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\begin{center}
\vskip 1cm{\LARGE\bf
Convolutions with the Bernoulli \\
\vskip .1in
and Euler Numbers}
\vskip 1cm
\large
Meryem Bouzeraib and Ali Boussayoud\\
LMAM Laboratory and Department of Mathematics\\
Mohamed Seddik Ben Yahia University \\
Jijel \\
Algeria\\
\href{mailto:meryembouzeraib@gmail.com}{\tt meryembouzeraib@gmail.com}\\
\href{mailto:aboussayoud@yahoo.fr}{\tt aboussayoud@yahoo.fr}\\
\ \\
Baghdadi Aloui\\
Higher Institute of Industrial Systems of Gabes\\
Department of Electromechanics\\
City of Erriadh\\
6072 Gabes \\
Tunisia\\
\href{mailto:Baghdadi.Aloui@fsg.rnu.tn}{\tt Baghdadi.Aloui@fsg.rnu.tn}\\
\end{center}

\vskip .2in

\begin{abstract}
We use exponential generating functions to study the
relationship between Bernoulli and Euler numbers with $k$-Jacobsthal numbers,
$k$-Jacobsthal-Lucas  numbers, and bivariate Fibonacci, Lucas,
Pell and Pell-Lucas polynomials.
\end{abstract}


\section{Introduction and preliminaries}\label{Sec1}

Recently, there has been some
study of special numbers and polynomials,
especially the Bernoulli numbers and Euler numbers,
by virtue of their applications in almost all branches of mathematics, computer algorithms, engineering, and other areas.
(For other numbers see~\cite{Chelgham, Dilcher, FA, Zhangg, Zerroug}).

Bernoulli and Euler polynomials are famous mathematical objects and are fairly
well understood. They are, respectively, defined by the exponential
generating functions as follows:
\begin{align}
\sum_{n=0}^{\infty}B_{n}\frac{z^{n}}{n!}&amp;=\frac{z}{\exp(z)-1}, \ |z|&lt;2\pi,\label{mk}\\
\sum_{n\geqslant 0}E_{n}\frac{z^{n}}{n!}&amp;=\frac{2\exp (z)}{\exp (2z)+1}, \ |z|&lt;\pi. \label{16}
\end{align}

 The Euler and  Bernoulli  numbers have appeared in many important results.
 There has been a growing interest in deriving new relations for these two pairs of sequences.
 In 1975, Byrd~\cite{Byrd3} derived the following identity relating Lucas numbers to Euler
 numbers:
 \begin{equation}
 \sum\limits_{l=0}^{\lfloor n/2\rfloor }{n\choose 2l}\Bigl( \frac{5
 }{4}\Bigr) ^{l}L_{n-2l}
 E_{2l}=2^{1-n}.\label{byard}
 \end{equation}

 Zhang and Ma~\cite{Zhang} proved a relation between Fibonacci polynomials and Bernoulli numbers.
 The following identity is a special case of their result:
$$\sum_{l=0}^{n}\binom{n}{l}5^{\frac{n-k}{2}}F_{k}B_{n-k}=n\bigl(\frac{1-\sqrt{5}}{2}\bigr) ^{n-1},$$
or equivalently
$$\sum\limits_{l=0}^{\lfloor n/2\rfloor }\binom{n}{2l}5^{l}
F_{n-2l}B_{2l}=\frac{nL_{n-1}}{2}.$$

 Patel et al.~\cite{Frontczak} obtained similar
identities when Fibonacci and Lucas numbers are replaced, respectively,
by the Pell and Lucas polynomials (defined below in Eqs.~\eqref{P} and
\eqref{Q}):
$$\sum\limits_{l=0}^{n}{n\choose l}
(2\sqrt{x^{2}+ 1})^{l}P_{n-l}(x)B_{l}=n(x-\sqrt{x^{2}+ 1})^{n-1}.$$
This can be stated as
$$\sum\limits_{l=0}^{\lfloor n/2\rfloor }{n\choose 2l}
4^{l}(x^{2}+1)^{l}P_{n-2l}(x)B_{2l}=\frac{nQ_{n-1}(x)}{2}.$$

 Still other relations are contained in the many articles (see~\cite{Frontczak2, FA, F}).
 The present paper is devoted to developing further relations between Bernoulli and Euler
 numbers with  famous number sequences. To achieve this goal, we use elementary methods, including
 exponential generating functions, to study the relationship between Bernoulli
 and Euler numbers and between the
 $k$-Jacobsthal numbers $k$-Jacobsthal-Lucas numbers and
 bivariate Fibonacci, Lucas,
Pell, and Pell-Lucas polynomials.

For every  positive real number $k$, the $k$-Jacobsthal numbers $(J_{n,k})_{n\in \mathbb{N}}$
are defined recursively by the relation~\cite{Jhala}
\begin{equation*}
	J_{k,n}=kJ_{k,n-1}+2J_{k,n-1},\text{\ for }n\geq 2,
\end{equation*}
with initial values $J_{k,0}=0$ and $J_{k,1}=1$.

Jhala et al.~\cite{Jhalad} defined the $k$-Jacobsthal-Lucas numbers as follows:
\begin{equation*}
j_{k,n}=kj_{k,n-1}+2j_{k,n-1},\text{\ for }n\geq 2,\text{ with }
j_{k,0}=2,\text{ } j_{k,1}=k.
\end{equation*}
The well-known Binet formulas for $k$-Jacobsthal  $(J_{n,k})_{n\in \mathbb{N}}$ and $k$-Jacobsthal-Lucas numbers $(j_{n,k})_{n\in \mathbb{N}}$
are given as follows:
\begin{align*}
J_{n,k}&amp;=\frac{\lambda_{1}^{n}-\lambda_{2}^{n}}{\lambda_{1}-\lambda_{2}}  \\
j_{n,k}&amp;=\lambda_{1}^{n}+\lambda_{2}^{n},
\end{align*}
where $\lambda_{1}=\frac{k + \sqrt{k^{2}+8}}{2}$ and $\lambda_{2}=\frac{k - \sqrt{k^{2}+8}}{2}$ are the roots of the characteristic equation $\lambda^{2}-k\lambda-2=0$.

If $k=1$, the classical Jacobsthal and Jacobsthal-Lucas numbers are
\begin{align*}
J_{n}&amp;=J_{n-1}+2J_{n-1},\text{\ for }n\geq 2,\text{ with }
J_{0}=0,\text{ }J_{1}=1,\\
j_{n}&amp;=j_{n-1}+2j_{n-1},\text{\ for }n\geq 2,\text{ with }
j_{0}=2,\text{ } j_{1}=1.
\end{align*}

The bivariate Fibonacci $(F_{n,k}(x,y))_{n\geq 0}$ and bivariate Lucas $(L_{n,k}(x,y))_{n\geq 0}$
polynomial sequences are, respectively, defined by the following recurrence relations~\cite{Catarino, Catarinoo}:
\begin{equation}
F_{n}(x,y)=xF_{n-1}(x,y)+yF_{n-2}(x,y),\text{\ for }n\geq 2,\label{F}
\end{equation}
with initial values $F_{0}(x,y)=0$ and $F_{1}(x,y)=1$, and
\begin{equation}
L_{n}(x,y)=xL_{n-1}(x,y)+yL_{n-2}(x,y), \text{\ for }n\geq 2,\label{L}
\end{equation}
with initial values $L_{0}(x,y)=2$ and $L_{1}(x,y)=x$.

The bivariate Pell and Pell-Lucas polynomials are, respectively, as follows~\cite{Catarinof}:
\begin{equation}
P_{n}(x,y)=2xyP_{n-1}(x,y)+yP_{n-2}(x,y), \ n\geq 2,\label{P}
\end{equation}
with initial values $P_{0}(x,y)=0$ and $P_{1}(x,y)=1$, and
\begin{equation}
Q_{n}(x,y)=2xyQ_{n-1}(x,y)+yQ_{n-2}(x,y), \ n\geq 2,\label{Q}
\end{equation}
with initial values $Q_{0}(x,y)=2$ and $Q_{1}(x,y)=2xy$.

