic $\mstc$ that is Eulerian on multiset permutations and that, together with $\invv$, is equidistributed with the pair $(\des,\maj)$.

To the author's knowledge, not much is known about factorisations of the bivariate polynomials defined by Euler-Mahonian distributions. More is known (or conjectured) about the Eulerian polynomial  $\sum_{\sigma \in S_n}x^{\des(\sigma)}$ and its generalisations to multiset permutations.
 Frobenius proved (see \cite{Frobenius/1910}) that the Eulerian polynomial  has real, simple, negative roots, and that $-1$ features as a root if and only if $n$ is even. Later, Simion proved that its analogues on permutations of any multiset are also real rooted with simple, negative roots (see \cite{Simion/84}). 

We use our first result of equidistribution to show that the polynomial of the joint distribution of $\des$ and $\maj$ admits, on the set of permutations of words in the alphabet $\{1^r, 2^r\}$,  for odd~$r$, a unique \emph{unitary} factor  (i.e.\ a factor which arises from a univariate polynomial, all of whose roots lie on the unit circle, by means of a monomial substitution; cf.~Section~\ref{stcmulti} for  a precise definition).   Together with the factorisation of Carlitz's $q$-Eulerian polynomial showed in \cite[Lemma 2.7]{CarnevaleVoll/16}, this result may be considered a refinement of Frobenius' and Simion's results. The previous results support a conjecture we made in \cite[Conjecture~B]{CarnevaleVoll/16} and  which we translate in Section~\ref{stcmulti} in terms of the joint distribution of $(\stc,\invv)$ on quotients of the symmetric group.

Our interest in detecting factorisations of this type is motivated by questions regarding analytic properties of certain Dirichlet series studied by the author and Voll in \cite{CarnevaleVoll/16}. These series are closely related to Hadamard products of subgroup zeta functions of free abelian groups of finite rank. Their numerators are exactly the polynomials of the joint distributions of $(\des,\maj)$ on appropriate sets of multiset permutations. This combinatorial description allows us to easily read off analytic properties such as abscissa of convergence, meromorphic continuation or the existence of a natural boundary.

Generalisations of MacMahon's  result \eqref{eq:mmpart} to signed permutations  were first obtained by Adin, Brenti and Roichman in \cite{AdinBrentiRoichman/01} and to even-signed permutations by Biagioli in \cite{Biagioli/03}.
In the last section of this note we define  Eulerian statistics $\nstc$ and $\dstc$ which, together with the length, are  equidistributed with the Euler-Mahonian pairs $(\ndes,\nmaj)$ on the hyperoctahedral group and $(\ddes,\dmaj)$ on the even hyperoctahedral group, respectively.





\section{Stc on quotients of the symmetric group and multiset permutations}\label{stcmulti}
 We start with some definitions and notation, for further notation and basic facts about Coxeter groups we refer the reader to~\cite{BjoernerBrenti/05}.

For $n,m \in \N,$ $m\leq n$ we denote with $[n]=\{1,\ldots,n\}$ and $[m,n]=\{m,m+1,\ldots,n\}$.  For a permutation $\sigma \in S_n$ we use the one-line notation or the disjoint cycle notation. 
For a (signed) permutation $\sigma\in S_n$ (respectively, $B_n$), we let \[\iset(\sigma)=\{(i,j)\in [n]\times[n] \mid i<j,\, \sigma(i)>\sigma(j)\} \mbox{ and } \invv(\sigma)=|\iset(\sigma)|.\] The symmetric group $S_n$ is a Coxeter group with respect to the set of generators $S=\{s_1,\ldots,s_{n-1}\}$, where $s_i$ is the simple transposition $(i,\,i+1)$.  It is well-known that the Coxeter length coincides with the inversion number. For  $J\subseteq [n-1]$ the corresponding quotient is defined as
 \[S_n ^J=\{w \in S_n \mid \Des(w)\subseteq [n-1]\setminus J\}.\] Moreover we denote with \[IS_n ^J=\{w \in S_n\mid \Des(w^{-1})\subseteq [n-1]\setminus J\} \]the inverse descent class corresponding to $J\subseteq [n-1]$. 


