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\begin{center}
\vskip 1cm{\LARGE\bf A Note on Fibonacci \&amp; Lucas and \\
\vskip .1in
Bernoulli \&amp; Euler Polynomials
}
\vskip 1cm
\large
Claudio de Jes\'{u}s Pita Ruiz Velasco \\
Universidad Panamericana\\
Mexico City, Mexico\\
\href{mailto:cpita@up.edu.mx}{\tt cpita@up.edu.mx}
\end{center}

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\begin{abstract}
We study certain polynomials $P_{m}\left( x,y;t\right) $ and $Q_{m}\left(
x,y;t\right) $ of the variable $t$ whose coefficients involve bivariate
Fibonacci polynomials $F_{j}\left( x,y\right) $ or bivariate Lucas
polynomials $L_{j}\left( x,y\right) $. By working with $P_{m}\left(
x,y;tx\right) $ and $Q_{m}\left( x,y;tx\right) $, together with the
generating functions for Bernoulli polynomials $B_{i}\left( t\right) $ and
Euler polynomials $E_{i}\left( t\right) $, we obtain a list of eight
identities connecting $F_{j}\left( x,y\right) $ or $L_{j}\left( x,y\right) $
with $B_{i}\left( t\right) $ or $E_{i}\left( t\right) $. We present also
some consequences of these results.
\end{abstract}

\section{Introduction}
\label{Sec1}

We use $\mathbb{N}$ for the natural numbers and $\mathbb{N}^{\prime }$ for $%
\mathbb{N}\cup \left\{ 0\right\} $.

We recall now some definitions and basic facts of the main mathematical
objects involved in this work, namely Fibonacci and Lucas numbers and
polynomials (see \cite{K} and \cite{V}), and Bernoulli and Euler numbers and
polynomials (see \cite{Di}).

We follow the standard notation $F_{n}\left( x,y\right) $ and $L_{n}\left(
x,y\right) $ for the sequences of bivariate Fibonacci and Lucas polynomials,
defined by the recurrences $F_{n+2}\left( x,y\right) =xF_{n-1}\left(
x,y\right) +yF_{n}\left( x,y\right) $, $F_{0}\left( x,y\right) =0$, $%
F_{1}\left( x,y\right) =1$, and $L_{n+2}\left( x,y\right) =xL_{n-1}\left(
x,y\right) +yL_{n}\left( x,y\right) $, $L_{0}\left( x,y\right) =2$, $%
L_{1}\left( x,y\right) =x$, respectively, and extended to $n\in \mathbb{Z}$
as $F_{-n}\left( x,y\right) =-\left( -y\right) ^{-n}F_{n}\left( x,y\right) $
and $L_{-n}\left( x,y\right) =\left( -y\right) ^{-n}L_{n}\left( x,y\right) $%
. Plainly we have $F_{n}\left( 1,1\right) =F_{n}$ and $L_{n}\left(
1,1\right) =L_{n}$, the Fibonacci and Lucas number sequences (\seqnum{A000045}
and \seqnum{A000032} of Sloane's \textit{Encyclopedia}).
Some bivariate Fibonacci
polynomials are $F_{2}\left( x,y\right) =x$, $F_{3}\left( x,y\right)
=x^{2}+y $, $F_{4}\left( x,y\right) =x^{3}+2xy$, $F_{5}\left( x,y\right)
=x^{4}+3x^{2}y+y^{2}$,\ldots , and some bivariate Lucas polynomials are $%
L_{2}\left( x,y\right) =x^{2}+2y$, $L_{3}\left( x,y\right) =x^{3}+3xy$, $%
L_{4}\left( y\right) =x^{4}+4x^{2}y+2y^{2}$, $L_{5}\left( y\right)
=x^{5}+5x^{3}y+5xy^{2}$, \ldots . We will use extensively Binet's formulas
(without further comments):%
\begin{equation}
F_{n}\left( x,y\right) =\frac{1}{\sqrt{x^{2}+4y}}\left( \alpha ^{n}\left(
x,y\right) -\beta ^{n}\left( x,y\right) \right) \ \ \ \text{and}\ \ \
L_{n}\left( x,y\right) =\alpha ^{n}\left( x,y\right) +\beta ^{n}\left(
x,y\right) ,  \label{1.1}
\end{equation}%
where 
\begin{equation}
\alpha \left( x,y\right) =\frac{1}{2}\left( x+\sqrt{x^{2}+4y}\right) \text{
\ \ \ and \ \ \ }\beta \left( x,y\right) =\frac{1}{2}\left( x-\sqrt{x^{2}+4y}%
\right) \text{,}  \label{1.2}
\end{equation}%
together with the basic facts $\alpha \left( x,y\right) +\beta \left(
x,y\right) =x$ and $\alpha \left( x,y\right) \beta \left( x,y\right) =-y$.
We will use also the following explicit formulas for bivariate Fibonacci and
Lucas polynomials:%
\begin{equation}
F_{n}\left( x,y\right) =\sum_{k=0}^{\left\lfloor \frac{n-1}{2}\right\rfloor }%
\binom{n-1-k}{k}x^{n-1-2k}y^{k}\text{ \ and \ }L_{n}\left( x,y\right)
=\sum_{k=0}^{\left\lfloor \frac{n}{2}\right\rfloor }\frac{n}{n-k}\binom{n-k}{%
k}x^{n-2k}y^{k},  \label{1.3}
\end{equation}%
(the first formula is valid for $n\in \mathbb{N}^{\prime }$, and the second
one is valid for $n\in \mathbb{N}$).

We will be working with Bernoulli and Euler polynomials, which can be
defined as%
\begin{equation}
B_{n}\left( t\right) =\sum_{j=0}^{n}\binom{n}{j}B_{j}t^{n-j}\text{ \ \ \ and
\ \ \ }E_{n}\left( t\right) =\sum_{j=0}^{n}\binom{n}{j}\frac{E_{j}}{2^{j}}%
\left( t-\frac{1}{2}\right) ^{n-j},  \label{1.4}
\end{equation}%
where $B_{j}$ and $E_{j}$ are the Bernoulli and Euler numbers, respectively,
defined by the generating functions%
\begin{equation}
\frac{z}{e^{z}-1}=\sum_{j=0}^{\infty }B_{j}\frac{z^{j}}{j!}\ \ \ \text{and}\
\ \ \frac{2e^{z}}{e^{2z}+1}=\sum_{j=0}^{\infty }E_{j}\frac{z^{j}}{j!},
\label{1.5}
\end{equation}

The corresponding generating functions for Bernoulli and Euler polynomials
are%
\begin{equation}
\frac{ze^{zt}}{e^{z}-1}=\sum_{j=0}^{\infty }B_{j}\left( t\right) \frac{z^{j}%
}{j!}\ \ \ \text{and}\ \ \ \frac{2e^{zt}}{e^{z}+1}=\sum_{n=0}^{\infty
}E_{n}\left( t\right) \frac{z^{n}}{n!}.  \label{1.6}
\end{equation}

It is not difficult to see that for $j\in \mathbb{N}$, one has $B_{2j+1}=0$
and $E_{2j-1}=0$ (odd Bernoulli numbers are zero, except $B_{1}=-\frac{1}{2}$%
, and odd Euler numbers are zero). Also we have $B_{0}=E_{0}=1$. Some other
values are $B_{2}=\frac{1}{6}$, $B_{4}=-\frac{1}{30}$, $B_{6}=\frac{1}{42}$, 
$B_{8}=-\frac{1}{30}$, \ldots\ and $E_{2}=-1$, $E_{4}=5$, $E_{6}=-61$, $%
E_{8}=1385$, \ldots . The first Bernoulli polynomials are $B_{0}\left(
t\right) =1$, $B_{1}\left( t\right) =t-\frac{1}{2}$, $B_{2}\left( t\right)
=t^{2}-t+\frac{1}{6}$, $B_{3}\left( t\right) =t^{3}-\frac{3}{2}t^{2}+\frac{1%
}{2}t$, $B_{4}\left( t\right) =t^{4}-2t^{3}+t^{2}-\frac{1}{30}$, \ldots ,
and the first Euler polynomials are $E_{0}\left( t\right) =1$, $E_{1}\left(
t\right) =t-\frac{1}{2}$, $E_{2}\left( t\right) =t^{2}-t$, $E_{3}\left(
t\right) =t^{3}-\frac{3}{2}t^{2}+\frac{1}{4}$, $E_{4}\left( t\right)
=t^{4}-2t^{3}+t$, \ldots . One can see easily that for $n\in \mathbb{N}$,
one has $B_{n}^{\prime }\left( t\right) =nB_{n-1}\left( t\right) $ and $%
E_{n}^{\prime }\left( t\right) =nE_{n-1}\left( t\right) $.