We present the Binet formulas of the bivariate Fibonacci, Lucas,
Pell, and Pell-Lucas polynomials in Table~\ref{k}.
\begin{table}[H]
	\centering
	\begin{tabular}{|l|l|l|}
		\hline
Bivariate polynomials &amp; Roots ($\lambda_{1},\lambda_{1}$)&amp; Binet formula \\
		\hline
	Bivariate Fibonacci polynomials &amp;$\lambda_{1,2}=\frac{x\pm\sqrt{x^{2}+4y}}{2}$&amp;$F_{n}(x,y)=\frac{\lambda_{1}^{n}-\lambda_{2}^{n}}{\lambda_{1}-\lambda_{2}}$\\
		\hline
		Bivariate Lucas polynomials &amp;$\lambda_{1,2}=\frac{x\pm\sqrt{x^{2}+4y}}{2}$&amp;$Q_{n}(x,y)=\lambda_{1}^{n}+\lambda_{2}^{n}$\\
		\hline
			Bivariate Pell polynomials &amp;$\lambda_{1,2}=xy\pm\sqrt{x^{2}y^{2}+y}$&amp;$P_{n}(x,y)=\frac{\lambda_{1}^{n}-\lambda_{2}^{n}}{\lambda_{1}-\lambda_{2}}$\\
		\hline
		Bivariate	Pell-Lucas polynomials &amp;$\lambda_{1,2}=xy\pm\sqrt{x^{2}y^{2}+y}$&amp;$Q_{n}(x,y)=\lambda_{1}^{n}+\lambda_{2}^{n}$\\
		\hline
	\end{tabular}
	\caption{Binet's formulas of the bivariate polynomials.}\label{k}
\end{table}

Putting $y=1$ in the Eqs~(\ref{F}),~(\ref{L}),~(\ref{P}) and~(\ref{Q}) yields,
respectively, the following four polynomials:
\begin{itemize}
\item
the Fibonacci polynomials defined by $F_{0}(x)=0$, $F_{1}(x)=1$, $F_{n}(x)=xF_{n-1}(x)+F_{n-2}(x)$, for $n\geq 2$,

\item
the Lucas polynomials defined by $L_{0}(x)=2$, $L_{1}(x)=x$, $L_{n}(x)=xL_{n-1}(x)+L_{n-2}(x)$, for $n\geq 2$,

\item
the Pell polynomials defined by $P_{0}(x)=0$, $P_{1}(x)=1$, $P_{n}(x)=2xP_{n-1}(x)+P_{n-2}(x)$, for $n\geq 2$,

\item
the Pell-Lucas polynomials defined by $Q_{0}(x)=2$, $Q_{1}(x)=2x$, $Q_{n}(x)=2xQ_{n-1}(x)+Q_{n-2}(x)$, for $n\geq 2$.
\end{itemize}

Putting $x=y=1$ in Eqs.~(\ref{F}),~(\ref{L}),~(\ref{P}) and~(\ref{Q}), we respectively get the Fibonacci, Lucas, Pell,
and Pell-Lucas numbers, which are given as follows:
\begin{itemize}
\item
$F_{n}=F_{n-1}+F_{n-2}, \ n\geq 2,$
with initial values $F_{0}=0$, $F_{1}=1$,

\item
$L_{n}=L_{n-1}+L_{n-2}, \ n\geq 2,$
with initial values $L_{0}=2$, $L_{1}=1$,

\item
$P_{n}=2P_{n-1}+P_{n-2}, \ n\geq 2,$
with initial values $P_{0}=0$, $P_{1}=1$,

\item
$Q_{n}=2Q_{n-1}+Q_{n-2}, \ n\geq 2,$
with initial values $Q_{0}(x)=2$, $Q_{1}=2$.
\end{itemize}

\section{New $k$-Jacobsthal-Bernoulli and $k$-Jacobsthal-Euler relations}\label{Sec2}

In this section, we present our first findings in three theorems, which provide some relations involving Bernoulli and Euler numbers  with $k$-Jacobsthal and $k$-Jacobsthal-Lucas numbers.
\begin{theorem}\label{1}
	Let $n$ be a positive integer. We have
\begin{align}
\sum_{l=0}^{n}\binom{n}{l}\Bigl(\sqrt{k^{2}+8}\Bigr)
		^{l}J_{k,n-l}B_{l}&amp;=n\Bigl( \frac{k - \sqrt{k^{2}+8}}{2}\Bigr)^{n-1}, \nonumber\\
\sum_{l=0}^{n}\binom{n}{l}\Bigl( \sqrt{k^{2}+8}\Bigr)
		^{l}(2^{l}-1)j_{k,n-l}B_{l}&amp;=-n\sqrt{k^{2}+8}\Bigl( \frac{k - \sqrt{k^{2}+8}}{2}\Bigr) ^{n-1}.\label{ff}
\end{align}
\end{theorem}
\begin{proof}
Using the change of variables $z=\sqrt{k^{2}+8}z$ in~(\ref{mk}), we obtain
	\begin{equation}
		\sum_{n=0}^{\infty}\Bigl( \sqrt{k^{2}+8}\Bigr) ^{n}B_{n}\frac{z^{n}}{n!}=\frac{\sqrt{k^{2}+8}z}{\exp (\sqrt{k^{2}+8}z)-1}.  \label{klo}
	\end{equation}
Multiplying~(\ref{klo}) by the exponential generating function for $(J_{k,n})_{n\in \mathbb{N}}$, we have
	\begin{align*}
		\Bigl( \sum_{n=0}^{\infty}J_{k,n}\frac{z^{n}}{n!}\Bigr) \Bigl(
		\sum_{l=0}^{\infty}\Bigl( \sqrt{k^{2}+8}\Bigr) ^{l}B_{l}\frac{z^{l}}{l!}\Bigr)&amp;=\sum_{n=0}^{\infty}\sum_{l=0}^{\infty}\Bigl( \sqrt{k^{2}+8}\Bigr) ^{l}B_{l}\frac{z^{l}}{l!}\frac{z^{n}}{n!}\\
		&amp;=\sum_{n=0}^{\infty}\sum_{l=0}^{n}{n\choose l}\Bigl( \sqrt{k^{2}+8}\Bigr) ^{l}B_{l}J_{k,n-l}\frac{z^{n}}{n!}.
	\end{align*}
Therefore
\begin{align*}
&amp;\frac{ \exp  \Bigl( \frac{k + \sqrt{k^{2}+8}}{2}z\Bigr) -\exp  \Bigl( \frac{k - \sqrt{k^{2}+8}}{2}z\Bigr)}{\sqrt{k^{2}+8}} \times  \frac{
			\sqrt{k^{2}+8}z}{\exp (\sqrt{k^{2}+8}z)-1}\\
&amp;\quad =\exp(\frac{k - \sqrt{k^{2}+8}}{2}z)(\exp (\sqrt{k^{2}+8}z)-1)\times\frac{z}{\exp (\sqrt{k^{2}+8}z)-1}\\
&amp;\quad =z\exp \Bigl( \frac{k - \sqrt{k^{2}+8}}{2}z\Bigr) \\
&amp;\quad =\sum_{n=0}^{\infty}n\Bigl( \frac{k - \sqrt{k^{2}+8}}{2}z\Bigr)^{n-1}\frac{z^{n}}{n!}.
\end{align*}
By comparing the coefficients of $\frac{z^{n}}{n!}$, we obtain the desired result.
	
