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\begin{center}
\vskip 1cm{\LARGE\bf On Some Properties of Bivariate Fibonacci \\
\vskip .1in and
Lucas Polynomials } \vskip 1cm \large Hac\`ene
Belbachir\footnote{This research is partially supported by the LAID3
Laboratory.} and Farid Bencherif\footnote{This research is partially
supported by the LATN Laboratory.}\\
USTHB/Faculty of Mathematics\\
P. O.~Box 32 \\
El Alia, 16111\\
Bab Ezzouar \\
Algeria \\
\href{mailto:hbelbachir@usthb.dz}{\tt hbelbachir@usthb.dz}  \\
\href{mailto:hacenebelbachir@gmail.com}{\tt hacenebelbachir@gmail.com} \\
\href{mailto:fbencherif@usthb.dz}{\tt fbencherif@usthb.dz} \\
\href{mailto:fbencherif@yahoo.fr}{\tt fbencherif@gmail.com}\\
\end{center}

\vskip .2 in

\begin{abstract}
In this paper we generalize to bivariate Fibonacci and Lucas
polynomials, properties obtained for Chebyshev polynomials. We prove
that the coordinates of the bivariate polynomials over appropriate
bases are families of integers satisfying remarkable recurrence
relations.
\end{abstract}

\newtheorem{theorem}{Theorem}[section]
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\newtheorem{corollary}{Corollary}[section]
\newtheorem{lemma}{Lemma}[section]


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%\date{\today}

\section{Introduction}


In \cite{bel10}, the authors established that Chebyshev polynomials
of the first and second kind admit remarkable integer coordinates in
a specific basis. It turns out that this property can be extended to
Jacobsthal polynomials \cite{djo3, hor2}, Vieta polynomials
\cite{2vie, 2hor12, 2rob1, 2sha1}, Morgan-Voyce polynomials
\cite{2mor, 2and3, 2hor4, 2lee, 2and1, 2swa2, 2che} and
quasi-Morgan-Voyce polynomials \cite{2hor2}, and more generally to
bivariate polynomials associated with recurrence sequences of order
two.

The bivariate polynomials of Fibonacci and Lucas, denoted respectively by $%
(U_{n})=(U_{n}(x,y))$ and $(V_{n})=(V_{n}(x,y))$, are polynomials
belonging
to $\mathbb{Z}[x,y]$ and defined by%
\begin{equation*}
\left\{
\begin{array}{l}
U_{0}=0,\text{ }U_{1}=1, \\
U_{n}=xU_{n-1}+yU_{n-2}\text{ \ }\left( n\geq 2\right) ,%
\end{array}%
\right. \ \ \ \text{ and}\ \ \ \left\{
\begin{array}{l}
V_{0}=2,\text{ }V_{1}=x, \\
V_{n}=xV_{n-1}+yV_{n-2}\text{  }\left( n\geq 2\right) .%
\end{array}%
\right.
\end{equation*}

It is established, see for example \cite{luc,2sha2,bel1}, that%
\begin{eqnarray}
U_{n+1} &=&\sum_{k=0}^{\left[ n/2\right] }{\binom{n-k}{k}}x^{n-2k}y^{k}, \\
V_{n} &=&\sum_{k=0}^{\left[ n/2\right] }\frac{n}{n-k}{\binom{n-k}{k}}%
x^{n-2k}y^{k}\text{  }\left( n\geq 1\right).
\end{eqnarray}

Let $\mathcal{E}_{n}$ be the $%
%TCIMACRO{\U{211a} }%
%BeginExpansion
\mathbb{Q}
%EndExpansion
$-vector space spanned by the free family $\mathcal{C}%
_{n}=(x^{n-2k}y^{k})_{k}$, $\left( 0\leq k\leq \left\lfloor
n/2\right\rfloor \right) $. Thus the relations $(1)$ and $(2)$
appear as the decompositions
of $U_{n+1}$ and $V_{n}$\ over the canonical basis $\mathcal{C}_{n}$ of $%
\mathcal{E}_{n}$.

The goal of this paper is to prove that the families $\mathfrak{U}%
_{n}:=\left( x^{k}U_{n+1-k}\right) _{k}$ and
$\mathfrak{V}_{n}:=\left( x^{k}V_{n-k}\right) _{k}$ for
$n-2\left\lfloor n/2\right\rfloor \leq k\leq
n-\left\lfloor n/2\right\rfloor $ constitute two other bases of $\mathcal{E}%
_{n}$ (Theorem 2.1) with respect to which, the polynomials $2U_{n+1}$ and $%
2V_{n}$ admit remarkable integer coordinates.

\section{Main results}

\begin{theorem}
For any $n\geq 1,$ $\mathfrak{U}_{n}$ and $\mathfrak{V}_{n}$ are bases of $%
\mathcal{E}_{n}$
\end{theorem}

As $U_{n+1}$ and $V_{n}$ belong to $\mathcal{E}_{n}$, the polynomials $%
U_{2n+1}$ and $V_{2n}$ are elements of $\mathcal{E}_{2n}$ with basis $%
\mathfrak{U}_{2n}$ or $\mathfrak{V}_{2n}.$ Similarly, $U_{2n}$ and
$V_{2n-1}$
belong to $\mathcal{E}_{2n-1}$ with basis $\mathfrak{U}_{2n-1}$ or $%
\mathfrak{V}_{2n-1}$.

