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\begin{center}
\vskip 1cm{\LARGE\bf
Some Determinantal Representations of\\
\vskip .1in
Eulerian Polynomials and Their $q$-Analogues}
\vskip 1cm
\large
Chak-On Chow\\
28A, Block 2, Banyan Garden\\
Cheung Sha Wan\\
Hong Kong\\
\href{mailto: cchow@alum.mit.edu}{\tt cchow@alum.mit.edu}
\end{center}

\vskip .2 in


\begin{abstract}
The author recently $q$-extended the approach of Qi, Wang, and Guo 
to derangement numbers,
and successfully applied this extended approach to
obtain expressions for certain types $B$ and $D$ of 
derangement polynomials as determinants of order $n+1$.
In this paper,
we demonstrate that the $q$-extended approach is applicable to other generating functions, including the Eulerian polynomials $A_n(t),B_n(t)$, and $D_n(t)$ 
for types $A,B$, and $D$, as well as the types $A$ and $B$ ``inv'' analogues of $q$-Eulerian polynomials $A_n^{\inv}(t,q)$ and $B_n^{\inv}(t,q)$. 
We also present a new recurrence relation for $B_n^{\inv}(t,q)$.
\end{abstract}


\section{Introduction}\label{sec:01}

Certain mathematical quantities are expressible as determinants. See \cite{k99} for useful tools for evaluating determinants, examples of evaluated determinants in context, and further references. As far as the present work is concerned, a notable example is the $n$th classical derangement number $d_n$.
Qi, Wang, and Guo \cite{qwg16}, by exploiting Lemma~\ref{lem:05} below, were able to express $d_n$ as a tridiagonal determinant of order $n+1$.
The involved generating function of $d_n$ is of the exponential type. See 
Section~\ref{sec:02} for undefined terms.

An inspection of the approach of Qi et al.\ reveals that it is $q$-generalizable. The author \cite{c23c}, by coming up with a $q$-version of Lemma~\ref{lem:05}, obtained  expressions for types $B$ and $D$ of derangement polynomials $d_n^B(q):=\sum_{\sigma\in\mathcal{D}_n^B}q^{\fmaj(\sigma)}$ and $d_n^D(q):=\sum_{\sigma\in\mathcal{D}_n^D}q^{\maj(\sigma)}$, studied previously in \cite{c06,c23a}, as determinants of order $n+1$, whose generating functions are of the factorial type.

The above mentioned derangement polynomials enumerate signed and even-signed derangements according to a certain statistic.
By extrapolating this idea, we can, in general, consider generating functions that enumerate other combinatorial objects according to tuples of statistics.
For instance, one can consider the classical Eulerian polynomial $A_n(x)$, which is a univariate generating function of $n$-permutations by their descent numbers, or $q$-Eulerian polynomials that are bivariate generating function of permutations by their descent numbers and some other statistics. Examples of the latter include the types $A$ and $B$ ``inv'' analogues of $q$-Eulerian polynomials, namely, $A_n^{\inv}(t,q):=\sum_{\sigma\in\mathfrak{S}_n}t^{\des(\sigma)+1}q^{\inv(\sigma)}$ and $B_n^{\inv}(t,q):=\sum_{\sigma\in B_n}t^{\des_B(\sigma)}q^{\inv_B(\sigma)}$.

The successes in \cite{qwg16,c23b,c23c} suggest that the original and extended approach of Qi et al.\ readily apply to generating functions of the exponential or of factorial type and whose closed form expressions are quotients, yielding determinantal expressions of them.

The purpose of this work is to show that the $q$-extended approach of Qi et al.\ applies to finding determinantal expressions of $k$-variate generating functions, with $k=1,2$.

The organization of this work is as follows. In Section~\ref{sec:02}, we gather notation and undefined terms that are needed in the rest of this paper.
In Section~\ref{sec:03}, we compute determinantal expressions of Eulerian polynomials of types $A,B$, and $D$.
In Section~\ref{sec:04}, we compute determinantal expressions of the $q$-Eulerian polynomials $A_n^{\inv}(t,q)$ and $B_n^{\inv}(t,q)$.
In Section~\ref{sec:05}, we explore the relationship between $A_n^{\inv}(t,q)$ and $B_n^{\inv}(t,q)$ and obtain a recurrence relation for $B_n^{\inv}(t,q)$.
In Section~\ref{sec:06}, we end this work by raising two questions.


\section{Notation and preliminaries}\label{sec:02}

Let $\N$, $\Q$, and $\R$ denote the sets of non-negative integers, rational numbers, and real numbers, respectively.
Let $a,b\in\N$. Define the interval of integers $[a,b]$ by $\{a,a+1,\ldots,b\}$. In case $b&lt;a$, we have $[a,b]=\varnothing$, the empty set.
Let $\#S$ denote the cardinality of a finite set $S$.

Let $n\geq 1$. A square matrix $A=(a_{ij})_{1\leq i,j\leq n}$ of order $n$ is lower Hessenberg if all entries above the superdiagonal are zeros.

Let $\mathfrak{S}_n$ denote the symmetric group of degree $n$, consisting of all permutations of $1,2,\ldots,n$.
An element $\sigma$ of $\mathfrak{S}_n$ is represented by the associated word $\sigma_1\sigma_2\cdots\sigma_n$, where $\sigma_i=\sigma(i)$ for $i=1,2,\ldots,n$.
An $n$-permutation $\sigma=\sigma_1\sigma_2\cdots\sigma_n$ is said to be a {\it derangement} if $\sigma_i\ne i$ for all $i=1,2,\ldots,n$. Let $\mathcal{D}_n$ denote the set of $n$-derangements. Define the major index, the descent number, and the inversion number of $\sigma\in\mathfrak{S}_n$ by

\begin{center}
\resizebox{.50\linewidth}{!}{
\begin{minipage}{.55\linewidth}
\begin{align*}
\maj(\sigma)&amp;:=\sum_{i=1}^{n-1}i\chi(\sigma_i&gt;\sigma_{i+1}), \\
\des(\sigma)&amp;:=\sum_{i=1}^{n-1}\chi(\sigma_i&gt;\sigma_{i+1}), \\
\inv(\sigma)&amp;:=\sum_{i=1}^{n-1}\sum_{j=i+1}^n\chi(\sigma_i&gt;\sigma_j),
\end{align*}
\end{minipage}
}
\end{center}

\bigskip

\noindent respectively, where $\chi(P)=1$ or $0$ depending on whether the statement $P$ is true or not. This notion of major index applies to any permutations, signed or not.
The $n$th classical Eulerian polynomial is defined by $A_n(x):=\sum_{\sigma\in\mathfrak{S}_n}x^{\des(\sigma)+1}$. It is well known that the exponential generating function of $(A_n(x))_{n\geqslant 0}$ has the following expression:
\begin{equation}\label{eqn:100}
A(x,z):=\sum_{n\geq 0}A_n(x)\frac{z^n}{n!}=\frac{1-x}{1-xe^{z(1-x)}},
\end{equation}
where $A_0(x):=1$ by convention.