	Similarly, we use the change of variables $z=2\sqrt{k^{2}+8}z$ in~(\ref{mk}),
	we obtain
	\begin{equation}
		\sum_{n=0}^{\infty}\Bigl( 2\sqrt{k^{2}+8}\Bigr) ^{n}B_{n}\frac{z^{n}}{n!}=\frac{
			2\sqrt{k^{2}+8}z}{\exp (2\sqrt{k^{2}+8}z)-1},  \label{mln}
	\end{equation}
	and multiplying~(\ref{mln}), by the exponential generating function for $(j_{n,k})_{n\in \mathbb{N}}$,  we get
	\begin{align*}
		\Bigl( \sum_{n=0}^{\infty}j_{k,n}\frac{z^{n}}{n!}\Bigr)
		\Bigl( \sum_{l=0}^{\infty}\Bigl( 2\sqrt{k^{2}+8}\Bigr) ^{l}B_{l}\frac{z^{l}}{l!}
		\Bigr)&amp;= \sum_{n=0}^{\infty}\sum_{l=0}^{\infty}
		\Bigl( 2\sqrt{k^{2}+8}\Bigr) ^{l}B_{l}j_{k,n}\frac{z^{l}}{l!}\frac{z^{n}}{n!}
		\\
		&amp;=\sum_{n=0}^{\infty}\sum\limits_{l=0}^{n}{n\choose l}
		2^{l}\Bigl( \sqrt{k^{2}+8}\Bigr) ^{l}j_{k,n-l} B_{l}\frac{z^{n} }{n!}.
	\end{align*}
Then we obtain
	\begin{align*}
&amp;\Bigl(\exp\Bigl(\frac{(k + \sqrt{k^{2}+8})z}{2}\Bigr)+\exp\Bigl( \frac{(k - \sqrt{k^{2}+8})z}{2}\Bigr)\Bigr)\frac{2\sqrt{k^{2}+8}z}{\exp (2\sqrt{k^{2}+8}z)-1}\\
&amp; \quad =\exp \Bigl( \frac{(k - \sqrt{k^{2}+8})z}{2}\Bigr)\frac{2\sqrt{k^{2}+8}z}{\exp(\sqrt{k^{2}+8}z)-1} \\
&amp; \quad =\Bigl( \sum_{n=0}^{\infty}\Bigl( \frac{k - \sqrt{k^{2}+8}}{2}\Bigr)^{n}\frac{z^{n}}{n!} \Bigr) \Bigl(\sum_{n=0}^{\infty}2\Bigl( \sqrt{k^{2}+8}\Bigr) ^{l}B_{l}\frac{z^{l}}{l!} \Bigr) \\
&amp; \quad =\sum_{n=0}^{\infty}\sum_{l=0}^{n}{n\choose l}
		2\Bigl( \sqrt{k^{2}+8}\Bigr)^{l}\Bigl(\frac{k - \sqrt{k^{2}+8}}{2}\Bigr)^{n-l}B_{l}\frac{z^{n}}{n!}.
\end{align*}
By comparing the coefficients of $\frac{z^{n}}{n!}$, we get
	\begin{equation*}
		\sum\limits_{l=0}^{n}{n\choose k}2^{l}\Bigl( \sqrt{k^{2}+8}\Bigr) ^{l}j_{k,n-l}
		B_{l}=\sum_{l=0}^{n}{n\choose l}
		2\Bigl( \sqrt{k^{2}+8}\Bigr)^{l}\Bigl(\frac{k - \sqrt{k^{2}+8}}{2}\Bigr)^{n-l}B_{l},
	\end{equation*}
which is equivalent to
\begin{equation*}
\sum\limits_{l=0}^{n}{n\choose k}2^{l}\Bigl( \sqrt{k^{2}+8}\Bigr) ^{l}j_{k,n-l}
B_{l}=\sum_{l=0}^{n}{n\choose l}
2\Bigl( \sqrt{k^{2}+8}\Bigr)^{l}\Bigl(\frac{
	j_{k,n-1} -\sqrt{k^{2}+8}J_{k,n-1}
}{2}\Bigr)B_{l}.
\end{equation*}
	Therefore
	\begin{align*}
	\sum\limits_{l=0}^{n}{n\choose l}\Bigl( \sqrt{k^{2}+8}\Bigr) ^{l}(2^{l}-1)j_{k,n-l}
	B_{l}&amp;=-\sqrt{k^{2}+8}\sum_{l=0}^{n}{n\choose l}\Bigl( \sqrt{k^{2}+8}\Bigr)
	^{l}J_{k,n-l}B_{l},\\
	&amp;=-n\sqrt{k^{2}+8}\Bigl( \frac{k - \sqrt{k^{2}+8}}{2}\Bigr)^{n-1}.
	\end{align*}
Hence the desired result.
\end{proof}
\begin{theorem}\label{2}
	Let $n$ be a positive integer. The following results hold:
\begin{align*}
&amp;\sum_{l=0}^{\lfloor n/2\rfloor}{n\choose 2l}\Bigl( k^{2}+8\Bigr)
		^{l}J_{k,n-2l}B_{2l}=\frac{nj_{k,n-1}}{2},\\
&amp;\sum_{l=0}^{\lfloor n/2\rfloor }{n\choose 2l}\Bigl( k^{2}+8\Bigr)
		^{l}(2^{2l}-1)j_{k,n-2l}B_{2l}=\frac{(k^{2}+8)nJ_{k,n-1}}{2}.
\end{align*}
\end{theorem}
\begin{proof}
	We have
	\begin{equation*}
		\sum_{l=0}^{n}{n\choose l}\Bigl(\sqrt{k^{2}+8}\Bigr)
		^{l}J_{k,n-l}B_{l}=n\Bigl( \frac{k - \sqrt{k^{2}+8}}{2}\Bigr)^{n-1}.
	\end{equation*}
	This gives
	\begin{equation*}
		\sum\limits_{l=0}^{n}{n\choose l}
		\Bigl( \sqrt{k^{2}+8}\Bigr) ^{n-l}J_{k,n-l}B_{l}=\frac{
			n(j_{k,n-1} -\sqrt{k^{2}+8}J_{k,n-1})
		}{2},
	\end{equation*}
	with the conditions $B_{0}=1$ and $B_{1}=-\frac{1}{2}$.
	
	For $l\geq0$, we have $B_{2l+1}=0$ and then
	\begin{equation*}
		\sum\limits_{l=0}^{\lfloor n/2\rfloor }{n\choose 2l}
		(k^{2}+8)^{l}J_{k,n-2l}B_{2l}=\frac{
			nj_{k,n-1}}{2}.
	\end{equation*}

	Similarly, from~(\ref{ff}), we get
	\begin{equation*}
		\sum\limits_{l=0}^{n}{n\choose l}\Bigl( \sqrt{k^{2}+8}\Bigr) ^{l}(2^{l}-1)
		j_{k,n-l}B_{l}=\frac{
			-\sqrt{k^{2}+8}n(j_{k,n-1} -\sqrt{k^{2}+8}J_{k,n-1})
		}{2},
	\end{equation*}
with $B_{0}=1$ and $B_{1}=-\frac{1}{2}$.
	
	For $l\geq0$, we have $B_{2l+1}=0$. Then we obtain
	\begin{equation*}
		\sum\limits_{l=0}^{\lfloor n/2\rfloor }\binom{n}{2l}
		(k^{2}+8)^{l}(2^{2l}-1)j_{k,n-2l} B_{2l}=\frac{
			(k^{2}+8)^{2}nJ_{k,n-1}}{2}.
	\end{equation*}
Which complete the proof.
\end{proof}
\begin{itemize}
	\item Putting $k=1$ in Theorem~\ref{1} and Theorem~\ref{2} we have the
	following Jacobsthal-Bernoulli and Jacobsthal-Lucas-Bernoulli identities
\begin{align*}
&amp;\sum\limits_{l=0}^{n}{n\choose l}3^{l}J_{n-l}B_{l}=(-1)^{n-1}n, \text{
			or}\sum\limits_{l=0}^{\lfloor n/2\rfloor }{n\choose 2l}9^{l}
		J_{n-2l}B_{2l}=\frac{nj_{n-1}}{2}.\\
&amp;\sum\limits_{l=0}^{n}{n\choose 2l}3^{l}(2^{l}-1)j_{n-l}B_{l}=(-1)^{n}n, \text{
			or}\sum\limits_{l=0}^{\lfloor n/2\rfloor }{n\choose 2l}
		9^{l}(2^{l}-1)j_{n-2l}B_{2l}=\frac{9nJ_{n-1}}{2}.
\end{align*}
\end{itemize}

