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\begin{center}
\vskip 1cm{\LARGE\bf 
Identities Involving Two Kinds of \\
\vskip .1in
$q$-Euler Polynomials and Numbers
}
\vskip 1cm
\large
Abdelmejid Bayad\\
D\'epartement de Math\'ematiques \\
Universit\'e d'Evry Val d'Essonne\\
B\^atiment I.B.G.B.I., 3\`eme \'etage \\
23 Bd.\ de France\\
91037 Evry Cedex \\
France \\
\href{mailto:abayad@maths.univ-evry.fr}{\tt abayad@maths.univ-evry.fr} \\
\ \\
Yoshinori Hamahata\\
Faculty of Engineering Science \\
Kansai University\\
3-3-35 Yamate-cho, Suita-shi \\
Osaka 564-8680 \\
Japan\\
\href{mailto:hamahata@fc.ritsumei.ac.jp}{\tt hamahata@fc.ritsumei.ac.jp}\\
\end{center}

\vskip .2 in

\begin{abstract}
We introduce  two kinds of $q$-Euler polynomials and numbers, 
and investigate many of their interesting properties.
In particular, we establish $q$-symmetry properties of these
$q$-Euler polynomials, from which we recover the so-called  
Kaneko-Momiyama identity for the ordinary Euler polynomials, 
discovered recently by  Wu, Sun, and Pan. Indeed, 
a $q$-symmetry and $q$-recurrence formulas among sum of product 
of these $q$-analogues Euler numbers and polynomials  are obtained. 
As an application, from these  $q$-symmetry formulas 
we deduce non-linear recurrence formulas for the product of the 
ordinary Euler numbers and polynomials.
\end{abstract}


\section{Introduction and preliminaries}
\subsection{The ordinary Euler numbers and polynomials: An analytic
  overview} 
Let $\mathbb{N}=\{0,1,2,\ldots\}$. The ordinary Euler polynomials
$E_n(x)$ are defined by the generating series
$$
\frac{2e^{xt}}{e^t+1}=
\sum_{n=0}^{\infty}E_n(x)
\frac{t^n}{n!}.
$$
The first few values are
\begin{eqnarray*}
E_0(x)=1,
E_1(x)=x-\frac{1}{2},
E_2(x)=x^2-x,
E_3(x)=x^3-\frac{3}{2}x^2+\frac{1}{4}.
\end{eqnarray*}
The ordinary Euler numbers $E_n$ ($n=0, 1, 2, \ldots$) are defined by the generating function
\begin{eqnarray}\label{Euler_numbers}
\frac{2}{e^t+1}=
\sum_{n=0}^{\infty}E_n
\frac{t^n}{n!}.
\end{eqnarray}
The $n^{\textrm{th}}$-Euler number and polynomial are connected by the equation: $E_n=E_n(0).$
The first few values are
$E_0=1$, $E_1=-1/2$, $E_2=0$, $E_3=1/4$, and it holds that
$E_{2k}=0$
($k=1, 2, 3, \ldots$).
\begin{remark}\label{schol1}

 Note that the Euler numbers $E_n$ which we consider in this paper are 
different from the Euler numbers defined by  M. Abramowitz and I. A. Stegun
\cite[Ch.23]{[1]}. 

\end{remark}
From the definition we can easily deduce the following well-known difference formula:
\begin{eqnarray}\label{differenceformula}
(-1)^nE_n(-x)+E_n(x)=2x^n, (n\in\mathbb{N}).
\end{eqnarray}
In 2004,  K.J. Wu, Z.W. Sun, and H. Pan \cite{WSP} proved the
following important formulae: 

\begin{eqnarray}
&amp;&amp;(-1)^m\sum_{k=0}^m
{m\choose k}E_{n+k}(x)=
 (-1)^n\sum_{k=0}^n
{n\choose k}E_{m+k}(-x),\label{Sun}\\
&amp;&amp;(-1)^m\sum_{k=0}^{m+1}
{m+1\choose k}(n+k+1)E_{n+k}(x)\notag\\
&amp;&amp;\quad +
 (-1)^n\sum_{k=0}^{n+1}
{n+1\choose k}(m+k+1)E_{m+k}(x)=0.\label{Kaneko-Momiyama-Sun}
\end{eqnarray}
The last identity (\ref{Kaneko-Momiyama-Sun}) is an Euler polynomial 
version of Kaneko-Momiyama relations among Bernoulli numbers. 
See M. Kaneko \cite{Kaneko}, H. Momiyama \cite{Momiyama}, 
I. M. Gessel \cite{Gessel2} 
and Wu-Sun-Pan \cite{WSP} for details. 

The identity (\ref{Sun})  
can be viewed as an integral version of  the Kaneko-Momiyama type 
identity for the Euler polynomials. 
In this present paper, we introduce and 
investigate two kinds of $q$-Euler polynomials and numbers. 
For instance, we find $q$-analogues for the identities (\ref{Sun}) 
and (\ref{Kaneko-Momiyama-Sun}). On the other hand, 
we also establish a relation between sums of products of our $q$-Euler polynomials.