Besides the trivial fact $B_{n}\left( 0\right) =B_{n}$, we will use that

\begin{equation}
B_{n}\left( \frac{1}{2}\right) =\left( 2^{1-n}-1\right) B_{n}\text{ \ \ \ \
, \ \ \ \ }E_{n}\left( \frac{1}{2}\right) =2^{-n}E_{n}.  \label{1.7}
\end{equation}%
\begin{equation}
E_{n}\left( 0\right) =-\frac{2\left( 2^{n+1}-1\right) }{n+1}B_{n+1}.
\label{1.8}
\end{equation}

There are interesting papers in the literature pursuing relations among
Bernoulli and Euler numbers and/or polynomials (see the 2006 works of Sun
and Pan \cite{Zhi1,Zhi2} and references therein). On the other hand, there
are certainly much research establishing relations among Fibonacci and Lucas
numbers or polynomials (see references in the books of Koshy \cite%
{K} and Vajda \cite{V}). But also there has been interest in finding
connections between the mathematics of Bernoulli and Euler and the
mathematics of Fibonacci and Lucas, and this interest is not new: in 1975 P.
F. Byrd \cite{By} obtains some nice formulas connecting Bernoulli, Fibonacci
and Lucas numbers. We also have to mention the 1957 work of Kelisky \cite%
{Kel}, the 2005 work of T. Zhang and Y. Ma \cite{Zha}, and the nice papers of
J. Cigler \cite{Ci,Ci2}, which contains some of the results of our
corollary \ref%
{Cor3.1} in section \ref{Sec3}. This article responds to the interest in
exploring more about these kind of connections. We work with certain kind of
Appel sequences of polynomials $P_{n}\left( x,y;t\right) $ and $Q_{n}\left(
x,y;t\right) $ in the variable $t$, whose coefficients are in turn bivariate
Fibonacci polynomials $F_{m}\left( x,y\right) $ or bivariate Lucas
polynomials $L_{m}\left( x,y\right) $. By working with generating functions
of Bernoulli and Euler polynomials, we establish some identities involving
the polynomials $P_{n}\left( x,y;xt\right) $ and $Q_{n}\left( x,y;xt\right) $
together with Bernoulli polynomials $B_{j}\left( t\right) $, Euler
polynomials $E_{j}\left( t\right) $, bivariate Fibonacci $F_{k}\left(
x,y\right) $ and bivariate Lucas $L_{k}\left( x,y\right) $ polynomials.
These identities are the main results of the work (proposition \ref{Prop2.1}%
). In section \ref{Sec3} we obtain some corollaries from identities of
section \ref{Sec2}.

\section{The main results}
\label{Sec2}

We begin with a lemma with two easy identities that we will need in the
proof of the main results of this work.

\begin{lemma}
The following identities hold%
\begin{equation}
\left( 1-e^{-xz}\right) \sum_{n=0}^{\infty }L_{n}\left( x,y\right) \frac{%
z^{n}}{n!}=2\sum_{n=0}^{\infty }L_{2n+1}\left( x,y\right) \frac{z^{2n+1}}{%
\left( 2n+1\right) !}.  \label{2.1}
\end{equation}%
\begin{equation}
\left( 1+e^{-xz}\right) \sum_{n=0}^{\infty }L_{n}\left( x,y\right) \frac{%
z^{n}}{n!}=2\sum_{n=0}^{\infty }L_{2n}\left( x,y\right) \frac{z^{2n}}{\left(
2n\right) !}.  \label{2.2}
\end{equation}
\end{lemma}

\begin{proof}
We have%
\begin{equation*}
e^{\alpha \left( x,y\right) z}+e^{\beta \left( x,y\right) z}=e^{\left(
x-\beta \left( x,y\right) \right) z}+e^{\left( x-\alpha \left( x,y\right)
\right) z}=e^{xz}\left( e^{-\alpha \left( x,y\right) z}+e^{-\beta \left(
x,y\right) z}\right) ,
\end{equation*}%
and then%
\begin{equation*}
e^{-xz}\sum_{n=0}^{\infty }L_{n}\left( x,y\right) \frac{z^{n}}{n!}%
=\sum_{n=0}^{\infty }L_{n}\left( x,y\right) \frac{\left( -z\right) ^{n}}{n!}.
\end{equation*}

Thus%
\begin{eqnarray*}
\left( 1-e^{-xz}\right) \sum_{n=0}^{\infty }\!L_{n}\left( x,y\right) \frac{%
z^{n}}{n!} &amp;=&amp;\sum_{n=0}^{\infty }\!L_{n}\left( x,y\right) \frac{z^{n}}{n!}%
-\sum_{n=0}^{\infty }\!L_{n}\left( x,y\right) \frac{\left( -z\right) ^{n}}{n!%
} \\
&amp;=&amp;2\sum_{n=0}^{\infty }\!L_{2n+1}\left( x,y\right) \frac{z^{2n+1}}{\left(
2n+1\right) !},
\end{eqnarray*}%
which proves (\ref{2.1}), and 
\begin{eqnarray*}
\left( 1+e^{-xz}\right) \sum_{n=0}^{\infty }L_{n}\left( x,y\right) \frac{%
z^{n}}{n!} &amp;=&amp;\sum_{n=0}^{\infty }L_{n}\left( x,y\right) \frac{z^{n}}{n!}%
+\sum_{n=0}^{\infty }L_{n}\left( x,y\right) \frac{\left( -z\right) ^{n}}{n!}
\\
&amp;=&amp;2\sum_{n=0}^{\infty }L_{2n}\left( x,y\right) \frac{z^{2n}}{\left(
2n\right) !},
\end{eqnarray*}%
which proves (\ref{2.2}).
\end{proof}

Let us consider the polynomials%
\begin{equation}
Q_{n}\left( x,y;t\right) =\sum_{k=0}^{n}\binom{n}{k}\left( -1\right)
^{k}L_{k}\left( x,y\right) t^{n-k},  \label{2.3}
\end{equation}%
which can be written as%
\begin{equation}
Q_{n}\left( x,y;t\right) =\left( t-\alpha \left( x,y\right) \right)
^{n}+\left( t-\beta \left( x,y\right) \right) ^{n}.  \label{2.4}
\end{equation}

Observe that%
\begin{eqnarray*}
\sum_{n=0}^{\infty }Q_{n}\left( x,y;tx\right) \frac{z^{n}}{n!}
&amp;=&amp;\sum_{n=0}^{\infty }\left( tx-\alpha \left( x,y\right) \right) ^{n}\frac{%
z^{n}}{n!}+\sum_{n=0}^{\infty }\left( tx-\beta \left( x,y\right) \right) ^{n}%
\frac{z^{n}}{n!} \\
&amp;=&amp;e^{\left( tx-\alpha \left( x,y\right) \right) z}+e^{\left( tx-\beta
\left( x,y\right) \right) z} \\
&amp;=&amp;e^{\left( tx-x+\beta \left( x,y\right) \right) z}+e^{\left( tx-x+\alpha
\left( x,y\right) \right) z} \\
&amp;=&amp;e^{\left( t-1\right) xz}\left( e^{\beta \left( x,y\right) z}+e^{\alpha
\left( x,y\right) z}\right) \\
&amp;=&amp;e^{\left( t-1\right) xz}\sum_{n=0}^{\infty }\left( \alpha ^{n}\left(
x,y\right) +\beta ^{n}\left( x,y\right) \right) \frac{z^{n}}{n!}.
\end{eqnarray*}