Let $B_n$ denote the $n$th hyperoctahedral group, consisting of all signed permutations of $1,2,\ldots,n$, represented by the associated words. The above  definition of derangements also applies to signed permutations. Let $\mathcal{D}_n^B$ denote the set of signed $n$-derangements. Three needed statistics on $B_n$ are the flag major index, the type $B$ descent number, and the type $B$ inversion number, defined respectively by
\begin{align*}
\fmaj(\sigma)&amp;:=2\maj(\sigma)+N(\sigma), \\
\des_B(\sigma)&amp;:=\sum_{i=0}^{n-1}\chi(\textrm{$\sigma_i&gt;\sigma_{i+1}$}),\\
\inv_B(\sigma)&amp;:=\inv(\sigma)+\sum_{i=1}^n|\sigma_i|\chi(\sigma_i&lt;0),
\end{align*}
where $N(\sigma):=\#\{i\colon\sigma_i&lt;0\}$ is the number of negative letters of $\sigma$, and $\sigma_0:=0$.
The above definition of flag major index is due to Adin-Brenti-Roichman \cite{abr01}, and those of $\des_B$ and $\inv_B$ are due to Brenti \cite{b94}, who had also shown that $\inv_B\equiv l_B$, where $l_B$ is the type $B$ length function.


The $n$th type $B$ Eulerian polynomial is defined by $B_n(x):=\sum_{\sigma\in B_n}x^{\des_B(\sigma)}$. It is known that the exponential generating function of $(B_n(x))_{n\geqslant 0}$ has the following expression:
\begin{equation}\label{eqn:200}
B(x,z):=\sum_{n\geq 0}B_n(x)\frac{z^n}{n!}=\frac{(1-x)e^{z(1-x)}}{1-xe^{2z(1-x)}},
\end{equation}
where $B_0(x):=1$ by convention.

Let $D_n$ denote the $n$th even-signed permutation group, consisting even-signed permutations of $1,2,\ldots,n$ of length $n$. An element $\sigma$ of $D_n$ is a signed $n$-permutation $\sigma_1\sigma_2\cdots\sigma_n$ with $N(\sigma)$ even. Being a subgroup of $B_n$, the definition of derangements applies to $D_n$. Let $\mathcal{D}_n^D$ denote the set of even-signed $n$-derangements. The type $D$ descent number of $\sigma\in D_n$ is defined by
$$
\des_D(\sigma):=\sum_{i=0}^{n-1}\chi(\sigma_i&gt;\sigma_{i+1}),
$$
where $\sigma_0:=-\sigma_2$. The $n$th type $D$ Eulerian polynomial is defined by $D_n(x):=\sum_{\sigma\in D_n}x^{\des_D(\sigma)}$. It is known that the exponential generating function of $(D_n(x))_{n\geqslant 0}$ has the following expression:
\begin{equation}\label{eqn:300}
D(x,z):=\sum_{n\geq 0}D_n(x)\frac{z^n}{n!}=\frac{(1-x)\bigl(e^{z(1-x)}-xze^{2z(1-x)}\bigr)}{1-xe^{2z(1-x)}},
\end{equation}
where $D_0(x):=1$ by convention.





Those two classes of derangement polynomials mentioned in the Introduction are
$$
d_n^B(q):=\sum_{\sigma\in\mathcal{D}_n^B}q^{\fmaj(\sigma)},\quad\textrm{and}\quad d_n^D(q):=\sum_{\sigma\in\mathcal{D}_n^D}q^{\maj(\sigma)},
$$
studied previously by the author \cite{c06,c23a}.

Let $x$ and $q$ be commuting indeterminates. Define
$$
(x;q)_n:=\begin{cases}
(1-x)(1-xq)\cdots(1-xq^{n-1})&amp; \textrm{if $n\geq 1$;}\\
1&amp; \textrm{otherwise.}\end{cases}
$$
$Q$-integers, $q$-factorials and $q$-binomial coefficients are defined by
\begin{equation*}
\begin{split}
[n]_q&amp;:=1+q+\cdots+q^{n-1},\\
[n]_q!&amp;:=[1]_q[2]_q\cdots[n]_q,\quad\textrm{and}\\
\left[\begin{matrix}
n\\
k\end{matrix}\right]_q&amp;:=\frac{[n]_q!}{[k]_q![n-k]_q!},\quad 0\leq k\leq n,
\end{split}
\end{equation*}
respectively.


The original approach of Qi et al.\ is based on Lemma~\ref{lem:05}, which is stated in \cite[p.~40]{b76} as an exercise.
A solution of this Bourbaki exercise can be found in \cite{c23b}.

\begin{lemma}\label{lem:05}
Let $u=u(x)$ and $v=v(x)$ be two real functions that are $n$ times differentiable on an interval $I\subset\R$. Then at every point where $v(x)\ne 0$, we have
$$
\frac{d^n}{dx^n}\biggl(\frac{u}{v}\biggr)=\frac{(-1)^n}{v^{n+1}}\left|\begin{matrix}
u&amp; v&amp; 0&amp; \cdots&amp; 0&amp; 0\\
u'&amp; v'&amp; v&amp; \cdots&amp; 0&amp; 0\\
u''&amp; v''&amp; 2v'&amp; \cdots&amp; 0&amp; 0\\
\vdots&amp; \vdots&amp; \vdots&amp; \ddots&amp; \vdots&amp; \vdots\\
u^{(n-1)}&amp; v^{(n-1)}&amp; \binom{n-1}{1}v^{(n-2)}&amp; \cdots&amp; \binom{n-1}{n-2}v'&amp; v\\
u^{(n)}&amp; v^{(n)}&amp; \binom{n}{1}v^{(n-1)}&amp; \cdots&amp; \binom{n}{n-2}v''&amp; \binom{n}{n-1}v'\end{matrix}\right|.
$$
\end{lemma}


Let $\{f_n\}$ be a sequence of numbers or $k$-variate formal power series independent of $x$, where $k\geq 1$.
If the exponential generating function of $\{f_n\}$,
$$
f(x)=\sum_{n\geq 0}f_n\frac{x^n}{n!},
$$
can be written as a quotient $u(x)/v(x)$, then with Lemma~\ref{lem:05}, the approach of Qi et al.\ allows us to express
$$
f_n=\left.\frac{\partial^n}{\partial x^n}f(x)\right|_{x=0}
$$
as a determinant of order $n+1$.

