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\begin{center}
\vskip 1cm{\LARGE\bf 
Sums of Products of $s$-Fibonacci \\
\vskip .1in
Polynomial Sequences
}
{
\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}

\vskip .2 in


\begin{abstract}
We consider $s$-Fibonacci polynomial sequences $\left( F_{0}\left( x\right)
,F_{s}\left( x\right) ,F_{2s}\left( x\right) ,\ldots \right) $, where $s\in 
\mathbb{N}$ is given. By studying certain $z$-polynomials involving $s$%
-polyfibonomials $\binom{n}{k}_{\!F_{s}\left( x\right) }\!\!=\frac{%
F_{sn}\left( x\right) \cdots F_{s\left( n-k+1\right) }\left( x\right) }{%
F_{s}\left( x\right) \cdots F_{ks}\left( x\right) }$ and $s$-Gibonacci
polynomial sequences $\left( G_{0}\left( x\right) ,G_{s}\left( x\right)
,G_{2s}\left( x\right) ,\ldots \right) $, we generalize some known results
(and obtain some new results) concerning sums of products and addition
formulas of Fibonacci numbers.
\end{abstract}

\section{\label{Sec1}Introduction}

We use $\mathbb{N}$ for the natural numbers and $\mathbb{N}^{\prime }$ for $%
\mathbb{N}\cup \left\{ 0\right\} $. Throughout this article $s$ will denote
a natural number.

Recall that Fibonacci polynomials $F_{n}\left( x\right) $ are defined as $%
F_{0}\left( x\right) =0$, $F_{1}\left( x\right) =1$, and $F_{n}\left(
x\right) =xF_{n-1}\left( x\right) +F_{n-2}\left( x\right) $, $n\geq 2$, and
extended for negative integers as $F_{-n}\left( x\right) =\left( -1\right)
^{n+1}F_{n}\left( x\right) $. Similarly, Lucas polynomials $L_{n}\left(
x\right) $ are defined as $L_{0}\left( x\right) =2$, $L_{1}\left( x\right)
=x $, and $L_{n}\left( x\right) =xL_{n-1}\left( x\right) +L_{n-2}\left(
x\right) $, $n\geq 2$, and extended for negative integers as $L_{-n}\left(
x\right) =\left( -1\right) ^{n}L_{n}\left( x\right) $. It is clear that $%
F_{n}\left( 1\right) $ and $L_{n}\left( 1\right) $ correspond to the
Fibonacci $F_{n}$ and Lucas $L_{n}$ number sequences, respectively (\seqnum{A000045}
and \seqnum{A000032} of Sloane's \textit{Encyclopedia}, respectively). A generalized
Fibonacci (Gibonacci) polynomial $G_{n}\left( x\right) $ is defined as $%
G_{n}\left( x\right) =xG_{n-1}\left( x\right) +G_{n-2}\left( x\right) $, $%
n\geq 2$, where $G_{0}\left( x\right) $ and $G_{1}\left( x\right) $ are
given (arbitrary) initial conditions. We will use also $H_{n}\left( x\right) 
$ to denote Gibonacci polynomials. It is easy to see that 
\begin{equation*}
G_{n}\left( x\right) =G_{0}\left( x\right) F_{n-1}\left( x\right)
+G_{1}\left( x\right) F_{n}\left( x\right) ,
\end{equation*}%
and that we can extend for negative integers as%
\begin{equation*}
G_{-n}\left( x\right) =\left( -1\right) ^{n+1}G_{n}\left( x\right) +\left(
-1\right) ^{n}G_{0}\left( x\right) L_{n}\left( x\right) .
\end{equation*}

We have Binet's formulas%
\begin{equation}
F_{n}\left( x\right) =\frac{1}{\sqrt{x^{2}+4}}\left( \alpha ^{n}\left(
x\right) -\beta ^{n}\left( x\right) \right) \text{ \ \ },\text{ \ \ \ }%
L_{n}\left( x\right) =\alpha ^{n}\left( x\right) +\beta ^{n}\left( x\right) ,
\label{1.01}
\end{equation}%
where%
\begin{equation}
\alpha \left( x\right) =\frac{x+\sqrt{x^{2}+4}}{2}\text{ \ \ , \ }\beta
\left( x\right) =\frac{x-\sqrt{x^{2}+4}}{2},  \label{1.011}
\end{equation}%
(the roots of $z^{2}-xz-1=0$).

For a Gibonacci polynomial $G_{n}\left( x\right) $ we have%
\begin{equation}
G_{n}\left( x\right) =\frac{1}{\sqrt{x^{2}+4}}\left( c_{1}\left( x\right)
\alpha ^{n}\left( x\right) -c_{2}\left( x\right) \beta ^{n}\left( x\right)
\right) ,  \label{1.03}
\end{equation}%
where $c_{1}\left( x\right) =G_{1}\left( x\right) -G_{0}\left( x\right)
\beta \left( x\right) $ and $c_{2}\left( x\right) =G_{1}\left( x\right)
-G_{0}\left( x\right) \alpha \left( x\right) $.

Some basic relations involving $\alpha \left( x\right) $ and $\beta \left(
x\right) $ (as $\alpha \left( x\right) \beta \left( x\right) =-1$) will be
used throughout the work, as well as some basic Fibonacci polynomial
identities (most of times without further comments). About this point we
would like to comment the following: some Fibonacci number identities are
valid as Fibonacci polynomial identities. For example, the well-known
identity 
\begin{equation*}
F_{n+m}=F_{n}F_{m+1}+F_{n-1}F_{m}
\end{equation*}
is just the case $x=1$ of 
\begin{equation}
F_{n+m}\left( x\right) =F_{n}\left( x\right) F_{m+1}\left( x\right)
+F_{n-1}\left( x\right) F_{m}\left( x\right) .  \label{1.04}
\end{equation}

However, it is more natural to accept the existence of Fibonacci number
identities that are not valid as identities with Fibonacci polynomials. An
example is the Gelin-Ces\`{a}ro identity 
\begin{equation*}
F_{n-2}F_{n-1}F_{n+1}F_{n+2}+1=F_{n}^{4},
\end{equation*}%
which is false for Fibonacci polynomials: for example, for $n=3$ we have
that the left-hand side is the polynomial%
\begin{equation*}
F_{1}\left( x\right) F_{2}\left( x\right) F_{4}\left( x\right) F_{5}\left(
x\right) +1=x^{8}+5x^{6}+7x^{4}+2x^{2}+1,
\end{equation*}%
while the right-hand side is 
\begin{equation*}
F_{3}^{4}\left( x\right) =x^{8}+4x^{6}+6x^{4}+4x^{2}+1.
\end{equation*}

What we can say in general about this example is that 
\begin{equation*}
F_{n-2}\left( x\right) F_{n-1}\left( x\right) F_{n+1}\left( x\right)
F_{n+2}\left( x\right) +1-F_{n}^{4}\left( x\right)
\end{equation*}%
is a polynomial that has a zero for $x=1$. For example, for $n=3$ this
polynomial is $x^{2}\left( x^{2}-1\right) \left( x^{2}+2\right) $.

For a given Gibonacci polynomial sequence $G_{n}\left( x\right) $, the
corresponding $s$-Gibonacci polynomial sequence $G_{sn}\left( x\right) $ is
given by $G_{sn}\left( x\right) =\left( G_{0}\left( x\right) ,G_{s}\left(
x\right) ,G_{2s}\left( x\right) ,\ldots \right) $. We will be dealing with $%
s $-\textit{Gibonacci polynomial factorials}, denoted as $\left( G_{n}\left(
x\right) !\right) _{s}$, where $n\in \mathbb{N}^{\prime }$, and defined as $%
\left( G_{0}\left( x\right) !\right) _{s}=1$ and $\left( G_{n}\left(
x\right) !\right) _{s}=G_{s}\left( x\right) G_{2s}\left( x\right) \cdots
G_{ns}\left( x\right) $ for $n\in \mathbb{N}$. Also we will be working with $%
s$-Gibonomials 
\begin{equation*}
\binom{n}{k}_{G_{s}}=\frac{\left( G_{n}!\right) _{s}}{\left( G_{k}!\right)
_{s}\left( G_{n-k}!\right) _{s}},
\end{equation*}%
(see \cite{P2}), where the $s$-Gibonacci sequences are replaced by $s$%
-Gibonacci polynomial sequences $G_{sn}\left( x\right) $. We will refer to
these objects as $s$-\textit{polygibonomials}, and we will use the natural
notation $\binom{n}{k}_{G_{s}\left( x\right) }$ (with $n,k\in \mathbb{N}%
^{\prime }$) for them. That is, we have that%
\begin{equation*}
\binom{n}{k}_{G_{s}\left( x\right) }=\frac{\left( G_{n}\left( x\right)
!\right) _{s}}{\left( G_{k}\left( x\right) !\right) _{s}\left( G_{n-k}\left(
x\right) !\right) _{s}},
\end{equation*}%
for $0\leq k\leq n$, and $\binom{n}{k}_{G_{s}\left( x\right) }=0$ otherwise.
In other words, if $0\leq k\leq n$ we have 
\begin{equation}
\binom{n}{k}_{G_{s}\left( x\right) }=\frac{G_{sn}\left( x\right) G_{s\left(
n-1\right) }\left( x\right) \cdots G_{s\left( n-k+1\right) }\left( x\right) 
}{G_{s}\left( x\right) G_{2s}\left( x\right) \cdots G_{ks}\left( x\right) }.
\label{1.05}
\end{equation}

Plainly it is valid the symmetry property $\binom{n}{k}_{G_{s}\left(
x\right) }=\binom{n}{n-k}_{G_{s}\left( x\right) }$, and then we have also

\begin{equation}
\sum_{k=0}^{2n-1}\left( -1\right) ^{k}\binom{2n-1}{k}_{G_{s}\left( x\right)
}=0.  \label{1.1}
\end{equation}

We call the attention to the fact that in the case $G_{n}\left( x\right)
=F_{n}\left( x\right) $, the $s$-polyfibonomials $\binom{n}{k}_{F_{s}\left(
x\right) }$ \emph{are indeed polynomials} (despite the polynomial quotients
in the definition). To see this we use the same argument that shows that $s$%
-Fibonomials $\binom{n}{k}_{F_{s}}$ are integers (see \cite{H}): from (\ref%
{1.04}) we have that%
\begin{equation*}
F_{s\left( n-k\right) +1}\left( x\right) F_{sk}\left( x\right)
+F_{sk-1}\left( x\right) F_{s\left( n-k\right) }\left( x\right)
=F_{sn}\left( x\right) ,
\end{equation*}%
and then we can write 
\begin{equation*}
\binom{n}{k}_{F_{s}\left( x\right) }=F_{s\left( n-k\right) +1}\left(
x\right) \binom{n-1}{k-1}_{F_{s}\left( x\right) }+F_{sk-1}\left( x\right) 
\binom{n-1}{k}_{F_{s}\left( x\right) }.
\end{equation*}

Thus, with a simple induction argument we obtain the desired conclusion. In
fact, it is easy to see that the degree of $\binom{n}{k}_{F_{s}\left(
x\right) }$ is $sk\left( n-k\right) $. (Of course, the number sequences $%
\binom{n}{k}_{F_{s}\left( 1\right) }$ correspond to $s$-Fibonomial sequences
(see \cite{P2}).) However, $s$-polygibonomials are in general rational
functions. For example, the $2$-polylucanomial $\binom{4}{2}_{L_{2}\left(
x\right) }$ is%
\begin{eqnarray*}
\binom{4}{2}_{L_{2}\left( x\right) } &amp;=&amp;\frac{L_{8}\left( x\right)
L_{6}\left( x\right) }{L_{2}\left( x\right) L_{4}\left( x\right) } \\
&amp;=&amp;\frac{\left( x^{4}+4x^{2}+1\right) \left(
x^{8}+8x^{6}+20x^{4}+16x^{2}+2\right) }{x^{4}+4x^{2}+2}.
\end{eqnarray*}

Two examples of $s$-polyfibonomials $\binom{n}{k}_{F_{s}\left( x\right) }$,
as triangular arrays, with $n$ for lines and $k=0,1,\ldots ,n$ for columns,
are the following:

For $s=1$ we have%
\begin{equation*}
\begin{array}{ccccccccccccc}
&amp;  &amp;  &amp;  &amp;  &amp;  &amp; ^{1} &amp;  &amp;  &amp;  &amp;  &amp;  &amp;  \\ 
&amp;  &amp;  &amp;  &amp;  &amp; ^{1} &amp;  &amp; ^{1} &amp;  &amp;  &amp;  &amp;  &amp;  \\ 
&amp;  &amp;  &amp;  &amp; ^{1} &amp;  &amp; ^{x} &amp;  &amp; ^{1} &amp;  &amp;  &amp;  &amp;  \\ 
&amp;  &amp;  &amp; ^{1} &amp;  &amp; ^{x^{2}+1} &amp;  &amp; ^{x^{2}+1} &amp;  &amp; ^{1} &amp;  &amp;  &amp;  \\ 
&amp;  &amp; ^{1} &amp;  &amp; ^{x^{3}+2x} &amp;  &amp; ^{x^{4}+3x^{2}+2} &amp;  &amp; ^{x^{3}+2x} &amp;  &amp; ^{1}
&amp;  &amp;  \\ 
&amp; ^{1} &amp;  &amp; ^{x^{4}+3x^{2}+1} &amp;  &amp; ^{x^{6}+5x^{4}+7x^{2}+2} &amp;  &amp; 
^{x^{6}+5x^{4}+7x^{2}+2} &amp;  &amp; ^{x^{4}+3x^{2}+1} &amp;  &amp; ^{1} &amp;  \\ 
&amp;  &amp; \vdots &amp;  &amp; \vdots &amp;  &amp; \vdots &amp;  &amp; \vdots &amp;  &amp; \vdots &amp;  &amp;  \\ 
&amp;  &amp;  &amp;  &amp;  &amp;  &amp;  &amp;  &amp;  &amp;  &amp;  &amp;  &amp; 
\end{array}%
\end{equation*}

For $s=2$ we have%
\begin{equation*}
\begin{array}[b]{ccccccccccc}
&amp;  &amp;  &amp;  &amp;  &amp; ^{1} &amp;  &amp;  &amp;  &amp;  &amp;  \\ 
&amp;  &amp;  &amp;  &amp; ^{1} &amp;  &amp; ^{1} &amp;  &amp;  &amp;  &amp;  \\ 
&amp;  &amp;  &amp; ^{1} &amp;  &amp; ^{x^{2}+2} &amp;  &amp; ^{1} &amp;  &amp;  &amp;  \\ 
&amp;  &amp; ^{1} &amp;  &amp; ^{\substack{ x^{4}+4x^{2}  \\ +3}} &amp;  &amp; ^{\substack{ %
x^{4}+4x^{2}  \\ +3}} &amp;  &amp; ^{1} &amp;  &amp;  \\ 
&amp; ^{1} &amp;  &amp; _{\substack{ +10x^{2}  \\ +4}}^{x^{6}+6x^{4}} &amp;  &amp; _{\substack{ %
+21x^{4}+20x^{2}  \\ +6}}^{x^{8}+8x^{6}} &amp;  &amp; _{\substack{ +10x^{2}  \\ +4}}%
^{x^{6}+6x^{4}} &amp;  &amp; ^{1} &amp;  \\ 
^{1} &amp;  &amp; ^{\substack{ x^{8}+8x^{6}  \\ +21x^{4}+20x^{2}+5}} &amp;  &amp; 
_{\substack{ +127x^{4}+60x^{2}  \\ +10}}^{\substack{ x^{12}+12x^{10}  \\ %
+55x^{8}+120x^{6}}} &amp;  &amp; _{\substack{ +127x^{4}+60x^{2}  \\ +10}}^{\substack{
x^{12}+12x^{10}  \\ +55x^{8}+120x^{6}}} &amp;  &amp; ^{\substack{ x^{8}+8x^{6}  \\ %
+21x^{4}+20x^{2}+5}} &amp;  &amp; ^{1} \\ 
&amp; \vdots &amp;  &amp; \vdots &amp;  &amp; \vdots &amp;  &amp; \vdots &amp;  &amp; \vdots &amp;  \\ 
&amp;  &amp;  &amp;  &amp;  &amp;  &amp;  &amp;  &amp;  &amp;  &amp; 
\end{array}%
\end{equation*}

(The case $x=1$ of the previous arrays corresponds to Fibonomials and $2$%
-Fibonomials, \seqnum{A010048} and \seqnum{A034801} of Sloane's \textit{Encyclopedia},
respectively.)