Therefore, there are a priori 8 possible decompositions:%
\begin{equation*}
\begin{array}{lllllll}
& \text{over }\mathfrak{U}_{2n} &
\begin{tabular}{|c|}
\hline 1 \\ \hline
\end{tabular}%
\rightarrow trivial, &  &  & \text{over }\mathfrak{U}_{2n} &
\begin{tabular}{|c|}
\hline 3 \\ \hline
\end{tabular}%
\rightarrow simple, \\
U_{2n+1}%
\begin{array}{l}
\nearrow  \\
\searrow
\end{array}
&  &  &  & V_{2n}%
\begin{array}{l}
\nearrow  \\
\searrow
\end{array}
&  &  \\
& \text{over }\mathfrak{V}_{2n} &
\begin{tabular}{|c|}
\hline 2 \\ \hline
\end{tabular}%
\rightarrow Th.\ A, &  &  & \text{over }\mathfrak{V}_{2n} &
\begin{tabular}{|c|}
\hline 4 \\ \hline
\end{tabular}%
\rightarrow trivial, \\
&  &  & \ \ \ \  &  &  &  \\
& \text{over }\mathfrak{U}_{2n-1}\ \  &
\begin{tabular}{|c|}
\hline 5 \\ \hline
\end{tabular}%
\rightarrow Th.\ B, &  &  & \text{over }\mathfrak{U}_{2n-1}\ \  &
\begin{tabular}{|c|}
\hline 7 \\ \hline
\end{tabular}%
\rightarrow Th.\ C, \\
U_{2n}%
\begin{array}{l}
\nearrow  \\
\searrow
\end{array}
&  &  &  & V_{2n-1}%
\begin{array}{l}
\nearrow  \\
\searrow
\end{array}
&  &  \\
& \text{over }\mathfrak{V}_{2n-1} &
\begin{tabular}{|c|}
\hline 6 \\ \hline
\end{tabular}%
\rightarrow Th.\ E, &  &  & \text{over }\mathfrak{V}_{2n-1} &
\begin{tabular}{|c|}
\hline 8 \\ \hline
\end{tabular}%
\rightarrow Th.\ D,%
\end{array}%
\end{equation*}%
where the cases 1 and 4 are obvious since $U_{2n+1}\in
\mathfrak{U}_{2n}$ and $V_{2n}\in \mathfrak{V}_{2n}$.

The decomposition of $V_{2n}$ in $\mathfrak{U}_{2n}$ is simple: we have $%
V_{2n}=2U_{2n+1}-xU_{2n}$.

\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \

The remaining cases are established by the five following results.

\begin{theorem}
(\textbf{A}). Decomposition of $2U_{2n+1}$ over the basis
$\mathfrak{V}_{2n}.$

For every integer $n\geq 0$, one has%
\begin{equation*}
2U_{2n+1}=\sum_{k=0}^{n}a_{n,k}x^{k}V_{2n-k},
\end{equation*}%
where%
\begin{equation*}
a_{n,k}=\sum_{j=0}^{n}(-1)^{j+k}(2-\delta _{n,j}){\binom{j}{k}}.
\end{equation*}

Moreover, $\left( a_{n,k}\right) _{n,k\geq 0}$ is a family of
integers
satisfying the following recurrence relation:%
\begin{equation*}
\left\{
\begin{array}{l}
a_{n,k}=-a_{n-1,k}+a_{n-1,k-1}\text{ \ }(n\geq 1,\text{ }k\geq 1); \\
a_{n,0}=1\text{ }\left( n\geq 0\right) ; \\
a_{0,k}=\delta _{k,0}\text{ }\left( k\geq 0\right) .%
\end{array}%
\right.
\end{equation*}%
($\delta _{i,j}$ being the Kronecker symbol).
\end{theorem}

The recurrence relation permits us to obtain the following table:%
\begin{equation*}
\begin{array}{cccccccccc}
\text{\textit{n}}\setminus \text{\textit{k}} & \text{\textit{0}} & \text{%
\textit{1}} & \text{\textit{2}} & \text{\textit{3}} &
\text{\textit{4}} & \text{\textit{5}} & \text{\textit{6}} &
\text{\textit{7}} & \text{\textit{8}}
\\
\text{\textit{0}} & 1 &  &  &  &  &  &  &  &  \\
\text{\textit{1}} & 1 & \text{ \ }1 &  &  &  &  &  &  &  \\
\text{\textit{2}} & 1 & \text{ \ }0 & \text{ \ }1 &  &  &  &  &  &  \\
\text{\textit{3}} & 1 & \text{ \ }1 & -1 & \text{ \ }1 &  &  &  &  &  \\
\text{\textit{4}} & 1 & \text{ \ }0 & \text{ \ }2 & -2 & \text{ \ \ }1 &  &  &  &  \\
\text{\textit{5}} & 1 & \text{ \ }1 & -2 & \text{ \ }4 & -3 & \text{ \ }1 &  &  &  \\
\text{\textit{6}} & 1 & \text{ \ }0 & \text{ \ }3 & -6 & \text{ \ \
}7 & -4 & \text{ \
}1 &  &  \\
\text{\textit{7}} & 1 & \text{ \ }1 & -3 & \text{ \ }9 & -13 &
\text{ }11 & -5 & \text{
\ }1 &  \\
\text{\textit{8}} & 1 & \text{ \ }0 & \text{ \ }4 & -12 & \text{ \
}22 & -24 & 16 & -6
& \text{ \ }1%
\end{array}%
\end{equation*}%
from which it follows that%
\begin{eqnarray*}
2U_{1} &=&V_{0}, \\
2U_{3} &=&V_{2}+xV_{1}, \\
2U_{5} &=&V_{4}+0V_{3}+x^{2}V_{2}, \\
2U_{7} &=&V_{6}+xV_{5}-x^{2}V_{4}+x^{3}V_{3}.
\end{eqnarray*}

\begin{theorem}
(\textbf{B}). Decomposition of $U_{2n}$ over the basis
$\mathfrak{U}_{2n-1}.$

For every integer $n\geq 1$, one has%
\begin{equation*}
U_{2n}=\sum_{k=1}^{n}b_{n,k}x^{k}U_{2n-k}
\end{equation*}%
where%
\begin{equation*}
b_{n,k}=(-1)^{k+1}{\binom{n}{k}}.
\end{equation*}