The author \cite{c23c} $q$-extended Lemma~\ref{lem:05} to become Lemma~\ref{lem:10}, where for formal power series $f(t;q)\in\Q[[t,q]]$ in $t$ and $q$ with rational coefficients, its $q$-derivative with respect to $t$ is defined as
\begin{equation}\label{eqn:500}
D_qf(t;q):=\frac{f(tq;q)-f(t;q)}{(q-1)t}.
\end{equation}
It is easy to see that $D_qt^n=[n]_qt^{n-1}$. See \cite[\S2.2]{a07} for details of $q$-calculus.


\begin{lemma}\label{lem:10}
Let $n\geq 1$ and let $u=u(t;q),v=v(t;q)\in\Q[[t,q]]$ be such that their $n$th $q$-derivatives $D_q^nu(t;q)$ and $D_q^nv(t;q)$ exist. Then at every point where $v(t;q)\ne 0$, we have
$$
D_q^n\biggl(\frac{u}{v}\biggr)(t;q)=\frac{(-1)^n}{v^{n+1}}\left|\begin{matrix}
u&amp; v&amp; 0&amp; \cdots&amp; 0&amp; 0\\
D_qu&amp; D_qv&amp; v&amp; \cdots&amp; 0&amp; 0\\
D_q^2u&amp; D_q^2v&amp; \bigl[\substack{2\\
\noalign{\vskip2pt}
1}\bigr]_qD_qv&amp; \cdots&amp; 0&amp; 0\\
\vdots&amp; \vdots&amp; \vdots&amp; \ddots&amp; \vdots&amp; \vdots\\
D_q^{n-1}u&amp; D_q^{n-1}v&amp; \bigl[\substack{n-1\\
\noalign{\vskip2pt}
1}\bigr]_qD_q^{n-2}v&amp; \cdots&amp; \bigl[\substack{n-1\\
\noalign{\vskip2pt}
n-2}\bigr]_qD_qv&amp; v\\
D_q^nu&amp; D_q^nv&amp; \bigl[\substack{n\\
\noalign{\vskip2pt}
1}\bigr]_qD_q^{n-1}v&amp; \cdots&amp; \bigl[\substack{n\\
\noalign{\vskip2pt}
n-2}\bigr]_qD_q^2v&amp; \bigl[\substack{n\\
\noalign{\vskip2pt}
n-1}\bigr]_qD_qv\\
\end{matrix}\right|,
$$
where all the formal power series on the right side are evaluated at $(tq^{-n};q)$.
\end{lemma}

To simplify the expressions, we write in the sequel $D_q^nu(tq^{-n};q)$ to mean the result of substituting the arguments of the $n$th $q$-derivative of $u(t;q)$ by $(tq^{-n};q)$, etc.

Let $(f_n)_{n\geqslant 0}$ be a sequence of formal power series dependent on $q$ and on some other indeterminates but independent of $x$.
If the factorial generating function of $(f_n)_{n\geqslant 0}$,
$$
f(x)=\sum_{n\geq 0}f_n\frac{x^n}{[n]_q!},
$$
can be written as a quotient $u(x)/v(x)$, then with Lemma~\ref{lem:10}, the $q$-extended approach of Qi et al.\ allows us to express
$$
f_n=D_q^nf(x)\bigr|_{x=0}
$$
as a determinant of order $n+1$, where $D_q^nf(x)$ on the right denotes the $n$th $q$-derivative of $f(x)$ with respect to $x$.



\section{Determinantal expressions of Eulerian polynomials}\label{sec:03}

The goal of this section is to present determinantal expressions of the Eulerian polynomials $A_n(x),B_n(x)$, and $D_n(x)$.


\begin{theorem}
Let $n\geq 1$. The Eulerian polynomial $A_n(x)$ is expressible as the following lower Hessenberg determinant of order $n+1$:
{\footnotesize
$$
A_n(x)=(-1)^n\left|\begin{matrix}
1&amp; 1&amp; 0&amp; \cdots&amp; 0&amp; 0\\
0&amp; -x&amp; 1&amp; \cdots&amp; 0&amp; 0\\
0&amp; -x(1-x)&amp; -2x&amp; \cdots&amp; 0&amp; 0\\
\vdots&amp; \vdots&amp; \vdots&amp; \ddots&amp; \vdots&amp; \vdots\\
0&amp; -x(1-x)^{n-2}&amp; -\binom{n-1}{1}x(1-x)^{n-3}&amp; \cdots&amp; -\binom{n-1}{n-2}x&amp; 1\\
0&amp; -x(1-x)^{n-1}&amp; -\binom{n}{1}x(1-x)^{n-2}&amp; \cdots&amp; -\binom{n}{n-2}x(1-x)&amp; -\binom{n}{n-1}x
\end{matrix}\right|.
$$}
\end{theorem}
\begin{proof}
By virtue of the exponential generating function \eqref{eqn:100} of $(A_n(x))_{n\geqslant 0}$, regarding $x$ as a parameter, we let $u(z;x)=1-x$ and $v(z;x)=1-xe^{z(1-x)}$. Then for $k\geq 1$, $\partial_z^ku(z;x)=0$ and $\partial_z^kv(z;x)=-x(1-x)^ke^{z(1-x)}$. Applying Lemma~\ref{lem:05}, and letting $z\to 0$, we have
{\footnotesize
\begin{equation*}
\begin{split}
A_n(x)&amp;=\left.\frac{\partial^n}{\partial z^n}A(x,z)\right|_{z=0}\\
&amp;=\frac{(-1)^n}{(1-x)^{n+1}}\left|\begin{matrix}
1-x&amp; 1-x&amp; 0&amp; \cdots&amp; 0&amp; 0\\
0&amp; -x(1-x)&amp; 1-x&amp; \cdots&amp; 0&amp; 0\\
0&amp; -x(1-x)^2&amp; -2x(1-x)&amp; \cdots&amp; 0&amp; 0\\
\vdots&amp; \vdots&amp; \vdots&amp; \ddots&amp; \vdots&amp; \vdots\\
0&amp; -x(1-x)^{n-1}&amp; -\binom{n-1}{1}x(1-x)^{n-2}&amp; \cdots&amp; -\binom{n-1}{n-2}x(1-x)&amp; 1-x\\
0&amp; -x(1-x)^n&amp; -\binom{n}{1}x(1-x)^{n-1}&amp; \cdots&amp; -\binom{n}{n-2}x(1-x)^2&amp; -\binom{n}{n-1}x(1-x)\\
\end{matrix}\right|\\[.1in]
&amp;=(-1)^n\left|\begin{matrix}
1&amp; 1&amp; 0&amp; \cdots&amp; 0&amp; 0\\
0&amp; -x&amp; 1&amp; \cdots&amp; 0&amp; 0\\
0&amp; -x(1-x)&amp; -2x&amp; \cdots&amp; 0&amp; 0\\
\vdots&amp; \vdots&amp; \vdots&amp; \ddots&amp; \vdots&amp; \vdots\\
0&amp; -x(1-x)^{n-2}&amp; -\binom{n-1}{1}x(1-x)^{n-3}&amp; \cdots&amp; -\binom{n-1}{n-2}x&amp; 1\\
0&amp; -x(1-x)^{n-1}&amp; -\binom{n}{1}x(1-x)^{n-2}&amp; \cdots&amp; -\binom{n}{n-2}x(1-x)&amp; -\binom{n}{n-1}x
\end{matrix}\right|,
\end{split}
\end{equation*}}
where the last equality follows from dividing each column by $1-x$.
\end{proof}



