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\begin{center}
\vskip 1cm{\LARGE\bf
Two-Parameter Identities\\ 
\vskip .1in
for $q$-Appell Polynomials
}
\vskip 1cm
\large
Emanuele Munarini \\
Dipartimento di Matematica \\
Politecnico di Milano \\
Piazza Leonardo da Vinci 32 \\
20133 Milano \\
Italy \\
\href{mailto:emanuele.munarini@polimi.it}{\tt emanuele.munarini@polimi.it} \\
\end{center}

\vskip .2 in

\begin{abstract}
 In this paper, by using the techniques of the $q$-exponential generating series,
 we extend a well-known two-parameter identity for the Appell polynomials
 to the $q$-Appell polynomials of type I and II.
 More precisely, we obtain two different $q$-analogues of such an identity.
 Then, we specialize these identities for some $q$-polynomials
 arising in combinatorics, in $q$-calculus or in the theory of orthogonal polynomials.
 In particular, we consider the generalized $q$-Bernoulli and $q$-Euler polynomials
 and then we deduce some further identities involving the Bernoulli and Euler numbers.
 In this way, as a byproduct, we derive the symmetric Carlitz identity for the Bernoulli numbers.
 Finally, we find a (non-symmetric) $q$-analogue of Carlitz's identity
 involving the $q$-Bernoulli numbers of type I and II.
\end{abstract}

\section{Introduction}

Appell sequences form an important and interesting class
containing many classical polynomials arising
in physics, in numerical analysis,
in the theory of orthogonal polynomials \cite{Chihara},
in analysis \cite{BoasBuck,Carlson,Scaravelli},
in the modern umbral calculus \cite{Roman,RomanRota,RotaKahanerOdlyzko}
and in the theory of Sheffer sequences \cite{Sheffer,Steffensen}
or, equivalently, in the theory of Sheffer matrices (or exponential Riordan arrays)
\cite{MunariniA,MunariniC}.
The ordinary powers, the generalized Hermite polynomials,
the generalized Bernoulli and Euler polynomials,
and the generalized rencontres polynomials are all examples of Appell polynomials.

An \emph{Appell sequence} $ \{ a_n(x) \}_{n\in\NN} $
can be characterized in several ways \cite{Appell,VeinDale}.
The following statements are equivalent.
\begin{enumerate}
 \item For every $ n \in \NN $, the polynomial $ a_n(x) $ has degree $ n $
  and $ a'_n(x) = n a_{n-1}(x) $.
 \item There exists a sequence $ \{ g_n \}_{n\in\NN} $, with $ g_0 \ne 0 $, such that
  $$ a_n(x) = \sum_{k=0}^n { n \choose k } g_{n-k}\, x^k $$
  or, equivalently,
  $$ g_n = \sum_{k=0}^n { n \choose k } (-1)^k x^k\, a_{n-k}(x)  . $$
 \item The sequence $ \{ a_n(x) \}_{n\in\NN} $ has exponential generating series
  $$ A(x;t) = \sum_{n\geq0} a_n(t)\, \frac{t^n}{n!} = g(t)\, \ee^{xt}, $$
  where $ g(t) = \sum_{n\geq0} g_n \frac{t^n}{n!} $
  is an exponential series with $ g_0 \ne 0 $.
 \item There exists a sequence $ \{ g_n \}_{n\in\NN} $, with $ g_0 \ne 0 $, such that
  $$ a_n(x) = \sum_{k\geq0} g_k\,\frac{\DD^k}{k!}\, x^n, $$
  where $ \DD $ is the derivative with respect to $ x $.
 \item The sequence $ \{ a_n(x) \}_{n\in\NN} $ satisfies the Appell identity
  $$ \sum_{k=0}^n { n \choose k } a_k(x)\, y^{n-k} = a_n(x+y)  . $$
\end{enumerate}

Furthermore, the Appell polynomials $ a_n(x) $ and the coefficients $ g_n $
are related by the following two-parameter binomial identity 
\cite[p.\ 316]{VeinDale} \cite[Formula ($16'$)]{Scaravelli}:
\begin{equation}\label{Id-Appell}
 \sum_{k=0}^n { n \choose k } (-1)^k x^k\, a_{m+n-k}(x) =
 \sum_{k=0}^m { m \choose k } g_{n+k}\, x^{m-k} \,  .
\end{equation}
It is easy to see that this is a further characterization of the Appell polynomials.
More precisely, we have that the polynomials $ a_n(x) $ form an Appell sequence if and only if
there exists a sequence $ \{ g_n \}_{n\in\NN} $, with $ g_0 \ne 0 $,
for which identity (\ref{Id-Appell}) is satisfied for every $ m, n \in \NN $.

In 1967, after a paper by Sharma and Chak \cite{SharmaChak},
Al-Salam \cite{Al-Salam} introduced the \emph{$q$-Appell polynomials} (of type I)
and, recently, Sadjang \cite{Sadjang} introduced the \emph{$q$-Appell polynomials of type II}.
In this paper, we extend identity (\ref{Id-Appell}) to the $q$-Appell polynomials of type I and II.
Specifically, through the techniques of the theory the $q$-exponential generating series,
we obtain two different $q$-analogues of this identity.
Then, we specialize these identities for some $q$-polynomials
arising in combinatorics, in $q$-calculus or in the theory of orthogonal polynomials,
such as the $q$-Hermite polynomials, the Gaussian polynomials,
the Al-Salam-Carlitz polynomials, the $q$-permutation polynomials,
the $q$-rencontres and $q$-arrangement polynomials (and some of their generalizations),
the $q$-Bernoulli and $q$-Euler polynomials.
In this last case, we further specialize our identities to the Bernoulli and Euler polynomials
and we deduce some other identities involving the Bernoulli and Euler numbers.
In particular, as a byproduct, we derive the Carlitz symmetric identity
for the Bernoulli numbers \cite{CarlitzBeB,MunariniA}.
Finally, we find a $q$-analogue of Carlitz's identity.
Such an identity, however, is not symmetric and involves the $q$-Bernoulli numbers of type I and II.

\section{$q$-Appell polynomials}

We start by recalling some definitions \cite{KacCheung}.
For every $ n \in \NN $, we have the \emph{$q$-natural number} $ [n]_q = 1+q+q^2+\cdots+q^{n-1} $
and the \emph{$q$-factorial number} $ [n]_q! = [n]_q[n-1]_q\cdots[2]_q[1]_q $.
Then, for every $ n, k \in \NN $,
the \emph{$q$-binomial coefficients} (or \emph{Gaussian coefficients}) are defined by
$$ { n \choose k }_{\!\!q} = \frac{[n]_q!}{[k]_q![n-k]_q!} \qquad \text{for } k = 0, 1, \ldots, n $$
and by 0 otherwise.
A \emph{$q$-polynomial} is a polynomial with coefficients in $ \mathbb{Q}_q = \mathbb{Q}(q) $,
i.e., the field of quotients of $ \mathbb{Q}[q] $,
where $ q $ is an indeterminate.
The algebra of the $q$-polynomials are denoted by $ \mathbb{Q}_q[x] $.
The \emph{$q$-Pochhammer symbol} is the $q$-polynomial defined by
\begin{equation}\label{Pochhammer}
 (x;q)_n = (1-x)(1-qx)\cdots(1-q^{n-1}x)
 = \sum_{k=0}^n { n \choose k }_{\!\!q} (-1)^k q^{{ k \choose 2 }} x^k  .
\end{equation}