\begin{theorem}\label{3}
For every positive integer $n$, we have
\begin{align*}
\sum\limits_{l=0}^{\lfloor n/2\rfloor }{n\choose 2l}
2(4)^{n-2l-1}\Bigl( \sqrt{k^{2}+8}\Bigr) ^{2l+1}J_{k,n-2l}E_{2l}&amp;=(2k+\sqrt{k^{2}+8})^{n}-(2k+\sqrt{k^{2}+8})^{n},\\
\sum\limits_{l=0}^{\lfloor n/2\rfloor }{n\choose 2l}\Bigl( \frac{k^{2}+8
}{4}\Bigr) ^{l}j_{k,n-2l}
E_{2l}&amp;=2\Bigl( \frac{k}{2}\Bigr)^{n}.
\end{align*}
\end{theorem}
\begin{proof}
	Using the change of variables $z=\sqrt{k^{2}+8}z$  in~(\ref{16}) and $z=4z$ in the exponential generating function for $(J_{k,n})_{n\in \mathbb{N}}$, we obtain
\begin{align}
\sum_{n=0}^{\infty}\Bigl( \sqrt{k^{2}+8}\Bigr)^{l}E_{l}\frac{z^{l}}{l!}&amp;=\frac{2\exp
(\sqrt{k^{2}+8}z)}{\exp (2\sqrt{k^{2}+8}z)+1},  \label{j}\\
\sum_{n=0}^{\infty}4^{n}J_{k,n}\frac{z^{n}}{n!}&amp;=\frac{\exp
(k-\sqrt{k^{2}+8})z)}{\sqrt{k^{2}+8}}(\exp (2\sqrt{k^{2}+8}z)-1)(\exp
(2\sqrt{k^{2}+8}z)+1). \label{b}
\end{align}
Multiplying~(\ref{b}) by~(\ref{j}), this gives
	\begin{equation*}
	\Bigl( \sum_{n=0}^{\infty}4^{n}J_{k,n}\frac{z^{n}}{n!}\Bigr)
	\Bigl(\sum_{l=0}^{\infty}\Bigl( \sqrt{k^{2}+8}\Bigr) ^{l}E_{l}\frac{z^{l}}{l!}\Bigr)
	=\sum_{n=0}^{\infty}\sum_{l=0}^{n}\binom{n}{l}
	4^{n-l}\Bigl( \sqrt{k^{2}+8}\Bigr) ^{l}J_{k,n-l}E_{l}\frac{z^{n}}{n!}.
	\end{equation*}
Then we get
	\begin{align*}
&amp;\frac{\exp (k-\sqrt{k^{2}+8})z)}{\sqrt{k^{2}+8}}(\exp (2\sqrt{k^{2}+8}z)-1)(\exp
(2\sqrt{k^{2}+8}z)+1)	\times\frac{2\exp (\sqrt{k^{2}+8}z)}{(\exp
		2\sqrt{k^{2}+8}z)+1}\\
	&amp; \quad =\frac{2}{\sqrt{k^{2}+8}}\Bigl(  \exp\Bigl((2k+\sqrt{k^{2}+8})z\Bigr)-\exp\Bigl((2k-\sqrt{k^{2}+8})z\Bigr) \Bigr)  \\
	&amp; \quad =\frac{2}{\sqrt{k^{2}+8}}\sum_{n=0}^{\infty}\Bigl( 2k+\sqrt{k^{2}+8}\Bigr) ^{n}-\Bigl( 2k-\sqrt{k^{2}+8}\Bigr) ^{n}\frac{z^{n}}{n!}.
\end{align*}
By comparing the coefficients of $\frac{z^{n}}{n!}$, we obtain
\begin{equation*}
	\sum\limits_{l=0}^{n}\binom{n}{l}
	2(4)^{n-l-1}\Bigl( \sqrt{k^{2}+8}\Bigr) ^{l+1}J_{k,n-l}E_{l}=\Bigl( 2k+\sqrt{k^{2}+8}\Bigr) ^{n}-\Bigl( 2k-\sqrt{k^{2}+8}\Bigr)^{n}.
	\end{equation*}
	For $l\geq0$, we have $E_{2l+1}=0$, which gives the desired result.
	
Similarly, we use the change of variables $z=\frac{\sqrt{k^{2}+8}}{2}z$ in~(\ref{16}). We have
	\begin{equation}
	\sum_{n=0}^{\infty}\Bigl( \frac{\sqrt{k^{2}+8}}{2}\Bigr) ^{n}E_{n}\frac{z^{n}}{n!}=\frac{2\exp \Bigl( \frac{\sqrt{k^{2}+8}}{2}z\Bigr) }{\exp (\sqrt{k^{2}+8}z)+1},
	\label{y}
	\end{equation}
and multiplying~(\ref{y}) by the exponential generating function for $(j_{k,n})_{n\in \mathbb{N}}$, we get
	\begin{equation*}
\Bigl( \sum_{n=0}^{\infty}j_{k,n}\dfrac{z^{n}}{n!}
\Bigr) \Bigl( \sum_{l=0}^{\infty}\Bigl( \frac{\sqrt{k^{2}+8}}{2}\Bigr)
^{l}E_{l}\dfrac{z^{l}}{l!}\Bigr) =	\sum_{l=0}^{\infty}\sum\limits_{l=0}^{n}{n\choose l}\Bigl(\frac{\sqrt{k^{2}+8}}{2}\Bigr) ^{l}
j_{k,n-l} E_{l}\frac{z^{n}}{n!},
\end{equation*}
and
	\begin{equation*}
\frac{2\exp (\frac{\sqrt{k^{2}+8}}{2}z)}{\exp (\sqrt{k^{2}+8}z)+1}\Bigl( \exp
\Bigl( \frac{k+\sqrt{k^{2}+8}}{2}z\Bigr) +\exp \Bigl( \frac{k-\sqrt{k^{2}+8}}{2}z\Bigr)\Bigr)
=2\sum_{k=0}^{\infty}\Bigl( \frac{k}{2}\Bigr) ^{n}\frac{z^{n}}{n!}.
	\end{equation*}
By comparing the coefficients of $\frac{z^{n}}{n!}$, we get
	\begin{equation*}
	\sum\limits_{l=0}^{n}{n\choose l}\Bigl( \frac{\sqrt{k^{2}+8}}{2}\Bigr)
	^{l}j_{k,n-l}
	E_{l}=2\Bigl( \frac{k}{
		2}\Bigr)^{n}.
	\end{equation*}
For $k\geq0$, we obtain $E_{2k+1}=0$; hence the desired result.
\end{proof}
\begin{itemize}
	\item   By taking $k=1$ in the Theorem~\ref{3},  we can state  the
	following identities involving Euler numbers with Jacobsthal and Jacobsthal-Lucas numbers
\begin{align*}
\sum\limits_{l=0}^{\lfloor n/2\rfloor }{n\choose 2l}
	2(4)^{n-2l-1}3^{2l+1}J_{n-2l}E_{2l}&amp;=5^{n}-(-1)^{n},\\
\sum\limits_{l=0}^{\lfloor n/2\rfloor }{n\choose 2l}\Bigl( \frac{9
	}{4}\Bigr) ^{l}j_{n-2l}
	E_{2l}&amp;=2^{1-n}.
\end{align*}
\end{itemize}
	\section{Bivariate polynomial identities}\label{Sec3}
	The aim of this section is to establish some new identities involving two variables $x$ and $y$. We start with three theorems involving Bernoulli and Euler numbers with bivariate Fibonacci and Lucas polynomials.
\begin{theorem}\label{4j}
	For every positive integer $n$, we have
\begin{align}
\sum\limits_{l=0}^{n}{n\choose l}
		\Bigl( \sqrt{x^{2}+ 4y}\Bigr) ^{l}F_{n-l}(x,y)B_{l} &amp;=n\Bigl( \frac{x-\sqrt{x^{2}+4y}}{2}\Bigr)^{n-1},  \label{mmk}\\
\sum\limits_{l=0}^{n}{n\choose l}\Bigl( \sqrt{x^{2}+ 4y}\Bigr) ^{l}(2^{l}-1)
		L_{n-l}(x,y) B_{l}&amp;=-2\sqrt{x^{2}y^{2}+y}n\Bigl( \frac{x-\sqrt{x^{2}+4y}}{2}\Bigr) ^{n-1}. \nonumber
\end{align}
\end{theorem}
\begin{proof}
	Using the change of variables $z=\sqrt{x^{2}+4y}z$ in~(\ref{mk}), we obtain
	\begin{equation}
		\sum_{n=0}^{\infty}\Bigl( \sqrt{x^{2}+4y}\Bigr)^{n}B_{n}\frac{z^{n}}{n!}=\frac{\sqrt{x^{2}+4y}z}{\exp (\sqrt{x^{2}+4y}z)-1}.  \label{klm}
	\end{equation}
	Multiplying~(\ref{klm}) by the exponential generating function for $(F_{n}(x,y))_{n\in \mathbb{N}}$, we get
	\begin{align*}
		\Bigl(\sum_{n=0}^{\infty}F_{n}(x,y)\frac{z^{n}}{n!}\Bigr) \Bigl(
	\sum_{l=0}^{\infty}\Bigl(\sqrt{x^{2}+4y}\Bigr) ^{l}B_{l}\frac{z^{l}}{l!}\Bigr)&amp;=\sum_{n=0}^{\infty}\sum_{l=0}^{\infty}\Bigl( \sqrt{x^{2}+4y}\Bigr) ^{l}B_{l}F_{n}(x,y)\frac{z^{l}}{l!}\frac{z^{n}}{n!}\\
		&amp;=\sum_{n=0}^{\infty}\sum_{l=0}^{n}\binom{n}{l}
		\Bigl( \sqrt{x^{2}+4y}\Bigr)^{l}B_{l}F_{n}(x,y)\frac{z^{n}}{n!}.
	\end{align*}
Then
\begin{align*}
&amp;\frac{\exp \Bigl(  \frac{x-\sqrt{x^{2}+4y}}{2}z\Bigr) -\exp \Bigl(  \frac{x-\sqrt{x^{2}+4y}}{2}z\Bigr)}{\sqrt{x^{2}+4y}} \times\frac{\sqrt{x^{2}+4y}z}{\exp (\sqrt{x^{2}+4y}z)-1}\\
&amp; \quad =z\exp \Bigl( \frac{x-\sqrt{x^{2}+4y}}{2}z\Bigr)=\sum_{n=0}^{\infty}n\Bigl( \frac{x-\sqrt{x^{2}+4y}}{2}\Bigr)^{n-1}\frac{z^{n}}{n!}.
\end{align*}
By comparing the coefficients of $\frac{z^{n}}{n!}$, we obtain the desired result.
	