It is well-known that the symmetric group $S_n$ is in bijection with the set of words $w=w_1\cdots w_n $ $\in E_n$ where 
\[E_n=\{w_1\cdots w_n \mid w_i \in [0,n-i] \mbox{ for  } i=1,.\ldots, n\}. \]
One of such bijections is the Lehmer code, defined as follows.
For  $\sigma \in S_n$,  $\code(\sigma)=c_1\cdots c_n \in E_n$  where $c_i= |\{j \in [i+1,n] \mid \sigma(i)>\sigma(j)\}|$.
The sum of the $c_i$s gives, for each permutation, the inversion number. The Eulerian  statistic  $\stc$ is defined as follows (cf. \cite[Definition 3.1]{Skandera/01}):
$\stc(\sigma)=\sta(\code(\sigma))$, where for a word $w\in E_n$,
%$$\sta(w)=\max\{ r \in [n] \mid\mbox{ there exists a subsequence } w_{i_1}\cdots w_{i_r} >(r-1)(r-2)\cdots 1 \, 0\}$$
$$\sta(w)=\max\{ r \in [n] \mid\exists\, 1\leq i_1,\cdots<i_r\leq n \mid w_{i_1}\cdots w_{i_r} >(r-1)(r-2)\cdots 1 \, 0\}.$$
Informally, $\sta(w)$ is the maximum $r$ for which there exists a subword of $w$ of length $r$  elementwise strictly greater than the \emph{r-staircase word}  $(r-1)(r-2)\cdots 1\,0$.

For example let $\sigma=452361 \in S_6$. Then $\code(\sigma) =331110$, $\invv(\sigma)=\sum_i c_i=9$, $\stc(\sigma)=\sta(\code(\sigma))=3$. The statistic $\stc$  constitutes an Eulerian partner for the inversions on $S_n$.
\begin{thm}[{\cite[Theorem 3.1]{Skandera/01}}]\label{thm:ska}
Let $n \in \N$. Then
	\begin{equation*}
	\sum_{w\in S_n}x^{\des(w)}q^{\maj(w)}=\sum_{w\in S_n}x^{\stc(w)}q^{\ell(w)}
	\end{equation*}
\end{thm}
Recall that given a composition $\rho$ of $N$, we denote by $\Sr$ the set  of permutations of the multiset $\{1^{\rho_1},\ldots,{m}^{\rho_m}\}$, i.e.\ rearrangements of the word $w_\rho=\underbrace{1\dots 1}_{\rho_1}\underbrace{2\dots 2}_{\rho_2}\ldots\underbrace{m \dots m}_{\rho_m}$. 

A natural way to associate a permutation to a multiset permutation is the standardisation, which we now describe; cf.\ also \cite[\textsection 1.7]{Stanley/12}. Given $\rho$ a composition of $N$ and a word $w$ in the alphabet $\{{1}^{\rho_1},\ldots,{m}^{\rho_m}\}$, $\stdr(w)$ is the element of $S_N$ obtained by $w$ substituting, in the order of appearance in $w$ from left to right,  the $\rho_1$ $1$s with the sequence $1\,2\ldots\rho_1$, the $\rho_2$ $2$s with the sequence $\rho_1+1 \ldots \rho_1+\rho_2$ and so on. So for example if $\rho=(2,3,2)$ and $w=1223132\in \Sr$, then $\stdr(w)=1346275 \in S_7$. Clearly the map $\stdr:\Sr \to S_N$ is not surjective. The set $\Sr$  is in bijection with certain quotients and inverse descent classes of $S_N$ depending on $\rho$. In particular, for $\rho=(\rho_1,\ldots,\rho_m)$ a composition of $N$, for $i=1\ldots, m-1$ we let 
\begin{equation}\label{eq:subset}
 r_i=\sum_{k=1}^i {\rho _k},\qquad {\bf r}=\{r_i \mid i\in [m-1]\}\subseteq [N-1] \qquad \mbox{and}\qquad R=[N-1]\setminus {\bf r}.
 \end{equation}
 The quotient and the inverse descent class that will play a role in our results of equidistribution are $\SR$ and $\ISR$, respectively.