That is, we have%
\begin{equation}
\sum_{n=0}^{\infty }Q_{n}\left( x,y;tx\right) \frac{z^{n}}{n!}=e^{\left(
t-1\right) xz}\sum_{n=0}^{\infty }L_{n}\left( x,y\right) \frac{z^{n}}{n!}.
\label{2.5}
\end{equation}

We can use (\ref{2.1}) to write (\ref{2.5}) as%
\begin{equation}
\sum_{n=0}^{\infty }Q_{n}\left( x,y;tx\right) \frac{z^{n}}{n!}=2\frac{e^{txz}%
}{e^{xz}-1}\sum_{n=0}^{\infty }L_{2n+1}\left( x,y\right) \frac{z^{2n+1}}{%
\left( 2n+1\right) !}.  \label{2.6}
\end{equation}

By using the generating function for Bernoulli polynomials (\ref{1.6}) we
can write (\ref{2.6}) as%
\begin{equation}
\sum_{n=0}^{\infty }Q_{n}\left( x,y;tx\right) \frac{z^{n}}{n!}%
=2\sum_{j=0}^{\infty }\sum_{n=0}^{\infty }\binom{2n+j}{j}B_{j}\left(
t\right) x^{j-1}\frac{L_{2n+1}\left( x,y\right) }{2n+1}\frac{z^{2n+j}}{%
\left( 2n+j\right) !}.  \label{2.7}
\end{equation}

By equating the coefficients of similar powers of $z$ in (\ref{2.7}) we get%
\begin{equation}
Q_{2m}\left( x,y;tx\right) =2\sum_{j=0}^{m}\binom{2m}{2j}B_{2j}\left(
t\right) \frac{x^{2j-1}}{2m+1-2j}L_{2m+1-2j}\left( x,y\right) ,  \label{2.8}
\end{equation}%
and%
\begin{equation}
Q_{2m+1}\left( x,y;tx\right) =2\sum_{j=0}^{m}\binom{2m+1}{2j+1}%
B_{2j+1}\left( t\right) \frac{x^{2j}}{2m+1-2j}L_{2m+1-2j}\left( x,y\right) .
\label{2.9}
\end{equation}

Similarly, observe that%
\begin{eqnarray*}
\sum_{n=0}^{\infty }Q_{n}\left( x,y;tx\right) \frac{\left( -z\right) ^{n}}{n!%
} &amp;=&amp;\sum_{n=0}^{\infty }\left( -tx+\alpha \left( x,y\right) \right) ^{n}%
\frac{z^{n}}{n!}+\sum_{n=0}^{\infty }\left( -tx+\beta \left( x,y\right)
\right) ^{n}\frac{z^{n}}{n!} \\
&amp;=&amp;e^{\left( -tx+\alpha \left( x,y\right) \right) z}+e^{\left( -tx+\beta
\left( x,y\right) \right) z} \\
&amp;=&amp;e^{-txz}\left( e^{\alpha \left( x,y\right) z}+e^{\beta \left( x,y\right)
z}\right) \\
&amp;=&amp;e^{-txz}\sum_{n=0}^{\infty }\left( \alpha ^{n}\left( x,y\right) +\beta
^{n}\left( x,y\right) \right) \frac{z^{n}}{n!}.
\end{eqnarray*}

That is, we have%
\begin{equation}
\sum_{n=0}^{\infty }Q_{n}\left( x,y;tx\right) \frac{\left( -z\right) ^{n}}{n!%
}=e^{-txz}\sum_{n=0}^{\infty }L_{n}\left( x,y\right) \frac{z^{n}}{n!}.
\label{2.10}
\end{equation}

We can use (\ref{2.2}) to write (\ref{2.10}) as%
\begin{equation}
\sum_{n=0}^{\infty }Q_{n}\left( x,y;tx\right) \frac{\left( -z\right) ^{n}}{n!%
}=\frac{2e^{-txz}}{1+e^{-xz}}\sum_{n=0}^{\infty }L_{2n}\left( x,y\right) 
\frac{z^{2n}}{\left( 2n\right) !}.  \label{2.11}
\end{equation}

By using the generating functions of Euler polynomials (\ref{1.6}), we can
write (\ref{2.11}) as%
\begin{equation}
\sum_{n=0}^{\infty }Q_{n}\left( x,y;tx\right) \frac{\left( -z\right) ^{n}}{n!%
}=\sum_{j=0}^{\infty }\sum_{n=0}^{\infty }\binom{2n+j}{j}E_{j}\left(
t\right) \left( -x\right) ^{j}L_{2n}\left( x,y\right) \frac{z^{2n+j}}{\left(
2n+j\right) !}.  \label{2.12}
\end{equation}

By equating the coefficients of similar powers of $z$ in (\ref{2.12}) we
obtain%
\begin{equation}
Q_{2m}\left( x,y;tx\right) =\sum_{j=0}^{m}\binom{2m}{2j}E_{2j}\left(
t\right) x^{2j}L_{2m-2j}\left( x,y\right) ,  \label{2.13}
\end{equation}%
and%
\begin{equation}
Q_{2m+1}\left( x,y;tx\right) =\sum_{j=0}^{m}\binom{2m+1}{2j+1}E_{2j+1}\left(
t\right) x^{2j+1}L_{2m-2j}\left( x,y\right) .  \label{2.14}
\end{equation}

Thus, (\ref{2.8}) and (\ref{2.13}) give us%
\begin{eqnarray}
Q_{2m}\left( x,y;tx\right) &amp;=&amp;2\sum_{j=0}^{m}\binom{2m}{2j}B_{2j}\left(
t\right) \frac{x^{2j-1}}{2m+1-2j}L_{2m+1-2j}\left( x,y\right)  \label{2.15}
\\
&amp;=&amp;\sum_{j=0}^{m}\binom{2m}{2j}E_{2j}\left( t\right) x^{2j}L_{2m-2j}\left(
x,y\right) ,  \notag
\end{eqnarray}%
and (\ref{2.9}) and (\ref{2.14}) give us%
\begin{eqnarray}
Q_{2m+1}\left( x,y;tx\right) &amp;=&amp;2\sum_{j=0}^{m}\binom{2m+1}{2j+1}%
B_{2j+1}\left( t\right) \frac{x^{2j}}{2m+1-2j}L_{2m+1-2j}\left( x,y\right)
\label{2.16} \\
&amp;=&amp;\sum_{j=0}^{m}\binom{2m+1}{2j+1}E_{2j+1}\left( t\right)
x^{2j+1}L_{2m-2j}\left( x,y\right) .  \notag
\end{eqnarray}

If we consider now the polynomials 
\begin{eqnarray}
P_{n}\left( x,y;t\right) &amp;=&amp;\sum_{k=0}^{n}\binom{n}{k}\left( -1\right)
^{k+1}F_{k}\left( x,y\right) t^{n-k}  \label{2.17} \\
&amp;=&amp;-\frac{1}{\sqrt{x^{2}+4y}}\left( \left( t-\alpha \left( x,y\right)
\right) ^{n}-\left( t-\beta \left( x,y\right) \right) ^{n}\right) ,  \notag
\end{eqnarray}%
it is possible to establish similar results to (\ref{2.15}) and (\ref{2.16})
(involving Fibonacci polynomials $F_{k}\left( x,y\right) $ instead of Lucas
polynomials $L_{k}\left( x,y\right) $). We describe the main steps of the
procedure to obtain these new results and leave the reader to complete the
details of the proofs. Firstly one proves that%
\begin{equation}
\left( 1-e^{-xz}\right) \sum_{n=0}^{\infty }F_{n}\left( x,y\right) \frac{%
z^{n}}{n!}=2\sum_{n=0}^{\infty }F_{2n}\left( x,y\right) \frac{z^{2n}}{\left(
2n\right) !}.  \label{2.18}
\end{equation}%
\begin{equation}
\left( 1+e^{-xz}\right) \sum_{n=0}^{\infty }F_{n}\left( x,y\right) \frac{%
z^{n}}{n!}=2\sum_{n=0}^{\infty }F_{2n+1}\left( x,y\right) \frac{z^{2n+1}}{%
\left( 2n+1\right) !}.  \label{2.19}
\end{equation}