Similarly, by using the change of variables $z=2\sqrt{x^{2}+4y}z$ in~(\ref{mk}),
	we obtain
	\begin{equation}
	\sum_{n=0}^{\infty}\Bigl( 2\sqrt{x^{2}+4y}\Bigr)^{n}B_{n}\frac{z^{n}}{n!}=\frac{2\sqrt{x^{2}+4y}z}{\exp (2\sqrt{x^{2}+4y}z)-1},  \label{mjn}
	\end{equation}
	and multiplying~(\ref{mjn}) by the exponential generating function for $(L_{n}(x,y))_{n\in \mathbb{N}}$, we get
\begin{align*}
\Bigl(\sum_{n=0}^{\infty}L_{n}(x,y)\frac{z^{n}}{n!}\Bigr)
		\Bigl(\sum_{l=0}^{\infty}\Bigl( 2\sqrt{x^{2}+4y}\Bigr)^{l}B_{l}\frac{z^{l}}{l!}
		\Bigr)&amp;=\sum_{n=0}^{\infty}\sum_{l=0}^{\infty}
		\Bigl(2\sqrt{x^{2}+4y}\Bigr)^{l}B_{l}L_{n}(x,y)\frac{z^{l}}{l!}\frac{z^{n}}{n!}\\
&amp;=\sum_{n=0}^{\infty}\sum\limits_{l=0}^{n}{n\choose l}
		2^{l}\Bigl( \sqrt{x^{2}+4y}\Bigr) ^{l}L_{n-l}(x,y)B_{l}\frac{
			z^{n}}{n!}.
\end{align*}
Therefore
	\begin{align*}
		&amp;\Bigl( \exp \Bigl( \frac{x+\sqrt{x^{2}+4y}}{2}z\Bigr)+ \exp \Bigl( \frac{x-\sqrt{x^{2}+4y}}{2}z\Bigr)\Bigr) \frac{2\sqrt{x^{2}+4y}z}{\exp (2\sqrt{x^{2}+4y}z)-1}\\		
&amp; \quad =\exp\Bigl( \frac{x-\sqrt{x^{2}+4y}}{2}z\Bigr) \frac{2\sqrt{x^{2}+4y}z}{\exp(\sqrt{x^{2}+4y}z)-1} \\
&amp; \quad =\sum_{n=0}^{\infty}\sum_{l=0}^{n}\binom{n}{l}
		2\Bigl( \sqrt{x^{2}+4y}\Bigr) ^{l}\Bigl( \frac{x-\sqrt{x^{2}+4y}}{2}\Bigr) ^{n-l}B_{l}\frac{z^{n}}{n!}.
	\end{align*}
By comparing the coefficients of $\frac{z^{n}}{n!}$, we obtain
	\begin{equation*}
		\sum\limits_{l=0}^{n}{n\choose k}2^{l}\Bigl( \sqrt{x^{2}+4y}\Bigr) ^{l}L_{n-l}(x,y)
		B_{l}=\sum_{l=0}^{n}{n\choose l}
		2\Bigl( \sqrt{x^{2}+4y}\Bigr)^{l}\Bigl( \frac{x-\sqrt{x^{2}+4y}}{2}\Bigr) ^{n-l}B_{l}.
	\end{equation*}
We know that
\begin{equation*}
\Bigl( \frac{x-\sqrt{x^{2}+4y}}{2}\Bigr) ^{n-l}=\frac{
	L_{n-l}(x,y)-\sqrt{x^{2}+4y}F_{n-l}(x,y)}{2},
\end{equation*}
so we get
	\begin{equation*}
		\sum\limits_{l=0}^{n}{n\choose l}\Bigl( \sqrt{x^{2}+4y}\Bigr) ^{l}(2^{l}-1)L_{n-l}(x,y)
		B_{l}=-\sqrt{x^{2}+4y}n\Bigl( \frac{x-\sqrt{x^{2}+4y}}{2}\Bigr)^{n-1}.
	\end{equation*}
Hence the result.
\end{proof}
\begin{theorem}\label{5j}
	For every positive integer $n$, we have
\begin{align}
\sum\limits_{l=0}^{\lfloor n/2\rfloor }{n\choose 2l}(x^{2}+4y)^{l}
	F_{n-2l}(x,y)B_{2l}&amp;=\frac{nL_{n-1}(x,y)}{2}, \nonumber\\
\sum\limits_{l=0}^{\lfloor n/2\rfloor }{n\choose 2l}
		(x^{2}+4y)^{l}(2^{2l}-1)L_{n-2l}(x,y) B_{2l}&amp;=\frac{n(x^{2}+4y)F_{n-1}(x,y)}{2}.\label{jh}
\end{align}
\end{theorem}
\begin{proof}
From~(\ref{mmk}) and on account of the identity
\begin{equation*}
\Bigl( \frac{x-\sqrt{x^{2}+4y}}{2}\Bigr) ^{n-l}=\frac{
	L_{n-l}(x,y)-\sqrt{x^{2}+4y}F_{n-l}(x,y)}{2},
\end{equation*}
 we get
\begin{equation*}
	\sum\limits_{l=0}^{n}{n\choose l}
	\big(\sqrt{x^{2}+4y}\big)^{n-l}F_{n-l}(x,y)B_{l}=\frac{
		n(L_{n-1}(x,y) -\sqrt{x^{2}+4y}F_{n-1}(x,y))
	}{2},
\end{equation*}
with $B_{0}=1$ and $B_{1}=-\frac{1}{2}$.