In this work we are concerned with sums of products of $s$-Fibonacci
polynomial sequences. The problem of finding closed formulas for these sums
in the case $s=x=1$ has a quite long story. Lots of results for different
particular cases are now available. Some examples are the works of Melham 
\cite{M1,M2}; Prodinger \cite{Pr}; Seibert \&amp; Trojovsk\'{y} \cite {S-T2};
and Witu\l a and S\l ota \cite{W}, among many others.

The motivation for this work came from identity (5) in \cite{T-F} (and all
the nice and important formulas that can be derived from it). A version of
this identity can be stated as follows: for $m\in \mathbb{N}$ and $%
a_{1},\ldots ,a_{m}\in \mathbb{Z}$ given, one has that

\begin{equation}
\sum_{r=0}^{m}\left( -1\right) ^{\frac{r\left( r+1\right) }{2}+r}\dbinom{m}{r%
}_{G}H_{n+m-r}\prod_{i=1}^{m}G_{a_{i}+m-r}=G_{m}!H_{n+a_{1}+\cdots +a_{m}+%
\frac{m\left( m+1\right) }{2}},  \label{1.2}
\end{equation}%
where $H_{n}$ and $G_{n}$ are two given Gibonacci sequences, with $G_{0}=0$.
What we present in this work are closed formulas for sums that resembles the
left-hand side of (\ref{1.2}), where the involved Gibonomial $\binom{m}{r}%
_{G}$ is replaced by the $s$-polyfibonomial $\binom{m}{r}_{F_{s}\left(
x\right) }$, and the remaining Gibonacci sequences $H_{n}$ and $G_{n}$ are
replaced for the corresponding $s$-Gibonacci polynomial sequences $%
H_{sn}\left( x\right) $ and $G_{sn}\left( x\right) $. (Thus, identity (\ref%
{1.2}) becomes the case $x=s=1$ of some of our results.) While the proof of (%
\ref{1.2}) presented in \cite{T-F} is an induction argument, in this work we
proceed explicitly to study certain polynomials, their factorizations and
the evaluations of them in certain powers of $\alpha \left( x\right) $ and $%
\beta \left( x\right) $, that eventually will lead us to the required proofs.

The main results presented in this work are the following polynomial
identities: for given $m\in \mathbb{N}$ and $k,l,p_{1},\ldots ,p_{2m+1}\in 
\mathbb{Z}$, we have

\begin{eqnarray}
&amp;&amp;\sum_{r=0}^{2m+1}\left( -1\right) ^{\frac{\left( sr+2\left( s+1\right)
\right) \left( r+1\right) }{2}}\dbinom{2m+1}{r}_{F_{s}\left( x\right)
}H_{s\left( 2m-r+k\right) +l}\left( x\right) \prod_{i=1}^{2m+1}G_{s\left(
n+r+p_{i}\right) }\left( x\right)  \label{1.3} \\
&amp;=&amp;\!\!\left( -1\right) ^{s\left( m+k\right) +l+1}\left( x^{2}+4\right) ^{-%
\frac{1}{2}}\!\left( F_{2m+1}\left( x\right) !\right) _{s}\!  \notag \\
&amp;&amp;\times \left( \!%
\begin{array}{c}
\left( H_{1}\left( x\right) -H_{0}\left( x\right) \alpha \left( x\right)
\right) \left( G_{1}\left( x\right) -G_{0}\left( x\right) \beta \left(
x\right) \right) ^{2m+1} \\ 
\times \alpha ^{s\left( \left( 2m+1\right) \left( n+m\right) +p_{1}+\cdots
+p_{2m+1}-k+1\right) -l}\left( x\right) \\ 
-\left( H_{1}\left( x\right) -H_{0}\left( x\right) \beta \left( x\right)
\right) \left( G_{1}\left( x\right) -G_{0}\left( x\right) \alpha \left(
x\right) \right) ^{2m+1} \\ 
\times \beta ^{s\left( \left( 2m+1\right) \left( n+m\right) +p_{1}+\cdots
+p_{2m+1}-k+1\right) -l}\left( x\right)%
\end{array}%
\!\right) ,  \notag
\end{eqnarray}%
and

\begin{eqnarray}
&amp;&amp;\sum_{r=0}^{2m}\left( -1\right) ^{\frac{\left( sr+2\left( s+1\right)
\right) \left( r+1\right) }{2}}\dbinom{2m}{r}_{F_{s}\left( x\right)
}H_{s\left( 2m+r+k\right) +l}\left( x\right) \prod_{i=1}^{2m}G_{s\left(
n+r+p_{i}\right) }\left( x\right)  \label{1.4} \\
&amp;=&amp;\left( -1\right) ^{s\left( m+1\right) +1}\left( x^{2}+4\right) ^{-\frac{1%
}{2}}\left( F_{2m}\left( x\right) !\right) _{s}  \notag \\
&amp;&amp;\times \left( 
\begin{array}{c}
\left( H_{1}\left( x\right) -H_{0}\left( x\right) \beta \left( x\right)
\right) \left( G_{1}\left( x\right) -G_{0}\left( x\right) \beta \left(
x\right) \right) ^{2m} \\ 
\times \alpha ^{s\left( 2m\left( m+n\right) +p_{1}+\cdots
+p_{2m}+3m+k\right) +l}\left( x\right) \\ 
-\left( H_{1}\left( x\right) -H_{0}\left( x\right) \alpha \left( x\right)
\right) \left( G_{1}\left( x\right) -G_{0}\left( x\right) \alpha \left(
x\right) \right) ^{2m} \\ 
\times \beta ^{s\left( 2m\left( m+n\right) +p_{1}+\cdots +p_{2m}+3m+k\right)
+l}\left( x\right)%
\end{array}%
\right) .  \notag
\end{eqnarray}

In section \ref{Sec3} we present the proofs of (\ref{1.3}) and (\ref{1.4}).
What we do to prove these formulas is studying the factorization of certain
polynomials involving $s$-polyfibonomials and $s$-Gibonacci polynomial
sequences, and the evaluation of these polynomials in certain powers of $%
\alpha \left( x\right) $ and $\beta \left( x\right) $ as well. This is done
in section \ref{Sec2}. In section \ref{Sec4} we give some examples of the
results proved in section \ref{Sec3}. Finally, in the appendix we give the
proofs of some identities used in section \ref{Sec2}.

\section{\label{Sec2}Preliminary results}

We begin this section by showing the factorization of a $z$-polynomial with $%
s$-polyfibonomials as coefficients.

\begin{proposition}
\label{Prop2.1}For $t,m\in \mathbb{N}^{\prime }$ we have that%
\begin{eqnarray}
&amp;&amp;\left( -1\right) ^{s\left( m+1\right) +1}\sum_{r=0}^{2m}\left( -1\right) ^{%
\frac{\left( sr+2\left( s+1\right) \right) \left( r+1\right) }{2}+sr\left(
t+1\right) }\binom{2m}{r}_{F_{s}\left( x\right) }z^{r}  \label{2.1} \\
&amp;=&amp;\prod\limits_{p=1}^{m}\left( z^{2}+\left( -1\right) ^{st+1}L_{s\left(
4p-2m-1\right) }\left( x\right) z+\left( -1\right) ^{s}\right) .  \notag
\end{eqnarray}
\end{proposition}

\begin{proof}
We proceed by induction on $m$. The case $m=0$ (and $m=1$) can be verified
easily. If we suppose the result is true for a given $m\in \mathbb{N}$, then
we consider the right-hand side of (\ref{2.1}) with $m+1$ replacing $m$ and
write%
\begin{eqnarray*}
&amp;&amp;\prod\limits_{p=1}^{m+1}\left( z^{2}+\left( -1\right) ^{st+1}L_{s\left(
4p-2m-3\right) }\left( x\right) z+\left( -1\right) ^{s}\right) \\
&amp;=&amp;\left( z^{2}+\left( -1\right) ^{st+1}L_{s\left( 2m+1\right) }\left(
x\right) z+\left( -1\right) ^{s}\right) \prod\limits_{p=1}^{m}\left(
z^{2}+\left( -1\right) ^{st+1}L_{s\left( 4p-2m-3\right) }\left( x\right)
z+\left( -1\right) ^{s}\right) .
\end{eqnarray*}%
Observe that%
\begin{eqnarray*}
&amp;&amp;\prod\limits_{p=1}^{m}\left( z^{2}+\left( -1\right) ^{st+1}L_{s\left(
4p-2m-3\right) }\left( x\right) z+\left( -1\right) ^{s}\right) \\
&amp;=&amp;\prod\limits_{p=1}^{m}\left( z^{2}+\left( -1\right) ^{st+1}L_{-s\left(
4p-2m-1\right) }\left( x\right) z+\left( -1\right) ^{s}\right) \\
&amp;=&amp;\prod\limits_{p=1}^{m}\left( z^{2}+\left( -1\right) ^{s\left( t+1\right)
+1}L_{s\left( 4p-2m-1\right) }\left( x\right) z+\left( -1\right) ^{s}\right)
.
\end{eqnarray*}%
Then, by using the induction hypothesis we have%
\begin{eqnarray*}
&amp;&amp;\prod\limits_{p=1}^{m+1}\left( z^{2}+\left( -1\right) ^{st+1}L_{s\left(
4p-2m-3\right) }\left( x\right) z+\left( -1\right) ^{s}\right) \\
&amp;=&amp;\left( z^{2}+\left( -1\right) ^{st+1}L_{s\left( 2m+1\right) }\left(
x\right) z+\left( -1\right) ^{s}\right) \prod\limits_{p=1}^{m}\left(
z^{2}+\left( -1\right) ^{st+1}L_{s\left( 4p-2m-3\right) }\left( x\right)
z+\left( -1\right) ^{s}\right) \\
&amp;=&amp;\left( -1\right) ^{s\left( m+1\right) +1}\left( z^{2}+\left( -1\right)
^{st+1}L_{s\left( 2m+1\right) }\left( x\right) z+\left( -1\right) ^{s}\right)
\\
&amp;&amp;\times \sum_{r=0}^{2m}\left( -1\right) ^{\frac{\left( sr+2\left(
s+1\right) \right) \left( r+1\right) }{2}+srt}\binom{2m}{r}_{F_{s}\left(
x\right) }z^{r} \\
&amp;=&amp;\left( -1\right) ^{s\left( m+1\right) +1}\left( 
\begin{array}{c}
\sum_{r=2}^{2m+2}\left( -1\right) ^{\frac{\left( s\left( r-2\right) +2\left(
s+1\right) \right) \left( r-1\right) }{2}+s\left( r-2\right) t}\binom{2m}{r-2%
}_{F_{s}\left( x\right) } \\ 
+\sum_{r=1}^{2m+1}\left( -1\right) ^{\frac{\left( s\left( r-1\right)
+2\left( s+1\right) \right) r}{2}+s\left( r-1\right) t+st+1}\binom{2m}{r-1}%
_{F_{s}\left( x\right) }L_{s\left( 2m+1\right) }\left( x\right) \\ 
+\sum_{r=0}^{2m}\left( -1\right) ^{\frac{\left( sr+2\left( s+1\right)
\right) \left( r+1\right) }{2}+srt+s}\binom{2m}{r}_{F_{s}\left( x\right) }%
\end{array}%
\right) z^{r} \\
&amp;=&amp;\left( -1\right) ^{s\left( m+1\right) +1}\sum_{r=0}^{2m+2}\left(
-1\right) ^{\frac{\left( sr+2\left( s+1\right) \right) \left( r+1\right) }{2}%
+sr\left( t+1\right) }\binom{2m+2}{r}_{F_{s}\left( x\right) }z^{r}\frac{1}{%
F_{s\left( 2m+2\right) }\left( x\right) F_{s\left( 2m+1\right) }\left(
x\right) } \\
&amp;&amp;\times \left( 
\begin{array}{c}
\left( -1\right) ^{s\left( r+1\right) }F_{sr}\left( x\right) F_{s\left(
r-1\right) }\left( x\right) +\left( -1\right) ^{s}F_{sr}\left( x\right)
F_{s\left( 2m+2-r\right) }\left( x\right) L_{s\left( 2m+1\right) }\left(
x\right) \\ 
+\left( -1\right) ^{s\left( r+1\right) }F_{s\left( 2m+2-r\right) }\left(
x\right) F_{s\left( 2m+1-r\right) }\left( x\right)%
\end{array}%
\right) \\
&amp;=&amp;\left( -1\right) ^{s\left( m+2\right) +1}\sum_{r=0}^{2m+2}\left(
-1\right) ^{\frac{\left( sr+2\left( s+1\right) \right) \left( r+1\right) }{2}%
+sr\left( t+1\right) }\binom{2m+2}{r}_{F_{s}\left( x\right) }z^{r},
\end{eqnarray*}%
as wanted. In the last step we used that%
\begin{eqnarray}
&amp;&amp;\left( -1\right) ^{s\left( r+1\right) }F_{sr}\left( x\right) F_{s\left(
r-1\right) }\left( x\right) +\left( -1\right) ^{s}F_{sr}\left( x\right)
F_{s\left( 2m+2-r\right) }\left( x\right) L_{s\left( 2m+1\right) }\left(
x\right)  \label{2.1.1} \\
&amp;&amp;+\left( -1\right) ^{s\left( r+1\right) }F_{s\left( 2m+2-r\right) }\left(
x\right) F_{s\left( 2m+1-r\right) }\left( x\right)  \notag \\
&amp;=&amp;\left( -1\right) ^{s}F_{s\left( 2m+2\right) }\left( x\right) F_{s\left(
2m+1\right) }\left( x\right) ,  \notag
\end{eqnarray}%
(see appendix for the proof).
\end{proof}