We will need the following result due to Foata and Han.
\begin{pro}[{\cite[Propriet\' e 2.2]{FoataHan/04}}]\label{thm:fh}
	Let $n \in \N$, $J\subseteq [n-1]$. Then
	\begin{equation}
	\sum_{ \substack{\{w\in S_n \mid\\ {\Des(w)=J}\}}}x^{\des(w^{-1})}q^{\maj(w^{-1})}=\sum_{ \substack{\{w\in S_n \mid\\ {\Des(w)=J}\}}}x^{\stc(w)}q^{\ell(w)}.
	\end{equation}
\end{pro}
Our first result is the following.
\begin{pro}\label{thm:equi}
Let $N \in \N$, $\rho$  a composition of $N$ and $R\subseteq [N-1]$ as in \eqref{eq:subset}. The pair $(\stc,\ell)$ on $\SR$ is equidistributed with $(\des,\maj)$ on $\Sr$:
\begin{equation}\label{equi}
C_{\rho}(x,q)=\sum_{w \in \Sr}{x^{\des(w)}q^{\maj(w)}}=\sum_{w \in \SR}{x^{\stc(w)}q^{\ell(w)}}.
\end{equation}
\end{pro}
\begin{proof}
The standardisation $\stdr$ is a bijection between $\Sr$ and $\ISR$, and preserves  $\des$ and $\maj$, so
\[\sum_{w\in \Sr}x^{\des(w)}q^{\maj(w)}=\sum_{w\in \Sr}x^{\des(\stdr(w))}q^{\maj(\stdr(w))}=\sum_{w\in \ISR}x^{\des(w)}q^{\maj(w)}=\sum_{w\in \SR}x^{\stc(w)}q^{\ell(w)},\]
where the last equality follows from Proposition \ref{thm:fh}.\qedhere
\end{proof}

As an application, we prove a result  about the bivariate factorisation of the polynomial $C_\rho(x,q)$, that in \cite{CarnevaleVoll/16} is used to  deduce analytic properties of some orbit Dirichlet series. We will need the following definition (see also \cite[Remark 2.9]{CarnevaleVoll/16}).

\begin{dfn} We say that a bivariate polynomial $f(x,y)\in \Z[x,y]$ is unitary if there exist integers $\alpha, \beta\geq 0$ and $g\in \Z[t]$ so that $f(x,y)=g(x^\alpha y^\beta)$ and all the complex roots of $g$ lie on the unit circle.\end{dfn} 
\begin{pro}\label{pro:fact}
Let $\rho=(r,r)$ where $r \equiv 1 \pmod 2$. Then
\begin{equation}\label{fact}
C_{\rho}(x,q)=(1+xq^{r})\widetilde{C}_{\rho}(x,q),\end{equation}
where $\widetilde{C}_{\rho}(x,q)$ has no
 unitary factor.
\end{pro}
Before we prove Proposition \ref{pro:fact}, we give a characterisation of the stc for permutations with at most one descent.
\begin{lem}\label{lem:stc}
Let $\rho=(\rho_1,\rho_2)$, $N=\rho_1 +\rho_2$. Let $w\in S^{\{\rho_1\}^c}_N$. Then
$$\stc(w)=|\{i\in[\rho_1]\mid w(i)>\rho_1\}|.$$