\begin{theorem}\label{thm:04}
Let $n\geq 1$. The type $B$ Eulerian polynomial $B_n(x)$ is expressible as the following lower Hessenberg determinant of order $n+1$:
{\scriptsize
$$
B_n(x)=(-1)^n\left|\begin{matrix}
1&amp; 1&amp; 0&amp; \cdots&amp; 0&amp; 0\\
1-x&amp; -2x&amp; 1&amp; \cdots&amp; 0&amp; 0\\
(1-x)^2&amp; -2^2x(1-x)&amp; -2^2x&amp; \cdots&amp; 0&amp; 0\\
\vdots&amp; \vdots&amp; \vdots&amp; \ddots&amp; \vdots&amp; \vdots\\
(1-x)^{n-1}&amp; -2^{n-2}x(1-x)^{n-2}&amp; -\binom{n-1}{1}2^{n-3}x(1-x)^{n-3}&amp; \cdots&amp; -\binom{n-1}{n-2}2x&amp; 1\\
(1-x)^n&amp; -2^nx(1-x)^{n-1}&amp; -\binom{n}{1}2^{n-2}x(1-x)^{n-2}&amp; \cdots&amp; -\binom{n}{n-2}2^2x(1-x)&amp; -\binom{n}{n-1}2x
\end{matrix}\right|.
$$}
\end{theorem}
\begin{proof}
By virtue of \eqref{eqn:200}, let $u(z;x)=(1-x)e^{z(1-x)}$ and $v(z;x)=1-xe^{2z(1-x)}$. Then for $k\geq 1$, $\partial_z^ku(z;x)=(1-x)^{k+1}e^{z(1-x)}$ and $\partial_z^kv(z;x)=-2^kx(1-x)^ke^{z(1-x)}$. Applying Lemma~\ref{lem:05}, letting $z\to 0$, and followed by dividing each column by $1-x$, we obtain
{\scriptsize
\begin{equation*}
\begin{split}
B_n(x)&amp;=\left.\frac{\partial^n}{\partial z^n}B(x,z)\right|_{z=0}\\
&amp;=(-1)^n\left|\begin{matrix}
1&amp; 1&amp; 0&amp; \cdots&amp; 0&amp; 0\\
1-x&amp; -2x&amp; 1&amp; \cdots&amp; 0&amp; 0\\
(1-x)^2&amp; -2^2x(1-x)&amp; -2^2x&amp; \cdots&amp; 0&amp; 0\\
\vdots&amp; \vdots&amp; \vdots&amp; \ddots&amp; \vdots&amp; \vdots\\
(1-x)^{n-1}&amp; -2^{n-2}x(1-x)^{n-2}&amp; -\binom{n-1}{1}2^{n-3}x(1-x)^{n-3}&amp; \cdots&amp; -\binom{n-1}{n-2}2x&amp; 1\\
(1-x)^n&amp; -2^nx(1-x)^{n-1}&amp; -\binom{n}{1}2^{n-2}x(1-x)^{n-2}&amp; \cdots&amp; -\binom{n}{n-2}2^2x(1-x)&amp; -\binom{n}{n-1}2x
\end{matrix}\right|.
\end{split}
\end{equation*}}
\end{proof}


\begin{theorem}
Let $n\geq 1$. The type $D$ Eulerian polynomial $D_n(x)$ is expressible as the following lower Hessenberg determinant of order $n+1$:
{\tiny
\begin{equation*}
\begin{split}
&amp;D_n(x)\\
&amp;=(-1)^n\left|\begin{matrix}
1&amp; 1&amp; 0&amp; \cdots&amp; 0&amp; 0\\
(1-2x)&amp; -2x&amp; 1&amp; \cdots&amp; 0&amp; 0\\
(1-x)(1-5x)&amp; -2^2x(1-x)&amp; -2^2x&amp; \cdots&amp; 0&amp; 0\\
\vdots&amp; \vdots&amp; \vdots&amp; \ddots&amp; \vdots&amp; \vdots\\
(1-x)^{n-2}(1-((n-1)2^{n-2}+1)x)&amp; -2^{n-1}x(1-x)^{n-2}&amp; -\binom{n-1}{1}2^{n-2}x(1-x)^{n-3}&amp; \cdots&amp; -\binom{n-1}{n-2}2x&amp; 1\\
(1-x)^{n-1}(1-(n2^{n-1}+1)x)&amp; -2^nx(1-x)^{n-1}&amp; -\binom{n}{1}2^{n-1}x(1-x)^{n-2}&amp; \cdots&amp; -\binom{n}{n-2}2^2x(1-x)&amp; -\binom{n}{n-1}2x
\end{matrix}\right|.
\end{split}
\end{equation*}}
\end{theorem}
\begin{proof}
By virtue of \eqref{eqn:300}, let $u(z;x)=(1-x)\bigl(e^{z(1-x)}-xze^{2z(1-x)}\bigr)$ and $v(z;x)=1-xe^{2z(1-x)}$. For $k\geq 1$,
\begin{equation*}
\begin{split}
\partial_z^ku(z;x)&amp;=(1-x)\bigl((1-x)^ke^{z(1-x)}-x(2^k(1-x)^kz+
\bigl(\substack{k\\
\noalign{\vskip2pt}
1}\bigr)2^{k-1}(1-x)^{k-1})e^{2z(1-x)}\bigr),\\
\partial_z^kv(z;x)&amp;=-2^kx(1-x)^ke^{z(1-x)}
\end{split}
\end{equation*}
so that
$$
\partial_z^ku(z;x)\bigr|_{z=0}=(1-x)^k\bigl(1-(k2^{k-1}+1)x)\bigr),\quad\textrm{and}\quad\partial_z^kv(z;x)\bigr|_{z=0}=-2^kx(1-x)^k.
$$
Applying Lemma~\ref{lem:05} with $z\to 0$, we have
{\tiny
\begin{equation*}
\begin{split}
&amp;D_n(x)=\left.\frac{\partial^n}{\partial z^n}D(x,z)\right|_{z=0}\\
&amp;=(-1)^n\left|\begin{matrix}
1&amp; 1&amp; 0&amp; \cdots&amp; 0&amp; 0\\
(1-2x)&amp; -2x&amp; 1&amp; \cdots&amp; 0&amp; 0\\
(1-x)(1-5x)&amp; -2^2x(1-x)&amp; -2^2x&amp; \cdots&amp; 0&amp; 0\\
\vdots&amp; \vdots&amp; \vdots&amp; \ddots&amp; \vdots&amp; \vdots\\
(1-x)^{n-2}(1-((n-1)2^{n-2}+1)x)&amp; -2^{n-1}x(1-x)^{n-2}&amp; -\binom{n-1}{1}2^{n-2}x(1-x)^{n-3}&amp; \cdots&amp; -\binom{n-1}{n-2}2x&amp; 1\\
(1-x)^{n-1}(1-(n2^{n-1}+1)x)&amp; -2^nx(1-x)^{n-1}&amp; -\binom{n}{1}2^{n-1}x(1-x)^{n-2}&amp; \cdots&amp; -\binom{n}{n-2}2^2x(1-x)&amp; -\binom{n}{n-1}2x
\end{matrix}\right|.
\end{split}
\end{equation*}}
\end{proof}