For $l\geq0$, we have $B_{2l+1}=0$, and then
\begin{equation*}
	\sum\limits_{l=0}^{\lfloor n/2\rfloor }{n\choose 2l}
	(x^{2}+4y)^{l}F_{n-2l}(x,y)B_{2l}=\frac{
		nL_{n-1}(x,y)}{2}.
\end{equation*}

Similarly, from~(\ref{jh}), we get
\begin{equation*}
	\sum\limits_{l=0}^{n}{n\choose l}\big(\sqrt{x^{2}+4y}\big)^{l}(2^{l}-1)
	L_{n-l}(x,y)B_{l}=\frac{
		-\sqrt{x^{2}+4y}n(L_{n-1}(x,y) -\sqrt{x^{2}+4y}F_{n-1}(x,y))
	}{2},
\end{equation*}
with $B_{0}=1$, $B_{1}=-\frac{1}{2}$.

For $l\geq0$, we have $B_{2l+1}=0$. Hence we obtain
\begin{equation*}
	\sum\limits_{l=0}^{\lfloor n/2\rfloor }{n\choose 2l}
	(x^{2}+4y)^{l}(2^{2l}-1)L_{n-2l}(x,y) B_{2l}=\frac{
		(x^{2}+4y)^{2}nF_{n-1}(x,y)}{2}.
\end{equation*}
Hence the desired result is proved.
\end{proof}
\begin{itemize}
	\item By putting $y=1$ in Theorem~\ref{4j} and Theorem~\ref{5j}, 
we obtain the following identities
\begin{align}
\sum\limits_{l=0}^{n}{n\choose l}
	\Bigl( \sqrt{x^{2}+ 4}\Bigr) ^{l}F_{n-l}(x)B_{l}&amp;=n\Bigl( \frac{x-\sqrt{x^{2}+4}}{2}\Bigr)^{n-1},  \label{mml}\\
\sum\limits_{l=0}^{\lfloor n/2\rfloor }{n\choose 2l}(x^{2}+4)^{l}
	F_{n-2l}(x)B_{2l}&amp;=\frac{nL_{n-1}(x)}{2},\label{gh}\\
\sum\limits_{l=0}^{n}{n\choose l}\Bigl(\sqrt{x^{2}+ 4}\Bigr) ^{l}(2^{l}-1)
	L_{n-l}(x)
	B_{l}&amp;=-2(\sqrt{x^{2}+4})n\Bigl( \frac{x-\sqrt{x^{2}+4}}{2}\Bigr) ^{n-1},  \label{nh}\\
\sum\limits_{l=0}^{\lfloor n/2\rfloor }{n\choose 2l}
	(x^{2}+4)^{l}(2^{2l}-1)L_{n-2l}(x) B_{2l}&amp;=\frac{n(x^{2}+4)F_{n-1}(x)}{2}.\label{h}
\end{align}
	  \item By setting $x=y=1$ in Theorem~\ref{4j} and Theorem~\ref{5j} or $x=1$
in the relationships~(\ref{mml}),~(\ref{gh}),~(\ref{nh}) and~(\ref{h})
we obtain the following identities involving Bernoulli numbers with Fibonacci
and Lucas numbers (see~\cite{Byrd2, Byrd3}):
\begin{align*}
\sum\limits_{l=0}^{n}{n\choose l}
\Bigl( \sqrt{5}\Bigr) ^{l}F_{n-l}B_{l}&amp;=n\Bigl( \frac{1-\sqrt{5}}{2}\Bigr)^{n-1},\\
\sum\limits_{l=0}^{\lfloor n/2\rfloor }{n\choose 2l}5^{l}
	F_{n-2l}B_{2l}&amp;=\frac{nL_{n-1}}{2},\\
\sum\limits_{l=0}^{n}{n\choose l}\Bigl( \sqrt{5}\Bigr) ^{l}(2^{l}-1)
	L_{n-l}
	B_{l}&amp;=-2\sqrt{5}n\Bigl( \frac{1-\sqrt{5}}{2}\Bigr) ^{n-1},\\
\sum\limits_{l=0}^{\lfloor n/2\rfloor }{n\choose 2l}
	5^{l}(2^{2l}-1)L_{n-2l} B_{2l}&amp;=\frac{n5F_{n-1}}{2}.
\end{align*}
\end{itemize}
\begin{theorem}\label{6}
		For every positive integer $n$, we have
\begin{align*}
\sum\limits_{l=0}^{\lfloor n/2\rfloor }{n\choose 2l}
	2(4)^{n-2k-1}\Bigl( \sqrt{x^{2}+4y}\Bigr) ^{2l+1}F_{n-2l}(x,y)E_{2l}&amp;=
\Bigl( 2x+\sqrt{x^{2}+4y}\Bigr)^{n}-\Bigl( 2x-\sqrt{x^{2}+4y}\Bigr)^{n},\\
\sum\limits_{l=0}^{\lfloor n/2\rfloor }{n\choose 2l}\Bigl( \frac{x^{2}+4y
	}{4}\Bigr) ^{2l}L_{n-2l}(x,y)
	E_{2l}&amp;=2\Bigl(\frac{x}{2}\Bigr)^{n}.
\end{align*}
\end{theorem}

\begin{proof}
	Using the change of variables $z=\sqrt{x^{2}+4y}z$
in~(\ref{16}), we obtain
	\begin{equation}
\sum_{l=0}^{\infty}\Bigl( \sqrt{x^{2}+4y}\Bigr) ^{l}E_{l}\frac{z^{l}}{l!}=\frac{2\exp
		(\sqrt{x^{2}+4y}z)}{\exp(2\sqrt{x^{2}+4y}z)+1}.  \label{jf}
	\end{equation}
By multiplying~(\ref{jf}) by the exponential generating function for the sequence $(4^{n}F_{n}(x,y))_{n\in \mathbb{N}}$, we obtain
\begin{align*}
\Bigl( \sum_{k=0}^{\infty}4^{n}F_{n}(x,y)\frac{z^{n}}{n!}\Bigr)
	\Bigl( \sum_{l=0}^{\infty}\Bigl(\sqrt{x^{2}+4y}\Bigr)^{l}E_{l}\frac{z^{l}}{l!}\Bigr)
=\sum_{n=0}^{\infty}\sum_{l=0}^{n}{n\choose l}
	4^{n-l}\Bigl( \sqrt{x^{2}+4y}\Bigr)^{l}F_{n-l}(x,y)E_{l}\frac{z^{n}}{n!}.
\end{align*}
	Then we get
\begin{align*}
&amp;\frac{\exp\Bigl( 2(x+\sqrt{x^{2}+4y})z\Bigr) -\exp\Bigl(2(x-\sqrt{x^{2}+4y})z\Bigr)}{\sqrt{x^{2}+4y}}\times\frac{2\exp(\sqrt{x^{2}+4y}z)}{\exp(2\sqrt{x^{2}+4y}z)+1}\\
&amp; \quad =\frac{2\exp\Bigl( 2(x-\sqrt{x^{2}+4y})z\Bigr)\exp(\sqrt{x^{2}+4y}z)
}{\sqrt{x^{2}+4y}\Bigl( \exp(2\sqrt{x^{2}+4y}z)+1\Bigr) }\Bigl( \exp(4\sqrt{x^{2}+4y}z)-1\Bigr) \\
&amp; \quad =\frac{2}{\sqrt{x^{2}+4y}}\Bigl( \exp\Bigl( (2x+\sqrt{x^{2}+4y})z\Bigr)+\exp\Bigl( (2x-\sqrt{x^{2}+4y})z\Bigr)\Bigr) \\
&amp; \quad =\frac{2}{\sqrt{x^{2}+4y}}\sum_{n=0}^{\infty}\Bigl( (2x+\sqrt{x^{2}+4y})^{n}-(2x-\sqrt{x^{2}+4y})^{n}\Bigr) \frac{z^{n}}{n!}.\\	
\end{align*}
	By comparing the coefficients of $\frac{z^{n}}{n!}$, we get
	\begin{equation*}
	\sum\limits_{l=0}^{n}{n\choose l}
	2(4)^{n-l-1}\Bigl(\sqrt{x^{2}+4y}\Bigr) ^{l+1}F_{n-l}(x,y)E_{l}=\Bigl( 2x+\sqrt{x^{2}+4y}\Bigr)^{n}-\Bigl( 2x-\sqrt{x^{2}+4y}\Bigr)^{n}.
	\end{equation*}
	For $l\geq0$, we have $E_{2l+1}=0$, which gives the result.
	