(Similar to (\ref{2.1}) and (\ref{2.2}).) Secondly one proves that%
\begin{equation}
\sum_{n=0}^{\infty }P_{n}\left( x,y;tx\right) \frac{z^{n}}{n!}=e^{\left(
t-1\right) xz}\sum_{n=0}^{\infty }F_{n}\left( x,y\right) \frac{z^{n}}{n!},
\label{2.20}
\end{equation}%
then uses (\ref{2.18}) to write this expression as%
\begin{equation}
\sum_{n=0}^{\infty }P_{n}\left( x,y;tx\right) \frac{z^{n}}{n!}=2\frac{e^{txz}%
}{e^{xz}-1}\sum_{n=0}^{\infty }F_{2n}\left( x,y\right) \frac{z^{2n}}{\left(
2n\right) !},  \label{2.21}
\end{equation}%
and finally uses the generating function for Bernoulli polynomials (\ref{1.6}%
) to write (\ref{2.21}) as%
\begin{equation}
\sum_{n=0}^{\infty }P_{n}\left( x,y;tx\right) \frac{z^{n}}{n!}%
=\sum_{n=0}^{\infty }\sum_{j=0}^{\infty }\binom{2n+1+j}{j}B_{j}\left(
t\right) \frac{F_{2n+2}\left( x,y\right) }{n+1}x^{j-1}\frac{z^{2n+j+1}}{%
\left( 2n+1+j\right) !}.  \label{2.21.5}
\end{equation}

By equating the coefficients of similar powers of $z$ in (\ref{2.21.5}) one
obtains that%
\begin{equation}
P_{2m+1}\left( x,y;tx\right) =\sum_{j=0}^{m}\binom{2m+1}{2j}B_{2j}\left(
t\right) \frac{x^{2j-1}}{m+1-j}F_{2m+2-2j}\left( x,y\right) ,  \label{2.22}
\end{equation}%
and%
\begin{equation}
P_{2m}\left( x,y;tx\right) =\sum_{j=0}^{m-1}\binom{2m}{2j+1}B_{2j+1}\left(
t\right) \frac{x^{2j}}{m-j}F_{2m-2j}\left( x,y\right) .  \label{2.23}
\end{equation}

On the other hand, one proves first that%
\begin{equation}
\sum_{n=0}^{\infty }P_{n}\left( x,y;tx\right) \frac{\left( -z\right) ^{n}}{n!%
}=-e^{-txz}\sum_{n=0}^{\infty }F_{n}\left( x,y\right) \frac{z^{n}}{n!},
\label{2.24}
\end{equation}%
then uses (\ref{2.19}) to write (\ref{2.24}) as%
\begin{equation}
\sum_{n=0}^{\infty }P_{n}\left( x,y;tx\right) \frac{\left( -z\right) ^{n}}{n!%
}=-\frac{2e^{-txz}}{1+e^{-xz}}\sum_{n=0}^{\infty }F_{2n+1}\left( x,y\right) 
\frac{z^{2n+1}}{\left( 2n+1\right) !},  \label{2.25}
\end{equation}%
and finally uses the generating function of Euler polynomials (\ref{1.6}) to
write (\ref{2.25}) as%
\begin{equation}
\sum_{n=0}^{\infty }P_{n}\left( x,y;tx\right) \frac{\left( -z\right) ^{n}}{n!%
}=-\sum_{j=0}^{\infty }\sum_{n=0}^{\infty }\binom{2n+1+j}{j}E_{j}\left(
t\right) \left( -x\right) ^{j}F_{2n+1}\left( x,y\right) \frac{z^{2n+j+1}}{%
\left( 2n+1+j\right) !}.  \label{2.26}
\end{equation}

By equating the coefficients of similar powers of $z$ in (\ref{2.26}) one
gets that%
\begin{equation}
P_{2m}\left( x,y;tx\right) =\sum_{j=0}^{m-1}\binom{2m}{2j+1}E_{2j+1}\left(
t\right) x^{2j+1}F_{2m-1-2j}\left( x,y\right) ,  \label{2.27}
\end{equation}%
and%
\begin{equation}
P_{2m+1}\left( x,y;tx\right) =\sum_{j=0}^{m}\binom{2m+1}{2j}E_{2j}\left(
t\right) x^{2j}F_{2m+1-2j}\left( x,y\right) .  \label{2.28}
\end{equation}

Thus, we have identities (\ref{2.23}) and (\ref{2.27}) similar to (\ref{2.15}%
), namely%
\begin{eqnarray}
P_{2m}\left( x,y;tx\right) &amp;=&amp;\sum_{j=0}^{m-1}\binom{2m}{2j+1}B_{2j+1}\left(
t\right) \frac{x^{2j}}{m-j}F_{2m-2j}\left( x,y\right)  \label{2.29} \\
&amp;=&amp;\sum_{j=0}^{m-1}\binom{2m}{2j+1}E_{2j+1}\left( t\right)
x^{2j+1}F_{2m-1-2j}\left( x,y\right) ,  \notag
\end{eqnarray}%
and identities (\ref{2.22}) and (\ref{2.28}) similar to (\ref{2.16}), namely%
\begin{eqnarray}
P_{2m+1}\left( x,y;tx\right) &amp;=&amp;\sum_{j=0}^{m}\binom{2m+1}{2j}B_{2j}\left(
t\right) \frac{x^{2j-1}}{m+1-j}F_{2m+2-2j}\left( x,y\right)  \label{2.30} \\
&amp;=&amp;\sum_{j=0}^{m}\binom{2m+1}{2j}E_{2j}\left( t\right)
x^{2j}F_{2m+1-2j}\left( x,y\right) .  \notag
\end{eqnarray}

In the following proposition we collect the main identities (\ref{2.15}), (%
\ref{2.16}), (\ref{2.29}) and (\ref{2.30}) obtained in this section, which
are the main results of this work.

\begin{proposition}
\label{Prop2.1}The following identities hold%
\begin{eqnarray}
\sum_{k=0}^{2m}\!\binom{2m}{k}\!\left( -1\right) ^{k+1}\!F_{k}\left(
x,y\right) \!\left( tx\right) ^{2m-k} &amp;=&amp;\sum_{j=0}^{m-1}\binom{2m}{2j+1}%
B_{2j+1}\left( t\right) \frac{x^{2j}}{m-j}F_{2m-2j}\left( x,y\right)
\label{2.31} \\
&amp;=&amp;\sum_{j=0}^{m-1}\binom{2m}{2j+1}E_{2j+1}\left( t\right)
x^{2j+1}F_{2m-1-2j}\left( x,y\right) .  \notag
\end{eqnarray}%
\begin{eqnarray}
\sum_{k=0}^{2m+1}\!\binom{2m\!+\!1}{k}\!\left( -1\right)
^{k+1}\!F_{k}\!\left( x,y\right) \!\left( tx\right) ^{2m+1-k}\!\!\!
&amp;=&amp;\!\!\sum_{j=0}^{m}\!\binom{2m\!+\!1}{2j}\frac{B_{2j}\!\left( t\right)
x^{2j\!-\!1}}{m\!+\!1\!-\!j}\!F_{2m+2-2j}\!\left( x,y\right)  \label{2.32} \\
&amp;=&amp;\sum_{j=0}^{m}\binom{2m+1}{2j}E_{2j}\left( t\right)
x^{2j}F_{2m+1-2j}\left( x,y\right) .  \notag
\end{eqnarray}%
\begin{eqnarray}
\sum_{k=0}^{2m}\!\binom{2m}{k}\!\left( -1\right) ^{k}\!L_{k}\!\left(
x,y\right) \!\left( tx\right) ^{2m-k} &amp;=&amp;2\sum_{j=0}^{m}\binom{2m}{2j}\frac{%
B_{2j}\left( t\right) x^{2j-1}}{2m+1-2j}L_{2m+1-2j}\left( x,y\right)
\label{2.33} \\
&amp;=&amp;\sum_{j=0}^{m}\binom{2m}{2j}E_{2j}\left( t\right) x^{2j}L_{2m-2j}\left(
x,y\right) .  \notag
\end{eqnarray}%
\begin{eqnarray}
\sum_{k=0}^{2m+1}\!\binom{2m\!+\!1}{k}\!\left( -1\right) ^{k}\!L_{k}\!\left(
x,y\right) \!\left( tx\right) ^{2m+1-k}\!\!\! &amp;=&amp;\!\!2\!\sum_{j=0}^{m}\!%
\binom{2m\!+\!1}{2j\!+\!1}\frac{B_{2j+1}\!\left( t\right) x^{2j}}{%
2m\!+\!1\!-\!2j}L_{2m+1-2j}\!\left( x,y\right)  \label{2.34} \\
&amp;=&amp;\sum_{j=0}^{m}\binom{2m+1}{2j+1}E_{2j+1}\left( t\right)
x^{2j+1}L_{2m-2j}\left( x,y\right) .  \notag
\end{eqnarray}
\end{proposition}