Moreover, $\left( b_{n,k}\right) _{n,k\geq 0}$ is a family of
integers
satisfying the following recurrence relation:%
\begin{equation*}
\left\{
\begin{array}{l}
b_{n,k}=b_{n-1,k}-b_{n-1,k-1}\text{ }(n\geq 1,\text{ }k\geq 1); \\
b_{n,0}=-1\text{ }\left( n\geq 0\right) ; \\
b_{0,k}=-\delta _{k,0}\text{ }(k\geq 0).%
\end{array}%
\right.
\end{equation*}
\end{theorem}

The latter recurrence relation permits us to obtain the following table:

\begin{equation*}
\begin{array}{cccccccccc}
\text{\textit{n}}\setminus \text{\textit{k}} & \text{\textit{0}} & \text{%
\textit{1}} & \text{\textit{2}} & \text{\textit{3}} &
\text{\textit{4}} & \text{\textit{5}} & \text{\textit{6}} &
\text{\textit{7}} & \text{\textit{8}}
\\
\text{\textit{0}} & \multicolumn{1}{l}{-1} & \multicolumn{1}{l}{} &
\multicolumn{1}{l}{} & \multicolumn{1}{l}{} & \multicolumn{1}{l}{} &
\multicolumn{1}{l}{} & \multicolumn{1}{l}{} & \multicolumn{1}{l}{} &
\multicolumn{1}{l}{} \\
\text{\textit{1}} & \multicolumn{1}{l}{-1} & \multicolumn{1}{l}{\ \
1} & \multicolumn{1}{l}{} & \multicolumn{1}{l}{} &
\multicolumn{1}{l}{} & \multicolumn{1}{l}{} & \multicolumn{1}{l}{} &
\multicolumn{1}{l}{} &
\multicolumn{1}{l}{} \\
\text{\textit{2}} & \multicolumn{1}{l}{-1} & \multicolumn{1}{l}{\ \
2} & \multicolumn{1}{l}{-1} & \multicolumn{1}{l}{} &
\multicolumn{1}{l}{} & \multicolumn{1}{l}{} & \multicolumn{1}{l}{} &
\multicolumn{1}{l}{} &
\multicolumn{1}{l}{} \\
\text{\textit{3}} & \multicolumn{1}{l}{-1} &
\multicolumn{1}{l}{\text{ \ }3} & \multicolumn{1}{l}{-3} &
\multicolumn{1}{l}{\ \ 1} & \multicolumn{1}{l}{} &
\multicolumn{1}{l}{} & \multicolumn{1}{l}{} & \multicolumn{1}{l}{} &
\multicolumn{1}{l}{} \\
\text{\textit{4}} & \multicolumn{1}{l}{-1} & \multicolumn{1}{l}{\ \
4} & \multicolumn{1}{l}{-6} & \multicolumn{1}{l}{\ \ 4} &
\multicolumn{1}{l}{-1} & \multicolumn{1}{l}{} & \multicolumn{1}{l}{}
& \multicolumn{1}{l}{} &
\multicolumn{1}{l}{} \\
\text{\textit{5}} & \multicolumn{1}{l}{-1} &
\multicolumn{1}{l}{\text{ \ }5} & \multicolumn{1}{l}{-10} &
\multicolumn{1}{l}{\text{ \ }10} & \multicolumn{1}{l}{-5} &
\multicolumn{1}{l}{\ \ 1} & \multicolumn{1}{l}{} &
\multicolumn{1}{l}{} & \multicolumn{1}{l}{} \\
\text{\textit{6}} & \multicolumn{1}{l}{-1} & \multicolumn{1}{l}{\ \
6} & \multicolumn{1}{l}{-15} & \multicolumn{1}{l}{\ \ 20} &
\multicolumn{1}{l}{-15 } & \multicolumn{1}{l}{\ \ 6} &
\multicolumn{1}{l}{-1} & \multicolumn{1}{l}{}
& \multicolumn{1}{l}{} \\
\text{\textit{7}} & \multicolumn{1}{l}{-1} & \multicolumn{1}{l}{\ \
7} & \multicolumn{1}{l}{-21} & \multicolumn{1}{l}{\ \ 35} &
\multicolumn{1}{l}{-35 } & \multicolumn{1}{l}{\ \ 21} &
\multicolumn{1}{l}{-7} & \multicolumn{1}{l}{
\ \ 1} & \multicolumn{1}{l}{} \\
\text{\textit{8}} & \multicolumn{1}{l}{-1} & \multicolumn{1}{l}{\ \
8} & \multicolumn{1}{l}{-28} & \multicolumn{1}{l}{\ \ 56} &
\multicolumn{1}{l}{-70 } & \multicolumn{1}{l}{\ \ 56} &
\multicolumn{1}{l}{-28} &
\multicolumn{1}{l}{\ \ 8} & \multicolumn{1}{l}{-1}%
\end{array}%
\end{equation*}%
from which it follows that%
\begin{eqnarray*}
U_{2} &=&\ \ xU_{1} \\
U_{4} &=&2xU_{3}-x^{2}U_{2} \\
U_{6} &=&3xU_{5}-3x^{2}U_{4}+x^{3}U_{3} \\
U_{8} &=&4xU_{7}-6x^{2}U_{6}+4x^{3}U_{5}-x^{4}U_{4}
\end{eqnarray*}

\begin{theorem}
(\textbf{C}). Decomposition of $V_{2n-1}$ over the basis
$\mathfrak{U}_{2n-1}.$

For every integer $n\geq 1,$ one has
\begin{equation*}
V_{2n-1}=\sum_{k=1}^{n}c_{n,k}x^{k}U_{2n-k}\text{ }
\end{equation*}%
where%
\begin{equation*}
c_{n,k}=2\left( -1\right) ^{k+1}\dbinom{n}{k}-\delta _{k,1}.
\end{equation*}