\subsection{$q$-shifted factorials} Let $a\in\mathbb{C}$.
The {\it $q$-shifted factorials} are defined by
$$
(a,q)_0=1,\quad
(a,q)_n=
\prod_{k=0}^{n-1}(1-aq^k)\quad (n=1,2,\ldots ).
$$
If $|q|&lt;1$, then we define
$$
(a,q)_{\infty}=\lim_{n\to\infty}(a,q)_n
=\prod_{k=0}^{\infty}(1-aq^k).
$$
We also denote
\begin{eqnarray*}
\left[x\right]_q &amp;=&amp;
\frac{1-q^x}{1-q},
\quad x\in\mathbb{C},\\
\left[n\right]_{q}! &amp;=&amp;
\frac{(q,q)_n}{(1-q)^n},\quad
n\in\mathbb{N},\\
\begin{bmatrix}n\\ k\end{bmatrix}_q
&amp;=&amp;
\frac{[n]_q!}{[k]_q![n-k]_q!}
,\quad
k,n\in\mathbb{N},\\
\left[\begin{array}{c}n\\
i_1, \ldots ,i_m\end{array}
\right]_q
&amp;=&amp;
\frac{[n]_q!}{[i_1]_q!\cdots [i_m]_q!}
,\quad n, i_1,\ldots , i_m\in
\mathbb{N}, \textrm{ with } i_1+\cdots + i_m=n .
\end{eqnarray*}

\subsection{$q$-Exponential functions}

The {\it $q$-exponential functions}
are given by
\begin{eqnarray}\label{exponential functions}
e_q(z):=\sum_{n=0}^{\infty}\frac{z^n}{[n]_q!},\textrm{ and
 }\quad e_{q^{-1}}(z):=\sum_{n=0}^{\infty} \frac{z^n}{[n]_{q^{-1}}!}.
\end{eqnarray}
It is easy to see that
$[n]_{q^{-1}}!=
q^{-{n\choose 2}}[n]_q!$.
Hence
$$
e_{q^{-1}}(z)=\sum_{n=0}^{\infty}
\frac{q^{n\choose 2}z^n}{[n]_{q}!}.
$$
Recently both $q$-exponential functions are intensively studied 
in $q$-calulus and and quatum theory. See I. M. Gessel \cite{Gessel1}, 
W. P. Johnson \cite{Johnson} for
related topics. As is well-known, these functions are related 
to the infinite product $ \left(z,q\right)_{\infty}$ by
$$
e_q(z)=\frac{1}{\left( (1-q)z,q\right)_{\infty}},
\qquad
e_{q^{-1}}(z)=\left( -(1-q)z,q\right)_{\infty}.
$$
This yields
$
e_q(z)e_{q^{-1}}(-z)=1.
$


\subsection{$q$-Euler polynomials and numbers}

\begin{definition}
We define two kinds of $q$-Euler polynomials
$E_n(x,q)$ and $F_n(x,q^{-1})$
($n=0, 1, 2, \ldots$)
by
\begin{eqnarray*}
\frac{2e_q(xt)}{e_q(t)+1} &amp;=&amp;
\sum_{n=0}^{\infty}E_n(x,q)\frac{t^n}{[n]_q!},\\
\frac{2e_q(xt)}{e_{q^{-1}}(t)+1} &amp;=&amp;
\sum_{n=0}^{\infty}F_n(x,q^{-1})\frac{t^n}{[n]_{q}!}.
\end{eqnarray*}
We call
$E_n(x,q)$ (resp. $F_n(x,q^{-1})$)
the {\it first} (resp. {\it second})
{\it $q$-Euler polynomials}.
In particular, we call
$E_n(0,q)$ (resp. $F_n(0,q^{-1})$)
the
{\it first} (resp. {\it second})
{\it $q$-Euler numbers}.
\end{definition}