\bigskip

We want to mention that (\ref{2.31}) and (\ref{2.32}) can be obtained as
consequences of (\ref{2.34}) and (\ref{2.33}), respectively. Indeed, if we
use the well-known fact that $\frac{\partial }{\partial y}L_{s}\left(
x,y\right) =sF_{s-1}\left( x,y\right) $ (see \cite{Yu}) and take the
derivative with respect to $y$ in (\ref{2.34}), we get%
\begin{eqnarray*}
&amp;&amp;\sum_{k=1}^{2m+1}\binom{2m+1}{k}\left( -1\right) ^{k}kF_{k-1}\left(
x,y\right) \left( tx\right) ^{2m+1-k} \\
&amp;=&amp;2\sum_{j=0}^{m-1}\binom{2m+1}{2j+1}B_{2j+1}\left( t\right)
x^{2j}F_{2m-2j}\left( x,y\right) \\
&amp;=&amp;\sum_{j=0}^{m-1}\binom{2m+1}{2j+1}E_{2j+1}\left( t\right) x^{2j+1}\left(
2m-2j\right) F_{2m-1-2j}\left( x,y\right) ,
\end{eqnarray*}%
which is (\ref{2.31}) (after some easy simplifications). Similarly, the
derivative with respect to $y$ of (\ref{2.33}) is%
\begin{eqnarray*}
\sum_{k=1}^{2m}\binom{2m}{k}\left( -1\right) ^{k}kF_{k-1}\left( x,y\right)
\left( tx\right) ^{2m-k} &amp;=&amp;2\sum_{j=0}^{m-1}\binom{2m}{2j}B_{2j}\left(
t\right) x^{2j-1}F_{2m-2j}\left( x,y\right) \\
&amp;=&amp;\sum_{j=0}^{m-1}\binom{2m}{2j}E_{2j}\left( t\right) x^{2j}\left(
2m-2j\right) F_{2m-1-2j}\left( x,y\right) ,
\end{eqnarray*}%
which, after some simplifications, becomes (\ref{2.32}).

\section{\label{Sec3}Some Corollaries}

In this section we obtain some other identities that are contained in our
main results (\ref{2.31}) to (\ref{2.34}).

\begin{corollary}
\label{Cor3.1}The following identities hold%
\begin{equation}
F_{2m}\left( x,y\right) =\sum_{j=0}^{m-1}\binom{2m}{2j+1}\frac{\left(
4^{j+1}-1\right) B_{2j+2}}{j+1}x^{2j+1}F_{2m-1-2j}\left( x,y\right) .
\label{3.1}
\end{equation}%
\begin{equation}
F_{2m+1}\left( x,y\right) =\sum_{j=0}^{m}\binom{2m+1}{2j}\frac{B_{2j}}{m+1-j}%
x^{2j-1}F_{2m+2-2j}\left( x,y\right) .  \label{3.2}
\end{equation}%
\begin{equation}
L_{2m}\left( x,y\right) =2\sum_{j=0}^{m}\binom{2m}{2j}\frac{B_{2j}}{2m+1-2j}%
x^{2j-1}L_{2m+1-2j}\left( x,y\right) .  \label{3.3}
\end{equation}%
\begin{equation}
L_{2m+1}\left( x,y\right) =\sum_{j=0}^{m}\binom{2m+1}{2j+1}\frac{\left(
4^{j+1}-1\right) B_{2j+2}}{j+1}x^{2j+1}L_{2m-2j}\left( x,y\right) .
\label{3.4}
\end{equation}
\end{corollary}

\begin{proof}
If we set $t=0$ in (\ref{2.31}) we obtain (by using (\ref{1.8}))%
\begin{eqnarray}
-F_{2m}\left( x,y\right) &amp;=&amp;\sum_{j=0}^{m-1}\binom{2m}{2j+1}B_{2j+1}\frac{%
x^{2j}}{m-j}F_{2m-2j}\left( x,y\right)  \notag \\
&amp;=&amp;-\sum_{j=0}^{m-1}\binom{2m}{2j+1}\frac{2\left( 2^{2j+2}-1\right) }{2j+2}%
x^{2j+1}F_{2m-1-2j}\left( x,y\right) .  \notag
\end{eqnarray}

The first equality is trivial. The second one is (\ref{3.1}). Similarly, by
setting $t=0$ in (\ref{2.32}), (\ref{2.33}) and (\ref{2.34}), and using (\ref%
{1.8}), we obtain (\ref{3.2}), (\ref{3.3}) and (\ref{3.4}), respectively.
\end{proof}

Formulas (\ref{3.1}) to (\ref{3.4}) express even (odd) indexed bivariate
Fibonacci and Lucas polynomials as linear combinations of odd (even,
respectively) indexed polynomials of the same kind. Some versions of these
results appear in \cite{Ci} (see also \cite{Ci2}).

\begin{corollary}
\label{Cor3.2}The following identities hold%
\begin{eqnarray}
&amp;&amp;\sum_{j=0}^{m}\binom{2m+1}{2j}\left( 2^{1-2j}-1\right) \frac{B_{2j}}{m+1-j}%
x^{2j-1}F_{2m+2-2j}\left( x,y\right)  \notag \\
&amp;=&amp;\sum_{j=0}^{m}\binom{2m}{2j}\left( 2^{1-2j}-1\right) \frac{B_{2j}}{2m+1-2j%
}x^{2j-1}L_{2m+1-2j}\left( x,y\right)  \notag \\
&amp;=&amp;\sum_{j=0}^{m}\binom{2m+1}{2j}2^{-2j}E_{2j}x^{2j}F_{2m+1-2j}\left(
x,y\right)  \notag \\
&amp;=&amp;\frac{1}{2}\sum_{j=0}^{m}\binom{2m}{2j}2^{-2j}E_{2j}x^{2j}L_{2m-2j}\left(
x,y\right)  \notag \\
&amp;=&amp;\frac{1}{2^{2m}}\left( x^{2}+4y\right) ^{m}.  \label{3.5}
\end{eqnarray}
\end{corollary}

\begin{proof}
Observe that (from (\ref{2.17})) 
\begin{eqnarray*}
P_{n}\left( x,y;\frac{x}{2}\right) &amp;=&amp;-\frac{1}{\sqrt{x^{2}+4y}}\left(
\left( \frac{x}{2}-\alpha \left( x,y\right) \right) ^{n}-\left( \frac{x}{2}%
-\beta \left( x,y\right) \right) ^{n}\right) \\
&amp;=&amp;\frac{1-\left( -1\right) ^{n}}{2^{n}}\left( x^{2}+4y\right) ^{\frac{n-1}{2%
}},
\end{eqnarray*}%
and (from (\ref{2.4}))%
\begin{eqnarray*}
Q_{n}\left( x,y;\frac{x}{2}\right) &amp;=&amp;\left( \frac{x}{2}-\alpha \left(
x,y\right) \right) ^{n}+\left( \frac{x}{2}-\beta \left( x,y\right) \right)
^{n} \\
&amp;=&amp;\frac{1+\left( -1\right) ^{n}}{2^{n}}\left( x^{2}+4y\right) ^{\frac{n}{2}%
}.
\end{eqnarray*}