Moreover, $\left( c_{n,k}\right) _{n\geq 1,k\geq 0}$ is a family of
integers
satisfying the following recurrence relation:%
\begin{equation*}
\left\{
\begin{array}{l}
c_{n,k}=c_{n-1,k}-c_{n-1,k-1}-\delta _{k,2}\text{ \ }(n\geq 2\text{,
}k\geq
1); \\
c_{n,0}=-2\text{ \ }\left( n\geq 1\right) ; \\
c_{1,k}=-2\delta _{k,0}+\delta _{k,1}\text{ \ }(k\geq 0).%
\end{array}%
\right.
\end{equation*}
\end{theorem}

The latter recurrence relation permits us to obtain the following table:%
\begin{equation*}
\begin{array}{cccccccccc}
\text{\textit{n}}\setminus \text{\textit{k}} & \text{\textit{0}} & \text{%
\textit{1}} & \text{\textit{2}} & \text{\textit{3}} &
\text{\textit{4}} & \text{\textit{5}} & \text{\textit{6}} &
\text{\textit{7}} & \text{\textit{8}}
\\
\text{\textit{1}} & \multicolumn{1}{l}{-2} & \multicolumn{1}{l}{\ \
1} & \multicolumn{1}{l}{} & \multicolumn{1}{l}{} &
\multicolumn{1}{l}{} & \multicolumn{1}{l}{} & \multicolumn{1}{l}{} &
\multicolumn{1}{l}{} &
\multicolumn{1}{l}{} \\
\text{\textit{2}} & \multicolumn{1}{l}{-2} & \multicolumn{1}{l}{\ \
3} & \multicolumn{1}{l}{-2} & \multicolumn{1}{l}{} &
\multicolumn{1}{l}{} & \multicolumn{1}{l}{} & \multicolumn{1}{l}{} &
\multicolumn{1}{l}{} &
\multicolumn{1}{l}{} \\
\text{\textit{3}} & \multicolumn{1}{l}{-2} & \multicolumn{1}{l}{\ \
5} & \multicolumn{1}{l}{-6} & \multicolumn{1}{l}{\ \ 2} &
\multicolumn{1}{l}{} & \multicolumn{1}{l}{} & \multicolumn{1}{l}{} &
\multicolumn{1}{l}{} &
\multicolumn{1}{l}{} \\
\text{\textit{4}} & \multicolumn{1}{l}{-2} & \multicolumn{1}{l}{\ \
7} & \multicolumn{1}{l}{-12} & \multicolumn{1}{l}{\ \ 8} &
\multicolumn{1}{l}{-2} & \multicolumn{1}{l}{} & \multicolumn{1}{l}{}
& \multicolumn{1}{l}{} &
\multicolumn{1}{l}{} \\
\text{\textit{5}} & \multicolumn{1}{l}{-2} & \multicolumn{1}{l}{\ \
9} & \multicolumn{1}{l}{-20} & \multicolumn{1}{l}{\ \ 20} &
\multicolumn{1}{l}{-10 } & \multicolumn{1}{l}{\ \ 2} &
\multicolumn{1}{l}{} & \multicolumn{1}{l}{}
& \multicolumn{1}{l}{} \\
\text{\textit{6}} & \multicolumn{1}{l}{-2} & \multicolumn{1}{l}{\ \
11} & \multicolumn{1}{l}{-30} & \multicolumn{1}{l}{\ \ 40} &
\multicolumn{1}{l}{-30 } & \multicolumn{1}{l}{\ \ 12} &
\multicolumn{1}{l}{-2} & \multicolumn{1}{l}{
} & \multicolumn{1}{l}{} \\
\text{\textit{7}} & \multicolumn{1}{l}{-2} & \multicolumn{1}{l}{\ \
13} & \multicolumn{1}{l}{-42} & \multicolumn{1}{l}{\ \ 70} &
\multicolumn{1}{l}{-70 } & \multicolumn{1}{l}{\ \ 42} &
\multicolumn{1}{l}{-14} &
\multicolumn{1}{l}{\ \ 2} & \multicolumn{1}{l}{} \\
\text{\textit{8}} & \multicolumn{1}{l}{-2} & \multicolumn{1}{l}{\ \
15} & \multicolumn{1}{l}{-56} & \multicolumn{1}{l}{\ \ 112} &
\multicolumn{1}{l}{ -140} & \multicolumn{1}{l}{\ \ 112} &
\multicolumn{1}{l}{-56} &
\multicolumn{1}{l}{\ \ 16} & \multicolumn{1}{l}{-2}%
\end{array}%
\end{equation*}%
from which we get
\begin{equation*}
\left\{
\begin{array}{l}
V_{1}=\ \ xU_{1} \\
V_{3}=3xU_{3}-2x^{2}U_{2} \\
V_{5}=\ 5xU_{5}-6x^{2}U_{4}+\ 2x^{3}U_{3} \\
V_{7}=7xU_{7}-12x^{2}U_{6}+8x^{3}U_{5}-2x^{4}U_{4}%
\end{array}%
\right.
\end{equation*}

\begin{theorem}
(\textbf{D}). Decomposition of $2V_{2n-1}$ over the basis $\mathfrak{V}_{2n-1}.\ $%

For every integer $n\geq 1,$ one has
\begin{equation*}
2V_{2n-1}=\sum_{k=1}^{n}d_{n,k}x^{k}V_{2n-1-k}\text{ }
\end{equation*}%
where%
\begin{equation*}
d_{n,k}=\left( -1\right) ^{k+1}\frac{2n-k}{n}{\binom{n}{k}}.
\end{equation*}
\end{theorem}