The following identities (factorizations of signed sums of $s$%
-polyfibonomials) can be obtained as corollaries of (\ref{2.1})

\begin{equation}
\sum_{r=0}^{2m}\left( -1\right) ^{\frac{r\left( r-1+2s+2t+2rs\right) }{2}}%
\binom{2m}{r}_{F_{2s-1}\left( x\right) }=\left( -1\right)
^{mt}\prod\limits_{p=1}^{m}L_{\left( 2s-1\right) \left( 4p-2m-1\right)
}\left( x\right) .  \label{2.2}
\end{equation}%
\begin{equation}
\sum_{r=0}^{2m}\left( -1\right) ^{r}\binom{2m}{r}_{F_{4s}\left( x\right)
}=\left( -x^{2}-4\right) ^{m}\prod\limits_{p=1}^{m}F_{2s\left(
4p-2m-1\right) }^{2}\left( x\right) .  \label{2.3}
\end{equation}%
\begin{equation}
\sum_{r=0}^{2m}\binom{2m}{r}_{F_{4s}\left( x\right)
}=\prod\limits_{p=1}^{m}L_{2s\left( 4p-2m-1\right) }^{2}\left( x\right) .
\label{2.4}
\end{equation}

We continue, in propositions (\ref{Prop2.2}), (\ref{Prop2.3}) and (\ref%
{Prop2.4}), with other algebraic properties of certain $z$-polynomials, that
will be used in the remaining results of this section (propositions (\ref%
{Prop2.7}) to (\ref{Prop2.10})).

\begin{proposition}
\label{Prop2.2}For $t,m\in \mathbb{N}^{\prime }$ and $k,l\in \mathbb{Z}$, we
have that%
\begin{eqnarray}
&amp;&amp;\sum_{r=0}^{2m+1}\left( -1\right) ^{\frac{\left( sr+2\left( s+1\right)
\right) \left( r+1\right) }{2}+sr\left( t+1\right) }\dbinom{2m+1}{r}%
_{F_{s}\left( x\right) }H_{s\left( 2m-r+k\right) +l}\left( x\right) z^{r}
\label{2.5} \\
&amp;=&amp;\left( -1\right) ^{sm+1}\left( \left( -1\right) ^{st+1}H_{s\left(
k-1\right) +l}\left( x\right) z+\left( -1\right) ^{s}H_{s\left( 2m+k\right)
+l}\left( x\right) \right)  \notag \\
&amp;&amp;\times \prod\limits_{p=1}^{m}\left( z^{2}+\left( -1\right)
^{st+1}L_{s\left( 4p-2m-1\right) }\left( x\right) z+\left( -1\right)
^{s}\right) .  \notag
\end{eqnarray}
\end{proposition}

\begin{proof}
According to (\ref{2.1}) it is sufficient to prove that%
\begin{eqnarray}
&amp;&amp;\sum_{r=0}^{2m+1}\left( -1\right) ^{\frac{\left( sr+2\left( s+1\right)
\right) \left( r+1\right) }{2}+sr\left( t+1\right) }\dbinom{2m+1}{r}%
_{F_{s}\left( x\right) }H_{s\left( 2m-r+k\right) +l}\left( x\right) z^{r}
\label{2.6} \\
&amp;=&amp;\!\left( \left( -1\right) ^{s\left( t+1\right) +1}\!H_{s\left( k-1\right)
+l}\left( x\right) z+H_{s\left( 2m+k\right) +l}\left( x\right) \right) 
\notag \\
&amp;&amp;\times \!\!\sum_{r=0}^{2m}\left( -1\right) ^{\frac{\left( sr+2\left(
s+1\right) \right) \left( r+1\right) }{2}+sr\left( t+1\right) }\!\binom{2m}{r%
}_{F_{s}\left( x\right) }\!z^{r}.  \notag
\end{eqnarray}%
We have%
\begin{eqnarray*}
&amp;&amp;\left( \left( -1\right) ^{s\left( t+1\right) +1}H_{s\left( k-1\right)
+l}\left( x\right) z+H_{s\left( 2m+k\right) +l}\left( x\right) \right) \\
&amp;&amp;\times \!\!\sum_{r=0}^{2m}\left( -1\right) ^{\frac{\left( sr+2\left(
s+1\right) \right) \left( r+1\right) }{2}+sr\left( t+1\right) }\binom{2m}{r}%
_{F_{s}\left( x\right) }z^{r} \\
&amp;=&amp;H_{s\left( k-1\right) +l}\left( x\right) \sum_{r=1}^{2m+1}\left(
-1\right) ^{\frac{\left( s\left( r-1\right) +2\left( s+1\right) \right) r}{2}%
+rs\left( t+1\right) +1}\binom{2m}{r-1}_{F_{s}\left( x\right) }z^{r} \\
&amp;&amp;+H_{s\left( 2m+k\right) +l}\left( x\right) \sum_{r=0}^{2m}\left( -1\right)
^{\frac{\left( sr+2\left( s+1\right) \right) \left( r+1\right) }{2}+sr\left(
t+1\right) }\binom{2m}{r}_{F_{s}\left( x\right) }z^{r} \\
&amp;=&amp;\sum_{r=0}^{2m+1}\left( -1\right) ^{\frac{\left( sr+2\left( s+1\right)
\right) \left( r+1\right) }{2}+sr\left( t+1\right) }\binom{2m+1}{r}%
_{F_{s}\left( x\right) } \\
&amp;&amp;\times \frac{1}{F_{s\left( 2m+1\right) }\left( x\right) }\left( \left(
-1\right) ^{s\left( r+1\right) }H_{s\left( k-1\right) +l}\left( x\right)
F_{sr}\left( x\right) +H_{s\left( 2m+k\right) +l}\left( x\right) F_{s\left(
2m+1-r\right) }\left( x\right) \right) z^{r}
\end{eqnarray*}

But%
\begin{eqnarray}
&amp;&amp;\left( -1\right) ^{s\left( r+1\right) }H_{s\left( k-1\right) +l}\left(
x\right) F_{sr}\left( x\right) +H_{s\left( 2m+k\right) +l}\left( x\right)
F_{s\left( 2m+1-r\right) }\left( x\right)  \label{2.6.1} \\
&amp;=&amp;F_{s\left( 2m+1\right) }\left( x\right) H_{s\left( 2m-r+k\right)
+l}\left( x\right) ,  \notag
\end{eqnarray}%
(see appendix). Then (\ref{2.6}) follows.
\end{proof}

\begin{proposition}
\label{Prop2.3}For given $t,m\in \mathbb{N}$ and $k,l\in \mathbb{Z}$ we have

(a)%
\begin{eqnarray}
&amp;&amp;\left( 
\begin{array}{c}
\left( -1\right) ^{s+1}H_{s\left( 4m+k\right) +l}\left( x\right) z \\ 
+\left( -1\right) ^{st}H_{s\left( 2m+k\right) +l}\left( x\right)%
\end{array}%
\right) \sum_{r=0}^{2m-1}\left( -1\right) ^{\frac{\left( sr+2\left(
s+1\right) \right) \left( r+1\right) }{2}+st\left( r+1\right) }\dbinom{2m-1}{%
r}_{F_{s}\left( x\right) }z^{r}  \notag \\
&amp;=&amp;\sum_{r=0}^{2m}\left( -1\right) ^{\frac{\left( sr+2\left( s+1\right)
\right) \left( r+1\right) }{2}+sr\left( t+1\right) }\dbinom{2m}{r}%
_{F_{s}\left( x\right) }H_{s\left( 2m+r+k\right) +l}\left( x\right) z^{r}.
\label{2.8}
\end{eqnarray}

(b)%
\begin{eqnarray}
&amp;&amp;\left( z^{2}-\left( -1\right) ^{s\left( t+1\right) }L_{2sm}\left( x\right)
z+1\right) \sum_{r=0}^{2m-1}\left( -1\right) ^{\frac{\left( sr+2\left(
s+1\right) \right) \left( r+1\right) }{2}+st\left( r+1\right) }\dbinom{2m-1}{%
r}_{F_{s}\left( x\right) }z^{r}  \notag \\
&amp;=&amp;\sum_{r=0}^{2m+1}\left( -1\right) ^{\frac{\left( sr+2\left( s+1\right)
\right) \left( r+1\right) }{2}+st\left( r+1\right) +sr}\dbinom{2m+1}{r}%
_{F_{s}\left( x\right) }z^{r}.  \label{2.9}
\end{eqnarray}
\end{proposition}

\begin{proof}
(a) We have that%
\begin{eqnarray*}
&amp;&amp;\left( 
\begin{array}{c}
\left( -1\right) ^{s+1}H_{s\left( 4m+k\right) +l}\left( x\right) z \\ 
+\left( -1\right) ^{st}H_{s\left( 2m+k\right) +l}\left( x\right)%
\end{array}%
\right) \sum_{r=0}^{2m-1}\left( -1\right) ^{\frac{\left( sr+2\left(
s+1\right) \right) \left( r+1\right) }{2}+st\left( r+1\right) }\dbinom{2m-1}{%
r}_{F_{s}\left( x\right) }z^{r} \\
&amp;=&amp;\sum_{r=1}^{2m}\left( -1\right) ^{\frac{\left( s\left( r-1\right)
+2\left( s+1\right) \right) r}{2}+str+s+1}\dbinom{2m-1}{r-1}_{F_{s}\left(
x\right) }H_{s\left( 4m+k\right) +l}\left( x\right) z^{r} \\
&amp;&amp;+\sum_{r=0}^{2m-1}\left( -1\right) ^{\frac{\left( sr+2\left( s+1\right)
\right) \left( r+1\right) }{2}+str}\dbinom{2m-1}{r}_{F_{s}\left( x\right)
}H_{s\left( 2m+k\right) +l}\left( x\right) z^{r} \\
&amp;=&amp;\sum_{r=0}^{2m}\left( -1\right) ^{\frac{\left( sr+2\left( s+1\right)
\right) \left( r+1\right) }{2}+sr\left( t+1\right) }\dbinom{2m}{r}%
_{F_{s}\left( x\right) }\frac{1}{F_{2sm}\left( x\right) } \\
&amp;&amp;\times \left( 
\begin{array}{c}
H_{s\left( 4m+k\right) +l}\left( x\right) F_{sr}\left( x\right) \\ 
+\left( -1\right) ^{sr}H_{s\left( 2m+k\right) +l}\left( x\right) F_{s\left(
2m-r\right) }\left( x\right)%
\end{array}%
\right) z^{r}.
\end{eqnarray*}%
But%
\begin{eqnarray}
&amp;&amp;H_{s\left( 4m+k\right) +l}\left( x\right) F_{sr}\left( x\right) +\left(
-1\right) ^{sr}H_{s\left( 2m+k\right) +l}\left( x\right) F_{s\left(
2m-r\right) }\left( x\right)  \label{2.9.1} \\
&amp;=&amp;F_{2sm}\left( x\right) H_{s\left( 2m+r+k\right) +l}\left( x\right) , 
\notag
\end{eqnarray}%
(see appendix). Then (\ref{2.8}) follows.

(b) We have that%
\begin{eqnarray*}
&amp;&amp;\left( z^{2}-\left( -1\right) ^{s\left( t+1\right) }L_{2sm}\left( x\right)
z+1\right) \sum_{r=0}^{2m-1}\left( -1\right) ^{\frac{\left( sr+2\left(
s+1\right) \right) \left( r+1\right) }{2}+st\left( r+1\right) }\dbinom{2m-1}{%
r}_{F_{s}\left( x\right) }z^{r} \\
&amp;=&amp;\sum_{r=2}^{2m+1}\left( -1\right) ^{\frac{\left( s\left( r-2\right)
+2\left( s+1\right) \right) \left( r-1\right) }{2}+st\left( r-1\right) }%
\dbinom{2m-1}{r-2}_{F_{s}\left( x\right) }z^{r} \\
&amp;&amp;-\sum_{r=1}^{2m}\left( -1\right) ^{\frac{\left( s\left( r-1\right)
+2\left( s+1\right) \right) r}{2}+str+s\left( t+1\right) }\dbinom{2m-1}{r-1}%
_{F_{s}\left( x\right) }L_{2sm}\left( x\right) z^{r} \\
&amp;&amp;+\sum_{r=0}^{2m-1}\left( -1\right) ^{\frac{\left( sr+2\left( s+1\right)
\right) \left( r+1\right) }{2}+st\left( r+1\right) }\dbinom{2m-1}{r}%
_{F_{s}\left( x\right) }z^{r} \\
&amp;=&amp;\sum_{r=0}^{2m+1}\left( -1\right) ^{\frac{\left( sr+2\left( s+1\right)
\right) \left( r+1\right) }{2}+st\left( r+1\right) +sr}\dbinom{2m+1}{r}%
_{F_{s}\left( x\right) } \\
&amp;&amp;\times \frac{1}{F_{s\left( 2m+1\right) }\left( x\right) F_{2sm}\left(
x\right) }\left( 
\begin{array}{c}
\left( -1\right) ^{s\left( r+1\right) }F_{sr}\left( x\right) F_{s\left(
r-1\right) }\left( x\right) \\ 
+L_{2sm}\left( x\right) F_{sr}\left( x\right) F_{s\left( 2m+1-r\right)
}\left( x\right) \\ 
+\left( -1\right) ^{rs}F_{s\left( 2m+1-r\right) }\left( x\right) F_{s\left(
2m-r\right) }\left( x\right)%
\end{array}%
\right) z^{r}.
\end{eqnarray*}%
Now we use the identity 
\begin{eqnarray}
&amp;&amp;\left( -1\right) ^{s\left( r+1\right) }F_{sr}\left( x\right) F_{s\left(
r-1\right) }\left( x\right) +L_{2sm}\left( x\right) F_{sr}\left( x\right)
F_{s\left( 2m+1-r\right) }\left( x\right)  \notag \\
&amp;&amp;+\left( -1\right) ^{rs}F_{s\left( 2m+1-r\right) }\left( x\right)
F_{s\left( 2m-r\right) }\left( x\right)  \notag \\
&amp;=&amp;F_{s\left( 2m+1\right) }\left( x\right) F_{2sm}\left( x\right) ,
\label{2.10.1}
\end{eqnarray}%
(see appendix) to obtain (\ref{2.9}).
\end{proof}