Then, by setting $t=\frac{1}{2}$ in (\ref{2.32}) and (\ref{2.33}) (and using
(\ref{1.7}) and (\ref{1.8})) we obtain that%
\begin{eqnarray*}
P_{2m+1}\left( x,y;\frac{x}{2}\right) &amp;=&amp;\frac{1}{2^{2m}}\left(
x^{2}+4y\right) ^{m} \\
&amp;=&amp;\sum_{j=0}^{m}\binom{2m+1}{2j}\left( 2^{1-2j}-1\right) \frac{B_{2j}}{m+1-j%
}x^{2j-1}F_{2m+2-2j}\left( x,y\right) \\
&amp;=&amp;\sum_{j=0}^{m}\binom{2m+1}{2j}2^{-2j}E_{2j}x^{2j}F_{2m+1-2j}\left(
x,y\right) ,
\end{eqnarray*}

and%
\begin{eqnarray*}
Q_{2m}\left( x,y;\frac{x}{2}\right) &amp;=&amp;\frac{2}{2^{2m}}\left(
x^{2}+4y\right) ^{m} \\
&amp;=&amp;2\sum_{j=0}^{m}\binom{2m}{2j}\left( 2^{1-2j}-1\right) \frac{B_{2j}}{%
2m+1-2j}x^{2j-1}L_{2m+1-2j}\left( x,y\right) \\
&amp;=&amp;\sum_{j=0}^{m}\binom{2m}{2j}2^{-2j}E_{2j}x^{2j}L_{2m-2j}\left( x,y\right)
,
\end{eqnarray*}%
respectively. These are identities (\ref{3.5}). (Note that identities (\ref%
{2.31}) and (\ref{2.34}) produce only trivial cases with $t=\frac{1}{2}$.)
\end{proof}

In the rest of this section we write $\xi \left( x,y\right) $ to denote any
of $\alpha \left( x,y\right) $ or $\beta \left( x,y\right) $. When we write
an expression involving $\xi \left( x,y\right) $ together with the
plus-minus sign $\pm $, we understand that the plus sign corresponds to the
case $\xi =\alpha $, and the minus sign corresponds to the case $\xi =\beta $%
.

\begin{corollary}
\label{Cor3.3}The following identities hold%
\begin{eqnarray}
&amp;&amp;\sum_{j=0}^{m}\binom{2m+1}{2j}B_{2j}\left( \frac{\xi \left( x,y\right) }{x}%
\right) \frac{x^{2j-1}}{m+1-j}F_{2m+2-2j}\left( x,y\right)  \notag \\
&amp;=&amp;2\sum_{j=0}^{m}\binom{2m}{2j}B_{2j}\left( \frac{\xi \left( x,y\right) }{x}%
\right) \frac{x^{2j-1}}{2m+1-2j}L_{2m+1-2j}\left( x,y\right)  \notag \\
&amp;=&amp;\sum_{j=0}^{m}\binom{2m+1}{2j}E_{2j}\left( \frac{\xi \left( x,y\right) }{x%
}\right) x^{2j}F_{2m+1-2j}\left( x,y\right)  \notag \\
&amp;=&amp;\sum_{j=0}^{m}\binom{2m}{2j}E_{2j}\left( \frac{\xi \left( x,y\right) }{x}%
\right) x^{2j}L_{2m-2j}\left( x,y\right)  \notag \\
&amp;=&amp;\left( x^{2}+4y\right) ^{m}.  \label{3.6}
\end{eqnarray}%
\begin{eqnarray}
&amp;&amp;\left( x^{2}+4y\right) \sum_{j=0}^{m-1}\binom{2m}{2j+1}B_{2j+1}\left( 
\frac{\xi \left( x,y\right) }{x}\right) \frac{x^{2j}}{m-j}F_{2m-2j}\left(
x,y\right)  \notag \\
&amp;=&amp;2\sum_{j=0}^{m}\binom{2m+1}{2j+1}B_{2j+1}\left( \frac{\xi \left(
x,y\right) }{x}\right) \frac{x^{2j}}{2m+1-2j}L_{2m+1-2j}\left( x,y\right) 
\notag \\
&amp;=&amp;\left( x^{2}+4y\right) \sum_{j=0}^{m-1}\binom{2m}{2j+1}E_{2j+1}\left( 
\frac{\xi \left( x,y\right) }{x}\right) x^{2j+1}F_{2m-1-2j}\left( x,y\right)
\notag \\
&amp;=&amp;\sum_{j=0}^{m}\binom{2m+1}{2j+1}E_{2j+1}\left( \frac{\xi \left(
x,y\right) }{x}\right) x^{2j+1}L_{2m-2j}\left( x,y\right)  \notag \\
&amp;=&amp;\pm \left( x^{2}+4y\right) ^{\frac{2m+1}{2}}.  \label{3.7}
\end{eqnarray}
\end{corollary}

\begin{proof}
If we set $t=\alpha \left( x,y\right) $ in (\ref{2.17}) and (\ref{2.4}) we
get%
\begin{equation*}
P_{n}\left( x,y;\alpha \left( x,y\right) \right) =\left( x^{2}+4y\right) ^{%
\frac{n-1}{2}},
\end{equation*}%
and%
\begin{equation*}
Q_{n}\left( x,y;\alpha \left( x,y\right) \right) =\left( x^{2}+4y\right) ^{%
\frac{n}{2}}.
\end{equation*}%
Similarly, with $t=\beta \left( x,y\right) $ we get%
\begin{equation*}
P_{n}\left( x,y;\beta \left( x,y\right) \right) =\left( -1\right)
^{n+1}\left( x^{2}+4y\right) ^{\frac{n-1}{2}},
\end{equation*}%
and%
\begin{equation*}
Q_{n}\left( x,y;\beta \left( x,y\right) \right) =\left( -1\right) ^{n}\left(
x^{2}+4y\right) ^{\frac{n}{2}}.
\end{equation*}

Thus, we have%
\begin{equation*}
P_{2m+1}\left( \alpha \left( x,y\right) \right) =P_{2m+1}\left( \beta \left(
x,y\right) \right) =Q_{2m}\left( \alpha \left( x,y\right) \right)
=Q_{2m}\left( \beta \left( x,y\right) \right) =\left( x^{2}+4y\right) ^{m},
\end{equation*}%
and then identity (\ref{3.6}) is obtained by setting $t=\frac{\alpha \left(
x,y\right) }{x}$ and $t=\frac{\beta \left( x,y\right) }{x}$ in (\ref{2.32})
and (\ref{2.33}).

In the same way we obtain (\ref{3.7}) by setting $t=\frac{\alpha \left(
x,y\right) }{x}$ and $t=\frac{\beta \left( x,y\right) }{x}$ in identities (%
\ref{2.31}) and (\ref{2.34}).
\end{proof}

\begin{corollary}
\label{Cor3.4}(a) For $r=0,1,\ldots ,m$, the following identities hold%
\begin{eqnarray}
&amp;&amp;\sum_{j=0}^{m-1-r}\binom{2m}{2j+1}\binom{2m-2j-1-r}{r}\frac{1}{m-j}%
B_{2j+1}\left( t\right)  \label{3.15} \\
&amp;=&amp;\sum_{j=0}^{m-1-r}\binom{2m}{2j+1}\binom{2m-2j-2-r}{r}E_{2j+1}\left(
t\right)  \notag \\
&amp;=&amp;\sum_{j=0}^{2m-1-r}\binom{2m}{j}\binom{2m-j-1-r}{r}\left( -1\right)
^{j+1}t^{j}.  \notag
\end{eqnarray}%
\begin{eqnarray}
&amp;&amp;\sum_{j=0}^{m-r}\binom{2m+1}{2j}\binom{2m-2j+1-r}{r}\frac{1}{m+1-j}%
B_{2j}\left( t\right)  \label{3.16} \\
&amp;=&amp;\sum_{j=0}^{m-r}\binom{2m+1}{2j}\binom{2m-2j-r}{r}E_{2j}\left( t\right) 
\notag \\
&amp;=&amp;\sum_{j=0}^{2m-r}\binom{2m+1}{j}\binom{2m-j-r}{r}\left( -1\right)
^{j}t^{j}.  \notag
\end{eqnarray}