Moreover, $\left( d_{n,k}\right) _{n\geq 1,\ k\geq 0}$ is a family
of
integers satisfying the following recurrence relation:%
\begin{equation*}
\left\{
\begin{array}{l}
d_{n,k}=d_{n-1,k}-d_{n-1,k-1}\text{ \ }(n\geq 2\text{, }k\geq 1); \\
d_{n,0}=-2\text{ \ }\left( n\geq 1\right) ; \\
d_{1,k}=-2\delta _{k,0}+\delta _{k,1}\text{ \ }(k\geq 0).%
\end{array}%
\right.
\end{equation*}

The latter recurrence relation permits us to obtain the following table:%
\begin{equation*}
\begin{array}{cccccccccc}
\text{\textit{n}}\setminus \text{\textit{k}} & \text{\textit{0}} & \text{%
\textit{1}} & \text{\textit{2}} & \text{\textit{3}} &
\text{\textit{4}} & \text{\textit{5}} & \text{\textit{6}} &
\text{\textit{7}} & \text{\textit{8}}
\\
\text{\textit{1}} & \multicolumn{1}{l}{-2} & \multicolumn{1}{l}{\ \
1} & \multicolumn{1}{l}{} & \multicolumn{1}{l}{} &
\multicolumn{1}{l}{} & \multicolumn{1}{l}{} & \multicolumn{1}{l}{} &
\multicolumn{1}{l}{} &
\multicolumn{1}{l}{} \\
\text{\textit{2}} & \multicolumn{1}{l}{-2} & \multicolumn{1}{l}{\ \
3} & \multicolumn{1}{l}{-1} & \multicolumn{1}{l}{} &
\multicolumn{1}{l}{} & \multicolumn{1}{l}{} & \multicolumn{1}{l}{} &
\multicolumn{1}{l}{} &
\multicolumn{1}{l}{} \\
\text{\textit{3}} & \multicolumn{1}{l}{-2} &
\multicolumn{1}{l}{\text{ \ }5} & \multicolumn{1}{l}{-4} &
\multicolumn{1}{l}{\ \ 1} & \multicolumn{1}{l}{} &
\multicolumn{1}{l}{} & \multicolumn{1}{l}{} & \multicolumn{1}{l}{} &
\multicolumn{1}{l}{} \\
\text{\textit{4}} & \multicolumn{1}{l}{-2} & \multicolumn{1}{l}{\ \
7} & \multicolumn{1}{l}{-9} & \multicolumn{1}{l}{\ \ 5} &
\multicolumn{1}{l}{-1} & \multicolumn{1}{l}{} & \multicolumn{1}{l}{}
& \multicolumn{1}{l}{} &
\multicolumn{1}{l}{} \\
\text{\textit{5}} & \multicolumn{1}{l}{-2} &
\multicolumn{1}{l}{\text{ \ }9} & \multicolumn{1}{l}{-16} &
\multicolumn{1}{l}{\text{ \ }14} & \multicolumn{1}{l}{-6} &
\multicolumn{1}{l}{\ \ 1} & \multicolumn{1}{l}{} &
\multicolumn{1}{l}{} & \multicolumn{1}{l}{} \\
\text{\textit{6}} & \multicolumn{1}{l}{-2} & \multicolumn{1}{l}{\ \
11} & \multicolumn{1}{l}{-25} & \multicolumn{1}{l}{\ \ 30} &
\multicolumn{1}{l}{-20 } & \multicolumn{1}{l}{\ \ 7} &
\multicolumn{1}{l}{-1} & \multicolumn{1}{l}{}
& \multicolumn{1}{l}{} \\
\text{\textit{7}} & \multicolumn{1}{l}{-2} & \multicolumn{1}{l}{\ \
13} & \multicolumn{1}{l}{-36} & \multicolumn{1}{l}{\ \ 55} &
\multicolumn{1}{l}{-50 } & \multicolumn{1}{l}{\ \ 27} &
\multicolumn{1}{l}{-8} & \multicolumn{1}{l}{
\ \ 1} & \multicolumn{1}{l}{} \\
\text{\textit{8}} & \multicolumn{1}{l}{-2} & \multicolumn{1}{l}{\ \
15} & \multicolumn{1}{l}{-49} & \multicolumn{1}{l}{\ \ 91} &
\multicolumn{1}{l}{ -105} & \multicolumn{1}{l}{\ \ 77} &
\multicolumn{1}{l}{-35} &
\multicolumn{1}{l}{\ \ 9} & \multicolumn{1}{l}{-1}%
\end{array}%
\end{equation*}%
from which we obtain
\begin{equation*}
\left\{
\begin{array}{l}
2V_{1}=\text{ }xV_{0} \\
2V_{3}=3xV_{2}-x^{2}V_{1} \\
2V_{5}=5xV_{4}\ -4x^{2}V_{3}+x^{3}V_{2} \\
2V_{7}=7xV_{6}-9x^{2}V_{5}+5x^{3}V_{4}-x^{4}V_{3}%
\end{array}%
\right.
\end{equation*}

\begin{theorem}
(\textbf{E}). Decomposition of $2U_{2n}$ over the basis
$\mathfrak{V}_{2n-1}.\ $

For every integer $n\geq 1,$ one has%
\begin{equation*}
2U_{2n}=\sum_{k=1}^{n}e_{n,k}x^{k}V_{2n-1-k}
\end{equation*}%
where%
\begin{equation*}
e_{n,k}=\left( -1\right) ^{k+1}\frac{2n-k}{2n}{\binom{n}{k}+}\delta _{k,0}+\frac{1}{2}%
\sum_{j=0}^{n-1}(-1)^{j+k-1}(2-\delta _{n-1,j}){\binom{j}{k-1}}.
\end{equation*}