\begin{example}
\begin{eqnarray*}
E_0(x,q)&amp;=&amp;1,
E_1(x,q)=x-\frac{1}{2},\\
E_2(x,q)&amp;=&amp;x^2-\frac{[2]_q}{2}x-\frac{1}{2}
+\frac{[2]_q}{4},\\
E_3(x,q)&amp;=&amp;x^3-\frac{[3]_q}{2}x^2
+\left(\frac{[3]_q[2]_q}{4}-\frac{[3]_q}{2}\right)
x-\frac{1}{2}-\frac{[3]_q[2]_q}{8}+\frac{[3]_q}{2}.
\end{eqnarray*}
\begin{eqnarray*}
F_0(x,q^{-1})&amp;=&amp;1,
F_1(x,q^{-1})=x-\frac{1}{2},\\
F_2(x,q^{-1})&amp;=&amp;x^2-\frac{[2]_q}{2}x-\frac{q}{2}
+\frac{[2]_q}{4},\\
F_3(x,q^{-1})&amp;=&amp;x^3-\frac{[3]_q}{2}x^2
+\left(\frac{[3]_q[2]_q}{4}-\frac{[3]_q}{2}q\right)
x-\frac{q^3}{2}-\frac{[3]_q[2]_q}{8}+\frac{[3]_q}{2}q.
\end{eqnarray*}
\end{example}
\begin{remark}\
\begin{enumerate}
\item     The reason for introducing both kinds of  
$q$-analogue Bernoulli polynomials $E_n(x,q)$ and $F_n(x,q^{-1})$ 
is that they are needed in the $q$-analogues of symmetry, difference, 
recurrence and complementary argument formulas.
\item The case $q=1$ corresponds to the ordinary Euler polynomials and numbers.
\item In the literature there are many $q$-analogues of the Euler
  numbers 
and polynomials.  The $q$-analogues which we consider here are closely
related to  
$q$-calculus, $q$-Jackson integral and hypergeometric series.
\item Various $q$-analoques of the Euler numbers and polynomials 
are studied by many mathematicians. For more details for example 
you can refer to T. Kim \cite{Kim1},C. S. Ryoo \cite{Ryoo}, Y. Simsek \cite{Simsek} and others.
\item It seems to be difficult to clarify the connections between all
  the 
$q$-analogues of the Euler numbers and polynomials.
\end{enumerate}
\end{remark}
\section{$q$-Recurrence, $q$-Addition, $q$-Derivative and $q$-integral formulae}
In this section, we establish a series of formulas involving the  
polynomials $E_n(x,q)$ and $F_n(x,q^{-1})$, like  $q$-addition,
$q$-derivative, 
$q$-integral and $q$-recurrence formulas.
\subsection{$q$-Recurrence formulae}
\begin{proposition}[$q$-Recurrence formula]
For any $n\geq 1$, we have
\begin{eqnarray*}
E_n(x,q)&amp;=&amp;x^n-\frac{1}{2}
\sum_{i=0}^{n-1}\begin{bmatrix}n\\ i\end{bmatrix}_q
E_i(x,q),\\
F_n(x,q^{-1})&amp;=&amp;x^n-\frac{1}{2}
\sum_{i=0}^{n-1}\begin{bmatrix}n\\ i\end{bmatrix}_q
F_i(x,q^{-1})q^{\binom{n-i}{2}}.
\end{eqnarray*}
\end{proposition}

\begin{proof} As for the first identity,
we make use of
$$
\left(\sum_{n=0}^{\infty}E_n(x,q)\frac{t^n}{[n]_q!}
\right)(e_q(t)+1)=2e_q(xt).
$$
We deduce from this identity
$$
\sum_{n=0}^{\infty}\left(
\sum_{i=0}^n\begin{bmatrix}n\\ i\end{bmatrix}_q
E_i(x,q)+E_n(x,q)\right)
\frac{t^n}{[n]_q!}=
\sum_{n=0}2x^n
\frac{t^n}{[n]_q!},
$$
which yields the result.
We get the second result in the similar way.
\end{proof}

\par
As $q\to 1$, one has a recurrence formula for
the ordinary Euler polynomials:
\begin{eqnarray*}
E_n(x)=x^n-\frac{1}{2}
\sum_{i=0}^{n-1}\binom{n}{i}
E_i(x)\qquad (n\geq 1),
\end{eqnarray*}
then for the Euler numbers $E_{2n+1}$ we recover the well-known recurrence formula
\begin{eqnarray}\label{Euler_recurrence_1}
E_{2n+1}(x)=-\frac{1}{2}
\sum_{k=0}^{2n}\binom{2n+1}{k}
E_k\qquad (n\geq 0).
\end{eqnarray}

\subsection{$q$-Derivative and $q$-integral}

The {\it $q$-derivative} of a function $f$ is given by
$$
D_qf(x):=
\frac{f(x)-f(qx)}{(1-q)x}\qquad
(x\ne 0, q\ne 1),
$$
where $x$ and $qx$ should be in the domain of $f$.
If $f$ is differentiable on an open set $I$, then
for all $x\in I$,
$$
\lim_{q\to 1}D_qf(x)=f'(x).
$$
Besides, for all $n\in\mathbb{N}$,
\begin{eqnarray*}
D_q(x^n)&amp;=&amp;
[n]_qx^{n-1},
D_q(x,q)_n=
-[n]_q(xq,q)_{n-1},\\
D_{q^{-1}}(x,q)_n&amp;=&amp;
-[n]_q(x,q)_{n-1},
D_q\left(
\frac{x^n}{[n]_q!}
\right) =
\frac{x^{n-1}}{[n-1]_q!}.
\end{eqnarray*}
From the last identity, for instance, we have
$D_qe_q(x)=e_q(x)$.\\
  Our $q$-Euler polynomials form ``$q$-Appell sequences'':
\begin{proposition}[$q$-Derivative formula]
For any $n\geq 0$, we have
\begin{eqnarray*}
D_qE_{n+1}(x,q)&amp;=&amp;[n+1]_qE_n(x,q),\\
D_qF_{n+1}(x,q^{-1})&amp;=&amp;[n+1]_qF_n(x,q^{-1}).
\end{eqnarray*}
\end{proposition}

\begin{proof}
Since
$$
\sum_{n=0}^{\infty}D_qE_n(x,q)
\frac{t^n}{[n]_q!}=
\frac{2te_q(xt)}{e_q(t)+1}
=\sum_{n=1}^{\infty}[n]_qE_{n-1}(x,q)
\frac{t^n}{[n]_q!},
$$
we have the first identity.
The second identity can be obtained similarly.
\end{proof}