A \emph{$q$-exponential series} is a formal series of the form
$ f(t) = \sum_{n\geq0} f_n \frac{t^n}{[n]_q!} $.
The sum and the multiplication by a scalar are defined componentwise.
The product of two $q$-exponential series
$ f(t) = \sum_{n\geq0} f_n \frac{t^n}{[n]_q!} $ and
$ g(t) = \sum_{n\geq0} g_n \frac{t^n}{[n]_q!} $ is defined by
\begin{equation}\label{series-qProd}
 f(t) g(t) = \sum_{n\geq0} \left( \sum_{k=0}^n { n \choose k }_{\!\!q} f_k g_{n-k} \right) \frac{t^n}{[n]_q!}.
\end{equation}

The \emph{$q$-derivative} (\emph{Jackson's derivative}) $ \DD_q $
of a $q$-exponential generating series $ f(t) = \sum_{n\geq0} f_n \frac{t^n}{[n]_q!} $
is defined \cite{Jackson,Jackson1908} by the formula
$$ \DD_q f(t) = \frac{f(qt)-f(t)}{(q-1)t} = \sum_{n\geq0} f_{n+1} \frac{t^n}{[n]_q!}  . $$

Given two $q$-exponential series $ f(t) $ and $ g(t) $, we have the 
\emph{$q$-Leibniz formula}
\begin{equation}\label{qLeibniz}
 \DD_q^m f(t)g(t)  = \sum_{k=0}^m { m \choose k }_{\!\!q} \DD_q^k f(t)\cdot \QQ_q^k \DD_q^{m-k} g(t),
\end{equation}
where $ \QQ_q $ is the operator defined by $ \QQ_q h(t) = h(qt) $.

The \emph{$q$-exponential series} (\emph{Jackson's $q$-exponential}) \cite{Jackson}
\begin{equation}\label{series-E}
 E_q(t) = \sum_{n\geq0} \frac{t^n}{[n]_q!} = \prod_{k\geq0} \frac{1}{1+(q-1)q^kt}
\end{equation}
is the eigenfunction of the $q$-derivative, that is,
\begin{equation}\label{DqEq}
 \DD_q E_q(\lambda t) = \lambda E_q(t)  .
\end{equation}
In particular, since $ \DD_q E_q(t) = E_q(t) $, we have the relation
\begin{equation}\label{EEqq}
 E_q(qt) = (1-(1-q)t)\,E_q(t)  .
\end{equation}
Consequently, for every $ m \in \NN $, we have
\begin{equation}\label{mEEqq}
 E_q(q^mt) = ((1-q)t;q)_m\, E_q(t)  .
\end{equation}

The inverse of the $q$-exponential series $ E_q(t) $ is
\begin{equation}\label{series-Einv}
 E_q(t)^{-1} = \sum_{n\geq0} (-1)^n q^{{ n \choose 2 }} \frac{t^n}{[n]_q!} .
\end{equation}

Given two $q$-exponential series
$ f(t) = \sum_{n\geq0} f_n \frac{t^n}{[n]_q!} $ and
$ g(t) = \sum_{n\geq0} g_n \frac{t^n}{[n]_q!} $,
we have $ f(t) = E_q(\alpha t) g(t) $ if and only if $ g(t) = E_q(\alpha t)^{-1} f(t) $.
This is equivalent to the $q$-binomial inversion theorem:
$$
 f_n = \sum_{k=0}^n { n \choose k }_{\!\!q} \alpha^k g_{n-k}
 \qquad\iff\qquad
 g_n = \sum_{k=0}^n { n \choose k }_{\!\!q} (-1)^k q^{{ k \choose 2 }} \alpha^k f_{n-k} .
$$

A \emph{$q$-Appell sequence} (of type I) \cite{Al-Salam}
is a sequence $ \{ a_n(x) \}_{n\in\NN}, $
where $ a_n(x) $ is a $q$-polynomial of degree $ n $
and $ \DD_q a_n(x) = [n]_q a_{n-1}(x) $ for every $ n \in \NN $.
Also the $q$-Appell sequences can be characterized in several ways, as in the ordinary case.
Indeed, the following statements are equivalent.
\begin{enumerate}
 \item $ \{ a_n(x) \}_{n\in\NN} $ is a $q$-Appell sequence (of type I).
 \item There exists a sequence $ \{ g_n \}_{n\in\NN} $, with $ g_0 \ne 0 $, such that
  $$ a_n(x) = \sum_{k=0}^n { n \choose k }_{\!\!q} g_{n-k}\, x^k $$
  or, equivalently,
  $$ g_n = \sum_{k=0}^n { n \choose k }_{\!\!q} (-1)^k q^{{ k \choose 2 }} x^k\, a_{n-k}(x)  . $$
 \item The sequence $ \{ a_n(x) \}_{n\in\NN} $ has $q$-exponential generating series
  \begin{equation}\label{series-qAppell}
   A(x;t) = \sum_{n\geq0} a_n(x) \frac{t^n}{[n]_q!} = g(t)\, E_q(xt),
  \end{equation}
  where $ g(t) = \sum_{n\geq0} g_n \frac{t^n}{[n]_q!} $
  is a $q$-exponential series with $ g_0 \ne 0 $.
 \item There exists a sequence $ \{ g_n \}_{n\in\NN} $, with $ g_0 \ne 0 $, such that
  $$ a_n(x) = \sum_{k\geq0} g_k\,\frac{\DD_q^k}{[k]_q!}\, x^n  . $$
\end{enumerate}

Similarly, a \emph{$q$-Appell sequence of type II} \cite{Sadjang}
is a sequence $ \{ a_n(x) \}_{n\in\NN}, $
where $ a_n(x) $ is a $q$-polynomial of degree $ n $
and $ \DD_q a_n(x) = [n]_q a_{n-1}(qx) $, for every $ n \in \NN $.
Also in this case, we have the following equivalent characterizations:
\begin{enumerate}
 \item $ \{ a_n(x) \}_{n\in\NN} $ is a $q$-Appell sequence of type II.
 \item There exists a sequence $ \{ g_n \}_{n\in\NN} $, with $ g_0 \ne 0 $, such that
  $$ a_n(x) = \sum_{k=0}^n { n \choose k }_{\!\!q} q^{{ k \choose 2 }} g_{n-k}\, x^k $$
  or, equivalently,
  $$ g_n = \sum_{k=0}^n { n \choose k }_{\!\!q} (-1)^k x^k\, a_{n-k}(x)  . $$
 \item The sequence $ \{ a_n(x) \}_{n\in\NN} $ has $q$-exponential generating series
  \begin{equation}\label{series-qAppell-II}
   A(x;t) = \sum_{n\geq0} a_n(x) \frac{t^n}{[n]_q!} = g(t)\, E_q(-xt)^{-1},
  \end{equation}
  where $ g(t) = \sum_{n\geq0} g_n \frac{t^n}{[n]_q!} $
  is a $q$-exponential series with $ g_0 \ne 0 $.
 \item There exists a sequence $ \{ g_n \}_{n\in\NN} $, with $ g_0 \ne 0 $, such that
  $$ a_n(x) = \sum_{k\geq0} g_k\,q^{{ n-k \choose 2 }}\frac{\DD_q^k}{[k]_q!}\, x^n  . $$
\end{enumerate}