\end{lem}
\begin{proof}
A permutation $w\in S_{N}^{\{\rho_1\}^c}$ has at most a descent at $\rho_1$, so its code is of the form $\code(w)=c_1\cdots c_{\rho_1} 0\cdots 0$, with  $0\leq c_1\leq\ldots\leq c_{\rho_1}$. 
The first (possibly) non-zero element of the code is exactly the number of elements of the second block for which the image is in the first block. This number coincides with the length of the longest possible subword of the code which is elementwise greater than a staircase word. 
\end{proof}

\begin{proof}[Proof of Proposition \ref{pro:fact}]
The polynomial $C_{\rho}(x,1)$, descent polynomial of $\Sr$, has all real, simple, negative roots (cf.\ \cite[Corollary 2]{Simion/84}). Thus a factorisation of the form \eqref{fact} implies that $\widetilde{C}_{\rho}(x,q)$ has no
 unitary factor.
To prove \eqref{fact} we define an involution $\vf$ on $\SR$  such that, for all $w\in \SR$, $|\ell(\vf (w))-\ell(w)|=r$ and $|\stc(\vf (w))-\stc(w)|=1$.

For $w \in \SR$ we  define \[M_w=\{i\in [r] \mid w\uu(i)\leq r \mbox{ and } w\uu(i+r)> r \mbox{ or } w\uu(i)> r \mbox{ and } w\uu(i+r)\leq r \},\]
the set of $i\in [r]$ for which $i$ and $i+r$ are not in the same ascending block. Since $r$ is odd, $M_w$ is non-empty for all $w\in \Sr$.
We then define $\vf(w)=((\iota,\iota + r)w)^{R}$, where $\iota=\min\{i \in M_w\}$ and, for $\sigma \in S_N$,  $\sigma^{R}$ denotes the unique minimal coset representative in the quotient $\SR$. By Lemma~\ref{lem:stc} clearly $\stc(\vf(w))=\stc(w)\pm 1$.
Suppose now that $w^{-1}(\iota)\leq r$ and $w^{-1}(\iota)>r$ (the other case is analogous). Then \begin{align*}\ell(\vf(w))&=\ell(w)+|\{i \in [r]\mid w(i)>\iota\}|+|\{i \in [r+1,2r]\mid w(i)<\iota+r\}|\\ &=\ell(w)+r-i+i\end{align*}
as desired.\qedhere
\end{proof}
We now reformulate \cite[Conjecture B]{CarnevaleVoll/16} in terms of the bivariate distribution of $(\stc,\ell)$ on quotients of the symmetric group.
\begin{con}
	Let $\rho$ be a composition of $N$ and $R\subseteq [N-1]$ constructed as in \eqref{eq:subset}. Then $C_\rho (x,q)=\sum_{w\in \SR}{x^{\stc(w)}q^{\ell(w)}}$ has a unitary factor if and only if $\rho=(\rho_1,\ldots,\rho_m)$ where $\rho_1=\ldots=\rho_m=r$ for some odd $r$ and even $m$. In this case
	\[\sum_{w\in \SR}{x^{\stc(w)}q^{\ell(w)}}=(1+xq^{\frac{rm}{2}})\widetilde{C}_\rho(x,q)\]
	for some $\widetilde{C}_\rho(x,q)\in \Z[x,q]$ with no unitary factors.
\end{con}

Proposition \ref{thm:equi} suggests a natural extension of the definition of the statistic $\stc$ to multiset permutations, thus answering a question raised by Skandera, see  \cite[Question~5.1]{Skandera/01}. 

For $w\in \Sr$, $\stdr (w) \in \ISR$. This yields a bijection between multiset permutations $\Sr$ and the quotient $\SR$
\[\istd: \Sr \rightarrow \SR,\quad \istd(w)=(\stdr (w))^{-1}\]
which is inversion preserving:  $\invv(w)=\invv(\istd(w))$.
\begin{dfn}
	Let $\rho$ be a composition of $N$. For a multiset permutation $w \in \Sr$ the \emph{multistc} is
	\[\mstc(w)=\stc(\istd(w)).\]
\end{dfn}
The pair $(\mstc,\invv)$ is equidistributed with $(\des,\maj)$ on $\Sr$, as
\[\sum_{w \in \Sr} x^{\mstc(w)}q^{\invv(w)}=\sum_{w \in \SR} x^{\stc(w)}q^{\invv(w)}=\sum_{w \in \Sr} x^{\des(w)}q^{\maj(w)},\]
which together with \eqref{eq:mmpart} proves the following theorem.
\begin{thm}
	Let $\rho$ be a composition of $N\in \N$. Then
\[ \sum_{k\geq 0}\left(\prod_{j=1}^m\binom{\rho_j +k}{k}_q\right) x^k=\frac{\sum_{w \in \Sr}{x^{\mstc(w)}q^{\invv(w)}}}{\prod_{i=0}^{N}(1-xq^i)} \in \Z[q]\llbracket x\rrbracket.\]
\end{thm}