\par
As $q\to 1$, we have the identities of
Appell sequences of the ordinary
Euler polynomials:
\begin{eqnarray*}
\frac{d}{dx}E_{n+1}(x)=(n+1)E_n(x).
\end{eqnarray*}

\par
For the product of two functions
$f$ and $g$, the following formula holds:
\begin{eqnarray*}
D_q(f\cdot g)(x)&amp;=&amp;
g(x)D_q(x)+f(qx)D_qg(x)\\
&amp;=&amp;
f(x)D_qg(x)+g(qx)D_qf(x).
\end{eqnarray*}
We next treat the composition of $f(x)$ and $g(x)$.
When $g(x)=-x$,
the following
chain rule for the $q$-derivative
is valid:
$$
D_q(f\circ g)(x)=D_qf(g(x))D_qg(x),
$$
which will be used in the proofs
of Theorems \ref{3.5} and \ref{3.9}.
However, in general, the rule above
does not hold.
If we modify the definition of
the composition of two functions,
then a new chain rule for the
$q$-derivative is gained.
We refer to I. M. Gessel \cite{Gessel1} for this
topic.


The $q$-{\it Jackson integral} from $0$
to $a$
is defined by
\begin{eqnarray*}
\int_0^af(x)d_qx &amp;:=&amp;
(1-q)a\sum_{n=0}^{\infty}f(aq^n)q^n
\end{eqnarray*}
provided the infinite sums converge absolutely.
The $q$-Jackson integral in the generic
interval $[a,b]$ is given by
$$
\int_a^bf(x)d_qx=
\int_0^bf(x)d_qx-
\int_0^af(x)d_qx.
$$
For any function $f$ we have
$$
D_q\int_0^xf(t)d_qt=
f(x).
$$

\begin{proposition}[$q$-Integral formula]
For any $n\geq 0$,
\begin{eqnarray*}
\int_a^xE_n(t,q)d_qt&amp;=&amp;
\frac{E_{n+1}(x,q)-E_{n+1}(a,q)}
{[n+1]_q},\\
\int_a^xF_n(t,q^{-1})d_qt&amp;=&amp;
\frac{F_{n+1}(x,q^{-1})-F_{n+1}(a,q^{-1})}
{[n+1]_q}.
\end{eqnarray*}
\end{proposition}
This result  follows from $q$-derivative
formula. As $q\to 1$, we have integral formula
for the classical Euler polynomials:

\begin{equation*}
\int_a^xE_n(t)dt=
\frac{E_{n+1}(x)-E_{n+1}(a)}
{n+1}.
\end{equation*}



\subsection{$q$-Binomial formula}

Let $q\in\mathbb{C}$, and take two
$q$-commuting variables $x$ and $y$
which satisfy the relation
$$
xy=q^{-1}yx.
$$
Let $\mathbb{C}_q[x,y]$
be the complex associative algebra with
$1$ generated by $x$ and $y$.
Then the following identity is valid
in the algebra
$\mathbb{C}_q[x,y]$:
$$
(x+y)^n=
\sum_{k=0}^n
\begin{bmatrix}n\\ k\end{bmatrix}_q
x^ky^{n-k},\qquad n\in\mathbb{N},
$$
or alternatively,
$$
(x+y)^n=
\sum_{k=0}^n
\begin{bmatrix}n\\ k\end{bmatrix}_{q^{-1}}
y^kx^{n-k},\qquad n\in\mathbb{N}.
$$
For details, we refer to 
Andrews-Askey-Roy
\cite{AAR}, Gasper-Rahman \cite{Gasper-Rahman}.


\subsection{$q$-Exponential identity}
Let $x, y$ be the
$q$-commuting variables
satisfying the relation
$xy=q^{-1}yx$.
Let $\mathbb{C}_q[[x,y]]$
be the complex associative algebra with $1$
of formal power series
$$
\sum_{m=0}^{\infty}\sum_{n=0}^{\infty}
a_{m,n}x^my^n
$$
with arbitrary complex coefficients
$a_{m,n}$. One knows in
Andrews-Askey-Roy \cite{AAR}, 
Gasper-Rahman \cite{Gasper-Rahman} that
in $\mathbb{C}_q[[x,y]]$,
we have the following identity
$$
e_q(x+y)=e_q(x)e_q(y).
$$