\section{Two-parameter identities}

To obtain a $q$-analogue of identity (\ref{Id-Appell}),
we apply the $q$-Leibniz rule to the $q$-exponential generating series of a $q$-Appell sequence.
By applying such a rule in two different ways, we obtain two different extensions of identity (\ref{Id-Appell}).
First, we have the following result:
\begin{theorem}
 Let $ \{ a_n(x) \}_{n\in\NN} $ be the $q$-Appell sequence (of type I)
 associated with the sequence $ \{ g_n \}_{n\in\NN} $.
 Then, for every $ m,n \in \NN $, we have the identity
 \begin{equation}\label{Id-qAppell01}
  \sum_{k=0}^n { n \choose k }_{\!\!q} (-1)^k q^{{ k \choose 2 }} x^k a_{m+n-k}(x) =
  \sum_{k=0}^m { m \choose k }_{\!\!q} q^{nk} g_{n+m-k}\, x^k \,  .
 \end{equation}
\end{theorem}
\begin{proof}
 By applying the $q$-Leibniz rule (\ref{qLeibniz}) to series (\ref{series-qAppell}), we get
 $$
  \DD_q^m A(x;t) = \DD_q^m E_q(xt) g(t)
  = \sum_{k=0}^m { m \choose k }_{\!\!q} \DD_q^k E_q(xt)\cdot \QQ_q^k \DD_q^{m-k} g(t).
 $$
 Then, by identity (\ref{DqEq}) and by setting $ g^{(h)}(t) = \DD_q^{h} g(t) $, we have
 $$ \DD_q^m A(x;t) = \sum_{k=0}^m { m \choose k }_{\!\!q} x^k E_q(xt)\, g^{(m-k)}(q^kt), $$
 that is,
 $$ E_q(xt)^{-1} \DD_q^m A(x;t) = \sum_{k=0}^m { m \choose k }_{\!\!q} x^k\, g^{(m-k)}(q^kt) .$$
 Recalling formulas (\ref{series-qProd}) and (\ref{series-Einv})
 and taking the coefficients of $ \frac{t^n}{[n]_q!} $ from the first and last series,
 we obtain identity (\ref{Id-qAppell01}).
\end{proof}

To state the next theorem, we need the following definition:
given the $q$-Appell polynomials $ a_n(x) $ associated with the $q$-numbers $ g_n $,
we define the shifted polynomials $ a_n^{[m]}(x) $
as the $q$-Appell polynomials associated with the $q$-numbers $ g_{m+n} $;
namely,
$$ a_n^{[m]}(x) = \sum_{k=0}^n { n \choose k }_{\!\!q} g_{m+n-k}\, x^k  . $$
This means that their $q$-exponential generating series is
$$ \sum_{n\geq0} a_n^{[m]}(x) \frac{t^n}{[n]_q!} = g^{(m)}(t)\, E_q(xt), $$
where $ g^{(m)}(t) = \DD_q^m g(t) $.

\begin{theorem}
 Let $ \{ a_n(x) \}_{n\in\NN} $ be the $q$-Appell sequence (of type I)
 associated with the sequence $ \{ g_n \}_{n\in\NN} $.
 Then, for every $ m, n \in \NN $, we have the identity
 \begin{equation}\label{Id-qAppell02}
  \sum_{k=0}^n { n \choose k }_{\!\!q} (-1)^k q^{{ k \choose 2 }} x^k a_{m+n-k}(x) =
  \sum_{k=0}^{m\wedge n} { m \choose k }_{\!\!q} { n \choose k }_{\!\!q} [k]!
   (q-1)^k q^{{ k \choose 2 }} x^k a_{m-k}^{[n]}(x),
 \end{equation}
 where $ m\wedge n $ denotes the minimum between $ m $ and $ n $.
\end{theorem}
\begin{proof}
 By applying the $q$-Leibniz rule (\ref{qLeibniz}) to series (\ref{series-qAppell}), we have
 $$
  \DD_q^m A(x;t) = \DD_q^m g(t) E_q(xt)
  = \sum_{k=0}^m { m \choose k }_{\!\!q} \DD_q^k g(t)\cdot \QQ_q^k \DD_q^{m-k} E_q(xt)  .
 $$
 Then, by identity (\ref{DqEq}), we have
 $$ \DD_q^m A(x;t) = \sum_{k=0}^m { m \choose k }_{\!\!q} g^{(k)}(t)\, x^{m-k} E_q(q^k xt) $$
 and, by Eq.~(\ref{mEEqq}), we have
 $$ \DD_q^m A(x;t) = \sum_{k=0}^m { m \choose k }_{\!\!q} g^{(k)}(t)\, x^{m-k} ((1-q)xt;q)_k E_q(xt), $$
 that is,
 $$ E_q(xt)^{-1} \DD_q^m A(x;t) = \sum_{k=0}^m { m \choose k }_{\!\!q} g^{(k)}(t)\, x^{m-k} ((1-q)xt;q)_k  . $$
 Hence, by Eq.~(\ref{Pochhammer}), we have
 \begin{align*}
  \MoveEqLeft
  E_q(xt)^{-1} \DD_q^m A(x;t)
   = \sum_{k=0}^m { m \choose k }_{\!\!q} g^{(k)}(t)\, x^{m-k} \sum_{i=0}^k { k \choose i }_{\!\!q} (q-1)^i q^{{ i \choose 2 }} x^i t^i \\
  &amp;= \sum_{i,k\geq0}^m { m \choose k }_{\!\!q} { k \choose i }_{\!\!q} (q-1)^i q^{{ i \choose 2 }} x^{i+m-k} t^i g^{(k)}(t) \\
  &amp;= \sum_{i,k\geq0}^m { m \choose i }_{\!\!q} { m-i \choose k-i }_{\!\!q} (q-1)^i q^{{ i \choose 2 }} x^{i+m-k}
     \sum_{n\geq0} { n \choose i }_{\!\!q} [i]_q! g_{n+k-i} \frac{t^n}{[n]_q!}  \\
  &amp;= \sum_{n\geq0} \left( \sum_{i,k\geq0} { m \choose i }_{\!\!q} { m-i \choose k-i }_{\!\!q} { n \choose i }_{\!\!q} [i]_q!
      (q-1)^i q^{{ i \choose 2 }} g_{n+k-i} x^{m-k+i} \right) \frac{t^n}{[n]_q!} \\
  &amp;= \sum_{n\geq0} \left( \sum_{i\geq0} { m \choose i }_{\!\!q} { n \choose i }_{\!\!q} [i]_q!
     (q-1)^i q^{{ i \choose 2 }} x^i \sum_{k=i}^m { m-i \choose k-i }_{\!\!q} g_{n+k-i} x^{m-k} \right) \frac{t^n}{[n]_q!} 
\end{align*}
\begin{align*}
  &amp;= \sum_{n\geq0} \left( \sum_{i\geq0} { m \choose i }_{\!\!q} { n \choose i }_{\!\!q} [i]_q!
     (q-1)^i q^{{ i \choose 2 }} x^i \sum_{k=0}^{m-i} { m-i \choose k }_{\!\!q} g_{n+k} x^{m-i-k} \right) \frac{t^n}{[n]_q!} \\
  &amp;= \sum_{n\geq0} \left( \sum_{i\geq0} { m \choose i }_{\!\!q} { n \choose i }_{\!\!q} [i]_q!
     (q-1)^i q^{{ i \choose 2 }} x^i \sum_{k=0}^{m-i} { m-i \choose k }_{\!\!q} g_{n+m-i-k} x^k \right) \frac{t^n}{[n]_q!} \\
  &amp;= \sum_{n\geq0} \left( \sum_{i\geq0} { m \choose i }_{\!\!q} { n \choose i }_{\!\!q} [i]_q!
     (q-1)^i q^{{ i \choose 2 }} x^i a_{m-i}^{[n]}(x) \right) \frac{t^n}{[n]_q!}  .
 \end{align*}
 Taking the coefficients of $ \frac{t^n}{[n]_q!} $ in the first and last series,
 we obtain identity (\ref{Id-qAppell02}).
\end{proof}