\begin{proposition}
\textbf{\label{Prop2.4}}For given $t,m\in \mathbb{N}$ and $k,l\in \mathbb{Z}$
we have that%
\begin{eqnarray}
&amp;&amp;\sum_{r=0}^{2m}\left( -1\right) ^{\frac{\left( sr+2\left( s+1\right)
\right) \left( r+1\right) }{2}+str}\dbinom{2m}{r}_{F_{s}\left( x\right)
}H_{s\left( 2m+r+k\right) +l}\left( x\right) z^{r}  \label{2.11} \\
&amp;=&amp;\left( -1\right) ^{sm+s+1}\left( z-\left( -1\right) ^{s\left( t+m\right)
}\right) \left( H_{s\left( 4m+k\right) +l}\left( x\right) z-\left( -1\right)
^{st}H_{s\left( 2m+k\right) +l}\left( x\right) \right)  \notag \\
&amp;&amp;\times \prod\limits_{p=1}^{m-1}\left( z^{2}-\left( -1\right) ^{s\left(
p+m+t\right) }L_{2sp}\left( x\right) z+1\right) .  \notag
\end{eqnarray}
\end{proposition}

\begin{proof}
We proceed by induction on $m$. To see that (\ref{2.11}) is true for $m=1$
we use the identity 
\begin{equation*}
H_{s\left( 2+k\right) +l}\left( x\right) +\left( -1\right) ^{s}H_{s\left(
4+k\right) +l}\left( x\right) =\left( -1\right) ^{s}L_{s}\left( x\right)
H_{s\left( 3+k\right) +l}\left( x\right) ,
\end{equation*}%
(easy to check). Let us suppose the result is valid for a given $m\in 
\mathbb{N}$, and let us see it is also valid for $m+1$. We begin by writing%
\begin{eqnarray*}
&amp;&amp;\left( -1\right) ^{s\left( m+1\right) +s+1}\left( z-\left( -1\right)
^{s\left( t+m+1\right) }\right) \left( H_{s\left( 4m+4+k\right) +l}\left(
x\right) z-\left( -1\right) ^{st}H_{s\left( 2m+2+k\right) +l}\left( x\right)
\right) \\
&amp;&amp;\times \prod\limits_{p=1}^{m}\left( z^{2}-\left( -1\right) ^{s\left(
p+m+1+t\right) }L_{2sp}\left( x\right) z+1\right) \\
&amp;=&amp;\left( -1\right) ^{s}\frac{H_{s\left( 4m+4+k\right) +l}\left( x\right)
z-\left( -1\right) ^{st}H_{s\left( 2m+2+k\right) +l}\left( x\right) }{%
H_{s\left( 4m+k\right) +l}\left( x\right) z-\left( -1\right) ^{s\left(
t+1\right) }H_{s\left( 2m+k\right) +l}\left( x\right) }\left( z^{2}-\left(
-1\right) ^{s\left( t+1\right) }L_{2sm}\left( x\right) z+1\right) \\
&amp;&amp;\times \left( -1\right) ^{sm+s+1}\left( z-\left( -1\right) ^{s\left(
t+m+1\right) }\right) \left( H_{s\left( 4m+k\right) +l}\left( x\right)
z-\left( -1\right) ^{s\left( t+1\right) }H_{s\left( 2m+k\right) +l}\left(
x\right) \right) \\
&amp;&amp;\times \prod\limits_{p=1}^{m-1}\left( z^{2}-\left( -1\right) ^{s\left(
p+m+1+t\right) }L_{2sp}\left( x\right) z+1\right) \\
&amp;=&amp;-\frac{H_{s\left( 4m+4+k\right) +l}\left( x\right) z-\left( -1\right)
^{st}H_{s\left( 2m+2+k\right) +l}\left( x\right) }{\left( -1\right)
^{s+1}H_{s\left( 4m+k\right) +l}\left( x\right) z+\left( -1\right)
^{st}H_{s\left( 2m+k\right) +l}\left( x\right) }\left( z^{2}-\left(
-1\right) ^{s\left( t+1\right) }L_{2sm}\left( x\right) z+1\right) \\
&amp;&amp;\times \sum_{r=0}^{2m}\left( -1\right) ^{\frac{\left( sr+2\left(
s+1\right) \right) \left( r+1\right) }{2}+s\left( t+1\right) r}\dbinom{2m}{r}%
_{F_{s}\left( x\right) }H_{s\left( 2m+r+k\right) +l}\left( x\right) z^{r}.
\end{eqnarray*}%
By using proposition (\ref{Prop2.3}), first (a), then (b), and finally again
(a) (this second time with $m$ replaced by $m+1$ and $t$ replaced by $t+1$),
we obtain that%
\begin{eqnarray*}
&amp;&amp;-\frac{H_{s\left( 4m+4+k\right) +l}\left( x\right) z-\left( -1\right)
^{st}H_{s\left( 2m+2+k\right) +l}\left( x\right) }{\left( -1\right)
^{s+1}H_{s\left( 4m+k\right) +l}\left( x\right) z+\left( -1\right)
^{st}H_{s\left( 2m+k\right) +l}\left( x\right) }\left( z^{2}-\left(
-1\right) ^{s\left( t+1\right) }L_{2sm}\left( x\right) z+1\right) \\
&amp;&amp;\times \sum_{r=0}^{2m}\left( -1\right) ^{\frac{\left( sr+2\left(
s+1\right) \right) \left( r+1\right) }{2}+s\left( t+1\right) r}\dbinom{2m}{r}%
_{F_{s}\left( x\right) }H_{s\left( 2m+r+k\right) +l}\left( x\right) z^{r} \\
&amp;=&amp;-\left( H_{s\left( 4m+4+k\right) +l}\left( x\right) z-\left( -1\right)
^{st}H_{s\left( 2m+2+k\right) +l}\left( x\right) \right) \left( z^{2}-\left(
-1\right) ^{s\left( t+1\right) }L_{2sm}\left( x\right) z+1\right) \\
&amp;&amp;\times \sum_{r=0}^{2m-1}\left( -1\right) ^{\frac{\left( sr+2\left(
s+1\right) \right) \left( r+1\right) }{2}+st\left( r+1\right) }\dbinom{2m-1}{%
r}_{F_{s}\left( x\right) }z^{r} \\
&amp;=&amp;-\left( H_{s\left( 4m+4+k\right) +l}\left( x\right) z-\left( -1\right)
^{st}H_{s\left( 2m+2+k\right) +l}\left( x\right) \right) \\
&amp;&amp;\times \sum_{r=0}^{2m+1}\left( -1\right) ^{\frac{\left( sr+2\left(
s+1\right) \right) \left( r+1\right) }{2}+st\left( r+1\right) +sr}\dbinom{%
2m+1}{r}_{F_{s}\left( x\right) }z^{r} \\
&amp;=&amp;\sum_{r=0}^{2m+2}\left( -1\right) ^{\frac{\left( sr+2\left( s+1\right)
\right) \left( r+1\right) }{2}+srt}\dbinom{2m+2}{r}_{F_{s}\left( x\right)
}H_{s\left( 2m+r+k\right) +l}\left( x\right) z^{r},
\end{eqnarray*}%
as wanted.
\end{proof}

The following identity is a corollary from (\ref{2.11}):%
\begin{equation}
\sum_{r=0}^{2m}\dbinom{2m}{r}_{F_{2s}\left( x\right) }H_{2s\left(
2m+r+k\right) +l}\left( x\right) =2\left( H_{2s\left( 4m+k\right) +l}\left(
x\right) +H_{2s\left( 2m+k\right) +l}\left( x\right) \right)
\prod\limits_{p=1}^{m-1}L_{2sp}^{2}\left( x\right) .  \label{2.12}
\end{equation}

Another immediate corollary from (\ref{2.11}) is the identity:%
\begin{equation}
\sum_{r=0}^{2m-1}\left( -1\right) ^{\frac{\left( sr+2\left( s+1\right)
\right) \left( r+1\right) }{2}+sr\left( m+1\right) }\dbinom{2m-1}{r}%
_{F_{s}\left( x\right) }=0.  \label{2.13}
\end{equation}

Observe that (\ref{2.13}) is of the same type of (\ref{1.1}). In (\ref{2.13}%
) we have a non-trivial distribution of signs in the list of $s$%
-polyfibonomials $\binom{2m-1}{r}_{F_{s}\left( x\right) },$ $r=0,1,\ldots
,2m-1$, that makes they cancel out in the sum, as happens in (\ref{1.1})
(where we have the trivial distribution of signs given by $\left( -1\right)
^{r}$).

\begin{proposition}
\label{Prop2.7}Let $m\in \mathbb{N}$, $k,l\in \mathbb{Z}$, and $t=1,2,\ldots
,2m$ be given. For $u=\alpha \left( x\right) $ or $u=\beta \left( x\right) $
we have%
\begin{equation}
\sum_{r=0}^{2m+1}\left( -1\right) ^{\frac{\left( sr+2\left( s+1\right)
\right) \left( r+1\right) }{2}+sr\left( t+1\right) }\binom{2m+1}{r}%
_{F_{s}\left( x\right) }H_{s\left( 2m-r+k\right) +l}\left( x\right)
u^{sr\left( 2t-2m-1\right) }=0.  \label{2.18}
\end{equation}
\end{proposition}

\begin{proof}
First of all observe that for given $1\leq t\leq 2m$ there exists $p$, $%
1\leq p\leq m\,$, such that%
\begin{equation}
u^{2s\left( 2t-2m-1\right) }+\left( -1\right) ^{st+1}L_{s\left(
4p-2m-1\right) }\left( x\right) u^{s\left( 2t-2m-1\right) }+\left( -1\right)
^{s}=0,  \label{2.18.1}
\end{equation}%
where $u=\alpha \left( x\right) $ or $u=\beta \left( x\right) $. Indeed,
observe we can write (\ref{2.18.1}) as 
\begin{equation*}
L_{s\left( 4p-2m-1\right) }\left( x\right) =\left( -1\right) ^{st}L_{s\left(
2t-2m-1\right) }\left( x\right) .
\end{equation*}
Thus, we see that if $t=2k$, $k=1,2,\ldots ,m$, we can take $p=k$, and if $%
t=2k-1$, $k=1,2,\ldots ,m$, we can take $p=m-k+1$. This fact, together with (%
\ref{2.5}) give us the desired conclusion.
\end{proof}

\begin{proposition}
\label{Prop2.8}Let $m\in \mathbb{N}$, $k,l\in \mathbb{Z}$, and $t=1,2,\ldots
,2m-1$ be given. For $u=\alpha \left( x\right) $ or $u=\beta \left( x\right) 
$ we have%
\begin{equation}
\sum_{r=0}^{2m}\left( -1\right) ^{\frac{\left( sr+2\left( s+1\right) \right)
\left( r+1\right) }{2}+str}\dbinom{2m}{r}_{F_{s}\left( x\right) }H_{s\left(
2m+r+k\right) +l}\left( x\right) u^{sr\left( 2t-2m\right) }=0.  \label{2.19}
\end{equation}
\end{proposition}

\begin{proof}
We begin our proof by claiming that for given $1\leq t\leq 2m-1$, $t\neq m$,
there exists $p$, $1\leq p\leq m-1$, such that%
\begin{equation}
u^{2s\left( 2t-2m\right) }-\left( -1\right) ^{s\left( p+m+t\right)
}L_{2sp}\left( x\right) u^{s\left( 2t-2m\right) }+1=0.  \label{2.19.1}
\end{equation}

In fact, observe that we can write (\ref{2.19.1}) as 
\begin{equation*}
L_{2sp}\left( x\right) =\left( -1\right) ^{s\left( p+m+t\right) }L_{s\left(
2t-2m\right) }\left( x\right) .
\end{equation*}
Thus, if $t=k$ or $t=2m-k$, $k=1,2,\ldots ,m-1$, take $p=m-k$ to see our
claim is true. On the other hand observe that the case $m=1$ of (\ref{2.19})
is the identity 
\begin{equation*}
\left( -1\right) ^{s+1}H_{s\left( 2+k\right) +l}\left( x\right) +L_{s}\left(
x\right) H_{s\left( 3+k\right) +l}\left( x\right) -H_{s\left( 4+k\right)
+l}\left( x\right) =0,
\end{equation*}%
which can be verified easily. So let us take $m\geq 2$. Consider (\ref{2.11}%
) with $z$ replaced by $u^{s\left( 2t-2m\right) }$. Observe that if $t=m$
the conclusion follows from the fact that $u^{s\left( 2t-2m\right) }-\left(
-1\right) ^{s\left( t+m\right) }=0$. For the remaining cases ($t\neq m$) use
the previous claim to obtain the desired conclusion.
\end{proof}

\begin{proposition}
\label{Prop2.9}Let $m\in \mathbb{N}^{\prime }$ and $k,l\in \mathbb{Z}$ be
given. For $u=\alpha \left( x\right) $ or $u=\beta \left( x\right) $ we have
the following identity%
\begin{eqnarray}
&amp;&amp;\sum_{r=0}^{2m+1}\left( -1\right) ^{\frac{\left( sr+2\left( s+1\right)
\right) \left( r+1\right) }{2}}\dbinom{2m+1}{r}_{F_{s}\left( x\right)
}H_{s\left( 2m-r+k\right) +l}\left( x\right) u^{sr\left( 2m+1\right) }
\label{2.20} \\
&amp;=&amp;\left( -1\right) ^{s\left( k+m\right) +l+1}\left( x^{2}+4\right)
^{m}\left( F_{2m+1}\left( x\right) !\right) _{s}\left( H_{1}\left( x\right)
-H_{0}\left( x\right) u\right) u^{s\left( m\left( 2m+1\right) -k+1\right)
-l}.  \notag
\end{eqnarray}
\end{proposition}