\begin{proposition}[$q$-Addition formula] Let $x, y$ be the
$q$-commuting variables
satisfying the relation
$xy=q^{-1}yx$.
For any $n\geq 0$, we have
\begin{eqnarray*}
E_n(x+y,q)&amp;=&amp;
\sum_{k=0}^n
\begin{bmatrix}n\\ k\end{bmatrix}_q
E_k(x,q)y^{n-k},\\
F_n(x+y,q^{-1})&amp;=&amp;
\sum_{k=0}^n
\begin{bmatrix}n\\ k\end{bmatrix}_q
F_k(x,q^{-1})y^{n-k}.
\end{eqnarray*}
Particularly, it follows that
\begin{eqnarray*}
E_n(y,q)&amp;=&amp;\sum_{k=0}^n
\begin{bmatrix}n\\ k\end{bmatrix}_q
E_k(0,q)y^{n-k},\\
F_n(y,q^{-1})&amp;=&amp;\sum_{k=0}^n
\begin{bmatrix}n\\ k\end{bmatrix}_q
F_k(0,q^{-1})y^{n-k}.
\end{eqnarray*}
\end{proposition}

\begin{proof}
 The first identity follows from
$$
\frac{2e_q((x+y)t)}{e_q(t)+1}=
\frac{2e_q(xt)}{e_q(t)+1}\cdot
e_q(yt).
$$
One can easily prove the remaining
identities.
\end{proof}

\par
As $q\to 1$, we have the classical formula:

\begin{equation*}
E_{n+1}(x+y)=
\sum_{k=0}^n
\binom{n}{k}
E_k(x)y^{n-k}.
\end{equation*}
Particularly, it holds that
\begin{equation*}
E_n(y)=\sum_{k=0}^n
\binom{n}{k}
E_ky^{n-k}.
\end{equation*}

At the end of this section,
we give a list of limit
of $q$-analogues.
\begin{eqnarray*}
\lim_{q\to 1}e_q(z) &amp;=&amp;
\lim_{q\to 1}e_{q^{-1}}(z)
=e^z,\\
\lim_{q\to 1}[n]_q &amp;=&amp; n,\\
\lim_{q\to 1}[n]_q! &amp;=&amp; n!,\\
\lim_{q\to 1}
\begin{bmatrix}n\\ k\end{bmatrix}_q
&amp;=&amp; {n\choose k},\\
\lim_{q\to 1}\left[
\begin{array}{c}n\\
i_1,\ldots ,i_m\end{array}
\right]_q
&amp;=&amp;
\left(
\begin{array}{c}n\\
i_1,\ldots ,i_m\end{array}
\right)
:=
\frac{n!}{i_1!\cdots i_m!},\\
\lim_{q\to 1}E_n(x,q) &amp;=&amp;
\lim_{q\to 1}F_n(x,q^{-1})=
E_n(x).
\end{eqnarray*}



\section{$q$-symmetry and $q$-analogues to Kaneko-Momiyama identities}

\subsection{$q$-symmetry fo Sums of products}

\begin{theorem}[Sums of products]\label{3.1}
Let $m$ be a given positive integer.
Then for any $n\geq 0$, we have
\begin{multline*}
(-1)^n
\sum_{i_1+\cdots +i_m=n}
\left[\begin{array}{c}n\\
i_1,\ldots ,i_m\end{array}\right]_q
F_{i_1}(-x,q^{-1})\cdots
F_{i_m}(-x,q^{-1})\\
=\sum_{j=0}^m(-1)^j2^{m-j}
\binom{m}{j}
\sum_{k_1+\cdots k_m=n}
\left[\begin{array}{c}n\\
k_1,\ldots ,k_m\end{array}\right]_q
E_{k_1}(x,q)\cdots E_{k_j}(x,q)
x^{n-(k_1+\cdots +k_j)}.
\end{multline*}
\end{theorem}

\begin{remark}\label{3.1bis} The above theorem implies the following results. 
\begin{enumerate}
\item If $m=1$, then
\begin{eqnarray}\label{E-F}
(-1)^nF_n(-x,q^{-1})+E_n(x,q)=2x^n.
\end{eqnarray}
This relation can be viewed as a $q$-difference formula. 
If $q\to 1$ we recover the usual difference formula for the Euler polynomials (\ref{differenceformula}).
\item If $m=2$, then
\begin{eqnarray*}
&amp;&amp;(-1)^n\sum_{i=0}^n
\begin{bmatrix}n\\ i\end{bmatrix}_q
F_i(-x,q^{-1})F_{n-i}(-x,q^{-1})\\
&amp;&amp;\hspace{2cm}=\sum_{i=0}^n
\begin{bmatrix}n\\ i\end{bmatrix}_q
E_i(x,q)E_{n-i}(x,q)
-4\sum_{i=0}^n
\begin{bmatrix}n\\ i\end{bmatrix}_q
E_i(x,q)x^{n-i}
+4x^n\sum_{i=0}^n
\begin{bmatrix}n\\ i\end{bmatrix}_q.
\end{eqnarray*}
\item It should be noted that Simsek \cite{Simsek} 
found formulae for sums of products of another kind of $q$-Euler polynomials.
\end{enumerate}
\end{remark}