(b) For $r=1,2,\ldots ,m$, the following identities hold%
\begin{eqnarray}
&amp;&amp;2\sum_{j=0}^{m-r}\binom{2m}{2j}\binom{2m+1-2j-r}{r}\frac{1}{2m+1-2j-r}%
B_{2j}\left( t\right)  \label{3.17} \\
&amp;=&amp;\sum_{j=0}^{m-r}\binom{2m}{2j}\binom{2m-2j-r}{r}\frac{2m-2j}{2m-2j-r}%
E_{2j}\left( t\right)  \notag \\
&amp;=&amp;\sum_{j=2r}^{2m}\binom{2m}{j}\binom{j-r}{r}\frac{\left( -1\right) ^{j}j}{%
j-r}t^{2m-j}.  \notag
\end{eqnarray}%
\begin{eqnarray}
&amp;&amp;2\sum_{j=0}^{m-r}\binom{2m+1}{2j+1}\binom{2m+1-2j-r}{r}\frac{1}{2m+1-2j-r}%
B_{2j+1}\left( t\right)  \label{3.18} \\
&amp;=&amp;\sum_{j=0}^{m-r}\binom{2m+1}{2j+1}\binom{2m-2j-r}{r}\frac{2m-2j}{2m-2j-r}%
E_{2j+1}\left( t\right)  \notag \\
&amp;=&amp;\sum_{j=2r}^{2m+1}\binom{2m+1}{j}\binom{j-r}{r}\frac{\left( -1\right)
^{j}j}{j-r}t^{2m+1-j}.  \notag
\end{eqnarray}
\end{corollary}

\begin{proof}
(a) First substitute the explicit formula (\ref{1.3}) for $F_{n}\left(
x,y\right) $ in (\ref{2.31}) and (\ref{2.32}), then equate the coefficients
of similar terms $x^{2m+1-2r}y^{r}$, $r=0,1,\ldots ,m$ to obtain (\ref{3.15}%
) and (\ref{3.16}), respectively.

(b) First substitute the explicit formula (\ref{1.3}) for $L_{n}\left(
x,y\right) $ in (\ref{2.33}) and (\ref{2.34}), then equate the coefficients
of similar terms $x^{2m+1-2r}y^{r}$, $r=1,2,\ldots ,m$ to obtain (\ref{3.17}%
) and (\ref{3.18}), respectively.
\end{proof}

Fibonacci and Lucas polynomials are hidden in identities (\ref{3.15}) to (%
\ref{3.18}). To let them appear we just have to write the right-hand side
polynomials of these identities in a special form. The following lemma tells
us how to do this.

\begin{lemma}
\label{Lem3.5}(a) For $r=0,1,\ldots ,m$ we have%
\begin{eqnarray}
&amp;&amp;\sum_{j=0}^{2m-1-r}\binom{2m}{j}\binom{2m-j-1-r}{r}\left( -1\right)
^{j+1}t^{j}  \label{3.21} \\
&amp;=&amp;\left( -1\right) ^{r+1}\sum_{j=0}^{r}\binom{2m}{j}\binom{2r-j}{r}\left(
\left( t-1\right) ^{2m-j}+\left( -1\right) ^{j+1}t^{2m-j}\right) ,  \notag
\end{eqnarray}%
and%
\begin{eqnarray}
&amp;&amp;\sum_{j=0}^{2m-r}\binom{2m+1}{j}\binom{2m-j-r}{r}\left( -1\right) ^{j}t^{j}
\label{3.22} \\
&amp;=&amp;\left( -1\right) ^{r+1}\sum_{j=0}^{r}\binom{2m+1}{j}\binom{2r-j}{r}\left(
\left( t-1\right) ^{2m+1-j}+\left( -1\right) ^{j+1}t^{2m+1-j}\right) . 
\notag
\end{eqnarray}

(b) For $r=1,2,\ldots ,m$ we have%
\begin{eqnarray}
&amp;&amp;\sum_{j=2r}^{2m}\binom{2m}{j}\binom{j-r}{r}\frac{\left( -1\right) ^{j}j}{%
j-r}t^{2m-j}  \label{3.23} \\
&amp;=&amp;\left( -1\right) ^{r}\sum_{j=1}^{r}\binom{2m}{j}\binom{2r-j-1}{r-1}\frac{j%
}{r}\left( \left( t-1\right) ^{2m-j}+\left( -1\right) ^{j}t^{2m-j}\right) , 
\notag
\end{eqnarray}%
and%
\begin{eqnarray}
&amp;&amp;\sum_{j=2r}^{2m+1}\binom{2m+1}{j}\binom{j-r}{r}\frac{\left( -1\right) ^{j}j%
}{j-r}t^{2m+1-j}  \label{3.24} \\
&amp;=&amp;\left( -1\right) ^{r}\sum_{j=1}^{r}\binom{2m+1}{j}\binom{2r-j-1}{r-1}%
\frac{j}{r}\left( \left( t-1\right) ^{2m+1-j}+\left( -1\right)
^{j}t^{2m+1-j}\right) .  \notag
\end{eqnarray}
\end{lemma}

\begin{proof}
We present only the proof of (\ref{3.21}). The corresponding proofs of (\ref%
{3.22}), (\ref{3.23}) and (\ref{3.24}) are similar and left to the reader.

We have%
\begin{eqnarray}
&amp;&amp;\sum_{j=0}^{r}\binom{2m}{j}\binom{2r-j}{r}\left( \left( t-1\right)
^{2m-j}+\left( -1\right) ^{j+1}t^{2m-j}\right)  \notag \\
&amp;=&amp;\sum_{j=0}^{r}\!\binom{2m}{j}\!\binom{2r-j}{r}\!\sum_{k=0}^{2m-j}\!\binom{%
2m-j}{k}\!\left( -1\right) ^{k}t^{2m-j-k}+\sum_{j=0}^{r}\!\binom{2m}{j}\!%
\binom{2r-j}{r}\!\left( -1\right) ^{j+1}t^{2m-j}  \notag \\
&amp;=&amp;\sum_{j=0}^{2m}\sum_{i=0}^{\min (r,j)}\!\binom{2m}{i}\!\binom{2r-i}{r}\!%
\binom{2m-i}{j-i}\!\left( -1\right) ^{j-i}t^{2m-j}+\sum_{j=0}^{r}\!\binom{2m%
}{j}\!\binom{2r-j}{r}\!\left( -1\right) ^{j+1}t^{2m-j}  \notag \\
&amp;=&amp;\sum_{j=0}^{r}\binom{2m}{j}\left( \sum_{i=0}^{j}\binom{2r-i}{r}\binom{j}{i%
}\left( -1\right) ^{i}-\binom{2r-j}{r}\right) \left( -1\right) ^{j}t^{2m-j} 
\notag \\
&amp;&amp;+\sum_{j=r+1}^{2m}\binom{2m}{j}\left( \sum_{i=0}^{r}\binom{2r-i}{r}\binom{j%
}{i}\left( -1\right) ^{i}\right) \left( -1\right) ^{j}t^{2m-j}.  \label{3.25}
\end{eqnarray}

For $j=0,1,\ldots ,r$ we have%
\begin{equation*}
\sum_{i=0}^{j}\binom{2r-i}{r}\binom{j}{i}\left( -1\right) ^{i}=\binom{2r-j}{r%
},
\end{equation*}%
and for $j=r+1,\ldots ,2m$ we have 
\begin{equation*}
\sum_{i=0}^{r}\binom{2r-i}{r}\binom{j}{i}\left( -1\right) ^{i}=\binom{j-1-r}{%
r}\left( -1\right) ^{r}.
\end{equation*}