Given a sequence $ \{ g_n \}_{n\in\NN} $,
we denote by $ a_n(x) $ the associated $q$-Appell polynomials of type I
and by $ a_n^*(x) $ the associated $q$-Appell polynomials of type II.
To establish a relation between these two sequences,
we define the umbral map $ \map{\psi}{\mathbb{Q}_q[x]}{\mathbb{Q}_q[x]} $ by setting
$$ \psi(x^n) = q^{{ n \choose 2 }} x^n $$
and by extending it by linearity.
Then, we immediately have the following first result:
\begin{lemma}
 The umbral map $ \psi $ transforms the $q$-Appell polynomials of type I
 into the $q$-Appell polynomials of type II, that is, for every $ n \in \NN $, we have
 $$ \psi(a_n(x)) = a_n^*(x)  . $$
\end{lemma}
This result can be extended as follows.
\begin{lemma}\label{lemma-main}
 For every $ m \in \NN $ and for every polynomial $ p(x) $, we have
 \begin{equation}\label{trans-psi}
  \psi(x^mp(x)) = q^{{ m \choose 2 }} x^m\, p^*(q^mx),
 \end{equation}
 where $ p^*(x) = \psi(p(x)) $.
\end{lemma}
\begin{proof}
 Suppose $ p(x) = \sum_{k=0}^n p_k x^k $.
 Then, by the linearity of the umbral map, we have
 \begin{align*}
  \psi(x^mp(x))
  &amp;= \sum_{k=0}^n { n \choose k }_{\!\!q} p_k\, \psi(x^{m+k})
   = \sum_{k=0}^n { n \choose k }_{\!\!q} p_k\, q^{{ m+k \choose 2 }} x^{m+k} \\
  &amp;= q^{{ m \choose 2 }} x^m \sum_{k=0}^n { n \choose i }_{\!\!q} q^{{ k \choose 2 }} p_k\, (q^mx)^k
   = q^{{ m \choose 2 }} x^m p^*(q^mx)  .
 \end{align*}
 This establishes identity (\ref{trans-psi}).
\end{proof}

By this lemma, we can easily find the following analogues
of identities (\ref{Id-qAppell01}) and (\ref{Id-qAppell02})
for the $q$-Appell polynomials of type II:
\begin{theorem}
 Let $ \{ a_n^*(x) \}_{n\in\NN} $ be the $q$-Appell sequence of type II
 associated with the sequence $ \{ g_n \}_{n\in\NN} $.
 Then, for all $ m,n \in \NN $, we have the identities
 \begin{align}
  &amp; \sum_{k=0}^n { n \choose k }_{\!\!q} (-1)^k q^{k^2-k} x^k a_{m+n-k}^*(q^kx) =
    \sum_{k=0}^m { m \choose k }_{\!\!q} q^{{ k \choose 2 }+nk} g_{n+m-k}\, x^k \label{Id-qAppell01-II}\\
  &amp; \sum_{k=0}^n { n \choose k }_{\!\!q} (-1)^k q^{k^2-k} x^k a_{m+n-k}^*(q^kx) =
    \sum_{k=0}^{m\wedge n} { m \choose k }_{\!\!q} { n \choose k }_{\!\!q} [k]!
    (q-1)^k q^{k^2-k} x^k a_{n+m-k}^{[n]*}(q^kx)  . \label{Id-qAppell02-II}
 \end{align}
\end{theorem}
\begin{proof}
 By Lemma \ref{lemma-main}, we have
 $ \psi\big(x^ma_n^{[r]}(x)\big) = q^{{ m \choose 2 }} x^m a_n^{[r]*}(q^mx) $.
 Hence, by applying $ \psi $ to identities (\ref{Id-qAppell01}) and (\ref{Id-qAppell02})
 we obtain at once identities (\ref{Id-qAppell01-II}) and (\ref{Id-qAppell02-II}).
\end{proof}

\section{Examples (of type I)}

\subsection{$q$-Hermite polynomials}

The \emph{$q$-Hermite polynomials} (or \emph{Rogers-Szeg\"o polynomials})
\cite{Carlitz,AlSalamCarlitz} \cite[p.\ 180]{Roman}
are the $q$-Appell polynomials associated with the numbers $ g_n = 1 $; namely,
\begin{equation}\label{qHermite}
 H_n(q;x) = \sum_{k=0}^n { n \choose k }_{\!\!q} x^k \, ,
\end{equation}
and have $q$-exponential generating series
$$ \sum_{n\geq0} H_n(q;x) \frac{t^n}{[n]_q!} = E_q(t)\, E_q(xt) . $$
Since $ H_n^{[m]}(q;x) = H_n(q;x) $,
identities (\ref{Id-qAppell01}) and (\ref{Id-qAppell02}) become
\begin{align}
 &amp; \sum_{k=0}^n { n \choose k }_{\!\!q} (-1)^k q^{{ k \choose 2 }} x^k H_{m+n-k}(q;x) =
   \sum_{k=0}^m { m \choose k }_{\!\!q} q^{nk}\, x^k, \\
 &amp; \sum_{k=0}^n { n \choose k }_{\!\!q} (-1)^k q^{{ k \choose 2 }} x^k H_{m+n-k}(q;x) =
   \sum_{k=0}^{m\wedge n} { m \choose k }_{\!\!q} { n \choose k }_{\!\!q} [k]!
    (q-1)^k q^{{ k \choose 2 }} x^k H_{m-k}(q;x)  .
\end{align}
In particular, for $ x = 1 $, we have the \emph{Galois numbers}
$ G_n(q) = H_n(q;1) = \sum_{k=0}^n { n \choose k }_{\!\!q} $,
\cite{GoldmanRota,NijenhuisSolowWilf},
and the two previous identities become
\begin{align}
 &amp; \sum_{k=0}^n { n \choose k }_{\!\!q} (-1)^k q^{{ k \choose 2 }} G_{m+n-k}(q) =
   \sum_{k=0}^m { m \choose k }_{\!\!q} q^{nk}, \\
 &amp; \sum_{k=0}^n { n \choose k }_{\!\!q} (-1)^k q^{{ k \choose 2 }} G_{m+n-k}(q) =
   \sum_{k=0}^{m\wedge n} { m \choose k }_{\!\!q} { n \choose k }_{\!\!q} [k]!
    (q-1)^k q^{{ k \choose 2 }} G_{m-k}(q)  .
\end{align}

\subsection{Gaussian polynomials}

The \emph{Gaussian polynomials} \cite{GoldmanRota,GoldmanRota70}
are the $q$-Appell polynomials associated with the $q$-numbers $ g_n = (-1)^n q^{{ n \choose 2 }} $; namely,
$$
 g_n(q;x) = (x-1)(x-q)\cdots(x-q^{n-1})
 = \sum_{k=0}^n { n \choose k }_{\!\!q} (-1)^{n-k} q^{{ n-k \choose 2 }} x^k\, ,
$$
and have $q$-exponential generating series
$$ \sum_{n\geq0} g_n(q;x) \frac{t^n}{[n]_q!} = E_q(t)^{-1} E_q(xt) . $$