\begin{proof}
In view of $e_q(t)e_{q^{-1}}(-t)=1$, we have
$$
\frac{1}{e_{q^{-1}}(-t)+1}=1-
\frac{1}{e_q(t)+1}.
$$
Hence for $m\geq 1$,
$$
\left(\frac{2e_q((-x)(-t))}{e_{q^{-1}}(-t)+1}
\right)^m
=\left(2e_q(xt)-
\frac{2e_q(xt)}{e_q(t)+1}\right)^m.
$$
The left-hand side of the identity is
$$
\sum_{n=0}^{\infty}(-1)^n
\sum_{i_1+\cdots +i_m=n}
\left[\begin{array}{c}n\\
i_1,\ldots ,i_m\end{array}\right]_q
F_{i_1}(-x,q^{-1})\cdots
F_{i_m}(-x,q^{-1})
\frac{t^n}{[n]_q!}.
$$
The right-hand side becomes
\begin{eqnarray*}
&amp;&amp;\sum_{j=0}^m(-1)^j
\binom{m}{j}
\left(\frac{2e_q(xt)}{e_q(t)+1}\right)^j
2^{m-j}e_q(xt)^{m-j}\\
&amp;&amp;=
\sum_{j=0}^m(-1)^j
2^{m-j}\binom{m}{j}
\sum_{n=0}^{\infty}
\sum_{k_1+\cdots +k_m=n}
\left[\begin{array}{c}n\\
k_1,\ldots ,k_m\end{array}\right]_q
E_{k_1}(x,q)\cdots
E_{k_j}(x,q)x^{k_{j+1}}\cdots
x^{k_m}
\frac{t^n}{[n]_q!}.
\end{eqnarray*}
\end{proof}


As $q\to 1$ in the formula of Theorem \ref{3.1}, we have

\begin{theorem}\label{3.2}
Let $m$ be a given positive integer.
Then for any $n\geq 0$,
\begin{multline*}
(-1)^n
\sum_{i_1+\cdots +i_m=n}
\left(\begin{array}{c}n\\
i_1,\ldots ,i_m\end{array}\right)
E_{i_1}(-x)\cdots
E_{i_m}(-x)\\
=\sum_{j=0}^m(-1)^j2^{m-j}
\binom{m}{j}
\sum_{k_1+\cdots k_m=n}
\left(\begin{array}{c}n\\
k_1,\ldots ,k_m\end{array}\right)
E_{k_1}(x)\cdots E_{k_j}(x)
x^{n-(k_1+\cdots +k_j)}.
\end{multline*}
\end{theorem}
\begin{corollary}\label{3.2bis}
Especially in the cases $m=1, 2$, the following
results hold:
\par
{\rm (1)} For any $n\geq 0$, we have
$(-1)^nE_n(-x)+E_n(x)=2x^n$.
\par
{\rm (2)} For any $k\geq 1$, $E_{2k}=0$.
\par
{\rm (3)} For any $n\geq 0$, we get the {\bf non-linear recurrence} formulae
$$
(-1)^n\sum_{i=0}^n\binom{n}{i}E_i(-x)E_{n-i}(-x)
=\sum_{i=0}^n\binom{n}{i}
E_i(x)E_{n-i}(x)-4\sum_{i=0}^n\binom{n}{i}
E_i(x)x^{n-i}+2^{n+2}x^n.
$$
\par
{\rm (4)} If $n$ is an odd positive integer, 
then we obtain the well-known  {\bf Euler non-linear recurrence} formula
$$
\sum_{i=0}^n\binom{n}{i}E_iE_{n-i}=2E_n.
$$
\end{corollary}


\subsection{$q$-Symmetry}

\begin{theorem}[$q$-Symmetry 1]\label{3.3}
For any $m, n\in\mathbb{N}$, we have
\begin{equation}\label{Momiyama}
(-1)^m\sum_{k=0}^m
\begin{bmatrix}m\\ k\end{bmatrix}_q
E_{n+k}(x,q)
q^{-kn+mn}
=
(-1)^n\sum_{k=0}^n
\begin{bmatrix}n\\ k\end{bmatrix}_{q^{-1}}
F_{m+k}(-x,q^{-1})
q^{\binom{n}{2}-\binom{k}{2}}.
\end{equation}
\end{theorem}
This identity (\ref{Momiyama}) can be viewed as a $q$-analogue to the  
polynomial version of the integral Kaneko-Momiyama's formulae on Euler polynomials (\ref{Sun}).\\

\begin{proof}
Let $x, y$ be two $q$-commuting
variables with $xy=q^{-1}yx$.
We compute the generating functions