\section{Signed and even-signed permutations}
MacMahon's result \eqref{eq:mmpart} for the symmetric group (i.e.\ for $\rho_1=\ldots\rho_m=1$) is often present in the literature as Carlitz's identity, satisfied by  Carlitz's $q$-Eulerian polynomial $A_n(x,q)=\sum_{\sigma \in S_n} x^{\des(\sigma)}q^{\maj(\sigma)}$. 

This result was extended, for suitable statistics, to the groups of signed and even-signed permutations. The major indices  defined in such extensions are in both cases equidistributed with the Coxeter length $\ell$. In this section we define type $B$ and type $D$ analogues of the statistic $\stc$, that together with the length satisfy these generalised Carlitz's identities.
\subsection{Eulerian companion for the length in type B}
Let $n\in \N$. The hyperoctahedral group $B_n$ is the group of permutations $\sigma=\sigma_1\cdots \sigma_n$ of $\{\pm 1,\ldots,\pm n\}$ for which $|\sigma|=|\sigma_1|\ldots |\sigma_n| \in S_n$.
For $\sigma \in B_n$, the negative set and negative statistic are  
\[\nset(\sigma)=\{i \in [n] \mid \sigma(i)<0\}\quad \nneg(\sigma)=|\nset(\sigma)|.\]
The Coxeter length $\ell$ for $\sigma$ in $B_n$ has the following combinatorial interpretation (see, for instance \cite{BjoernerBrenti/05}):
\begin{equation*}
\ell(\sigma)=\invv(\sigma)+\nneg(\sigma)+\nsp(\sigma),
\end{equation*}
where $\invv$ is the usual inversion number and $\nsp(\sigma)=|\{(i,j)\in [n]\times[n] \mid i<j,\,\sigma(i)+\sigma(j)<0\}|$ is the number of negative sum pairs.

 In  \cite{AdinBrentiRoichman/01} an Euler-Mahonian pair of the negative type was defined as follows.
The negative descent and negative major index are, respectively,
\begin{equation}\label{eq:nstat}
\ndes(\sigma)=\des(\sigma)+\nneg(\sigma), \qquad \nmaj(\sigma)=\maj(\sigma)-\sum_{i\in\nset(\sigma)}\sigma(i).\end{equation} 
The pair $(\ndes,\nmaj)$ satisfies the following generalised Carlitz's identity.
\begin{thm}[{\cite[Theorem 3.2]{AdinBrentiRoichman/01}}]
Let $n\in \N$. Then
\begin{equation}
\sum_{r\geq 0} [r+1]^n _q x^r = \frac{\sum\limits_{\sigma \in B_n}{x^{\ndes(\sigma)}q^{\nmaj(\sigma)}}}{(1-x) \prod\limits_{i=1}^{n}{(1-x^2 q^{2i})}} \mbox{ in }\Z[q]\llbracket x\rrbracket.\end{equation}	
\end{thm}