\section{Determinantal expressions of $q$-Eulerian polynomials}\label{sec:04}

Bivariate extensions of $A_n(x)$ and $B_n(x)$ are available in the literature, namely, $A_n^{\inv}(t,q):=\sum_{\sigma\in\mathfrak{S}_n}t^{\des(\sigma)+1}q^{\inv(\sigma)}$, whose factorial generating function was computed by Stanley \cite{s76}:
\begin{equation}\label{eqn:400}
A^{\inv}(x;t,q):=\sum_{n\geq 0}A_n^{\inv}(t,q)\frac{x^n}{[n]_q!}=\frac{1-t}{1-te(x(1-t);q)},
\end{equation}
where $e(x;q):=\sum_{n\geq 0}\frac{x^n}{[n]_q!}$ is a $q$-exponential, and $B_n^{\inv}(t,q):=\sum_{\sigma\in B_n}t^{\des_B(\sigma)}q^{\inv_B(\sigma)}$.

D\'esarm\'enien and Foata \cite{df92} showed that the following ``semi'' $q$-recurrence relation holds:
\begin{equation}\label{eqn:1100}
A_n^{\inv}(t,q)=\sum_{k=0}^{n-1}\left[\begin{matrix}
n\\
k\end{matrix}\right]_qA_k^{\inv}(t,q)t(1-t)^{n-1-k},
\end{equation}
``semi'' in the sense that the summands on the right involve two factors one of which depends on $q$ and the other does not.
The author \cite{c10} showed that $A_n^{\inv}(t,q)$ satisfies the following ``fully'' $q$-recurrence relation:
$$
A_{n+1}^{\inv}(t,q)=(1+tq^n)A_n^{\inv}(t,q)+\sum_{k=1}^{n-1}\left[\begin{matrix}
n\\
k\end{matrix}\right]_qq^kA_{n-k}^{\inv}(t,q)A_k^{\inv}(t,q). 
$$

As far as coefficient extraction is concerned, $A^{\inv}(x;t,q)$ enjoys the following property:
$$
\left.D_q^nA^{\inv}(x;t,q)\right|_{x=0}=A_n^{\inv}(t,q),
$$
where the $q$-derivative is taken with respect to $x$. The key to the applicability of the approach of Qi et al., the original or the $q$-extended one, is that the exponential or factorial generating function be expressible as a quotient. That of $(A_n^{\inv}(t,q))_{n\geqslant 0}$, i.e., \eqref{eqn:400}, is definitely a quotient. Applying the $q$-extended approach of Qi et al.\ readily yields Theorem~\ref{thm:06}.