(See \cite{Sp}, identity (3.50), p. 65.) Thus (\ref{3.25}) can be written as%
\begin{equation*}
\sum_{j=0}^{r}\!\binom{2m}{j}\!\binom{2r-j}{r}\!\!\left( \left( t-1\right)
^{2m-j}+\left( -1\right) ^{j+1}t^{2m-j}\right) =\sum_{j=r+1}^{2m}\!\!\binom{%
2m}{j}\!\binom{j-1-r}{r}\!\left( -1\right) ^{r+j}t^{2m-j}\!,
\end{equation*}%
and finally we can write the right-hand side of (\ref{3.21}) as%
\begin{eqnarray*}
&amp;&amp;\left( -1\right) ^{r+1}\sum_{j=0}^{r}\binom{2m}{j}\binom{2r-j}{r}\left(
\left( t-1\right) ^{2m-j}+\left( -1\right) ^{j+1}t^{2m-j}\right) \\
&amp;=&amp;\left( -1\right) ^{r+1}\sum_{j=r+1}^{2m}\binom{2m}{j}\binom{j-1-r}{r}%
\left( -1\right) ^{r}\left( -1\right) ^{j}t^{2m-j} \\
&amp;=&amp;\sum_{j=r+1}^{2m}\binom{2m}{j}\binom{j-1-r}{r}\left( -1\right)
^{j+1}t^{2m-j} \\
&amp;=&amp;\sum_{j=0}^{2m-1-r}\binom{2m}{j}\binom{2m-j-1-r}{r}\left( -1\right)
^{j+1}t^{j},
\end{eqnarray*}%
as wanted.
\end{proof}

\begin{corollary}
\label{Cor3.6}(a) For $r=0,1,\ldots ,m$ we have that%
\begin{eqnarray}
&amp;&amp;\sum_{j=0}^{m-1-r}\binom{2m}{2j+1}\binom{2m-2j-1-r}{r}\frac{1}{m-j}%
B_{2j+1}\left( \frac{\xi \left( x,y\right) }{x}\right)  \label{3.26} \\
&amp;=&amp;\sum_{j=0}^{m-1-r}\binom{2m}{2j+1}\binom{2m-2j-2-r}{r}E_{2j+1}\left( 
\frac{\xi \left( x,y\right) }{x}\right)  \notag \\
&amp;=&amp;\pm \left( -1\right) ^{r+1}\sqrt{x^{2}+4y}\sum_{j=0}^{r}\binom{2m}{j}%
\binom{2r-j}{r}\frac{\left( -1\right) ^{j+1}}{x^{2m-j}}F_{2m-j}\left(
x,y\right) ,  \notag
\end{eqnarray}%
and%
\begin{eqnarray}
&amp;&amp;\sum_{j=0}^{m-r}\binom{2m+1}{2j}\binom{2m-2j+1-r}{r}\frac{1}{m+1-j}%
B_{2j}\left( \frac{\xi \left( x,y\right) }{x}\right)  \label{3.27} \\
&amp;=&amp;\sum_{j=0}^{m-r}\binom{2m+1}{2j}\binom{2m-2j-r}{r}E_{2j}\left( \frac{\xi
\left( x,y\right) }{x}\right)  \notag \\
&amp;=&amp;\left( -1\right) ^{r+1}\sum_{j=0}^{r}\binom{2m+1}{j}\binom{2r-j}{r}\frac{%
\left( -1\right) ^{j+1}}{x^{2m+1-j}}L_{2m+1-j}\left( x,y\right) .  \notag
\end{eqnarray}

(b) For $r=1,2,\ldots ,m$ we have that%
\begin{eqnarray}
&amp;&amp;2\sum_{j=0}^{m-r}\binom{2m}{2j}\binom{2m+1-2j-r}{r}\frac{1}{2m+1-2j-r}%
B_{2j}\left( \frac{\xi \left( x,y\right) }{x}\right)  \label{3.28} \\
&amp;=&amp;\sum_{j=0}^{m-r}\binom{2m}{2j}\binom{2m-2j-r}{r}\frac{2m-2j}{2m-2j-r}%
E_{2j}\left( \frac{\xi \left( x,y\right) }{x}\right)  \notag \\
&amp;=&amp;\frac{\left( -1\right) ^{r}}{r}\sum_{j=1}^{r}\binom{2m}{j}\binom{2r-j-1}{%
r-1}\frac{j\left( -1\right) ^{j}}{x^{2m-j}}L_{2m-j}\left( x,y\right) , 
\notag
\end{eqnarray}%
and%
\begin{eqnarray}
&amp;&amp;2\sum_{j=0}^{m-r}\binom{2m+1}{2j+1}\binom{2m+1-2j-r}{r}\frac{1}{2m+1-2j-r}%
B_{2j+1}\left( \frac{\xi \left( x,y\right) }{x}\right)  \label{3.29} \\
&amp;=&amp;\sum_{j=0}^{m-r}\binom{2m+1}{2j+1}\binom{2m-2j-r}{r}\frac{2m-2j}{2m-2j-r}%
E_{2j+1}\left( \frac{\xi \left( x,y\right) }{x}\right)  \notag \\
&amp;=&amp;\pm \frac{\left( -1\right) ^{r}}{r}\sqrt{x^{2}+4y}\sum_{j=1}^{r}\binom{%
2m+1}{j}\binom{2r-j-1}{r-1}\frac{j\left( -1\right) ^{j}}{x^{2m+1-j}}%
F_{2m+1-j}\left( x,y\right) .  \notag
\end{eqnarray}
\end{corollary}

\begin{proof}
By using (\ref{3.21}), (\ref{3.22}), (\ref{3.23}) and (\ref{3.24}), rewrite
identities (\ref{3.15}), (\ref{3.16}), (\ref{3.17}) and (\ref{3.18}),
respectively. Set $t=\frac{\alpha \left( x,y\right) }{x}$ and $t=\frac{\beta
\left( x,y\right) }{x}$ in the resulting identities, to obtain (\ref{3.26}),
(\ref{3.27}), (\ref{3.28}) and (\ref{3.29}), respectively.
\end{proof}

\section{Acknowledgements}

The first version of this work contained some versions of identities (\ref%
{2.31}) to (\ref{2.34}), which were demonstrated by means of Fourier
expansions of certain periodic extensions of certain restrictions of the
polynomials $P_{n}\left( x,1;t\right) $ and $Q_{n}\left( x,1;t\right) $. In
this version we present a larger list of identities, and the main ideas of
the corresponding proofs are different (now the proofs follow to Cigler \cite%
{Ci}). I thank the referee for call my attention to this way of proving
those identities. I also thank him/her for the additional comments in
his/her report, that certainly helped me to present a more readable version
of the article.

\bigskip

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\bibitem{Ci} Johann Cigler, Fibonacci polynomials, generalized Stirling
numbers, and Bernoulli, Genocchi and tangent numbers, 
\url{http://arxiv.org/abs/1103.2610}.

\bibitem{Ci2} Johann Cigler, q-Fibonacci polynomials and q-Genocchi numbers,
\url{http://arxiv.org/abs/0908.1219}.

\bibitem{Di} Karl Dilcher, \textit{Bernoulli and Euler Polynomials,} Digital
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\bibitem{Kel} Richard P. Kelisky, Congruences involving combinations of the
Bernoulli and Fibonacci numbers, \textit{Proc. Nat. Acad. Sci. USA},
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\bibitem{K} Thomas Koshy, \textit{Fibonacci and Lucas Numbers with
Applications,} John Wiley \&amp; Sons, Inc., 2001.

\bibitem{Sp} Renzo Sprugnoli, \textit{Riordan Array Proofs of Identities in
Gould's Book,} 2006.  Available at
\url{http://www.dsi.unifi.it/~resp/GouldBK.pdf}.

\bibitem{V} S. Vajda, \textit{Fibonacci and Lucas Numbers, and the Golden
Section,} Dover, 1989.

\bibitem{Yu} Hongquan Yu and Chuanguang Liang, Identities involving partial
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\end{thebibliography}

\bigskip
\hrule
\bigskip

\noindent 2010 {\it Mathematics Subject Classification}:
Primary 11B39; Secondary 11B68.

\noindent \emph{Keywords: } 
Fibonacci polynomial, Lucas polynomial, Bernoulli polynomial,
Euler polynomial.

\bigskip
\hrule
\bigskip

\noindent (Concerned with sequences
\seqnum{A000032} and
\seqnum{A000045}.)

\bigskip
\hrule
\bigskip

\vspace*{+.1in}
\noindent
Received September 9 2011;
revised versions received  December 8 2011; January 13 2012.
Published in {\it Journal of Integer Sequences}, January 14 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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