Moreover, $\left( e_{n,k}\right) _{n,k\geq 0}$ is a family of
integers
satisfying the following recurrence relation:%
\begin{equation*}
\left\{
\begin{array}{l}
e_{n,k}=e_{n-2,k}-2e_{n-2,k-1}+e_{n-2,k-2}\text{ \ }(n\geq 3\text{, }k\geq 2)%
\text{;} \\
e_{n,0}=0\ \ \text{and}\ \ e_{n,1}=n\ \ \left( n\geq 1\right) ; \\
e_{1,k}=\delta _{k,1}\ \ \text{and \ }e_{2,k}=2\delta _{k,1}\ \ (k\geq 0).%
\end{array}%
\right.
\end{equation*}
\end{theorem}

The latter recurrence relation permits us to obtain the following table:
\begin{equation*}
\begin{array}{cccccccccc}
\text{\textit{n}}\setminus \text{\textit{k}} & \text{\textit{0}} & \text{%
\textit{1}} & \text{\textit{2}} & \text{\textit{3}} &
\text{\textit{4}} & \text{\textit{5}} & \text{\textit{6}} &
\text{\textit{7}} & \text{\textit{8}}
\\
\text{\textit{1}} & \multicolumn{1}{l}{\ \ 0} & \multicolumn{1}{l}{\
\ 1} & \multicolumn{1}{l}{} & \multicolumn{1}{l}{} &
\multicolumn{1}{l}{} & \multicolumn{1}{l}{} & \multicolumn{1}{l}{} &
\multicolumn{1}{l}{} &
\multicolumn{1}{l}{} \\
\text{\textit{2}} & \multicolumn{1}{l}{\ \ 0} & \multicolumn{1}{l}{\
\ 2} & \multicolumn{1}{l}{\ \ 0} & \multicolumn{1}{l}{} &
\multicolumn{1}{l}{} & \multicolumn{1}{l}{} & \multicolumn{1}{l}{} &
\multicolumn{1}{l}{} &
\multicolumn{1}{l}{} \\
\text{\textit{3}} & \multicolumn{1}{l}{\ \ 0} & \multicolumn{1}{l}{\text{ \ }%
3} & \multicolumn{1}{l}{-2} & \multicolumn{1}{l}{\ \ 1} &
\multicolumn{1}{l}{ } & \multicolumn{1}{l}{} & \multicolumn{1}{l}{}
& \multicolumn{1}{l}{} &
\multicolumn{1}{l}{} \\
\text{\textit{4}} & \multicolumn{1}{l}{\ \ 0} & \multicolumn{1}{l}{\
\ 4} & \multicolumn{1}{l}{-4} & \multicolumn{1}{l}{\ \ 2} &
\multicolumn{1}{l}{\ \ 0 } & \multicolumn{1}{l}{} &
\multicolumn{1}{l}{} & \multicolumn{1}{l}{} &
\multicolumn{1}{l}{} \\
\text{\textit{5}} & \multicolumn{1}{l}{\ \ 0} & \multicolumn{1}{l}{\text{ \ }%
5} & \multicolumn{1}{l}{-8} & \multicolumn{1}{l}{\ \ 8} &
\multicolumn{1}{l}{ -4} & \multicolumn{1}{l}{\ \ 1} &
\multicolumn{1}{l}{} & \multicolumn{1}{l}{}
& \multicolumn{1}{l}{} \\
\text{\textit{6}} & \multicolumn{1}{l}{\ \ 0} & \multicolumn{1}{l}{\
\ 6} & \multicolumn{1}{l}{-12} & \multicolumn{1}{l}{\ \ 14} &
\multicolumn{1}{l}{-8} & \multicolumn{1}{l}{\ \ 2} &
\multicolumn{1}{l}{\ \ 0} & \multicolumn{1}{l}{
} & \multicolumn{1}{l}{} \\
\text{\textit{7}} & \multicolumn{1}{l}{\ \ 0} & \multicolumn{1}{l}{\
\ 7} & \multicolumn{1}{l}{-18} & \multicolumn{1}{l}{\ \ 29} &
\multicolumn{1}{l}{-28 } & \multicolumn{1}{l}{\ \ 17} &
\multicolumn{1}{l}{-6} & \multicolumn{1}{l}{
\ \ 1} & \multicolumn{1}{l}{} \\
\text{\textit{8}} & \multicolumn{1}{l}{\ \ 0} & \multicolumn{1}{l}{\
\ 8} & \multicolumn{1}{l}{-24} & \multicolumn{1}{l}{\ \ 44} &
\multicolumn{1}{l}{-48 } & \multicolumn{1}{l}{\ \ 32} &
\multicolumn{1}{l}{-12} &
\multicolumn{1}{l}{\ \ 2} & \multicolumn{1}{l}{\ \ 0}%
\end{array}%
\end{equation*}%
from which, we have
\begin{equation*}
\left\{
\begin{array}{l}
2U_{2}=\ xV_{0} \\
2U_{4}=2xV_{2}+0x^{2}V_{0} \\
2U_{6}=3xV_{4}-2x^{2}V_{3}+x^{3}V_{2} \\
2U_{8}=4xV_{6}-4x^{2}V_{5}+2x^{3}V_{4}+0x^{4}V_{3}%
\end{array}%
\right.
\end{equation*}

\section{Proof of Theorems}

Theorem 1 follows from the following lemma.

\begin{lemma}
For any integer $n\geq 0,$ by setting $m=\left\lfloor
n/2\right\rfloor ,$ we
have%
\begin{equation*}
\det\nolimits_{\mathcal{C}_{n}}(\mathfrak{U}_{n})=(-1)^{m(m+1)/2}\ \ \text{%
and }\
\det\nolimits_{\mathcal{C}_{n}}(\mathfrak{V}_{n})=2(-1)^{m(m+1)/2}.
\end{equation*}
\end{lemma}

\begin{proof}
Let us prove only the first equality as the proof of the other one
is similar.