\begin{theorem}\label{thm:06}
Let $n\geq 1$. The $q$-Eulerian polynomial $A_n^{\inv}(t,q)$ is expressible as the following lower Hessenberg determinant of order $n+1$:
{\small
$$
A^{\inv}(t,q)=(-1)^n\left|\begin{matrix}
1&amp; 1&amp; 0&amp; \cdots&amp; 0&amp; 0\\
0&amp; -t&amp; 1&amp; \cdots&amp; 0&amp; 0\\
0&amp; -t(1-t)&amp; -\bigl[\substack{2\\
\noalign{\vskip2pt}
1}\bigr]_qt&amp; \cdots&amp; 0&amp; 0\\
\vdots&amp; \vdots&amp; \vdots&amp; \ddots&amp; \vdots&amp; \vdots\\
0&amp; -t(1-t)^{n-2}&amp; -\bigl[\substack{n-1\\
\noalign{\vskip2pt}
1}\bigr]_qt(1-t)^{n-3}&amp; \cdots&amp; -\bigl[\substack{n-1\\
\noalign{\vskip2pt}
n-2}\bigr]_qt&amp; 1\\
0&amp; -t(1-t)^{n-1}&amp; \bigl[\substack{n\\
\noalign{\vskip2pt}
1}\bigr]_qt(1-t)^{n-2}&amp; \cdots&amp; -\bigl[\substack{n\\
\noalign{\vskip2pt}
n-2}\bigr]_qt(1-t)&amp; -\bigl[\substack{n\\
\noalign{\vskip2pt}
n-1}\bigr]_qt\\
\end{matrix}\right|.
$$}
\end{theorem}
\begin{proof}
Let $u(x;t,q):=1-t$ and $v(x;t,q):=1-te(x(1-t);q)$. It is clear that $u(0;t,q)=v(0;t,q)=1-t$. For $k\geq 1$, the $k$th $q$-derivatives of $u(x;t,q)$ and $v(x;t,q)$ with respect to $x$ are, respectively, $D_q^ku(x;t,q)=0$ and
$$
D_q^kv(x;t,q)=-t\sum_{n\geq k}\frac{(1-t)^nx^{n-k}}{[n-k]_q!}=-t(1-t)^k\sum_{n\geq 0}\frac{[x(1-t)]^n}{[n]_q!}.
$$
Evaluating them at $(xq^{-n};t,q)$ followed by letting $x\to 0$, we have $D_q^ku(xq^{-n};t,q)\bigr|_{x=0}=0$ and $D_q^kv(xq^{-n};t,q)\bigr|_{x=0}=-t(1-t)^k$. Applying Lemma~\ref{lem:10} with $x\to 0$, we obtain
{\small
\begin{equation*}
\begin{split}
&amp;A^{\inv}(t,q)\\
&amp;=\frac{(-1)^n}{(1-t)^{n+1}}\left|\begin{matrix}
1-t&amp; 1-t&amp; 0&amp; \cdots&amp; 0&amp; 0\\
0&amp; -t(1-t)&amp; 1-t&amp; \cdots&amp; 0&amp; 0\\
0&amp; -t(1-t)^2&amp; -\bigl[\substack{2\\
\noalign{\vskip2pt}
1}\bigr]_qt(1-t)&amp; \cdots&amp; 0&amp; 0\\
\vdots&amp; \vdots&amp; \vdots&amp; \ddots&amp; \vdots&amp; \vdots\\
0&amp; -t(1-t)^{n-1}&amp; -\bigl[\substack{n-1\\
\noalign{\vskip2pt}
1}\bigr]_qt(1-t)^{n-2}&amp; \cdots&amp; -\bigl[\substack{n-1\\
\noalign{\vskip2pt}
n-2}\bigr]_qt(1-t)&amp; 1-t\\
0&amp; -t(1-t)^n&amp; \bigl[\substack{n\\
\noalign{\vskip2pt}
1}\bigr]_qt(1-t)^{n-1}&amp; \cdots&amp; -\bigl[\substack{n\\
\noalign{\vskip2pt}
n-2}\bigr]_qt(1-t)^2&amp; -\bigl[\substack{n\\
\noalign{\vskip2pt}
n-1}\bigr]_qt(1-t)\\
\end{matrix}\right|\\[.1in]
&amp;=(-1)^n\left|\begin{matrix}
1&amp; 1&amp; 0&amp; \cdots&amp; 0&amp; 0\\
0&amp; -t&amp; 1&amp; \cdots&amp; 0&amp; 0\\
0&amp; -t(1-t)&amp; -\bigl[\substack{2\\
\noalign{\vskip2pt}
1}\bigr]_qt&amp; \cdots&amp; 0&amp; 0\\
\vdots&amp; \vdots&amp; \vdots&amp; \ddots&amp; \vdots&amp; \vdots\\
0&amp; -t(1-t)^{n-2}&amp; -\bigl[\substack{n-1\\
\noalign{\vskip2pt}
1}\bigr]_qt(1-t)^{n-3}&amp; \cdots&amp; -\bigl[\substack{n-1\\
\noalign{\vskip2pt}
n-2}\bigr]_qt&amp; 1\\
0&amp; -t(1-t)^{n-1}&amp; \bigl[\substack{n\\
\noalign{\vskip2pt}
1}\bigr]_qt(1-t)^{n-2}&amp; \cdots&amp; -\bigl[\substack{n\\
\noalign{\vskip2pt}
n-2}\bigr]_qt(1-t)&amp; -\bigl[\substack{n\\
\noalign{\vskip2pt}
n-1}\bigr]_qt\\
\end{matrix}\right|,
\end{split}
\end{equation*}}
where the second equality follows from dividing each column by $1-t$.
\end{proof}


Let us now look at $B_n^{\inv}(t,q)$, whose first three members are as follows:
\begin{equation*}
\begin{split}
B_1^{\inv}(t,q)&amp;=1+qt,\\
B_2^{\inv}(t,q)&amp;=1+(2q+2q^2+2q^3)t+q^4t^2,\\
B_3^{\inv}(t,q)&amp;=1+(3q+4q^2+5q^3+4q^4+4q^5+2q^6+q^7)t\\
&amp;\quad+(q^2+2q^3+4q^4+4q^5+5q^6+4q^7+3q^8)t^2+q^9t^3,\\
\end{split}
\end{equation*}

Reiner \cite[Corollary 4.7]{r93} showed, in our terminology, that
$$
\sum_{n\geq 0}\frac{x^n\sum_{\pi\in B_n}q^{\inv_B(\pi)}t^{n-\des_B(\pi)}}{[2]_q[4]_q\cdots[2n]_q}=\biggl(1-\sum_{n\geq 1}\frac{(t-1)^{n-1}x^n}{[n]_q!}\biggr)^{-1}\sum_{n\geq 0}\frac{(t-1)^nx^n}{[2]_q[4]_q\cdots[2n]_q}.
$$
Note that $[2k]_q=(1+q^k)[k]_q=(1+q)[k]_{q^2}$ and that $(1+q)(1+q^2)\cdots(1+q^n)=(-q;q)_n$. The replacements $x\mapsto xt$ and $t\mapsto t^{-1}$ in the preceding factorial generating function identity then yield
\begin{align}
B^{\inv}(x;t,q)&amp;:=\sum_{n\geq 0}B_n^{\inv}(t,q)\frac{x^n}{(-q;q)_n[n]_q!}\nonumber\\
&amp;=\biggl(1-\frac{t}{1-t}\sum_{n\geq 1}\frac{(1-t)^nx^n}{[n]_q!}\biggr)^{-1}\sum_{n\geq 0}\frac{(1-t)^nx^n}{(-q;q)_n[n]_q!},\label{eqn:700}
\end{align}
which enjoys the following property:
$$
\left.D_q^nB^{\inv}(x;t,q)\right|_{x=0}=\frac{B_n^{\inv}(t,q)}{(-q;q)_n}.
$$
The $q$-extended approach of Qi et al.\ readily applies to $B^{\inv}(x;t,q)$, yielding Theorem~\ref{thm:07}.