Motivated by \eqref{eq:nstat} and the well-known fact that the length in type $B$ may be also written as
\begin{equation}
\ell(\sigma)=\invv(\sigma)-\sum_{i \in \nset(\sigma)} \sigma(i),
\end{equation} we define the analogue of the statistic $\stc$ for signed permutations as follows.
\begin{dfn}
	Let $\sigma\in B_n$. Then 
\[\nstc(\sigma)=\stc (\sigma)+\nneg(\sigma).\]
\end{dfn}
\begin{thm}\label{thm:B}
	Let $n\in \N$. Then
\[\sum_{\sigma\in B_n}{x^{\nstc(\sigma)}q^{\ell(\sigma)}}=\sum_{\sigma\in B_n}{x^{\ndes(\sigma)}q^{\nmaj(\sigma)}}.\]
\end{thm}
\begin{proof}
	We use essentially the same argument as in the proof of \cite[Theorem 3]{LaiPet/11}. There, the following decomposition of $B_n$ is used. 
Every permutation $\tau \in S_n$ is associated with $2^n$ elements of $B_n$, via the choice of the $n$ signs. More precisely, given a signed permutation $\sigma \in B_n$ one can consider the ordinary permutation in which the elements are in the same relative positions as in $\sigma$. We write $\sy(\sigma)=\tau$. Then
\[B_n=\bigcup_{\tau \in S_n} B(\tau),\]
where $B(\tau)=\{\sigma \in B_n \mid \sy(\sigma)=\tau\}$. So every $\sigma\in B_n$ is uniquely identified by the permutation $\tau=\sy(\sigma)$ and the choice of signs $J(\sigma)=\{\sigma(j) \mid j \in \nset(\sigma)\}.$ 