In this case, we have
\begin{align*}
 g_n^{[m]}(q;x)
 &amp;= \sum_{k=0}^n { n \choose k }_{\!\!q} (-1)^{m+n-k} q^{{ m+n-k \choose 2 }} x^k \\
 &amp;= (-1)^m q^{{ m \choose 2 }} q^{mn}
    \sum_{k=0}^n { n \choose k }_{\!\!q} (-1)^{n-k} q^{{ n-k \choose 2 }} \Big(\frac{x}{q^m}\Big)^k \\
 &amp;= (-1)^m q^{{ m \choose 2 }} q^{mn} g_n(q;x/q^m)
  = (-1)^m q^{{ m \choose 2 }} \frac{g_{m+n}(q;x)}{g_n(q;x)}  .
\end{align*}
Hence, identities (\ref{Id-qAppell01}) and (\ref{Id-qAppell02}) become
\begin{align}
 &amp; \sum_{k=0}^n { n \choose k }_{\!\!q} (-1)^k q^{{ k \choose 2 }} x^k g_{m+n-k}(q;x) =
   \sum_{k=0}^m { m \choose k }_{\!\!q} (-1)^{n+m-k} q^{{ n+m-k \choose 2 }+nk} x^k, \\
 &amp; \begin{aligned}
    \MoveEqLeft
    g_n(q;x) \sum_{k=0}^n { n \choose k }_{\!\!q} (-1)^k q^{{ k \choose 2 }} x^k g_{m+n-k}(q;x) = \\
    &amp;= (-1)^n q^{{ n \choose 2 }} \sum_{k=0}^{m\wedge n} { m \choose k }_{\!\!q} { n \choose k }_{\!\!q} [k]!
       (q-1)^k q^{{ k \choose 2 }} x^k g_{n+m-k}(q;x)  .
   \end{aligned}
\end{align}

These two identities can be rewritten in terms of the Pochhammer symbol.
More precisely, since $ (x;q)_n = x^n g_n(q;x^{-1}) $, we have the identities
\begin{align}
 &amp; \sum_{k=0}^n { n \choose k }_{\!\!q} (-1)^k q^{{ k \choose 2 }} (x;q)_{m+n-k} =
   \sum_{k=0}^m { m \choose k }_{\!\!q} (-1)^{n+m-k} q^{{ n+m-k \choose 2 }+nk} x^{n+m-k}, \\
 &amp; \begin{aligned}
    \MoveEqLeft
    (x;q)_n \sum_{k=0}^n 
    { n \choose k }_{\!\!q} (-1)^{n-k} q^{{ k \choose 2 }} (x;q)_{m+n-k}  \\
    &amp;= q^{{ n \choose 2 }} x^n \sum_{k=0}^{m\wedge n} { m \choose k }_{\!\!q} { n \choose k }_{\!\!q} [k]!
       (q-1)^k q^{{ k \choose 2 }} (x;q)_{n+m-k}  .
   \end{aligned}
\end{align}

\subsection{Al-Salam-Carlitz polynomials}

The \emph{Al-Salam-Carlitz polynomials} (or \emph{$q$-Carlitz polynomials})
(\cite{AlSalamCarlitzU} \cite[p.\ 195]{Chihara} \cite{Ismail})
are defined by the formula
$$ U_n^{(\alpha)}(q;x) = \sum_{k=0}^n { n \choose k }_{\!\!q} (-\alpha)^{n-k} g_k(x) $$
and consequently they have $q$-exponential generating series
$$
 U_q^{(\alpha)}(x;t)
 = \sum_{n\geq0} U_n^{(\alpha)}(q;x) \frac{t^n}{[n]_q!}
 = \frac{E_q(xt)}{E_q(t)E_q(\alpha t)}
 = E_q(t)^{-1} E_q(\alpha t)^{-1} E_q(xt)  . \label{series-U}
$$
This means that they form the $q$-Appell sequence
associated with the $q$-numbers
$$
 u_n^{(\alpha)}(q)
 = (-1)^n \sum_{k=0}^n { n \choose k }_{\!\!q} q^{{ k \choose 2 }} q^{{ n-k \choose 2 }} \alpha^k
$$
defined by the $q$-exponential generating series
$$
 u_q^{(\alpha)}(t)
 = \sum_{n\geq0} u_n^{(\alpha)}(q) \frac{t^n}{[n]_q!}
 = E_q(t)^{-1} E_q(\alpha t)^{-1}  .
$$
In this case, identity (\ref{Id-qAppell01}) become
\begin{equation}
 \sum_{k=0}^n { n \choose k }_{\!\!q} (-1)^k q^{{ k \choose 2 }} x^k  U^{(\alpha)}_{m+n-k}(q;x) =
 \sum_{k=0}^m { m \choose k }_{\!\!q} q^{nk} u^{(\alpha)}_{n+m-k}(q)\, x^k \,  .
\end{equation}

\subsection{$q$-Factorial polynomials}

The \emph{$q$-factorial polynomials} are the $q$-Appell polynomials
associated with the $q$-factorial numbers $ g_n = [n]_q! $; namely,
$$ F_n(q;x) = \sum_{k=0}^n { n \choose k }_{\!\!q} [n-k]_q!\, x^k . $$
Hence, they have $q$-exponential generating series
$$ F_q(x,t) = \sum_{n\geq0} P_n(q;x) \frac{t^n}{[n]_q!} = \frac{E_q(xt)}{1-t}  . $$
Moreover, the polynomials
$$ F_n^{[m]}(q;x) = \sum_{k=0}^n { n \choose k }_{\!\!q} [m+n-k]_q!\, x^k $$
have $q$-exponential generating series
$$ F_q^{[m]}(x,t) = \sum_{n\geq0} P_n^{[m]}(q;x) \frac{t^n}{[n]_q!} = F_q^{[m]}(t) E_q(xt), $$
where
$$
 F_q^{[m]}(t) = \sum_{n\geq0} [m+n]_q! \frac{t^n}{[n]_q!}
 = [m]_q! \sum_{n\geq0} \frac{[m+n]_q!}{[m]_q![n]_q!}t^n
 = [m]_q! \sum_{n\geq0} { m+n \choose m }_{\!\!q} t^n,
$$
that is,
$$ F_q^{[m]}(t) = \frac{[m]_q!}{(t;q)_{m+1}} = \frac{[m]_q!}{(1-t)(1-qt)\cdots(1-q^mt)} . $$
Hence, we have
$$ F_q^{[m]}(x,t) = \frac{[m]_q!\,E_q(xt)}{(1-t)(1-qt)\cdots(1-q^mt)} . $$

Identities (\ref{Id-qAppell01}) and (\ref{Id-qAppell02}) become
\begin{align}
 &amp; \sum_{k=0}^n { n \choose k }_{\!\!q} (-1)^k q^{{ k \choose 2 }} x^k F_{m+n-k}(x) =
   \sum_{k=0}^m { m \choose k }_{\!\!q} q^{nk} [n+m-k]_q!\, x^k, \\
 &amp; \sum_{k=0}^n { n \choose k }_{\!\!q} (-1)^k q^{{ k \choose 2 }} x^k F_{m+n-k}(x) =
   \sum_{k=0}^{m\wedge n} { m \choose k }_{\!\!q} { n \choose k }_{\!\!q} [k]!
   (q-1)^k q^{{ k \choose 2 }} x^k F_{n+m-k}^{[n]}(x) .
\end{align}