\begin{theorem}\label{thm:07}
Let $n\geq 1$. The type $B$ $q$-Eulerian polynomial $B_n^{\inv}(t,q)$ can be expressed as the following lower Hessenberg determinant of order $n+1$:
{\small
\begin{equation*}
\begin{split}
&amp;B_n^{\inv}(t,q)\\
&amp;=(-1)^n(-q;q)_n\left|\begin{matrix}
1&amp; 1&amp; 0&amp; \cdots&amp; 0&amp; 0\\
\frac{(1-t)}{(-q;q)_1}&amp; -t&amp; 1&amp; \cdots&amp; 0&amp; 0\\
\noalign{\vskip2pt}
\frac{(1-t)^2}{(-q;q)_2}&amp; -t(1-t)&amp; -\bigl[\substack{2\\
\noalign{\vskip2pt}
1}\bigr]_qt&amp; \cdots&amp; 0&amp; 0\\
\noalign{\vskip2pt}
\vdots&amp; \vdots&amp; \vdots&amp; \ddots&amp; \vdots&amp; \vdots\\
\noalign{\vskip2pt}
\frac{(1-t)^{n-1}}{(-q;q)_{n-1}}&amp; -t(1-t)^{n-2}&amp; -\binom{n-1}{1}t(1-t)^{n-3}&amp; \cdots&amp; -\bigl[\substack{n-1\\
\noalign{\vskip2pt}
n-2}\bigr]_qt&amp; 1\\  
\noalign{\vskip2pt}
\frac{(1-t)^n}{(-q;q)_n}&amp; -t(1-t)^{n-1}&amp; -\bigl[\substack{n\\
\noalign{\vskip2pt}
1}\bigr]_q t(1-t)^{n-2}&amp; \cdots&amp; -\bigl[\substack{n\\
\noalign{\vskip2pt}
n-2}\bigr]_qt(1-t)&amp; -\bigl[\substack{n\\
\noalign{\vskip2pt}
n-1}\bigr]_qt\end{matrix}\right|.
\end{split}
\end{equation*}}
\end{theorem}
\begin{proof}
Let
$$
u(x;t,q):=\sum_{n\geq 0}\frac{(1-t)^nx^n}{(-q;q)_n[n]_q!}\quad\textrm{and}\quad v(x;t,q):=1-\frac{t}{1-t}\sum_{n\geq 1}\frac{(1-t)^nx^n}{[n]_q!}.
$$
It is easy to see that $u(0;t,q)=1$ and $v(0;t,q)=1$. For $1\leq k\leq n$, the $k$th $q$-derivatives of $u(x;t,q)$ and $v(x;t,q)$ with respect to $x$ are, respectively,
\begin{equation*}
\begin{split}
D_q^ku(x;t,q)&amp;=\sum_{n\geq k}\frac{(1-t)^nx^{n-k}}{(-q;q)_n[n-k]_q!}=(1-t)^k\sum_{n\geq 0}\frac{(1-t)^nx^n}{(-q;q)_{n+k}[n]_q!},\\
D_q^kv(x;t,q)&amp;=-\frac{t}{1-t}\sum_{n\geq k}\frac{(1-t)^nx^{n-k}}{[n-k]_q!}=-t(1-t)^{k-1}\sum_{n\geq 0}\frac{(1-t)^nx^n}{[n]_q!}.
\end{split}
\end{equation*}
It follows that
\begin{equation*}
\begin{split}
D_q^ku(xq^{-n};t,q)\bigr|_{x=0}&amp;=\lim_{x\to 0}(1-t)^k\sum_{n\geq 0}\frac{(1-t)^n(xq^{-n})^n}{(-q;q)_{n+k}[n]_q!}=\frac{(1-t)^k}{(-q;q)_k},\\
D_q^kv(xq^{-n};t,q)\bigr|_{x=0}&amp;=-\lim_{x\to 0}t(1-t)^{k-1}\sum_{n\geq 0}\frac{(1-t)^n(xq^{-n})^n}{[n]_q!}=-t(1-t)^{k-1}.
\end{split}
\end{equation*}
Since letting $x\to 0$ after evaluating at $(xq^{-n};t,q)$ makes no difference to letting $x\to 0$ directly, it now follows from Lemma~\ref{lem:10} that
{\small
\begin{equation*}
\begin{split}
&amp;B_n^{\inv}(t,q)=(-q;q)_nD_q^nB^{\inv}(xq^{-n};t,q)\bigr|_{x=0}\\
&amp;=(-1)^n(-q;q)_n\left|\begin{matrix}
1&amp; 1&amp; 0&amp; \cdots&amp; 0&amp; 0\\
\frac{(1-t)}{(-q;q)_1}&amp; -t&amp; 1&amp; \cdots&amp; 0&amp; 0\\
\noalign{\vskip2pt}
\frac{(1-t)^2}{(-q;q)_2}&amp; -t(1-t)&amp; -\bigl[\substack{2\\
\noalign{\vskip2pt}
1}\bigr]_qt&amp; \cdots&amp; 0&amp; 0\\
\noalign{\vskip2pt}
\vdots&amp; \vdots&amp; \vdots&amp; \ddots&amp; \vdots&amp; \vdots\\
\noalign{\vskip2pt}
\frac{(1-t)^{n-1}}{(-q;q)_{n-1}}&amp; -t(1-t)^{n-2}&amp; -\binom{n-1}{1}t(1-t)^{n-3}&amp; \cdots&amp; -\bigl[\substack{n-1\\
\noalign{\vskip2pt}
n-2}\bigr]_qt&amp; 1\\  
\noalign{\vskip2pt}
\frac{(1-t)^n}{(-q;q)_n}&amp; -t(1-t)^{n-1}&amp; -\bigl[\substack{n\\
\noalign{\vskip2pt}
1}\bigr]_q t(1-t)^{n-2}&amp; \cdots&amp; -\bigl[\substack{n\\
\noalign{\vskip2pt}
n-2}\bigr]_qt(1-t)&amp; -\bigl[\substack{n\\
\noalign{\vskip2pt}
n-1}\bigr]_qt\end{matrix}\right|.
\end{split}
\end{equation*}}
\end{proof}


It is clear that $B_n^{\inv}(t,1)=B_n(t)$. On the other hand, one can move the multiplicative factor $(-q;q)_n$ of the determinantal expression in Theorem~\ref{thm:07} into the determinant by multiplying each entry of the first column by $(-q;q)_n$, the resulting quantities $(-q;q)_n/(-q;q)_k$ are $q$-analogues of powers of $2$. In Theorem~\ref{thm:04}, powers of $2$ are are distributed over the last $n$ columns of the determinant. When $q\to 1$, the determinantal expression in Theorem~\ref{thm:07} does not reduce to become that in Theorem~\ref{thm:04}.