\begin{eqnarray*}
L(w,x,y) &amp;=&amp;
\sum_{m=0}^{\infty}\sum_{n=0}^{\infty}
(-1)^m\sum_{k=0}^m
\begin{bmatrix}m\\ k\end{bmatrix}_q
E_{n+k}(w,q)q^{-kn+mn}
\frac{x^m}{[m]_q!}\frac{y^n}{[n]_q!},\\
R(w,x,y) &amp;=&amp;
\sum_{m=0}^{\infty}\sum_{n=0}^{\infty}
(-1)^n\sum_{k=0}^n
\begin{bmatrix}n\\ k\end{bmatrix}_q
F_{m+k}(-w,q^{-1})q^{\binom{n}{2}-\binom{k}{2}}
\frac{x^m}{[m]_{q}!}\frac{y^n}{[n]_{q}!},
\end{eqnarray*}
where $w$ is a commuting variable
with $x$ and $y$.
\begin{eqnarray*}
L(w,x,y)
&amp;=&amp;
\sum_{m=0}^{\infty}\sum_{n=0}^{\infty}
(-1)^m\sum_{k=0}^m
\begin{bmatrix}m\\ k\end{bmatrix}_q
E_{n+k}(w,q)q^{-kn}
\frac{y^n}{[n]_q!}\frac{x^m}{[m]_q!}\\
&amp;=&amp;
\sum_{m=0}^{\infty}\sum_{n=0}^{\infty}
(-1)^m\sum_{k=0}^m
E_{n+k}(w,q)q^{-kn}
\frac{y^n}{[n]_q!}
\frac{x^k}{[k]_q!}
\frac{x^{m-k}}{[m-k]_q!}\\
&amp;=&amp;
\sum_{j=0}^{\infty}
\sum_{k=0}^{\infty}\sum_{n=0}^{\infty}
E_{n+k}(w,q)
\frac{(-x)^k}{[k]_q!}
\frac{y^n}{[n]_q!}
\frac{(-x)^{j}}{[j]_q!}\\
&amp;=&amp;
\left(
\sum_{i=0}^{\infty}E_i(x,q)
\sum_{k=0}^i
\frac{(-x)^k}{[k]_q!}
\frac{y^{i-k}}{[i-k]_q!}
\right)
e_q(-x)\\
&amp;=&amp;
\frac{2e_q(w(y-x))}{e_q(y-x)+1}
e_q(-x).
\end{eqnarray*}
\begin{eqnarray*}
R(w,x,y)&amp;=&amp;
\sum_{m=0}^{\infty}\sum_{n=0}^{\infty}
(-1)^n\sum_{k=0}^n
F_{m+k}(-w,q^{-1})
\frac{x^m}{[m]_q!}\frac{y^k}{[k]_q!}
\frac{q^{\binom{n-k}{2}}y^{n-k}}{[n-k]_q!}\\
&amp;=&amp;
\sum_{j=0}^{\infty}\sum_{k=0}^{\infty}
\sum_{m=0}^{\infty}(-1)^{j+k}
F_{m+k}(-w,q^{-1})
\frac{x^m}{[m]_{q}!}
\frac{y^k}{[k]_{q}!}
\frac{y^j}{[j]_{q^{-1}}!}\\
&amp;=&amp;
\left(
\sum_{m=0}^{\infty}\sum_{k=0}^{\infty}
F_{m+k}(-w,q^{-1})
\frac{x^m}{[m]_{q}!}
\frac{(-y)^k}{[k]_{q}!}
\right)
e_q(-y)\\
&amp;=&amp;
\frac{2e_q(-w(y-x))}{e_{q^{-1}}(x-y)+1}
e_{q^{-1}}(-y).
\end{eqnarray*}
Hence it follows that
\begin{eqnarray*}
R(w,x,y)e_q(y) &amp;=&amp;
\frac{2e_q(w(y-x))}{e_{q^{-1}}(x-y)+1}\\
&amp;=&amp;
\frac{2e_q(w(y-x))}{e_q(y-x)+1}
e_q(y-x)\\
&amp;=&amp;L(w,x,y)e_q(y),
\end{eqnarray*}
which provides $R(w,x,y)=L(w,x,y)$.
Therefore we can complete the proof.
\end{proof}

As $q\to 1$ in (\ref{Momiyama}) of Theorem \ref{3.3}, we have a symmetric relation
for the ordinary Euler polynomials:

\begin{theorem}\label{3.4}
For any $m, n\in\mathbb{N}$, we have
\begin{equation*}
(-1)^m\sum_{k=0}^m
\begin{pmatrix}m\\ k\end{pmatrix}
E_{n+k}(x)=
(-1)^n\sum_{k=0}^n
\begin{pmatrix}n\\ k\end{pmatrix}
E_{m+k}(-x).
\end{equation*}
\end{theorem}

\begin{theorem}[$q$-Symmetry 2]\label{3.5}
For any $m, n\in\mathbb{N}$, we have
\begin{multline}\label{(2)}
(-1)^m\sum_{k=0}^{m+1}
\begin{bmatrix}m+1\\ k\end{bmatrix}_q
[n+k+1]_q
E_{n+k}(x,q)
q^{-k(n+1)-{n\choose 2}+mn+1}\\
+
(-1)^n\sum_{k=0}^{n+1}
\begin{bmatrix}n+1\\ k\end{bmatrix}_{q^{-1}}
[m+k+1]_{q^{-1}}
F_{m+k}(-x,q^{-1})
q^{k(m+1)+{m\choose 2}}=0.
\end{multline}
\end{theorem}

\begin{proof}
Applying $q$-derivative formula to
the identity (\ref{Momiyama}) in Theorem \ref{3.3}
replaced $m$, $n$ by
$m+1$, $n+1$, respectively, we have the result.
\end{proof}