Let $r=n-2m,$ $W_{k}^{\left( m\right)}=x^{k}U_{2m+1-k}$ $\left( 0\leq k\leq m\right) $\ and $%
\Delta _{m}=\det\nolimits_{\mathcal{C}_{2m}}(W_{0}^{\left( m\right)
},W_{1}^{\left( m\right) },\ldots ,W_{m}^{\left( m\right) }),$ we have%
\begin{equation*}
\det\nolimits_{\mathcal{C}_{n}}(\mathfrak{U}_{n})=\det\nolimits_{\mathcal{C}%
_{2m+r}}\left( x^{r+k}U_{2m+1-k}\right) _{0\leq k\leq m}=\Delta
_{m}\text{.}
\end{equation*}

The result follows by noticing that $\Delta _{0}=1$ and $\Delta
_{m}=(-1)^{m}\Delta _{m-1}$ for $m\geq 1.$ Indeed, for $m\geq 1,$ we
have
\begin{equation*}
W_{k+1}^{\left( m\right) }-W_{k}^{\left( m\right)
}=x^{k}(xU_{2m-k}-U_{2m-k+1})=-yW_{k}^{\left( m-1\right) }\text{ \
}\left( 0\leq k\leq m-1\right) .
\end{equation*}

Thus,%
\begin{eqnarray*}
\Delta _{m} &=&\det\nolimits_{\mathcal{C}_{2m}}(W_{0}^{\left(
m\right) },W_{1}^{\left( m\right) }-W_{0}^{\left( m\right)
},...,W_{m-1}^{\left( m\right) }-W_{m-2}^{\left( m\right)
},W_{m}^{\left( m\right)
}-W_{m-1}^{\left( m\right) }) \\
&=&\det\nolimits_{\mathcal{C}_{2m}}(W_{0}^{\left( m\right)
},-yW_{0}^{\left( m-1\right) },-yW_{1}^{\left( m-1\right) },\ldots
,-yW_{m-1}^{\left( m-1\right) }).
\end{eqnarray*}

The ``component" of $W_{0}^{\left( m\right) }=U_{2m+1}$ over $x^{2m}$
is equal to $1$.

The ``component" of $-yW_{k}^{\left( m-1\right) }$ over $x^{2n},$ is
equal to $0,$ for $1\leq k\leq m$ , so we have

$\ \ \ \ \ \ \ \ \ \ \ \Delta _{m}=\det\nolimits_{\mathcal{C}%
_{2m-2}}(-W_{0}^{\left( m-1\right) },-W_{1}^{\left( m-1\right)
},\ldots ,-W_{m-1}^{\left( m-1\right) })=(-1)^{m}\Delta _{m-1}.$
\end{proof}

\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \

Let $A_{m},$ $B_{m},C_{m},$ $D_{m}$ and $E_{m}$ be the operators on
$\left(
\mathbb{Q}\left[ x,y\right] \right) ^{\mathbb{N}}$ defined by%
\begin{eqnarray*}
A_{m} &=&-\left( x-E\right) ^{m}+2\sum_{k=0}^{m}E^{k}\left( x-E\right) ^{m-k}%
\text{\ \ }\left( m\geq 0\right) , \\
B_{m} &=&-\left( E-x\right) ^{m}\text{\ \ }\left( m\geq 0\right) , \\
C_{m} &=&2E^{m}+2B_{m}-xE^{m-1}\text{\ \ }\left( m\geq 1\right) , \\
D_{m} &=&\left( E-x\right) ^{m-1}\left( x-2E\right) \text{\ \
}\left( m\geq
1\right) , \\
E_{m} &=&\frac{1}{2}(xA_{m-1}+D_{m})+E^{m}\text{\ \ }\left( m\geq
1\right) ,
\end{eqnarray*}%
where $E$ is the forward shift operator given by
\begin{equation*}
E\left( \left( W_{n}\right) _{n}\right) =\left( W_{n+1}\right) _{n}
\end{equation*}

Then, we have%
\begin{eqnarray*}
A_{m} &=&\sum_{k=0}^{m}a_{m,k}x^{k}E^{m-k}\qquad \text{with}\qquad
a_{m,k}=\sum_{j=0}^{m}(-1)^{j+k}(2-\delta _{m,j}){\binom{j}{k}} \\
B_{m} &=&\sum_{k=0}^{m}b_{m,k}x^{k}E^{m-k}\qquad \text{with}\qquad
b_{m,k}=\left( -1\right) ^{k+1}{\binom{m}{k}} \\
C_{m} &=&\sum_{k=1}^{m}c_{m,k}x^{k}E^{m-k}\qquad \text{with}\qquad
c_{m,k}=2\left( -1\right) ^{k+1}{\binom{m}{k}}-\delta _{k,1} \\
D_{m} &=&\sum_{k=0}^{m}d_{m,k}x^{k}E^{m-k}\qquad \text{with}\qquad
d_{m,k}=\left( -1\right) ^{k+1}\frac{2m-k}{m}{\binom{m}{k}} \\
E_{m} &=&\sum_{k=1}^{m}e_{m,k}x^{k}E^{m-k}\qquad \text{with}\qquad e_{m,k}=%
\frac{1}{2}\left( d_{m,k}{+}a_{m-1,k-1}\right) +\delta _{k,0}.
\end{eqnarray*}