\subsection{$q$-rencontres and $q$-arrangement polynomials}

The \emph{$q$-derangement numbers} $ d_n(q) $ \cite{Wachs,ChenRota,MunariniqD}
and the \emph{$q$-arrangement numbers} $ a_n(q) $
are respectively defined by the formulas
\begin{align*}
 d_n(q) = \sum_{k=0}^n { n \choose k }_{\!\!q} (-1)^k q^{{ k \choose 2 }} [n-k]_q!
 \qquad\text{and}\qquad
 a_n(q) = \sum_{k=0}^n { n \choose k }_{\!\!q} [n-k]_q!
\end{align*}
and have $q$-exponential generating series
\begin{align*}
 D_q(t) = \sum_{n\geq0} d_n(q) \frac{t^n}{[n]_q!} = \frac{E_q(t)^{-1}}{1-t}
 \qquad\text{and}\qquad
 A_q(t) = \sum_{n\geq0} a_n(q) \frac{t^n}{[n]_q!} = \frac{E_q(t)}{1-t} .
\end{align*}
For $ q = 1 $, we recover the derangement numbers $ d_n $
(\cite[p.\ 65]{Riordan} and \seqnum{A000166}
and the arrangement numbers $ a_n $ (\cite[p.\ 16]{Riordan} and
\seqnum{A000522}).

The \emph{$q$-rencontres polynomials} $ D_q(x;t) $
and the \emph{$q$-arrangement polynomials} $ A_q(x;t) $
are the $q$-Appell polynomials associated respectively with
the $q$-numbers $ d_n(q) $ and $ a_n(q) $; namely,
\begin{align*}
 D_n(q;x) = \sum_{k=0}^n { n \choose k }_{\!\!q} d_{n-k}(q)\, x^k
 \qquad\text{and}\qquad
 A_n(q;x) = \sum_{k=0}^n { n \choose k }_{\!\!q} a_{n-k}(q)\, x^k,
\end{align*}
and have $q$-exponential generating series
\begin{align*}
 &amp; D_q(x;t) = \sum_{n\geq0} D_n(q;x) \frac{t^n}{[n]_q!} = \frac{E_q(t)^{-1}}{1-t}\, E_q(xt),\\
 &amp; A_q(x;t) = \sum_{n\geq0} A_n(q;x) \frac{t^n}{[n]_q!} = \frac{E_q(t)}{1-t}\, E_q(xt) .
\end{align*}
Then, by identity (\ref{Id-qAppell01}), we have
\begin{align}
 &amp; \sum_{k=0}^n { n \choose k }_{\!\!q} (-1)^k q^{{ k \choose 2 }} x^k D_{m+n-k}(q;x) =
   \sum_{k=0}^m { m \choose k }_{\!\!q} q^{nk} d_{n+m-k}(q)\, x^k, \\
 &amp; \sum_{k=0}^n { n \choose k }_{\!\!q} (-1)^k q^{{ k \choose 2 }} x^k A_{m+n-k}(q;x) =
   \sum_{k=0}^m { m \choose k }_{\!\!q} q^{nk} a_{n+m-k}(q)\, x^k  .
\end{align}

More generally, given $ m \in \NN $,
we can consider the \emph{generalized $q$-derangement numbers}
$ d^{(m)}_n(q) $
and the \emph{generalized $q$-arrangement numbers} $ a^{(m)}_n(q) $ defined by
\begin{align*}
 d^{(m)}_n(q) = \sum_{k=0}^n { n \choose k }_{\!\!q} (-1)^k q^{{ k \choose 2 }} [m+n-k]_q!
 \qquad\text{and}\qquad
 a^{(m)}_n(q) = \sum_{k=0}^n { n \choose k }_{\!\!q} [m+n-k]_q!
\end{align*}
and having $q$-exponential generating series
\begin{align*}
 &amp; D^{(m)}_q(t) = \sum_{n\geq0} d^{(m)}_n(q) \frac{t^n}{[n]_q!}
                = \frac{[m]_q!\,E_q(t)^{-1}}{(1-t)(1-qt)\cdots(1-q^mt)}, \\
 &amp; A^{(m)}_q(t) = \sum_{n\geq0} a^{(m)}_n(q) \frac{t^n}{[n]_q!}
                = \frac{[m]_q!\,E_q(t)}{(1-t)(1-qt)\cdots(1-q^mt)} .
\end{align*}
Then, the associated $q$-Appell polynomials
are the \emph{generalized $q$-rencontres polynomials} $ D^{(m)}_q(x;t) $
and the \emph{generalized $q$-arrangement polynomials} $ A^{(m)}_q(x;t) $
defined by
\begin{align*}
 D^{(m)}_n(q;x) = \sum_{k=0}^n { n \choose k }_{\!\!q} d^{(m)}_{n-k}(q)\, x^k
 \qquad\text{and}\qquad
 A^{(m)}_n(q;x) = \sum_{k=0}^n { n \choose k }_{\!\!q} a^{(m)}_{n-k}(q)\, x^k
\end{align*}
and with $q$-exponential generating series
\begin{align*}
 &amp; D^{(m)}_q(x;t) = \sum_{n\geq0} D^{(m)}_n(q;x) \frac{t^n}{[n]_q!}
                  = \frac{[m]_q!\,E_q(t)^{-1}}{(1-t)(1-qt)\cdots(1-q^mt)}\, E_q(xt), \\
 &amp; A^{(m)}_q(x;t) = \sum_{n\geq0} A^{(m)}_n(q;x) \frac{t^n}{[n]_q!}
                  = \frac{[m]_q!\,E_q(t)}{(1-t)(1-qt)\cdots(1-q^mt)}\, E_q(xt) .
\end{align*}
In this way, when $ q = 1 $, we recover the generalized derangement numbers $ d^{(m)}_n $
and the generalized arrangement numbers $ a^{(m)}_n $,
and the associated Appell polynomials $ D^{(m)}_n(x) $ and $ A^{(m)}_n(x) $,
\cite{CFMunariniZ,FerrariMunarini,FerrariMunariniZagaglia}.

For these generalized polynomials, by identity (\ref{Id-qAppell01}), we have
\begin{align}
 &amp; \sum_{k=0}^n { n \choose k }_{\!\!q} (-1)^k q^{{ k \choose 2 }} x^k D^{(\mu)}_{m+n-k}(q;x) =
   \sum_{k=0}^m { m \choose k }_{\!\!q} q^{nk} d^{(\mu)}_{n+m-k}(q)\, x^k, \\
 &amp; \sum_{k=0}^n { n \choose k }_{\!\!q} (-1)^k q^{{ k \choose 2 }} x^k A^{(\mu)}_{m+n-k}(q;x) =
   \sum_{k=0}^m { m \choose k }_{\!\!q} q^{nk} a^{(\mu)}_{n+m-k}(q)\, x^k  .
\end{align}
In particular, for $ x = 1 $, we have $ D^{(\mu)}_n(q;1) = [\mu+n]_q! $.
Hence, the first identity becomes
\begin{equation}
 \sum_{k=0}^n { n \choose k }_{\!\!q} (-1)^k q^{{ k \choose 2 }} [\mu+m+n-k]_q! =
 \sum_{k=0}^m { m \choose k }_{\!\!q} q^{nk} d^{(\mu)}_{n+m-k}(q)  .
\end{equation}