As $q\to 1$ in (\ref{(2)}) of Theorem \ref{3.5},
we have another symmetric formula
for the ordinary Euler polynomials:

\begin{theorem}\label{3.6}
For any $m, n\in\mathbb{N}$, we have
\begin{equation}\label{(3)}
(-1)^m\sum_{k=0}^{m+1}
\begin{pmatrix}m+1\\ k\end{pmatrix}
(n+k+1)E_{n+k}(x)
+
(-1)^n\sum_{k=0}^{n+1}
\begin{pmatrix}n+1\\ k\end{pmatrix}
(m+k+1)E_{m+k}(-x)=0.
\end{equation}
\end{theorem}
This can be regarded as an Euler polynomial version
of Kaneko-Momiyama formulae for Bernoulli numbers.
To be precise, put $m=n$ and $x=0$ in (\ref{(3)}). Then
we have an analogue of Kaneko's formula:

\begin{theorem}\label{3.7} For any $n\in\mathbb{N}$,
$$
\sum_{k=0}^{n+1}\binom{n+1}{k}
(n+k+1)E_{n+k}=0.
$$
\end{theorem}

Then we obtain the nice formula
\begin{eqnarray}\label{Euler_recurrence_2}
E_{2n+1}=-\frac1{n+1}\sum_{k=0}^{n}
{n+1\choose k}(n+k+1)E_{n+k}.
\end{eqnarray}
\begin{remark}
The formula (\ref{Euler_recurrence_2}) has a strong resemblance to the
usual recurrence
 (\ref{Euler_recurrence_1}), ($0\leq k\leq ~2n$).  
Using  the formula (\ref{Euler_recurrence_1})  
and according to the fact that $E_k=0$ for $k$ even positive
integers, 
we need the $n$ first terms with odd indexes $k$  to compute
$E_{2n+1}$. 
But the recurrence formula (\ref{Euler_recurrence_2}) needs only 
half the number of those terms ($E_k$ with $n\leq k\leq 2n$ with $k$ odd) to calculate $E_{2n+1}$.
\end{remark}
Put $x=0$ in (\ref{(3)}). Then we have an analogue of
Momiyama's formula:

\begin{theorem}\label{3.8}
For any $m, n\in\mathbb{N}$, we have
\begin{equation*}
(-1)^m\sum_{k=0}^{m+1}
\begin{pmatrix}m+1\\ k\end{pmatrix}
(n+k+1)E_{n+k}
+
(-1)^n\sum_{k=0}^{n+1}
\begin{pmatrix}n+1\\ k\end{pmatrix}
(m+k+1)E_{m+k}=0.
\end{equation*}
\end{theorem}

\par
Using
$q$-integral formula to (\ref{Momiyama}) in Theorem \ref{3.3}, we have

\begin{theorem}[$q$-Symmetry 3]\label{3.9}
For any $m, n\in\mathbb{N}$ and $a, b\in\mathbb{R}$,
\begin{multline*}
(-1)^m\sum_{k=0}^m
\begin{bmatrix}m\\ k\end{bmatrix}_q
\frac{E_{n+k+1}(a,q)-E_{n+k+1}(b,q)}{[n+k+1]_q}
q^{-kn+mn}\\
+
(-1)^n\sum_{k=0}^n
\begin{bmatrix}n\\ k\end{bmatrix}_{q^{-1}}
\frac{F_{m+k+1}(-a,q^{-1})-F_{m+k+1}(-b,q^{-1})}{[m+k+1]_q}
q^{\binom{n}{2}-\binom{k}{2}}=0.
\end{multline*}
\end{theorem}

As $q\to 1$, we get

\begin{theorem}\label{3.10}
For any $m, n\in\mathbb{N}$ and $a, b\in\mathbb{R}$,
\begin{equation*}
(-1)^m\sum_{k=0}^m
\binom{m}{k}
\frac{E_{n+k+1}(a)-E_{n+k+1}(b)}{n+k+1}
+
(-1)^n\sum_{k=0}^n
\binom{n}{k}
\frac{E_{m+k+1}(-a)-E_{m+k+1}(-b)}{m+k+1}
=0.
\end{equation*}
\end{theorem}




\section{Acknowledgement}

The second named author was supported by
Grant-in-Aid for Scientific Research
(No.\ 20540026),
Japan Society for the Promotion of
Science.


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\end{thebibliography}

\bigskip
\hrule
\bigskip

\noindent 2010 {\it Mathematics Subject Classification}:
Primary 11B68; Secondary 05A30.

\noindent \emph{Keywords: } 
$q$-analogue, Euler number, Euler polynomial, 
$q$-symmetry, Keneko-Momiyama identity.

\bigskip
\hrule
\bigskip

\vspace*{+.1in}
\noindent
Received  February 15 2012;
revised version received  March 31 2012.
Published in {\it Journal of Integer Sequences}, April 9 2012.

\bigskip
\hrule
\bigskip

\noindent
Return to
\htmladdnormallink{Journal of Integer Sequences home page}{http://www.cs.uwaterloo.ca/journals/JIS/}.
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