\section{A recurrence relation for $B_n^{\inv}(t,q)$}\label{sec:05}

Unlike its type $A$ counterpart,
a recurrence relation for $B_n^{\inv}(t,q)$ was, up to now,
apparently not in the literature. The goal of this final section is to fill in this gap. Before doing so, it is important to note that $A_n^{\inv}(t,q)$ and $B_n^{\inv}(t,q)$ are closely related with each other.


\begin{proposition}
For $n\geq 1$, we have
\begin{align}
B_n^{\inv}(t,q)&amp;=\sum_{k=0}^n\left[\begin{matrix}
n\\
k\end{matrix}\right]_q\frac{(-q;q)_n(1-t)^{n-k}A_k^{\inv}(t,q)}{(-q;q)_{n-k}},\label{eqn:900}\\
&amp;=\sum_{k=0}^n\left[\begin{matrix}
n\\
k\end{matrix}\right]_{q^2}(-q;q)_k(1-t)^{n-k}A_k^{\inv}(t,q).\label{eqn:1500}
\end{align}
\end{proposition}
\begin{proof}
An inspection of the denominator of $B^{\inv}(x;t,q)$ reveals that
$$
1-\frac{t}{1-t}\sum_{n\geq 1}\frac{(1-t)^nx^n}{[n]_q!}=\frac{1-te(x(1-t);q)}{1-t}
$$
so that
\begin{equation}\label{eqn:1000}
B^{\inv}(x;t,q)=A^{\inv}(x;t,q)\sum_{n\geq 0}\frac{(1-t)^nx^n}{(-q;q)_n[n]_q!}.
\end{equation}
Extracting the coefficients of $x^n$ on both sides, we have
$$
\frac{B_n^{\inv}(t,q)}{(-q;q)_n[n]_q!}=\sum_{k=0}^n\frac{A_k^{\inv}(t,q)(1-t)^{n-k}}{[k]_q!(-q;q)_{n-k}[n-k]_q!},
$$
whence \eqref{eqn:900}. Since $[2k]_q!=(1+q)^k[k]_{q^2}!=(-q;q)_k[k]_q!$, we have
\begin{equation}\label{eqn:1400}
\left[\begin{matrix}
n\\
k\end{matrix}\right]_q=\frac{(-q;q)_k(-q;q)_{n-k}}{(-q;q)_n}\left[\begin{matrix}
n\\
k\end{matrix}\right]_{q^2}.
\end{equation}
Substituting \eqref{eqn:1400} into \eqref{eqn:900}, equality \eqref{eqn:1500} follows.
\end{proof}


Based on Stanley's factorial generating function \eqref{eqn:400}, Bj\"orner and Brenti \cite[\S7.2]{bb05} generalized the factorial generating functions by descent numbers and lengths to other Coxeter families of which \eqref{eqn:1000} in the preceding proof is a concrete type $B$ realization. Compare \eqref{eqn:1000} with \cite[Exercise 7.2.5(b)]{bb05}.


\begin{theorem}
For $n\geq 1$, $B_n^{\inv}(t,q)$ satisfies the following recurrence relation:
$$
B_{n+1}^{\inv}(t,q)=\sum_{k=0}^n\left[\begin{matrix}
n+1\\
k\end{matrix}\right]_{q^2}(-q;q)_k[(-q;q)_{n+1-k}t+(1-t)](1-t)^{n-k}A_k^{\inv}(t,q).
$$
\end{theorem}
\begin{proof}
Starting from \eqref{eqn:1500} with $n+1$ in place of $n$, followed by using \eqref{eqn:1100}, we obtain
\begin{equation*}
\begin{split}
B_{n+1}^{\inv}(t,q)&amp;=(-q;q)_{n+1}A_{n+1}^{\inv}(t,q)+\sum_{k=0}^n\left[\begin{matrix}
n+1\\
k\end{matrix}\right]_{q^2}(-q;q)_k(1-t)^{n+1-k}A_k^{\inv}(t,q)\\
&amp;=(-q;q)_{n+1}\sum_{k=0}^n\left[\begin{matrix}
n+1\\
k\end{matrix}\right]_qt(1-t)^{n-k}A_k^{\inv}(t,q)\\
&amp;\quad+\sum_{k=0}^n\left[\begin{matrix}
n+1\\
k\end{matrix}\right]_{q^2}(-q;q)_k(1-t)^{n+1-k}A_k^{\inv}(t,q)\\
&amp;=(-q;q)_{n+1}\sum_{k=0}^n\left[\begin{matrix}
n+1\\
k\end{matrix}\right]_{q^2}\frac{(-q;q)_k(-q;q)_{n+1-k}}{(-q;q)_{n+1}}t(1-t)^{n-k}A_k^{\inv}(t,q)\\
&amp;\quad+\sum_{k=0}^n\left[\begin{matrix}
n+1\\
k\end{matrix}\right]_{q^2}(-q;q)_k(1-t)^{n+1-k}A_k^{\inv}(t,q)\\
&amp;=\sum_{k=0}^n\left[\begin{matrix}
n+1\\
k\end{matrix}\right]_{q^2}(-q;q)_k[(-q;q)_{n+1-k}t+(1-t)](1-t)^{n-k}A_k^{\inv}(t,q).
\end{split}
\end{equation*}
It is not apparent that the sum on the right can be written as a combination of $B_k^{\inv}(t,q)$.
\end{proof}


\section{Two questions}\label{sec:06}

In the preceding sections, we obtained expressions of Eulerian polynomials of types $A,B$, and $D$, as well as the ``inv'' analogues of types $A$ and $B$ $q$-Eulerian polynomials, as lower Hessenberg determinants.


Besides representing Eulerian polynomials, determinants also play important roles in the theory of symmetric functions.
For instance, the Jacobi-Trudi identity is a determinantal representation of skew Schur functions in terms of homogeneous symmetric functions \cite[\S7.16]{s99}.
In view of the present work, it is natural to ask

\begin{question}\label{qu:10}
Can a given determinant be interpreted in terms of symmetric functions?
\end{question}

Another occurrence of determinants is in the classical definition of Schur functions \cite{s99}:
$$
a_{\lambda+\delta}/a_\delta=s_\lambda(x_1,\ldots,x_n),
$$
where $\lambda$ is a partition of $n$ with at most $n$ parts, $\delta=(n-1,n-2,\ldots,0)$ and $a_{\lambda+\delta}=\det\bigl(x_i^{\lambda_i+n-j}\bigr)_{i,j=1}^n$. Note that neither $a_{\lambda+\delta}$ nor $a_\delta$ is symmetric in $x_1,\ldots,x_n$, but $a_{\lambda+\delta}/a_\delta$ is.
Extending Question~\ref{qu:10}, more generally, one can ask


\begin{question}\label{qu:11}
Can a given quotient of two determinants be interpreted in terms of symmetric functions?
\end{question}


\section{Acknowledgments}

The author thanks the anonymous referee for his invaluable comments/suggestions that have led to much improved presentation of this work.


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\end{thebibliography}

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\noindent 2020 {\it Mathematics Subject Classification}: Primary 05A15. Secondary 05A05.

\noindent \emph{Keywords:} derangement polynomial,
determinant, lower Hessenberg determinant.

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\vspace*{+.1in}
\noindent
Received February 7 2023;
revised versions received  June 24 2023; June 26 2023.
Published in {\it Journal of Integer Sequences}, July 24 2023.

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