\begin{proof}
By using (\ref{2.5}) with $t=1$ and $z=u^{s\left( 2m+1\right) }$ we can write%
\begin{eqnarray*}
&amp;&amp;\sum_{r=0}^{2m+1}\left( -1\right) ^{\frac{\left( sr+2\left( s+1\right)
\right) \left( r+1\right) }{2}}\dbinom{2m+1}{r}_{F_{s}\left( x\right)
}H_{s\left( 2m-r+k\right) +l}\left( x\right) u^{sr\left( 2m+1\right) } \\
&amp;=&amp;\left( -1\right) ^{sm+1}\left( \left( -1\right) ^{s+1}H_{s\left(
k-1\right) +l}\left( x\right) u^{s\left( 2m+1\right) }+\left( -1\right)
^{s}H_{s\left( 2m+k\right) +l}\left( x\right) \right) \\
&amp;&amp;\times \prod\limits_{p=1}^{m}\left( u^{2s\left( 2m+1\right) }+\left(
-1\right) ^{s+1}L_{s\left( 4p-2m-1\right) }\left( x\right) u^{s\left(
2m+1\right) }+\left( -1\right) ^{s}\right) .
\end{eqnarray*}%
Use the identities (straightforward proofs)%
\begin{eqnarray*}
&amp;&amp;u^{2s\left( 2m+1\right) }+\left( -1\right) ^{s+1}L_{s\left( 4p-2m-1\right)
}\left( x\right) u^{s\left( 2m+1\right) }+\left( -1\right) ^{s} \\
&amp;=&amp;\left( x^{2}+4\right) F_{2sp}\left( x\right) F_{s\left( 2m-2p+1\right)
}\left( x\right) u^{s\left( 2m+1\right) },
\end{eqnarray*}%
and%
\begin{eqnarray*}
&amp;&amp;\left( -1\right) ^{sk+l+s+1}F_{s\left( 2m+1\right) }\left( x\right) \left(
H_{1}\left( x\right) -H_{0}\left( x\right) u\right) u^{s\left( -k+1\right)
-l} \\
&amp;=&amp;H_{s\left( k-1\right) +l}\left( x\right) u^{s\left( 2m+1\right)
}-H_{s\left( 2m+k\right) +l}\left( x\right) ,
\end{eqnarray*}%
to obtain%
\begin{eqnarray*}
&amp;&amp;\left( -1\right) ^{sm+1}\sum_{r=0}^{2m+1}\left( -1\right) ^{\frac{\left(
sr+2\left( s+1\right) \right) \left( r+1\right) }{2}}\dbinom{2m+1}{r}%
_{F_{s}\left( x\right) }H_{s\left( 2m-r+k\right) +l}\left( x\right)
u^{sr\left( 2m+1\right) } \\
&amp;=&amp;\left( -1\right) ^{sm+1}\left( -1\right) ^{sk+l}F_{s\left( 2m+1\right)
}\left( x\right) \left( H_{1}\left( x\right) -H_{0}\left( x\right) u\right)
u^{s\left( -k+1\right) -l} \\
&amp;&amp;\times \prod\limits_{p=1}^{m}\left( \left( x^{2}+4\right) F_{2sp}\left(
x\right) F_{s\left( 2m-2p+1\right) }\left( x\right) u^{s\left( 2m+1\right)
}\right) \\
&amp;=&amp;\left( -1\right) ^{s\left( k+m\right) +l+1}F_{s\left( 2m+1\right) }\left(
x\right) \left( H_{1}\left( x\right) -H_{0}\left( x\right) u\right)
u^{s\left( -k+1\right) -l}\left( x^{2}+4\right) ^{m} \\
&amp;&amp;\times \left( F_{2m}\left( x\right) !\right) _{s}u^{sm\left( 2m+1\right) }
\\
&amp;=&amp;\left( -1\right) ^{s\left( k+m\right) +l+1}\left( x^{2}+4\right)
^{m}\left( F_{2m+1}\left( x\right) !\right) _{s}\left( H_{1}\left( x\right)
-H_{0}\left( x\right) u\right) u^{s\left( m\left( 2m+1\right) -k+1\right)
-l},
\end{eqnarray*}%
as expected.
\end{proof}

\begin{proposition}
\label{Prop2.10}Let $m\in \mathbb{N}$ be given. For $u=\alpha \left(
x\right) $ and $u=\beta \left( x\right) $ we have%
\begin{eqnarray}
&amp;&amp;\pm \sum_{r=0}^{2m}\left( -1\right) ^{\frac{\left( sr+2\left( s+1\right)
\right) \left( r+1\right) }{2}}\dbinom{2m}{r}_{F_{s}\left( x\right)
}H_{s\left( 2m+r+k\right) +l}\left( x\right) u^{2msr}  \label{2.21} \\
&amp;=&amp;\left( -1\right) ^{sm+s+1}\left( x^{2}+4\right) ^{m-\frac{1}{2}}\left(
F_{2m}\left( x\right) !\right) _{s}\left( H_{1}\left( x\right) -H_{0}\left(
x\right) \overline{u}\right) u^{s\left( 2m^{2}+3m+k\right) +l},  \notag
\end{eqnarray}%
where the plus sign of the left-hand side corresponds to $u=\alpha \left(
x\right) $ and $\overline{u}=\beta \left( x\right) $, and the minus sign
corresponds to $u=\beta \left( x\right) $ and $\overline{u}=\alpha \left(
x\right) $.
\end{proposition}

\begin{proof}
We use (\ref{2.11}) with $t=2$ and $z=u^{2ms}$ to write%
\begin{eqnarray*}
&amp;&amp;\sum_{r=0}^{2m}\left( -1\right) ^{\frac{\left( sr+2\left( s+1\right)
\right) \left( r+1\right) }{2}}\dbinom{2m}{r}_{F_{s}\left( x\right)
}H_{s\left( 2m+r+k\right) +l}\left( x\right) u^{2msr} \\
&amp;=&amp;\left( -1\right) ^{sm+s+1}\left( u^{2ms}-\left( -1\right) ^{sm}\right)
\left( H_{s\left( 4m+k\right) +l}\left( x\right) u^{2ms}-H_{s\left(
2m+k\right) +l}\left( x\right) \right) \\
&amp;&amp;\times \prod\limits_{p=1}^{m-1}\left( u^{4ms}-\left( -1\right) ^{s\left(
p+m\right) }L_{2sp}\left( x\right) u^{2ms}+1\right) .
\end{eqnarray*}%
Use now the identities (straightforward proofs)%
\begin{eqnarray*}
u^{4ms}-\left( -1\right) ^{s\left( p+m\right) }L_{2sp}\left( x\right)
u^{2ms}+1 &amp;=&amp;\left( x^{2}+4\right) u^{2sm}F_{s\left( m+p\right) }\left(
x\right) F_{s\left( m-p\right) }\left( x\right) , \\
H_{s\left( 4m+k\right) +l}\left( x\right) u^{2ms}-H_{s\left( 2m+k\right)
+l}\left( x\right) &amp;=&amp;\left( H_{1}\left( x\right) -H_{0}\left( x\right) 
\overline{u}\right) F_{2sm}\left( x\right) u^{s\left( 4m+k\right) +l}, \\
u^{2ms}-\left( -1\right) ^{sm} &amp;=&amp;\pm \left( x^{2}+4\right) ^{\frac{1}{2}%
}F_{sm}\left( x\right) u^{sm},
\end{eqnarray*}%
(in the last identity we have $+$ if $u=\alpha \left( x\right) $ and $-$ if $%
u=\beta \left( x\right) $) to obtain%
\begin{eqnarray*}
&amp;&amp;\pm \sum_{r=0}^{2m}\left( -1\right) ^{\frac{\left( sr+2\left( s+1\right)
\right) \left( r+1\right) }{2}}\dbinom{2m}{r}_{F_{s}\left( x\right)
}H_{s\left( 2m+r+k\right) +l}\left( x\right) u^{2msr} \\
&amp;=&amp;\left( -1\right) ^{sm+s+1}\left( x^{2}+4\right) ^{\frac{1}{2}%
}F_{sm}\left( x\right) u^{sm}\left( H_{1}\left( x\right) -H_{0}\left(
x\right) \overline{u}\right) F_{2sm}\left( x\right) u^{s\left( 4m+k\right)
+l} \\
&amp;&amp;\times \prod\limits_{p=1}^{m-1}\left( \left( x^{2}+4\right)
u^{2sm}F_{s\left( m+p\right) }\left( x\right) F_{s\left( m-p\right) }\left(
x\right) \right) \\
&amp;=&amp;\left( -1\right) ^{sm+s+1}\left( x^{2}+4\right) ^{m-\frac{1}{2}}\left(
F_{2m}\left( x\right) !\right) _{s}\left( H_{1}\left( x\right) -H_{0}\left(
x\right) \overline{u}\right) u^{s\left( 2m^{2}+3m+k\right) +l},
\end{eqnarray*}%
as wanted.
\end{proof}

\section{\label{Sec3}The main results}

Let $m\in \mathbb{N}$, $p_{1},\ldots ,p_{m}\in \mathbb{Z}$ be given. For
every $t=0,1,\ldots ,m$, let $J_{t,m}$ be the family of subsets of $\left\{
1,2,\ldots ,m\right\} $ containing $t$ elements. (Thus $J_{t,m}$ contains $%
\binom{m}{t}$ subsets.) For each subset $A_{t}$ belonging to $J_{t,m}$,
define $p_{A_{t}}$ as $p_{A_{t}}=\sum_{\omega \in A_{t}}p_{\omega }$.

Let us consider the $s$-Gibonacci polynomial sequences 
\begin{equation*}
G_{s\left( n+p_{i}\right) }\left( x\right) =\left( x^{2}+4\right) ^{-\frac{1%
}{2}}\left( c_{1}\left( x\right) \alpha ^{s\left( n+p_{i}\right) }\left(
x\right) -c_{2}\left( x\right) \beta ^{s\left( n+p_{i}\right) }\left(
x\right) \right) ,
\end{equation*}%
where $i=1,2,\ldots ,M$, and 
\begin{eqnarray*}
c_{1}\left( x\right) &amp;=&amp;G_{1}\left( x\right) -G_{0}\left( x\right) \beta
\left( x\right) , \\
c_{2}\left( x\right) &amp;=&amp;G_{1}\left( x\right) -G_{0}\left( x\right) \alpha
\left( x\right) .
\end{eqnarray*}

It is not difficult to establish the following formulas for the products $%
\prod_{i=1}^{2m+1}G_{s\left( n+p_{i}\right) }\left( x\right) $ and $%
\prod_{i=1}^{2m}G_{s\left( n+p_{i}\right) }\left( x\right) $%
\begin{eqnarray}
&amp;&amp;\prod_{i=1}^{2m+1}G_{s\left( n+p_{i}\right) }\left( x\right)  \label{3.1}
\\
&amp;=&amp;\left( x^{2}+4\right) ^{-m-\frac{1}{2}}\sum_{t=0}^{2m+1}\left( -1\right)
^{t\left( sn+1\right) }c_{1}^{2m+1-t}\left( x\right) c_{2}^{t}\left( x\right)
\notag \\
&amp;&amp;\times \sum_{J_{t,2m+1}}\alpha ^{s\left( \left( 2m+1-2t\right)
n+p_{1}+\cdots +p_{2m+1}-p_{A_{t}}\right) }\!\left( x\right) \beta
^{sp_{A_{t}}}\!\left( x\right) .  \notag
\end{eqnarray}%
\begin{eqnarray}
&amp;&amp;\prod_{i=1}^{2m}G_{s\left( n+p_{i}\right) }\left( x\right)  \label{3.2} \\
&amp;=&amp;\left( x^{2}+4\right) ^{-m}\sum_{t=0}^{2m}\left( -1\right) ^{t\left(
sn+1\right) }c_{1}^{2m-t}\left( x\right) c_{2}^{t}\left( x\right)  \notag \\
&amp;&amp;\times \sum_{J_{t,2m}}\alpha ^{s\left( \left( 2m-2t\right) n+p_{1}+\cdots
+p_{2m}-p_{A_{t}}\right) }\left( x\right) \beta ^{sp_{A_{t}}}\left( x\right)
.  \notag
\end{eqnarray}

(The summation $\sum_{J_{t,m}}$ in (\ref{3.1}) and (\ref{3.2}) is over all
the subsets contained in $J_{t,m}$.)

Now we are ready to establish the two main results of this article.

\begin{theorem}
\label{Th3.1}Let $m\in \mathbb{N}^{\prime }$ and $k,l,p_{1},\ldots
,p_{2m+1}\in \mathbb{Z}$ be given. We have that%
\begin{eqnarray}
&amp;&amp;\sum_{r=0}^{2m+1}\left( -1\right) ^{\frac{\left( sr+2\left( s+1\right)
\right) \left( r+1\right) }{2}}\dbinom{2m+1}{r}_{F_{s}\left( x\right)
}H_{s\left( 2m-r+k\right) +l}\left( x\right) \prod_{i=1}^{2m+1}G_{s\left(
n+r+p_{i}\right) }\left( x\right)  \label{3.3} \\
&amp;=&amp;\!\!\left( -1\right) ^{s\left( m+k\right) +l+1}\left( x^{2}+4\right) ^{-%
\frac{1}{2}}\!\left( F_{2m+1}\left( x\right) !\right) _{s}\!  \notag \\
&amp;&amp;\times \left( \!%
\begin{array}{c}
\left( H_{1}\left( x\right) -H_{0}\left( x\right) \alpha \left( x\right)
\right) \left( G_{1}\left( x\right) -G_{0}\left( x\right) \beta \left(
x\right) \right) ^{2m+1} \\ 
\times \alpha ^{s\left( \left( 2m+1\right) \left( n+m\right) +p_{1}+\cdots
+p_{2m+1}-k+1\right) -l}\left( x\right) \\ 
-\left( H_{1}\left( x\right) -H_{0}\left( x\right) \beta \left( x\right)
\right) \left( G_{1}\left( x\right) -G_{0}\left( x\right) \alpha \left(
x\right) \right) ^{2m+1} \\ 
\times \beta ^{s\left( \left( 2m+1\right) \left( n+m\right) +p_{1}+\cdots
+p_{2m+1}-k+1\right) -l}\left( x\right)%
\end{array}%
\!\right) .  \notag
\end{eqnarray}
\end{theorem}

\begin{proof}
After inserting (\ref{3.1}) in the left-hand side of (\ref{3.3}), we obtain
an expression of the form 
\begin{eqnarray}
&amp;&amp;\sum_{r=0}^{2m+1}\left( -1\right) ^{\frac{\left( sr+2\left( s+1\right)
\right) \left( r+1\right) }{2}}\dbinom{2m+1}{r}_{F_{s}\left( x\right)
}H_{s\left( 2m-r+k\right) +l}\left( x\right) \prod_{i=1}^{2m+1}G_{s\left(
n+r+p_{i}\right) }\left( x\right)  \label{3.4} \\
&amp;=&amp;\sum_{t=0}^{2m+1}\Psi \left( x,t,m,s\right) ,  \notag
\end{eqnarray}%
where 
\begin{eqnarray}
&amp;&amp;\Psi \left( x,t,m,s\right)  \label{3.5} \\
&amp;=&amp;\left( x^{2}+4\right) ^{-m-\frac{1}{2}}\left( -1\right) ^{t\left(
sn+1\right) }c_{1}^{2m+1-t}\left( x\right) c_{2}^{t}\left( x\right)  \notag
\\
&amp;&amp;\times \sum_{J_{t,2m+1}}\alpha ^{s\left( \left( 2m+1-2t\right)
n-p_{A_{t}}+p_{1}+\cdots +p_{2m+1}\right) }\left( x\right) \beta
^{sp_{A_{t}}}\left( x\right)  \notag \\
&amp;&amp;\times \sum_{r=0}^{2m+1}\left( -1\right) ^{\frac{\left( sr+2\left(
s+1\right) \right) \left( r+1\right) }{2}+tsr}\dbinom{2m+1}{r}_{F_{s}\left(
x\right) }H_{s\left( 2m-r+k\right) +l}\left( x\right) \alpha ^{sr\left(
2m+1-2t\right) }\left( x\right) ,  \notag
\end{eqnarray}%
(with $c_{1}\left( x\right) =G_{1}\left( x\right) -G_{0}\left( x\right)
\beta \left( x\right) $ and $c_{2}\left( x\right) =G_{1}\left( x\right)
-G_{0}\left( x\right) \alpha \left( x\right) $). By writing the right-hand
side of (\ref{3.4}) as 
\begin{equation*}
\sum_{t=0}^{2m+1}\Psi \left( x,t,m,s\right) =\Psi \left( x,0,m,s\right)
+\Psi \left( x,2m+1,m,s\right) +\sum_{t=1}^{2m}\Psi \left( x,t,m,s\right) ,
\end{equation*}%
the proof ends if we show that 
\begin{equation*}
\sum_{t=1}^{2m}\Psi \left( x,t,m,s\right) =0,
\end{equation*}%
and that the remaining sum 
\begin{equation*}
\Psi \left( x,0,m,s\right) +\Psi \left( x,2m+1,m,s\right) ,
\end{equation*}%
is equal to the right-hand side of (\ref{3.3}).