\subsection{Generalized $q$-Bernoulli and $q$-Euler polynomials}

The \emph{generalized $q$-Bernoulli numbers} $ B_n^{(\nu)}(q) $
and the \emph{generalized $q$-Bernoulli polynomials} $ B_n^{(\nu)}(q;x) $, \cite{Al-SalamBE},
are defined, respectively, by the $q$-exponential generating series
\begin{align*}
 &amp; B_q^{(\nu)}(t) = \sum_{n\geq0} B_n^{(\nu)}(q)\, \frac{t^n}{[n]_q!}
 = \left(\frac{t}{E_q(t)-1}\right)^{\!\!\nu}, \\
 &amp; B_q^{(\nu)}(x;t) = \sum_{n\geq0} B_n^{(\nu)}(q;x)\, \frac{t^n}{[n]_q!}
 = \left(\frac{t}{E_q(t)-1}\right)^{\!\!\nu} E_q(xt) .
\end{align*}

Similarly, the $q$-numbers $ \widetilde{E}_n^{(\nu)}(q) $
and the \emph{generalized $q$-Euler polynomials} $ E_n^{(\nu)}(q;x) $
are defined, respectively, by the $q$-exponential generating series
\begin{align*}
 &amp; \widetilde{E}_q^{(\nu)}(t)
 = \sum_{n\geq0} \widetilde{E}_n^{(\nu)}(q)\, \frac{t^n}{[n]_q!}
 = \left(\frac{2}{E_q(t)+1}\right)^{\!\!\nu}, \\
 &amp; E_q^{(\nu)}(x;t) = \sum_{n\geq0} E_n^{(\nu)}(q;x)\, \frac{t^n}{[n]_q!}
 = \left(\frac{2}{E_q(t)+1}\right)^{\!\!\nu} E_q(xt) .
\end{align*}
Hence, by identity (\ref{Id-qAppell01}), we have
\begin{align}
 &amp; \sum_{k=0}^n { n \choose k }_{\!\!q} (-1)^k q^{{ k \choose 2 }} x^k B_{m+n-k}^{(\nu)}(q;x) =
   \sum_{k=0}^m { m \choose k }_{\!\!q} q^{nk} B_{n+m-k}^{(\nu)}(q)\, x^k, \label{Id-qBernoulli01} \\
 &amp; \sum_{k=0}^n { n \choose k }_{\!\!q} (-1)^k q^{{ k \choose 2 }} x^k E_{m+n-k}^{(\nu)}(q;x) =
   \sum_{k=0}^m { m \choose k }_{\!\!q} q^{nk} \widetilde{E}_{n+m-k}^{(\nu)}(q)\, x^k  . \label{Id-qBernoulli02}
\end{align}

Clearly, when $ \nu = 1 $ and $ q = 1 $,
we have the \emph{Bernoulli polynomials} $ B_n(x) $
and the \emph{Euler polynomials} $ E_n(x) $,
\cite{MilneThomson,Norlund}.
Moreover, we have the \emph{Bernoulli numbers} $ B_n = B_n(0) $
and $ \widetilde{E}_n = E_n(0) = (2-2^{n+2})\frac{B_{n+1}}{n+1} $,
the \emph{Euler numbers} $ E_n = 2^n E_n(1/2) $,
(\cite[p.\ 49]{Comtet} and \seqnum{A122045}, 
\seqnum{A000364}, \seqnum{A028296})
and the \emph{Springer numbers} 
$ S_n = (-1)^{\lceil n/2 \rceil} 4^n E_n(1/4) $
(\cite{Springer} and \seqnum{A001586}).
Hence, the above identities can be rewritten as
\begin{align}
 &amp; \sum_{k=0}^n { n \choose k } (-1)^{n-k} x^{n-k} B_{m+k}(x) =
   \sum_{k=0}^m { m \choose k } B_{n+k}\, x^{m-k} \label{Id-EuB}, \\
 &amp; \sum_{k=0}^n { n \choose k } (-1)^{n-k} x^{n-k} E_{m+k}(x) =
   \sum_{k=0}^m { m \choose k } \widetilde{E}_{n+k}\, x^{m-k}  . \label{Id-EuE}
\end{align}

In particular, when $ x = 1/2 $, we have $ B_n(1/2) = (2^{1-n}-1) B_n $ and
identities (\ref{Id-EuB}) and (\ref{Id-EuE}) become
\begin{align}
 &amp; \sum_{k=0}^n { n \choose k } (-1)^{n-k} (2-2^{m+k})\, B_{m+k} =
   \sum_{k=0}^m { m \choose k } 2^{n+k}\,B_{n+k} \label{IdBB12}, \\
 &amp; \sum_{k=0}^n { n \choose k } (-1)^{n-k} E_{m+k} =
   \sum_{k=0}^m { m \choose k } 2^{n+k} (2-2^{n+k+2})\,\frac{B_{n+k+1}}{n+k+1} .
\end{align}
Similarly, when $ x = 1/4 $,
we have $ B_n(1/4) = \frac{1}{2^n} \left( \frac{1}{2^{n-1}} - 1 \right) B_n - \frac{1}{4^n} n E_{n-1} $
and identities (\ref{Id-EuB}) and (\ref{Id-EuE}) become
\begin{align}
 &amp; \sum_{k=0}^n { n \choose k } (-1)^{n-k} \Big( (2-2^{m+k})\, B_{m+k} - (m+k)\, E_{m+k-1} \Big) =
   \sum_{k=0}^m { m \choose k } 4^{n+k}\,B_{n+k}, \label{IdBB14} \\
 &amp; \sum_{k=0}^n { n \choose k } (-1)^{n-k+\lceil\frac{m+k}{2}\rceil} S_{m+k} =
   \sum_{k=0}^m { m \choose k } 4^{n+k} (2-2^{n+k+2})\,\frac{B_{n+k+1}}{n+k+1} .
\end{align}
From identities (\ref{IdBB12}) and (\ref{IdBB14}), we also have
\begin{equation}
 \sum_{k=0}^n { n \choose k } (-1)^{n-k} (m+k)\, E_{m+k-1} =
 \sum_{k=0}^m { m \choose k } 2^{n+k}(1-2^{n+k})\,B_{n+k}  .
\end{equation}

Finally, when $ x = 1 $,
we have $ B_n(1) = (-1)^n B_n $ and $ E_n(1) = (-1)^n E_n(0) = (-1)^n \widetilde{E}_n $
and identities (\ref{Id-EuB}) and (\ref{Id-EuE}) become
\begin{gather}
 (-1)^n \sum_{k=0}^n { n \choose k } B_{m+k} = (-1)^m \sum_{k=0}^m { m \choose k } B_{n+k} \label{IdBB1} \\
 (-1)^n \sum_{k=0}^n { n \choose k } (2-2^{m+k+2})\,\frac{B_{m+k+1}}{m+k+1}, =
 (-1)^m \sum_{k=0}^m { m \choose k } (2-2^{n+k+2})\,\frac{B_{n+k+1}}{n+k+1} . \label{IdEE1}
\end{gather}

The first identity (\ref{IdBB1}) is the symmetric Carlitz identity
for the Bernoulli numbers \cite{CarlitzBeB}.
Such an identity can be proved (and generalized) in several ways
\cite{ChenSun,Gessel,Prodinger,Sun,VassilevMissana}.
In \cite{MunariniA}, we proved it by using a general method based on umbral calculus.
Here, we proved that it is a simple consequence
of the general two-parameter identity for the Appell polynomials
(specialized to the Bernoulli polynomials).
Similarly, the second identity (\ref{IdEE1}) is the analogue symmetric identity
for the numbers $ \widetilde{E}_n $.