	Similarly, using the change of variables $z=\frac{\sqrt{x^{2}+4y}}{2}z$ in~(\ref{16}), we obtain
	\begin{equation}
	\sum_{n=0}^{\infty}\Bigl( \frac{\sqrt{x^{2}+4y}}{2}\Bigr) ^{n}E_{n}\frac{z^{n}}{ n!}=\frac{2\exp (\frac{\sqrt{x^{2}+4y}}{2}z)}{\exp (\sqrt{x^{2}+4y}z)+1},
	\label{jv}
	\end{equation}
	and multiplying~(\ref{jv}) by the exponential generating function for $(L_{n}(x,y))_{n\in \mathbb{N}}$, we get
	\begin{equation*}
	\Bigl( \sum_{n=0}^{\infty}L_{n}(x,y)\frac{z^{n}}{n!}
	\Bigr) \Bigl(\sum_{l=0}^{\infty}\Bigl( \frac{\sqrt{x^{2}+4y}}{2}\Bigr)
	^{l}E_{l}\frac{z^{l}}{l!}\Bigr) =\sum_{n=0}^{\infty}\sum\limits_{l=0}^{n}{n\choose l}\Bigl(\frac{\sqrt{x^{2}+4y}}{2}\Bigr) ^{l}
	L_{n-l}(x,y) E_{l}\frac{z^{n}
	}{n!},
	\end{equation*}
	and
\begin{align*}
\Bigl(\exp	\Bigl( \frac{x+\sqrt{x^{2}+4y}}{2}z\Bigr) +\exp \Bigl( \frac{x-\sqrt{x^{2}+4y}}{2}z\Bigr) \Bigr)\times\frac{2\exp\Bigl( \frac{\sqrt{x^{2}+4y}}{2}z\Bigr)}{\exp (\sqrt{x^{2}+4y}z)+1}
&amp;=2\exp(\frac{x}{ 2}z) \\
&amp; =2\sum_{n=0}^{\infty}\Bigl( \frac{x}{
		2}\Bigr) ^{n}\frac{z^{n}}{n!}.
\end{align*}
	Comparing the coefficients of $\frac{z^{n}}{n!}$, we obtain
	\begin{equation*}
	\sum\limits_{l=0}^{n}{n\choose l}\Bigl( \frac{\sqrt{x^{2}+4y}}{2}\Bigr)
	^{l}L_{n-l}(x,y)
	E_{l}=2\Bigl( \frac{x}{
		2}\Bigr)^{n},
	\end{equation*}
	For $k\geq0$, we have $E_{2k+1}=0$, and the result follows.
\end{proof}
\begin{itemize}
	\item Putting $y=1$ in Theorem~\ref{6}, we obtain the
	following  identities  linking  Euler numbers to Fibonacci and Lucas polynomials
	\begin{align}
\sum\limits_{l=0}^{\lfloor n/2\rfloor }{n\choose 2l}
2(4)^{n-2k-1}(\sqrt{x^{2}+4})^{2l+1}F_{n-2l}(x)E_{2l}&amp;=\Bigl( 2x+\sqrt{x^{2}+4}\Bigr)^{n}-\Bigl( 2x-\sqrt{x^{2}+4}\Bigr)^{n},\label{4}\\
\sum\limits_{l=0}^{\lfloor n/2\rfloor }{n\choose 2l}\Bigl( \frac{x^{2}+4
}{4}\Bigr) ^{l}L_{n-2l}(x)
E_{2l}&amp;=2\Bigl(\frac{x}{2}\Bigr)^{n}.\label{5}
\end{align}
\item Putting $y=x=1$ in Theorem~\ref{6} or $x=1$ in the relationships~(\ref{4}) and~(\ref{5}), we obtain the
following identities linking  Euler numbers to Fibonacci and Lucas numbers
\begin{align}
\sum\limits_{l=0}^{\lfloor n/2\rfloor }{n\choose 2l}
2(4)^{n-2l-1}(\sqrt{5})^{2l+1}F_{n-2l}E_{2l}&amp;=(2+\sqrt{5})^{n}-(2-\sqrt{5})^{n}, \nonumber\\
\sum\limits_{l=0}^{\lfloor n/2\rfloor }{n\choose 2l}\Bigl( \frac{5
}{4}\Bigr) ^{l}L_{n-2l}
E_{2l}&amp;=2^{1-n}. \label{qw}
\end{align}
Note that Eq.~(\ref{qw}) is Byrd's result~(\ref{byard}).
\end{itemize}

Now we present the analogue results for the relation between Bernoulli numbers with bivariate Pell and bivariate Pell-Lucas polynomials.
\begin{theorem}\label{7}
	For every positive integer $n$, we have
\begin{align}
&amp;\sum\limits_{l=0}^{n}{n\choose l}
		\Bigl( 2\sqrt{x^{2}y^{2}+ y}\Bigr)^{l}P_{n-l}(x,y)B_{l}=n\Bigl(xy-\sqrt{x^{2}y^{2}+y}\Bigr)^{n-1},\label{mm}\\
&amp;\sum\limits_{l=0}^{n}\binom{n}{l}\Bigl(2\sqrt{x^{2}y^{2}+y}\Bigr) ^{l}(2^{l}-1)
		Q_{n-l}(x,y)
		B_{l}=-2\sqrt{x^{2}y^{2}+y}n\Bigl(xy-\sqrt{x^{2}y^{2}+ y}\Bigr)^{n-1}.  \label{nn}
\end{align}
\end{theorem}
\begin{proof}
	The proof is similar to the proof of Theorem~\ref{4j}, by using the 
	change of variables
$z=2\sqrt{x^{2}y^{2}+y}z$ in~(\ref{mk}) to prove Eq.~(\ref{mm}) and
	 $z=4\sqrt{x^{2}y^{2}+y}z$ in~(\ref{mk}) to 
	 prove the relationship~(\ref{nn}).
\end{proof}
\begin{theorem}\label{8}
	For every positive integer $n$, we have
\begin{align*}
\sum\limits_{l=0}^{\lfloor n/2\rfloor }{n\choose 2l}
		4^{l}(x^{2}y^{2}+y)^{l}P_{n-2l}(x,y)B_{2l}&amp;=\frac{nQ_{n-1}(x,y)}{2},\\
\sum\limits_{l=0}^{\lfloor n/2\rfloor }{n\choose 2l}
		4^{l}(x^{2}y^{2}+y)^{l}(2^{2l}-1)Q_{n-2l}(x,y) B_{2l}&amp;=
		2n(x^{2}y^{2}+y)P_{n-1}(x,y).
\end{align*}
\end{theorem}
\begin{proof}
	We have
	\begin{equation*}
	\sum_{l=0}^{n}{n\choose l}\Bigl( 2\sqrt{x^{2}y^{2}+y}\Bigr)
	^{l}P_{n-l}(x,y)B_{l}=n\Bigl(xy-\sqrt{x^{2}y^{2}+y}\Bigr)^{n-1},
	\end{equation*}
	which gives
	\begin{equation*}
	\sum\limits_{l=0}^{n}{n\choose l}
	\Bigl(2\sqrt{x^{2}y^{2}+y}\Bigr)^{n-l}P_{n-l}(x,y)B_{l}=\frac{n\Bigl( Q_{n-1}(x,y)-2\sqrt{x^{2}y^{2}+y}P_{n-1}(x,y)\Bigr)
	}{2},
	\end{equation*}
	with $B_{0}=1$ and $B_{1}=-\frac{1}{2}$.
	