With these notations, relations stated by Theorems $A,$ $B,$ $C,$
$D$ and $E$
may be expressed by means of the following relations%
\begin{eqnarray*}
\mathbf{a}.\text{ }\forall n &\in &\mathbb{N}\ \ \ \ \ \
A_{n}V_{n}=2U_{2n+1}
\\
\mathbf{b}.\text{ }\forall n &\in &\mathbb{N}^{\ast }\ \ \ \ \ B_{n}U_{n}=0 \\
\mathbf{c}.\text{ }\forall n &\in &\mathbb{N}^{\ast }\ \ \ \ \
C_{n}U_{n}=V_{2n-1} \\
\mathbf{d}.\text{ }\forall n &\in &\mathbb{N}^{\ast }\ \ \ \ \
D_{n}V_{n-1}=0
\\
\mathbf{e}.\text{ }\forall n &\in &\mathbb{N}^{\ast }\ \ \ \ \
E_{n}V_{n-1}=2U_{n}
\end{eqnarray*}

which are to be proven. For this, the following lemma will be useful
for us.

\begin{lemma}
For every integers $n$ and $m,$ we have

\begin{enumerate}

\item $V_{n}=2U_{n+1}-xU_{n}$ \ $\left( n\geq 0\right)$ \ and $V_{n}=U_{n+1}+yU_{n-1}$ \ $\left( n\geq 1\right) ,$

\item $\left( E-x\right) ^{n}U_{m}=y ^{n}U_{m-n}$ \ and $\
\left( E-x\right) ^{n}V_{m}=y ^{n}V_{m-n}$ \ $\left( m\geq n\geq
0\right) ,$

\item $\sum_{k=1}^{n}\left( -y\right) ^{n-k}V_{2k}=U_{2n+1}-\left( -y\right)
^{n}$ \ $\left( n\geq 0\right) .$
\end{enumerate}
\end{lemma}

\begin{proof}
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \

\begin{enumerate}

\item See relation (2.9) and (2.8) in \cite{2sha2}.

\item We proceed by induction on $n$.

\item For every integer $n\in \mathbb{N}$, put $T_{n}:=U_{2n+1}-%
\sum_{k=1}^{n}\left( -y\right) ^{n-k}V_{2k}.$ The relation to be
proven is equivalent to $T_{n}=\left( -y\right) ^{n}$ \ $\left(
n\geq 0\right) .$ Then, we remark that from the first relation of
this lemma, we have for
every integer $n\geq 1$%
\begin{equation*}
T_{n}+yT_{n-1}=U_{2n+1}+yU_{2n-1}-V_{2n}=0,
\end{equation*}%
$\left( T_{n}\right) _{n\geq 0}$ is then a geometric sequence with
multiplier $\left( -y\right) $ and of first term $T_{0}=1$. It
follows that for every integer $n\in \mathbb{N}$, $T_{n}=\left(
-y\right) ^{n}.$
\end{enumerate}
\end{proof}

\begin{proof}[Proof of relations a., b., c., d. and e]
Using the above Lemma, we have

\textbf{a}. $A_{n}V_{n}=\left( -y\right)
^{n}V_{0}+2\sum_{k=1}^{n}\left( -y\right) ^{n-k}V_{2k}=2U_{2n+1}.$

\textbf{b}. $B_{n}U_{n}=-\left( E-x\right)^{n}U_{n}=-y^{n}U_{0}=0.$

\textbf{c}. $C_{n}U_{n}=\left( 2E^{n}+2B_{n}-xE^{n-1}\right)
U_{n}=2U_{2n}-xU_{2n-1}=V_{2n-1}.$

\textbf{d}. $D_{n}V_{n-1}=\left( E-x\right)^{n-1} \left(
xV_{n-1}-2V_{n}\right) =y^{n-1}\left( xV_{0}-2V_{1}\right) =0.$

\textbf{e}. $E_{n}V_{n-1}=\left( \frac{1}{2}xA_{n-1}+\frac{1}{2}%
D_{n}+E^{n}\right) V_{n-1}=\frac{1}{2}xA_{n-1}V_{n-1}+\frac{1}{2}%
D_{n}V_{n-1}+V_{2n-1}.$ Using $A_{n-1}V_{n-1}=2U_{2n-1}$ and
$D_{n}V_{n-1}=0$, it follows that
$E_{n}V_{n-1}=xU_{2n-1}+V_{2n-1}=2U_{2n}$
\end{proof}

\begin{rem}{\rm
Theorems A, B, C, D and E generalize results obtained for the
Chebyshev polynomials \cite{bel10}, Indeed,
\begin{eqnarray*}
\frac{1}{2}V_{n}(2x,-1) &=&T_{n}(x)\text{ is the Chebyshev
polynomials of the
first kind, } \\
U_{n+1}(2x,-1) &=&U_{n}(x)\text{ is the Chebyshev polynomials of the
second kind,}
\end{eqnarray*}%
with%
\begin{equation*}
\left\{
\begin{array}{l}
T_{n}(x)=2xT_{n-1}-T_{n-2}, \\
T_{0}=1,T_{1}=x,%
\end{array}%
\right. \text{ \ and \ }\left\{
\begin{array}{l}
U_{n}(x)=2xU_{n-1}-U_{n-2}, \\
U_{0}(x)=1,U_{1}=2x.%
\end{array}%
\right.
\end{equation*}
}
\end{rem}

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\bibitem{2che} R.\ X. F. Chen and L.\ W.\ Shapiro, On sequences $G_{n}$\
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\end{thebibliography}

\bigskip
\hrule
\bigskip

\noindent 2000 {\it Mathematics Subject Classification}: Primary
11B39; Secondary 11B37 .

\noindent \emph{Keywords: } bivariate Fibonacci and Lucas
polynomials, linear recurrences.

\bigskip
\hrule
\bigskip

\noindent (Concerned with sequences 
\seqnum{A007318},
\seqnum{A029653}, and
\seqnum{A112468}.)

\bigskip
\hrule
\bigskip

\vspace*{+.1in}
\noindent
Received October 8 2007;
revised version received June 3 2008.
Published in {\it Journal of Integer Sequences}, June
27 2008.

\bigskip
\hrule
\bigskip

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


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