We conclude by noticing that the previous symmetric identities
can be extended to the generalized Bernoulli numbers.
Indeed, we have
$ B^{(\nu)}_n(\nu) = (-1)^n B^{(\nu)}_n $ and $ E^{(\nu)}_n(\nu) = (-1)^n \widetilde{E}^{(\nu)}_n $.
So, setting $ q = 1 $ and $ x = \nu $, we obtain
\begin{align}
 &amp; (-1)^n \sum_{k=0}^n { n \choose k } \nu^{n-k} B^{(\nu)}_{m+k} =
   (-1)^m \sum_{k=0}^m { m \choose k } \nu^{m-k} B^{(\nu)}_{n+k}, \\
 &amp; (-1)^n \sum_{k=0}^n { n \choose k } \nu^{n-k} \widetilde{E}^{(\nu)}_{m+k} =
   (-1)^m \sum_{k=0}^m { m \choose k } \nu^{m-k} \widetilde{E}^{(\nu)}_{n+k}  .
\end{align}

However, Carlitz's identity (\ref{IdBB1}) cannot be immediately extended to the $q$-Bernoulli numbers.
To obtain such an extension, we need to consider the $q$-Bernoulli numbers $ B^*_n(q) $ of type II
defined by the $q$-exponential generating series
$$ B^*_q(t) =
\sum_{n\geq0} B^*_n(q) \frac{t^n}{[n]_q!} = \frac{t}{E^*_q(t)-1}, $$
where
$$ E^*_q(t) = E_q(-t)^{-1} = \sum_{n\geq0} q^{{ n \choose 2 }} \frac{t^n}{[n]_q!}  . $$
Since the $q$-exponential generating series of the $q$-Bernoulli numbers can be written as follows
$$ B_q(t) = \frac{t}{E_q(t)-1} = \frac{-t\,E_q(t)^{-1}}{E_q(t)^{-1}-1} = B^*_q(-t) \,E_q(t)^{-1}\, , $$
we have $ B_q(1;t) = B^*_q(-t) $, that is, $ B_n(q;1) = (-1)^n B^*_n(q) $.
Therefore, for $ \nu = 1 $ and $ x = 1 $,
identity (\ref{Id-qBernoulli01}) can be rewritten as
\begin{equation}\label{Carlitz-qBernoulli01}
 (-1)^n \sum_{k=0}^n { n \choose k }_{\!\!q} q^{{ n-k \choose 2 }} B^*_{m+k}(q) =
 (-1)^m \sum_{k=0}^m { m \choose k }_{\!\!q} q^{n(m-k)} B_{n+k}(q)  .
\end{equation}
This is the $q$-analogue of Carlitz's identity (\ref{IdBB1}),
even though it is not symmetric
and involves the $q$-Bernoulli numbers of type I and II.
The $q$-Bernoulli numbers of type II, however,
can be expressed in terms of the $q$-Bernoulli numbers of type I.
Indeed, since
$$
 B^*_q(t) = \frac{t}{E_q(-t)^{-1}-1} = \frac{-t\,E_q(-t)}{E_q(-t)-1}
 = \frac{-t\,(1+E_q(-t)-1)}{E_q(-t)-1} = B_q(-t) - t\, ,
$$
we have $ B^*_n(q) = (-1)^n B_n(q) - \delta_{n,1} = (-1)^n ( B_n(q) + \delta_{n,1}) $.
Moreover, by the identity \cite{Ubriaco}
$$
 E_q(-t)\, E_{q^{-1}}(t) = 1\, ,
 \quad\text{or}\qquad
 E_q(-t)^{-1} = E_{q^{-1}}(t)\, ,
$$
we also have
$$ B^*_q(t) = \frac{t}{E_q(-t)^{-1}-1} = \frac{t}{E_{q^{-1}}(t)-1} = B_{q^{-1}}(t) . $$
Since $\; [n]_{q^{-1}}! = q^{-{ n \choose k }} [n]_q! \,$, we have
$$
 B^*_q(t) = \sum_{n\geq0} B_n(q^{-1}) \frac{t^n}{[n]_{q^{-1}}!}
          = \sum_{n\geq0} q^{{ n \choose k }} B_n(q^{-1}) \frac{t^n}{[n]_q!},
$$
that is,
$$ B^*_n(q) = q^{{ n \choose k }} B_n(q^{-1})  . $$
So, replacing this expression in \eqref{Carlitz-qBernoulli01}, we obtain the identity
\begin{equation}
 (-1)^n q^{{ n \choose 2 }} \sum_{k=0}^n { n \choose k }_{\!\!q} q^{k(m-n+k)} B_{m+k}(q^{-1}) =
 (-1)^m q^{-{ m \choose k }} \sum_{k=0}^m{ m \choose k }_{\!\!q} q^{n(m-k)} B_{n+k}(q)  .
\end{equation}

In a completely similar way, defining the $q$-numbers $ \widetilde{E}^*_n(q) $ of type II
by the $q$-exponential generating series
$$
 \widetilde{E}^*_q(t) = \sum_{n\geq0} \widetilde{E}^*_n(q) \frac{t^n}{[n]_q!}
 = \frac{2}{E^*_q(t)+1} = \frac{2}{E_{q^{-1}}(t)+1} = \widetilde{E}_{q^{-1}}(t) \, ,
$$
we can derive 
the following identity
from identity (\ref{Id-qBernoulli02}):
\begin{equation}
 (-1)^n \sum_{k=0}^n { n \choose k }_{\!\!q} q^{{ n-k \choose 2 }} \widetilde{E}^*_{m+k}(q) =
 (-1)^m \sum_{k=0}^m { m \choose k }_{\!\!q} q^{n(m-k)} \widetilde{E}_{n+k}(q)  .
\end{equation}
Finally, since $ \widetilde{E}^*_n(q) = q^{{ n \choose k }} \widetilde{E}_n(q^{-1}) $, we also have
\begin{equation}
 (-1)^n q^{{ n \choose 2 }} \sum_{k=0}^n { n \choose k }_{\!\!q} q^{k(m-n+k)} \widetilde{E}_{m+k}(q^{-1}) =
 (-1)^m q^{-{ m \choose k }} \sum_{k=0}^m{ m \choose k }_{\!\!q} q^{n(m-k)} \widetilde{E}_{n+k}(q)  .
\end{equation}

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\bigskip

\noindent 2010 {\it Mathematics Subject Classification}:
Primary 05A19;
Secondary 05A30, 05A40, 11B65, 11B68, 33C45, 33D45.
	
\noindent \emph{Keywords:}
combinatorial sum, $q$-binomial sum,
$q$-Appell polynomial,
Bernoulli and $q$-Bernoulli number,
Euler number, Springer number,
Carlitz's identity, $q$-exponential generating series.

\bigskip \hrule \bigskip

\noindent (Concerned with sequences
\seqnum{A000166},
\seqnum{A000364},
\seqnum{A000522},
\seqnum{A001586},
\seqnum{A028296}, and
\seqnum{A122045}.)

\bigskip
\hrule
\bigskip

\vspace*{+.1in}
\noindent
Received December 3 2021;
revised version received  March 7 2023.
Published in {\it Journal of Integer Sequences}, March 8 2023.

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\noindent
Return to
\htmladdnormallink{Journal of Integer Sequences home page}{https://cs.uwaterloo.ca/journals/JIS/}.
\vskip .1in


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