Clearly, for $\sigma \in B_n$ we have $\iset(\sigma)=\iset(\sy(\sigma))$, and thus $\stc(\sigma)=\stc(\sy(\sigma))$. So, for $\tau=\sy(\sigma)$
\[x^{\nstc(\sigma)}q^{\ell(\sigma)}=x^{\stc(\tau)}q^{\invv(\tau)}\prod_{j\in J(\sigma)} x q^j.\] The claim follows, as
\[\sum_{\sigma \in B_n} {x^{\nstc(\sigma)}q^{\ell(\sigma)}}=\sum_{\sigma \in B(\tau)}\sum_{\tau \in S_n}x^{\stc(\tau)}q^{\invv(\tau)}\sum_{J\subseteq [n]}\prod_{j\in J} x q^j=A_n(x,q)\prod_{i=1}^n {(1+xq^i)}.\]\qedhere
\end{proof}
\begin{cor}Let $n\in \N$. Then
\begin{equation}
	\sum_{r\geq 0} [r+1]^n _q x^r = \frac{\sum\limits_{\sigma \in B_n}{x^{\nstc(\sigma)}q^{\ell(\sigma)}}}{(1-x) \prod\limits_{i=1}^{n}{(1-x^2 q^{2i})}} \mbox{ in }\Z[q]\llbracket x\rrbracket.\end{equation}
\end{cor}
\subsection{Eulerian companion for the length in type D}
The even hyperoctahedral group $D_n$ is the subgroup of $B_n$ of signed permutations for which the negative statistic is even:
\[D_n=\{\sigma \in B_n \mid \nneg(\sigma) \equiv 0 \pmod 2\}.\]
Also for $\sigma$ in $D_n$ the Coxeter length can be computed in terms of the following statistics:
\begin{equation}
\ell(\sigma)=\invv(\sigma)+\nsp(\sigma).
\end{equation}
The problem of finding an analogue, on the group $D_n$ of even signed permutations, was solved in \cite{Biagioli/03}, 
where type $D$  statistics $\des$ and $\maj$ were defined, as follows. For $\sigma \in D_n$
\begin{equation}
\ddes(\sigma)=\des(\sigma)+|\DNeg(\sigma)|,\qquad
\dmaj(\sigma)=\maj(\sigma)-\sum_{i \in \DNeg(\sigma)}( \sigma(i)+1),
\end{equation}
where $\DNeg(\sigma)=\{i \in [n]|\sigma(i)<-1\}$.
The following holds.
\begin{thm}[{\cite[Theorem 3.4]{Biagioli/03}}]
Let $n \in \N$. Then
\begin{equation}
\sum_{r\geq 0} [r+1]^n _q x^r = \frac{\sum\limits_{\sigma \in D_n}{x^{\ddes(\sigma)}q^{\dmaj(\sigma)}}}{(1-x)(1-xq^n) \prod\limits_{i=1}^{n-1}{(1-x^2 q^{2i})}} \mbox{ in }\Z[q]\llbracket x\rrbracket.\end{equation}	
\end{thm}
\begin{dfn}
Let $\sigma \in D_n$. We set
$$\dstc(\sigma)=\stc (\sigma)+|\DNeg(\sigma)|=\stc(\sigma)+\nneg(\sigma)+\epsilon(\sigma),$$
where $$\epsilon(\sigma)=\begin{cases}
-1 \mbox{ if }\sigma^{-1}(1)<0, \\
0\:\:\; \mbox{ otherwise }.
\end{cases}$$
\end{dfn}
We now show that the statistic just defined
constitutes an Eulerian partner for the length on $D_n$, that is, the following holds.
\begin{thm}\label{thm:D}
Let $n\in \N$. Then
\[\sum_{\sigma\in D_n}{x^{\dstc(\sigma)}q^{\ell(\sigma)}}=\sum_{\sigma\in D_n}{x^{\ddes(\sigma)}q^{\dmaj(\sigma)}}.\]
\end{thm}
\begin{proof}
We use, as in \cite{Biagioli/03} the following decomposition of $D_n$. Let \begin{equation}\label{eq:asc}
T_n=\{\alpha \in D_n \mid \des(\alpha)=0\}=\{\alpha\in D_n \mid \iset(\alpha)=\varnothing\}
\end{equation}
then $D_n$ can be rewritten as the following disjoint union:
\begin{equation}\label{eq:ddec}
D_n=\bigcup_{\tau \in S_n}\{\alpha \tau \mid \alpha \in T_n\}.
\end{equation}
For $\alpha \in T_n$ and $\tau \in S_n$ the following hold:
\begin{align*}
\ell(\alpha \tau)&=\ell(\alpha)+\ell(\tau)=\nsp(\alpha)+\invv(\tau),\\ \nsp(\alpha \tau)&=\nsp(\alpha),\\ \dstc(\alpha \tau)&=\stc(\tau)+\nneg(\alpha)+\epsilon(\sigma).
\end{align*}
The last equation follows from the second equality in \eqref{eq:asc}. Thus
\begin{align*}
\sum_{\sigma\in D_n}{x^{\dstc(\sigma)}q^{\ell(\sigma)}}&=\sum_{\alpha \in T_n}\sum_{\tau \in S_n}x^{\stc(\tau)+\nneg(\alpha)+\epsilon(\alpha)}q^{\ell(\alpha)+\ell(\tau)}\\
&=\sum_{\alpha \in T_n}x^{\nneg(\alpha)+\epsilon(\alpha)}q^{\nsp(\alpha)} \sum_{\tau \in S_n}x^{\stc(\tau)}q^{\invv(\tau)} \\
&=\prod_{i=1}^{n-1}(1+x q^i)  A_n(x,q)\end{align*}
for the last equality see \cite[Lemma 3.3]{Biagioli/03}.
The result follows, as
\[\sum_{\sigma\in D_n}{x^{\ddes(\sigma)}q^{\dmaj(\sigma)}}=\prod_{i=1}^{n-1}(1+x q^i)  A_n(x,q).\]\qedhere
\end{proof}
\begin{cor}Let $n \in \N$. Then
	\begin{equation}
	\sum_{r\geq 0} [r+1]^n _q x^r = \frac{\sum\limits_{\sigma \in D_n}{x^{\dstc(\sigma)}q^{\ell(\sigma)}}}{(1-x)(1-xq^n) \prod\limits_{i=1}^{n-1}{(1-x^2 q^{2i})}} \mbox{ in }\Z[q]\llbracket x\rrbracket.\end{equation}
\end{cor}
\subsection*{Acknowledgements}
 The author would like to thank Christopher Voll for his comments. 




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