	For $l\geq0$, we have $B_{2l+1}=0$. Then we get
	\begin{equation*}
	\sum\limits_{l=0}^{\lfloor n/2\rfloor }{n\choose 2l}
	4^{l}(x^{2}y^{2}+y)^{l}P_{n-2l}(x,y)B_{2l}=\frac{
		nQ_{n-1}(x,y)}{2}.
	\end{equation*}
	Similarly, from~(\ref{nn}), we get
	\begin{equation*}
	\sum\limits_{l=0}^{n}{n\choose l}4^{l}(x^{2}y^{2}+y)^{l}(2^{l}-1)
	Q_{n-l}(x,y)B_{l}=\frac{
		-2\sqrt{x^{2}y^{2}+y}n\Bigl( Q_{n-1}(x,y) -2\sqrt{x^{2}y^{2}+y}P_{n-1}(x,y)\Bigr)
	}{2},
	\end{equation*}
	with $B_{0}=1$ and $B_{1}=-\frac{1}{2}$.
	
	For $l\geq0$, we have $B_{2l+1}=0$. Then we obtain
	\begin{equation*}
	\sum\limits_{l=0}^{\lfloor n/2\rfloor }{n\choose 2l}
	4^{l}(x^{2}y^{2}+y)^{l}(2^{2l}-1)Q_{n-2l}(x,y) B_{2l}=\frac{
		(x^{2}y^{2}+y)nF_{n-1}(x,y)}{2},
	\end{equation*}
	and the result follows.
\end{proof}
	\begin{itemize}
		\item 	By putting $y=1$ in Theorem~\ref{7} and Theorem~\ref{8},
we start with the following results involving Bernoulli numbers and  Pell, Pell-Lucas polynomials:
\begin{align}
\sum\limits_{l=0}^{n}{n\choose l}
		\Bigl( 2\sqrt{x^{2}+ 1}\Bigr) ^{l}P_{n-l}(x)B_{l}&amp;=n(x-\sqrt{x^{2}+ 1})^{n-1},  \label{tm}\\
\sum\limits_{l=0}^{\lfloor n/2\rfloor }{n\choose 2l}
		4^{l}(x^{2}+1)^{l}P_{n-2l}(x)B_{2l}&amp;=\frac{nQ_{n-1}(x)}{2},\label{nb}\\
\sum\limits_{l=0}^{n}{n\choose l}\Bigl(2\sqrt{x^{2}+1}\Bigr)^{l}(2^{l}-1)
		Q_{n-l}(x)
		B_{l}&amp;=-2n\sqrt{x^{2}+1}(x-\sqrt{x^{2}+ 1})^{n-1},  \label{en}\\
\sum\limits_{l=0}^{\lfloor n/2\rfloor }{n\choose 2l}
		4^{l}(x^{2}+1)^{l}(2^{2l}-1)Q_{n-2l}(x) B_{2l}&amp;=
		2n(x^{2}+1)P_{n-1}(x).\label{fc}
\end{align}
		\item By setting $x=y=1$ in Theorem~\ref{7} and Theorem~\ref{8} or $x=1$
in the relationships~(\ref{tm}),~(\ref{nb}),~(\ref{en}) and~(\ref{fc}) we deduce the following
		Bernoulli-Pell and the Bernoulli-Pell-Lucas identities:
\begin{align*}
\sum\limits_{l=0}^{n}{n\choose l}
	(2\sqrt{2})^{l}P_{n-l}B_{l}&amp;=n(1-\sqrt{2})^{n-1},\\
\sum\limits_{l=0}^{\lfloor n/2\rfloor }{n\choose 2l}
	8^{l}P_{n-2l}B_{2l}&amp;=\frac{nQ_{n-1}}{2},\\
\sum\limits_{l=0}^{n}{n\choose l}\Bigl(2\sqrt{2}\Bigr) ^{l}(2^{l}-1)
	Q_{n-l}
	B_{l}&amp;=-2n\sqrt{2}(1-\sqrt{2})^{n-1},\\
\sum\limits_{l=0}^{\lfloor n/2\rfloor }{n\choose 2l}
	8^{l}(2^{2l}-1)Q_{n-2l} B_{2l}&amp;= 2n(1+1)P_{n-1}.
\end{align*}
	\end{itemize}
	\begin{theorem}\label{9}
		For every positive integer $n$, we have
\begin{align}
\sum\limits_{l=0}^{\lfloor n/2\rfloor }{n\choose 2l}2^{n-2l}
		\Bigl(\sqrt{x^{2}y^{2}+y}\Bigr) ^{2l+1}P_{n-2l}(x,y)E_{2l}&amp;= \Bigl(2xy+\sqrt{x^{2}y^{2}+y}\Bigr)^{n}-\Bigl( 2xy-\sqrt{x^{2}y^{2}+y}\Bigr)^{n},\label{f}\\
\sum\limits_{l=0}^{\lfloor n/2\rfloor }{n\choose 2l}\Bigl(x^{2}y^{2}+y\Bigr)^{l}	Q_{n-2l}(x,y)E_{2l}&amp;=2(xy)^{n}.\label{qx}
\end{align}
	\end{theorem}
	\begin{proof}
	The proof is similar to the proof of Theorem~\ref{6},
	by using the change of variables $z=2\sqrt{x^{2}y^{2}+y}z$ in~(\ref{16}) to prove~(\ref{f}) and 
	$z=4\sqrt{x^{2}y^{2}+y}z$ in~(\ref{16}) to prove~(\ref{qx}).
	\end{proof}
	\begin{itemize}
	\item By putting $y=1$ in Theorem~\ref{9}, we obtain some
relations between the Euler numbers and the Pell and Pell-Lucas polynomials:
\begin{align}
\sum\limits_{l=0}^{\lfloor n/2\rfloor }{n\choose 2l}2^{n-2l}
\Bigl(\sqrt{x^{2}+1}\Bigr) ^{2l+1}P_{n-2l}(x)E_{2l}&amp;= \Bigl( 2x+\sqrt{x^{2}+1}\Bigr)^{n}-\Bigl(2x-\sqrt{x^{2}+1}\Bigr)^{n},\label{sx}\\
\sum\limits_{l=0}^{\lfloor n/2\rfloor }{n\choose 2l}\Bigl( x^{2}+1\Bigr)^{l}Q_{n-2l}(x)E_{2l}&amp;=2x^{n}.\label{fd}
\end{align}
	\item By putting $x=y=1$ in Theorem~\ref{9} or $x=1$ in the relationships~(\ref{sx}) and~(\ref{fd}), we can state the following Pell-Euler
and Pell-Lucas-Euler identities:
\begin{align*}
\sum\limits_{l=0}^{\lfloor n/2\rfloor }{n\choose 2l}(\sqrt{2})2^{n-l}P_{n-2l}E_{2l} &amp;=
\Bigl( 2+\sqrt{2}\Bigr)^{n}-\Bigl(2-\sqrt{2}\Bigr)^{n},\\
\sum\limits_{l=0}^{\lfloor n/2\rfloor }{n\choose 2l}2^{l}	P_{n-2l}E_{2l}&amp;=2.
\end{align*}
\end{itemize}

\section{Acknowledgments}

The authors would like to thank the anonymous referees for their
valuable comments and suggestions for improving the original
version of this manuscript.



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\noindent 2020 {\it Mathematics Subject Classification}:
Primary 05E05; Secondary 11B39.

\noindent \emph{Keywords: }
Bernoulli number, Bernoulli polynomial, Euler number, Euler polynomial, convolution, generating function.

\bigskip
\hrule
\bigskip

\noindent (Concerned with sequences
\seqnum{A000032},
\seqnum{A000045},
\seqnum{A000364},
\seqnum{A027641}, and
\seqnum{A027642}.)

\bigskip
\hrule
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\vspace*{+.1in}
\noindent
Received December 12 2022;
revised versions December 13 2022; February 17 2023; February 20 2023.
Published in {\it Journal of Integer Sequences}, February 24 2023.

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