According to proposition \ref{Prop2.7}, for $t=1,2,\ldots ,2m$ we have%
\begin{eqnarray*}
&amp;&amp;\sum_{r=0}^{2m+1}\left( -1\right) ^{\frac{\left( sr+2\left( s+1\right)
\right) \left( r+1\right) }{2}+tsr}\dbinom{2m+1}{r}_{F_{s}\left( x\right)
}H_{s\left( 2m-r+k\right) +l}\left( x\right) \alpha ^{sr\left(
2m+1-2t\right) }\left( x\right) \\
&amp;=&amp;\sum_{r=0}^{2m+1}\left( -1\right) ^{\frac{\left( sr+2\left( s+1\right)
\right) \left( r+1\right) }{2}+sr\left( t+1\right) }\dbinom{2m+1}{r}%
_{F_{s}\left( x\right) }H_{s\left( 2m-r+k\right) +l}\left( x\right) \beta
^{sr\left( 2t-2m-1\right) }\left( x\right) \\
&amp;=&amp;0,
\end{eqnarray*}%
and then $\sum_{t=1}^{2m}\Psi \left( x,t,m,s\right) =0$. On the other hand,
by using proposition \ref{Prop2.9} we have%
\begin{eqnarray*}
&amp;&amp;\Psi \left( x,0,m,s\right) +\Psi \left( x,2m+1,m,s\right) \\
&amp;=&amp;\left( x^{2}+4\right) ^{-m-\frac{1}{2}}c_{1}^{2m+1}\left( x\right) \alpha
^{s\left( \left( 2m+1\right) n+p_{1}+\cdots +p_{2m+1}\right) }\left( x\right)
\\
&amp;&amp;\times \sum_{r=0}^{2m+1}\left( -1\right) ^{\frac{\left( sr+2\left(
s+1\right) \right) \left( r+1\right) }{2}}\dbinom{2m+1}{r}_{F_{s}\left(
x\right) }H_{s\left( 2m-r+k\right) +l}\left( x\right) \alpha ^{sr\left(
2m+1\right) }\left( x\right) \\
&amp;&amp;-\left( x^{2}+4\right) ^{-m-\frac{1}{2}}c_{2}^{2m+1}\left( x\right) \beta
^{s\left( \left( 2m+1\right) n+p_{1}+\cdots +p_{2m+1}\right) }\left( x\right)
\\
&amp;&amp;\times \sum_{r=0}^{2m+1}\left( -1\right) ^{\frac{\left( sr+2\left(
s+1\right) \right) \left( r+1\right) }{2}}\dbinom{2m+1}{r}_{F_{s}\left(
x\right) }H_{s\left( 2m-r+k\right) +l}\left( x\right) \beta ^{sr\left(
2m+1\right) }\left( x\right) \\
&amp;=&amp;\left( x^{2}+4\right) ^{-m-\frac{1}{2}}c_{1}^{2m+1}\left( x\right) \alpha
^{s\left( \left( 2m+1\right) n+p_{1}+\cdots +p_{2m+1}\right) }\left(
x\right) \left( -1\right) ^{s\left( k+m\right) +l+1}\left( x^{2}+4\right)
^{m} \\
&amp;&amp;\times \left( F_{2m+1}\left( x\right) !\right) _{s}\left( H_{1}\left(
x\right) -H_{0}\left( x\right) \alpha \left( x\right) \right) \alpha
^{s\left( m\left( 2m+1\right) +\left( -k+1\right) \right) -l}\left( x\right)
\\
&amp;&amp;-\left( x^{2}+4\right) ^{-m-\frac{1}{2}}c_{2}^{2m+1}\left( x\right) \beta
^{s\left( \left( 2m+1\right) n+p_{1}+\cdots +p_{2m+1}\right) }\left(
x\right) \left( -1\right) ^{s\left( k+m\right) +l+1}\left( x^{2}+4\right)
^{m} \\
&amp;&amp;\times \left( F_{2m+1}\left( x\right) !\right) _{s}\left( H_{1}\left(
x\right) -H_{0}\left( x\right) \beta \left( x\right) \right) \beta ^{s\left(
m\left( 2m+1\right) +\left( -k+1\right) \right) -l}\left( x\right) \\
&amp;=&amp;\left( -1\right) ^{s\left( m+k\right) +l+1}\left( x^{2}+4\right) ^{-\frac{%
1}{2}}\!\left( F_{2m+1}\left( x\right) !\right) _{s}\! \\
&amp;&amp;\times \left( \!%
\begin{array}{c}
\left( H_{1}\left( x\right) -H_{0}\left( x\right) \alpha \left( x\right)
\right) c_{1}^{2m+1}\left( x\right) \alpha ^{s\left( \left( 2m+1\right)
\left( n+m\right) +p_{1}+\cdots +p_{2m+1}-k+1\right) -l}\left( x\right) \\ 
-\left( H_{1}\left( x\right) -H_{0}\left( x\right) \beta \left( x\right)
\right) c_{2}^{2m+1}\left( x\right) \beta ^{s\left( \left( 2m+1\right)
\left( n+m\right) +p_{1}+\cdots +p_{2m+1}-k+1\right) -l}\left( x\right)%
\end{array}%
\!\right) ,
\end{eqnarray*}%
as wanted.
\end{proof}

\begin{theorem}
\textbf{\label{Th3.2}}Let $m\in \mathbb{N}$ and $k,l,p_{1},\ldots
,p_{2m+1}\in \mathbb{Z}$ be given. We have that%
\begin{eqnarray}
&amp;&amp;\sum_{r=0}^{2m}\left( -1\right) ^{\frac{\left( sr+2\left( s+1\right)
\right) \left( r+1\right) }{2}}\dbinom{2m}{r}_{F_{s}\left( x\right)
}H_{s\left( 2m+r+k\right) +l}\left( x\right) \prod_{i=1}^{2m}G_{s\left(
n+r+p_{i}\right) }\left( x\right)  \label{3.6} \\
&amp;=&amp;\left( -1\right) ^{sm+s+1}\left( x^{2}+4\right) ^{-\frac{1}{2}}\left(
F_{2m}\left( x\right) !\right) _{s}  \notag \\
&amp;&amp;\times \left( 
\begin{array}{c}
\left( H_{1}\left( x\right) -H_{0}\left( x\right) \beta \left( x\right)
\right) \left( G_{1}\left( x\right) -G_{0}\left( x\right) \beta \left(
x\right) \right) ^{2m} \\ 
\times \alpha ^{s\left( 2m\left( m+n\right) +p_{1}+\cdots
+p_{2m}+3m+k\right) +l}\left( x\right) \\ 
-\left( H_{1}\left( x\right) -H_{0}\left( x\right) \alpha \left( x\right)
\right) \left( G_{1}\left( x\right) -G_{0}\left( x\right) \alpha \left(
x\right) \right) ^{2m} \\ 
\times \beta ^{s\left( 2m\left( m+n\right) +p_{1}+\cdots +p_{2m}+3m+k\right)
+l}\left( x\right)%
\end{array}%
\right) .  \notag
\end{eqnarray}
\end{theorem}

\begin{proof}
We follow the same strategy of the proof of theorem \ref{Th3.1}.
Substituting (\ref{3.2}) in the left-hand side of (\ref{3.6}) we can write%
\begin{eqnarray*}
&amp;&amp;\sum_{r=0}^{2m}\left( -1\right) ^{\frac{\left( sr+2\left( s+1\right)
\right) \left( r+1\right) }{2}}\dbinom{2m}{r}_{F_{s}\left( x\right)
}H_{s\left( 2m+r+k\right) +l}\left( x\right) \prod_{i=1}^{2m}G_{s\left(
n+r+p_{i}\right) }\left( x\right) \\
&amp;=&amp;\Phi \left( x,0,m,s\right) +\Phi \left( x,2m,m,s\right)
+\sum_{t=1}^{2m-1}\Phi \left( x,t,m,s\right) ,
\end{eqnarray*}%
where 
\begin{eqnarray*}
\Phi \left( x,t,m,s\right) &amp;=&amp;\left( x^{2}+4\right) ^{-m}\left( -1\right)
^{t\left( sn+1\right) }c_{1}^{2m-t}\left( x\right) c_{2}^{t}\left( x\right)
\\
&amp;&amp;\times \sum_{J_{t,2m}}\alpha ^{s\left( \left( 2m-2t\right) n+p_{1}+\cdots
+p_{2m}-p_{A_{t}}\right) }\left( x\right) \beta ^{sp_{A_{t}}}\left( x\right)
\\
&amp;&amp;\times \sum_{r=0}^{2m}\left( -1\right) ^{\frac{\left( sr+2\left(
s+1\right) \right) \left( r+1\right) }{2}}\dbinom{2m}{r}_{F_{s}\left(
x\right) }H_{s\left( 2m+r+k\right) +l}\left( x\right) \alpha ^{sr\left(
2m-2t\right) }\left( x\right) ,
\end{eqnarray*}%
(with $c_{1}\left( x\right) =G_{1}\left( x\right) -G_{0}\left( x\right)
\beta \left( x\right) $ and $c_{2}\left( x\right) =G_{1}\left( x\right)
-G_{0}\left( x\right) \alpha \left( x\right) $). According to proposition %
\ref{Prop2.8}, for $t=1,2,\ldots ,2m-1$ we have%
\begin{eqnarray*}
&amp;&amp;\sum_{r=0}^{2m}\left( -1\right) ^{\frac{\left( sr+2\left( s+1\right)
\right) \left( r+1\right) }{2}+tsr}\dbinom{2m}{r}_{F_{s}\left( x\right)
}H_{s\left( 2m+r+k\right) +l}\left( x\right) \alpha ^{sr\left( 2m-2t\right)
}\left( x\right) \\
&amp;=&amp;\sum_{r=0}^{2m}\left( -1\right) ^{\frac{\left( sr+2\left( s+1\right)
\right) \left( r+1\right) }{2}+str}\dbinom{2m}{r}_{F_{s}\left( x\right)
}H_{s\left( 2m+r+k\right) +l}\left( x\right) \alpha ^{sr\left( 2t-2m\right)
}\left( x\right) \\
&amp;=&amp;0,
\end{eqnarray*}%
and then 
\begin{equation*}
\sum_{t=1}^{2m-1}\Phi \left( x,t,m,s\right) =0.
\end{equation*}

On the other hand, by using proposition \ref{Prop2.10} we get%
\begin{eqnarray*}
&amp;&amp;\Phi \left( x,0,m,s\right) +\Phi \left( x,2m,m,s\right) \\
&amp;=&amp;\left( x^{2}+4\right) ^{-m}c_{1}^{2m}\left( x\right) \alpha ^{s\left(
2mn+p_{1}+\cdots +p_{2m}\right) }\left( x\right) \\
&amp;&amp;\times \sum_{r=0}^{2m}\left( -1\right) ^{\frac{\left( sr+2\left(
s+1\right) \right) \left( r+1\right) }{2}}\dbinom{2m}{r}_{F_{s}\left(
x\right) }G_{s\left( 2m+r+k\right) +l}\left( x\right) \alpha ^{2smr}\left(
x\right) \\
&amp;&amp;+\left( x^{2}+4\right) ^{-m}c_{2}^{2m}\left( x\right) \beta ^{s\left(
2mn+p_{1}+\cdots +p_{2m}\right) }\left( x\right) \\
&amp;&amp;\times \sum_{r=0}^{2m}\left( -1\right) ^{\frac{\left( sr+2\left(
s+1\right) \right) \left( r+1\right) }{2}}\dbinom{2m}{r}_{F_{s}\left(
x\right) }G_{s\left( 2m+r+k\right) +l}\left( x\right) \beta ^{2msr}\left(
x\right) \\
&amp;=&amp;\left( x^{2}+4\right) ^{-m}\alpha ^{s\left( 2mn+p_{1}+\cdots
+p_{2m}\right) }\left( x\right) \left( -1\right) ^{sm+s+1}\left(
x^{2}+4\right) ^{m-\frac{1}{2}}\left( F_{2m}\left( x\right) !\right) _{s} \\
&amp;&amp;\times \left( H_{1}\left( x\right) -H_{0}\left( x\right) \beta \left(
x\right) \right) c_{1}^{2m}\left( x\right) \alpha ^{s\left(
2m^{2}+3m+k\right) +l}\left( x\right) \\
&amp;&amp;-\left( x^{2}+4\right) ^{-m}\beta ^{s\left( 2mn+p_{1}+\cdots
+p_{2m}\right) }\left( x\right) \left( -1\right) ^{sm+s+1}\left(
x^{2}+4\right) ^{m-\frac{1}{2}}\left( F_{2m}\left( x\right) !\right) _{s} \\
&amp;&amp;\times \left( H_{1}\left( x\right) -H_{0}\left( x\right) \alpha \left(
x\right) \right) c_{2}^{2m}\left( x\right) \beta ^{s\left(
2m^{2}+3m+k\right) +l}\left( x\right) \\
&amp;=&amp;\left( -1\right) ^{sm+s+1}\left( x^{2}+4\right) ^{-\frac{1}{2}}\left(
F_{2m}\left( x\right) !\right) _{s} \\
&amp;&amp;\times \left( 
\begin{array}{c}
\left( H_{1}\left( x\right) -H_{0}\left( x\right) \beta \left( x\right)
\right) c_{1}^{2m}\left( x\right) \alpha ^{s\left( 2m\left( m+n\right)
+p_{1}+\cdots +p_{2m}+3m+k\right) +l}\left( x\right) \\ 
-\left( H_{1}\left( x\right) -H_{0}\left( x\right) \alpha \left( x\right)
\right) c_{2}^{2m}\left( x\right) \beta ^{s\left( 2m\left( m+n\right)
+p_{1}+\cdots +p_{2m}+3m+k\right) +l}\left( x\right)%
\end{array}%
\right)
\end{eqnarray*}%
as wanted.
\end{proof}

\section{\label{Sec4}Some examples}

If in (\ref{3.3}) we take $H_{sn}\left( x\right) =cL_{sn}\left( x\right)
+dF_{sn}\left( x\right) $ and $G_{sn}\left( x\right) =aL_{sn}\left( x\right)
+bF_{sn}\left( x\right) $ (with $a,b,c,d\in \mathbb{R}$), we obtain%
\begin{eqnarray*}
&amp;&amp;\sum_{r=0}^{2m+1}\left( -1\right) ^{\frac{\left( sr+2\left( s+1\right)
\right) \left( r+1\right) }{2}}\dbinom{2m+1}{r}_{F_{s}\left( x\right)
}\left( cL_{s\left( 2m-r+k\right) +l}\left( x\right) +dF_{s\left(
2m-r+k\right) +l}\left( x\right) \right) \\
&amp;&amp;\times \prod_{i=1}^{2m+1}\left( aL_{s\left( n+r+p_{i}\right) }\left(
x\right) +bF_{s\left( n+r+p_{i}\right) }\left( x\right) \right) \\
&amp;=&amp;\!\!\left( -1\right) ^{s\left( m+k\right) +l+1}\left( x^{2}+4\right) ^{-%
\frac{1}{2}}\!\left( F_{2m+1}\left( x\right) !\right) _{s}\! \\
&amp;&amp;\times \left( \!%
\begin{array}{c}
\left( d-c\sqrt{x^{2}+4}\right) \left( b+a\sqrt{x^{2}+4}\right)
^{2m+1}\alpha ^{s\left( \left( 2m+1\right) \left( n+m\right) +p_{1}+\cdots
+p_{2m+1}-k+1\right) -l}\left( x\right) \\ 
-\left( d+c\sqrt{x^{2}+4}\right) \left( b-a\sqrt{x^{2}+4}\right)
^{2m+1}\beta ^{s\left( \left( 2m+1\right) \left( n+m\right) +p_{1}+\cdots
+p_{2m+1}-k+1\right) -l}\left( x\right)%
\end{array}%
\!\right) .
\end{eqnarray*}

By expanding the binomials $\left( b\pm a\sqrt{x^{2}+4}\right) ^{2m+1}$ we
can write this expression as%
\begin{eqnarray*}
&amp;&amp;\sum_{r=0}^{2m+1}\left( -1\right) ^{\frac{\left( sr+2\left( s+1\right)
\right) \left( r+1\right) }{2}}\dbinom{2m+1}{r}_{F_{s}\left( x\right)
}H_{s\left( 2m-r+k\right) +l}\left( x\right) \prod_{i=1}^{2m+1}G_{s\left(
n+r+p_{i}\right) }\left( x\right) \\
&amp;=&amp;\!\!\left( -1\right) ^{s\left( m+k\right) +l+1}\left( F_{2m+1}\left(
x\right) !\right) _{s}\! \\
&amp;&amp;\times \sum_{j=0}^{m}a^{2j}\left( x^{2}+4\right) ^{j}b^{2m-2j} \\
&amp;&amp;\times \left( \!%
\begin{array}{c}
\alpha ^{s\left( \left( 2m+1\right) \left( n+m\right) +p_{1}+\cdots
+p_{2m+1}-k+1\right) -l}\left( x\right) \left( 
\begin{array}{c}
\binom{2m+1}{2j}bd\left( x^{2}+4\right) ^{-\frac{1}{2}}+\binom{2m+1}{2j+1}ad
\\ 
-\binom{2m+1}{2j}bc-\binom{2m+1}{2j+1}ac\left( x^{2}+4\right) ^{\frac{1}{2}}%
\end{array}%
\right) \\ 
-\beta ^{s\left( \left( 2m+1\right) \left( n+m\right) +p_{1}+\cdots
+p_{2m+1}-k+1\right) -l}\left( x\right) \left( 
\begin{array}{c}
\binom{2m+1}{2j}bd\left( x^{2}+4\right) ^{-\frac{1}{2}}-\binom{2m+1}{2j+1}ad
\\ 
+\binom{2m+1}{2j}bc-\binom{2m+1}{2j+1}ac\left( x^{2}+4\right) ^{\frac{1}{2}}%
\end{array}%
\right)%
\end{array}%
\!\right) .
\end{eqnarray*}

A simple further simplification gives us finally%
\begin{eqnarray}
&amp;&amp;\sum_{r=0}^{2m+1}\left( -1\right) ^{\frac{\left( sr+2\left( s+1\right)
\right) \left( r+1\right) }{2}}\dbinom{2m+1}{r}_{F_{s}\left( x\right)
}\left( cL_{s\left( 2m-r+k\right) +l}\left( x\right) +dF_{s\left(
2m-r+k\right) +l}\left( x\right) \right)  \label{3.7} \\
&amp;&amp;\times \prod_{i=1}^{2m+1}\left( aL_{s\left( n+r+p_{i}\right) }\left(
x\right) +bF_{s\left( n+r+p_{i}\right) }\left( x\right) \right)  \notag \\
&amp;=&amp;\!\!\left( -1\right) ^{s\left( m+k\right) +l+1}\left( F_{2m+1}\left(
x\right) !\right) _{s}\!\!\text{ }\sum_{j=0}^{m}a^{2j}b^{2m-2j}\left(
x^{2}+4\right) ^{j}  \notag \\
&amp;&amp;\times \left( 
\begin{array}{c}
\left( \binom{2m+1}{2j+1}ad-\binom{2m+1}{2j}bc\right) L_{s\left( \left(
2m+1\right) \left( n+m\right) +p_{1}+\cdots +p_{2m+1}-k+1\right) -l}\left(
x\right) \\ 
+\left( \binom{2m+1}{2j}bd-\binom{2m+1}{2j+1}ac\left( x^{2}+4\right) \right)
F_{s\left( \left( 2m+1\right) \left( n+m\right) +p_{1}+\cdots
+p_{2m+1}-k+1\right) -l}\left( x\right)%
\end{array}%
\right) ,  \notag
\end{eqnarray}

Similarly, beginning with (\ref{3.6}) and taking $H_{sn}\left( x\right)
=cL_{sn}\left( x\right) +dF_{sn}\left( x\right) $ and $G_{sn}\left( x\right)
=aL_{sn}\left( x\right) +bF_{sn}\left( x\right) $, we can obtain%
\begin{eqnarray}
&amp;&amp;\sum_{r=0}^{2m}\left( -1\right) ^{\frac{\left( sr+2\left( s+1\right)
\right) \left( r+1\right) }{2}}\dbinom{2m}{r}_{F_{s}\left( x\right) }\left(
cL_{s\left( 2m+r+k\right) +l}\left( x\right) +dF_{s\left( 2m+r+k\right)
+l}\left( x\right) \right)  \label{3.8} \\
&amp;&amp;\times \prod_{i=1}^{2m}\left( aL_{s\left( n+r+p_{i}\right) }\left(
x\right) +bF_{s\left( n+r+p_{i}\right) }\left( x\right) \right)  \notag \\
&amp;=&amp;\left( -1\right) ^{sm+s+1}\left( F_{2m}\left( x\right) !\right) _{s} 
\notag \\
&amp;&amp;\times \sum_{j=0}^{m-1}a^{2j}b^{2m-2j-1}\!\left( x^{2}+4\right)
^{j}\!\left( \!\!%
\begin{array}{c}
\left( \binom{2m}{2j+1}ad+\binom{2m}{2j}bc\right) \!L_{s\left( 2m\left(
m+n\right) +p_{1}+\cdots +p_{2m}+3m+k\right) +l}\left( x\right) \\ 
+\left( \binom{2m}{2j}bd+\binom{2m}{2j+1}ac\right) \!F_{s\left( 2m\left(
m+n\right) +p_{1}+\cdots +p_{2m}+3m+k\right) +l}\left( x\right)%
\end{array}%
\!\!\right)  \notag \\
&amp;&amp;+a^{2m}\!\left( x^{2}+4\right) ^{m}\!\left( 
\begin{array}{c}
cL_{s\left( 2m\left( m+n\right) +p_{1}+\cdots +p_{2m}+3m+k\right)
+l}\!\left( x\right) \\ 
+dF_{s\left( 2m\left( m+n\right) +p_{1}+\cdots +p_{2m}+3m+k\right)
+l}\!\left( x\right)%
\end{array}%
\right) \!.  \notag
\end{eqnarray}

More concretely, if in (\ref{3.7}) we set $c=0$, $p_{i}=q_{i}-n-m$, $k=1$
and $l=0$, we can write the resulting expression as an addition identity,
namely%
\begin{eqnarray}
&amp;&amp;\sum_{r=0}^{2m}\frac{\left( -1\right) ^{\frac{\left( sr+2\left( s+1\right)
\right) \left( r+1\right) }{2}+s\left( m+1\right) +1}}{\left( F_{r}\left(
x\right) !\right) _{s}\!\left( F_{2m-r}\left( x\right) !\right) _{s}\!}%
\prod_{i=1}^{2m+1}\left( aL_{s\left( q_{i}+r-m\right) }\left( x\right)
+bF_{s\left( q_{i}+r-m\right) }\left( x\right) \right)  \label{3.9} \\
&amp;=&amp;\!\!\sum_{j=0}^{m}a^{2j}b^{2m-2j}\left( x^{2}+4\right) ^{j}\!\left( 
\binom{2m+1}{2j+1}aL_{s\left( i_{1}+\cdots +i_{2m+1}\right) }\left( x\right)
+\binom{2m+1}{2j}bF_{s\left( i_{1}+\cdots +i_{2m+1}\right) }\left( x\right)
\right) \!.  \notag
\end{eqnarray}

Similarly, if in (\ref{3.8}) we set $c=0$, $p_{i}=q_{i}-n-m+1$, $k=-4m$ and $%
l=0$, we obtain%
\begin{eqnarray}
&amp;&amp;\sum_{r=0}^{2m-1}\frac{\left( -1\right) ^{\frac{\left( sr+2\left(
s+1\right) \right) \left( r+1\right) }{2}+s\left( r+m+1\right) }}{\left(
F_{r}\left( x\right) !\right) _{s}\left( F_{2m-1-r}\left( x\right) !\right)
_{s}}\prod_{i=1}^{2m}\left( aL_{s\left( q_{i}+r+1-m\right) }\left( x\right)
+bF_{s\left( q_{i}+r+1-m\right) }\left( x\right) \right)  \label{3.10} \\
&amp;=&amp;\sum_{j=0}^{m-1}a^{2j}b^{2m-2j-1}\!\left( x^{2}+4\right) ^{k}\!\left( 
\binom{2m}{2j+1}aL_{s\left( i_{1}+\cdots +i_{2m}+m\right) }\left( x\right) +%
\binom{2m}{2j}bF_{s\left( i_{1}+\cdots +i_{2m}+m\right) }\left( x\right)
\right)  \notag \\
&amp;&amp;+a^{2m}\left( x^{2}+4\right) ^{m}F_{s\left( i_{1}+\cdots +i_{2m}+m\right)
}\left( x\right) .  \notag
\end{eqnarray}

Some examples are the following:%
\begin{eqnarray}
&amp;&amp;\frac{\left( -1\right) ^{s+1}}{F_{s}\left( x\right) }\left(
aL_{sq_{1}}\left( x\right) +bF_{sq_{1}}\left( x\right) \right) \left(
aL_{sq_{2}}\left( x\right) +bF_{sq_{2}}\left( x\right) \right)  \label{3.11}
\\
&amp;&amp;+\frac{1}{F_{s}\left( x\right) }\left( aL_{s\left( q_{1}+1\right) }\left(
x\right) +bF_{s\left( q_{1}+1\right) }\left( x\right) \right) \left(
aL_{s\left( q_{2}+1\right) }\left( x\right) +bF_{s\left( q_{2}+1\right)
}\left( x\right) \right)  \notag \\
&amp;=&amp;2abL_{s\left( q_{1}+q_{2}+1\right) }\left( x\right) +\left(
b^{2}+a^{2}\left( x^{2}+4\right) \right) F_{s\left( q_{1}+q_{2}+1\right)
}\left( x\right) .  \notag
\end{eqnarray}

\begin{eqnarray}
&amp;&amp;\frac{\left( -1\right) ^{s}\left( 
\begin{array}{c}
aL_{s\left( q_{1}-1\right) }\left( x\right) \\ 
+bF_{s\left( q_{1}-1\right) }\left( x\right)%
\end{array}%
\right) \left( 
\begin{array}{c}
aL_{s\left( q_{2}-1\right) }\left( x\right) \\ 
+bF_{s\left( q_{2}-1\right) }\left( x\right)%
\end{array}%
\right) \left( 
\begin{array}{c}
aL_{s\left( q_{3}-1\right) }\left( x\right) \\ 
+bF_{s\left( q_{3}-1\right) }\left( x\right)%
\end{array}%
\right) }{F_{s}\left( x\right) F_{2s}\left( x\right) }  \label{3.12-1} \\
&amp;&amp;+\frac{\left( -1\right) ^{s+1}\left( 
\begin{array}{c}
aL_{sq_{1}}\left( x\right) \\ 
+bF_{sq_{1}}\left( x\right)%
\end{array}%
\right) \left( 
\begin{array}{c}
aL_{sq_{2}}\left( x\right) \\ 
+bF_{sq_{2}}\left( x\right)%
\end{array}%
\right) \left( 
\begin{array}{c}
aL_{sq_{3}}\left( x\right) \\ 
+bF_{sq_{3}}\left( x\right)%
\end{array}%
\right) }{F_{s}^{2}\left( x\right) }  \notag \\
&amp;&amp;+\frac{\left( 
\begin{array}{c}
aL_{s\left( q_{1}+1\right) }\left( x\right) \\ 
+bF_{s\left( q_{1}+1\right) }\left( x\right)%
\end{array}%
\right) \left( 
\begin{array}{c}
aL_{s\left( q_{2}+1\right) }\left( x\right) \\ 
+bF_{s\left( q_{2}+1\right) }\left( x\right)%
\end{array}%
\right) \left( 
\begin{array}{c}
aL_{s\left( q_{3}+1\right) }\left( x\right) \\ 
+bF_{s\left( q_{3}+1\right) }\left( x\right)%
\end{array}%
\right) }{F_{s}\left( x\right) F_{2s}\left( x\right) }  \notag \\
&amp;=&amp;\left( 3b^{2}+a^{2}\left( x^{2}+4\right) \right) aL_{s\left(
q_{1}+q_{2}+q_{3}\right) }\left( x\right) +\left( b^{2}+3a^{2}\left(
x^{2}+4\right) \right) bF_{s\left( q_{1}+q_{2}+q_{3}\right) }\left( x\right)
\!.  \notag
\end{eqnarray}

Formula (\ref{3.11}) includes $\left( -1\right) ^{s+1}F_{sn}^{2}\left(
x\right) +F_{s\left( n+1\right) }^{2}\left( x\right) =F_{s}\left( x\right)
F_{s\left( 2n+1\right) }\left( x\right) $ (the case $s=x=1$ is a famous
identity). Formula (\ref{3.12-1}) includes%
\begin{eqnarray*}
F_{s}^{2}\left( x\right) F_{s\left( a+b+c\right) }\left( x\right) &amp;=&amp;\frac{1%
}{L_{s}\left( x\right) }F_{s\left( a+1\right) }\left( x\right) \!F_{s\left(
b+1\right) }\left( x\right) \!F_{s\left( c+1\right) }\left( x\right) +\left(
-1\right) ^{s+1}\!F_{sa}\left( x\right) \!F_{sb}\left( x\right)
\!F_{sc}\left( x\right) \\
&amp;&amp;+\frac{\left( -1\right) ^{s}}{L_{s}\left( x\right) }F_{s\left( a-1\right)
}\left( x\right) F_{s\left( b-1\right) }\left( x\right) F_{s\left(
c-1\right) }\left( x\right) .
\end{eqnarray*}%
(The case $s=x=1$ is also a known identity. See \cite[identity 45, p.\ 89]{K} and
\cite[p.\ 5]{Jh}.)

Finally, we want to call the attention to the fact that (\ref{3.7}) and (\ref%
{3.8}) include identities that express $s$-Fibonacci and $s$-Lucas
polynomial sequences as linear combinations of $s$-polyfibonomial sequences.

Consider (\ref{3.7}) with $a=0$, $k=1$, $l=0$, $p_{i}=i$, $i=1,2,\ldots
,2m+1 $, and $n$ shifted to $n-2m-1$. If $c=0$ and $bd\neq 0$ we can write
the identity as%
\begin{eqnarray}
&amp;&amp;F_{\left( 2m+1\right) sn}\left( x\right)  \label{5.10.1} \\
&amp;=&amp;\left( -1\right) ^{s\left( m+1\right) +1}F_{\left( 2m+1\right) s}\left(
x\right) \sum_{r=0}^{2m}\left( -1\right) ^{\frac{\left( sr+2\left(
s+1\right) \right) \left( r+1\right) }{2}}\dbinom{2m}{r}_{F_{s}\left(
x\right) }\dbinom{n+r}{2m+1}_{F_{s}\left( x\right) },  \notag
\end{eqnarray}%
and if $d=0$ and $bc\neq 0$ we can write 
\begin{eqnarray}
&amp;&amp;L_{\left( 2m+1\right) sn}\left( x\right)  \label{5.11.1} \\
&amp;=&amp;\left( -1\right) ^{s\left( m+1\right) }\sum_{r=0}^{2m+1}\left( -1\right)
^{\frac{\left( sr+2\left( s+1\right) \right) \left( r+1\right) }{2}%
}L_{s\left( 2m+1-r\right) }\left( x\right) \!\dbinom{2m+1}{r}_{F_{s}\left(
x\right) }\!\dbinom{n+r}{2m+1}_{F_{s}\left( x\right) }.  \notag
\end{eqnarray}

Similarly, if in (\ref{3.8}) we set $a=0$, $k=-4m$, $l=0$, $p_{i}=i$, $%
i=1,2,\ldots ,2m$, and shift $n$ to $n-2m$, we obtain if $c=0$ and $bd\neq 0$%
\begin{equation}
F_{2smn}\left( x\right) =\left( -1\right) ^{s\left( m+1\right)
}F_{2ms}\left( x\right) \sum_{r=0}^{2m-1}\left( -1\right) ^{\frac{\left(
sr+2\left( s+1\right) \right) \left( r+1\right) }{2}+sr}\dbinom{2m-1}{r}%
_{F_{s}\left( x\right) }\dbinom{n+r}{2m}_{F_{s}\left( x\right) },
\label{5.14.2}
\end{equation}%
and if $d=0$ and $bc\neq 0$ 
\begin{equation}
L_{2smn}\left( x\right) =\left( -1\right) ^{s\left( m+1\right)
+1}\sum_{r=0}^{2m}\left( -1\right) ^{\frac{\left( sr+2\left( s+1\right)
\right) \left( r+1\right) }{2}+sr}L_{s\left( 2m-r\right) }\left( x\right) 
\dbinom{2m}{r}_{F_{s}\left( x\right) }\dbinom{n+r}{2m}_{F_{s}\left( x\right)
}.  \label{5.15}
\end{equation}

(Formulas (\ref{5.10.1}), (\ref{5.11.1}), (\ref{5.14.2}) and (\ref{5.15})
when $x=1$, were obtained in \cite{P2} by using different methods.)

Some examples are%
\begin{equation*}
F_{2sn}\left( x\right) =F_{2s}\left( x\right) \left( \dbinom{n+1}{2}%
_{F_{s}\left( x\right) }+\left( -1\right) ^{s+1}\dbinom{n}{2}_{F_{s}\left(
x\right) }\right) .
\end{equation*}%
\begin{equation}
F_{3sn}\left( x\right) =F_{3s}\left( x\right) \left( \dbinom{n+2}{3}%
_{F_{s}\left( x\right) }+\left( -1\right) ^{s+1}L_{s}\left( x\right) \dbinom{%
n+1}{3}_{F_{s}\left( x\right) }+\left( -1\right) ^{s}\dbinom{n}{3}%
_{F_{s}\left( x\right) }\right) .  \notag
\end{equation}%
\begin{eqnarray*}
L_{3sn}\left( x\right) &amp;=&amp;2\dbinom{n+3}{3}_{F_{s}\left( x\right) }-\frac{%
F_{3s}\left( x\right) }{F_{s}\left( x\right) }L_{s}\left( x\right) \dbinom{%
n+2}{3}_{F_{s}\left( x\right) } \\
&amp;&amp;+\left( -1\right) ^{s}\frac{F_{3s}\left( x\right) }{F_{s}\left( x\right) }%
L_{2s}\left( x\right) \dbinom{n+1}{3}_{F_{s}\left( x\right) }+\left(
-1\right) ^{s+1}L_{3s}\left( x\right) \dbinom{n}{3}_{F_{s}\left( x\right) }.
\end{eqnarray*}%
\begin{eqnarray*}
L_{4sn}\left( x\right) &amp;=&amp;2\dbinom{n+4}{4}_{F_{s}\left( x\right)
}-L_{2s}\left( x\right) L_{s}^{2}\left( x\right) \dbinom{n+3}{4}%
_{F_{s}\left( x\right) } \\
&amp;&amp;+\left( -1\right) ^{s}\frac{F_{3s}\left( x\right) }{F_{s}\left( x\right) }%
L_{2s}^{2}\left( x\right) \dbinom{n+2}{4}_{F_{s}\left( x\right) } \\
&amp;&amp;+\left( -1\right) ^{s+1}L_{s}\left( x\right) L_{2s}\left( x\right)
L_{3s}\left( x\right) \dbinom{n+1}{4}_{F_{s}\left( x\right) }+L_{4s}\left(
x\right) \dbinom{n}{4}_{F_{s}\left( x\right) }.
\end{eqnarray*}

\section{Appendix. Proofs of some identities}

In this appendix we present proofs of identities (\ref{2.1.1}) and (\ref%
{2.6.1}) used in propositions (\ref{Prop2.1}) and (\ref{Prop2.2}),
respectively. We will use some identities related to the so-called \textit{%
index-reduction formula}%
\begin{equation}
F_{a}\left( x\right) F_{b}\left( x\right) -F_{c}\left( x\right) F_{d}\left(
x\right) =\left( -1\right) ^{r}\left( F_{a-r}\left( x\right) F_{b-r}\left(
x\right) -F_{c-r}\left( x\right) F_{d-r}\left( x\right) \right) ,  \label{A1}
\end{equation}%
where $a,b,c,d,r\in \mathbb{Z}$ and $a+b=c+d$. Johnson \cite{Jh} proves (\ref%
{A1}) in the case $x=1$, but the argument can be adapted to our case (of
Fibonacci polynomials) by replacing the matrix 
\begin{equation*}
Q=\left[ 
\begin{array}{cc}
1 &amp; 1 \\ 
1 &amp; 0%
\end{array}%
\right] ,
\end{equation*}%
by the matrix%
\begin{equation*}
Q\left( x\right) =\left[ 
\begin{array}{cc}
x &amp; 1 \\ 
1 &amp; 0%
\end{array}%
\right] ,
\end{equation*}%
for which we have%
\begin{equation*}
\left[ 
\begin{array}{c}
F_{k}\left( x\right)  \\ 
F_{k-1}\left( x\right) 
\end{array}%
\right] =\left[ 
\begin{array}{cc}
x &amp; 1 \\ 
1 &amp; 0%
\end{array}%
\right] ^{k-1}\left[ 
\begin{array}{c}
1 \\ 
0%
\end{array}%
\right] .
\end{equation*}

We will use also the identity%
\begin{equation}
F_{a+b}\left( x\right) -\left( -1\right) ^{b}F_{a-b}\left( x\right)
=L_{a}\left( x\right) F_{b}\left( x\right) ,  \label{A2}
\end{equation}%
which proof is trivial (by using Binet's formulas).

\textit{Proof of (\ref{2.1.1}).}\textbf{\ }We can write (\ref{2.1.1}) as%
\begin{eqnarray}
&amp;&amp;\left( -1\right) ^{sr}F_{sr}\left( x\right) F_{s\left( r-1\right) }\left(
x\right) +F_{sr}\left( x\right) F_{s\left( 2m+2-r\right) }\left( x\right)
L_{s\left( 2m+1\right) }\left( x\right)  \label{A3} \\
&amp;&amp;+\left( -1\right) ^{sr}F_{s\left( 2m+2-r\right) }\left( x\right)
F_{s\left( 2m+1-r\right) }\left( x\right)  \notag \\
&amp;=&amp;F_{s\left( 2m+2\right) }\left( x\right) F_{s\left( 2m+1\right) }\left(
x\right) .  \notag
\end{eqnarray}

Let us consider the left-hand side of (\ref{A3})%
\begin{eqnarray*}
LHS\left( s,r,m,x\right) &amp;=&amp;\left( -1\right) ^{sr}F_{sr}\left( x\right)
F_{s\left( r-1\right) }\left( x\right) +F_{sr}\left( x\right) F_{s\left(
2m+2-r\right) }\left( x\right) L_{s\left( 2m+1\right) }\left( x\right) \\
&amp;&amp;+\left( -1\right) ^{sr}F_{s\left( 2m+2-r\right) }\left( x\right)
F_{s\left( 2m+1-r\right) }\left( x\right) .
\end{eqnarray*}

Observe that the right-hand side of (\ref{A3}) does not depend on $r$, and
that (\ref{A3}) is trivial for $r=0$ and $r=2m+2$. Thus, to prove this
identity it is sufficient to prove that $LHS\left( s,r,m,x\right) =LHS\left(
s,r+1,m,x\right) $.

We have%
\begin{eqnarray*}
&amp;&amp;LHS\left( s,r,m,x\right) -LHS\left( s,r+1,m,x\right) \\
&amp;=&amp;\left( -1\right) ^{sr}F_{sr}\left( x\right) F_{s\left( r-1\right) }\left(
x\right) +F_{sr}\left( x\right) F_{s\left( 2m+2-r\right) }\left( x\right)
L_{s\left( 2m+1\right) }\left( x\right) \\
&amp;&amp;+\left( -1\right) ^{sr}F_{s\left( 2m+2-r\right) }\left( x\right)
F_{s\left( 2m+1-r\right) }\left( x\right) \\
&amp;&amp;-\left( -1\right) ^{s\left( r+1\right) }F_{s\left( r+1\right) }\left(
x\right) F_{sr}\left( x\right) -F_{s\left( r+1\right) }\left( x\right)
F_{s\left( 2m+1-r\right) }\left( x\right) L_{s\left( 2m+1\right) }\left(
x\right) \\
&amp;&amp;-\left( -1\right) ^{s\left( r+1\right) }F_{s\left( 2m+1-r\right) }\left(
x\right) F_{s\left( 2m-r\right) }\left( x\right) \\
&amp;=&amp;-\left( -1\right) ^{s\left( r+1\right) }F_{sr}\left( x\right) \left(
F_{s\left( r+1\right) }\left( x\right) -\left( -1\right) ^{s}F_{s\left(
r-1\right) }\left( x\right) \right) \\
&amp;&amp;+\left( -1\right) ^{sr}F_{s\left( 2m+1-r\right) }\left( x\right) \left(
F_{s\left( 2m+2-r\right) }\left( x\right) -\left( -1\right) ^{s}F_{s\left(
2m-r\right) }\left( x\right) \right) \\
&amp;&amp;-L_{s\left( 2m+1\right) }\left( x\right) \left( F_{s\left( r+1\right)
}\left( x\right) F_{s\left( 2m+1-r\right) }\left( x\right) -F_{sr}\left(
x\right) F_{s\left( 2m+2-r\right) }\left( x\right) \right) .
\end{eqnarray*}

Now use (\ref{A2}) to write%
\begin{eqnarray*}
F_{s\left( r+1\right) }\left( x\right) -\left( -1\right) ^{s}F_{s\left(
r-1\right) }\left( x\right) &amp;=&amp;L_{sr}\left( x\right) F_{s}\left( x\right) ,
\\
F_{s\left( 2m+2-r\right) }\left( x\right) -\left( -1\right) ^{s}F_{s\left(
2m-r\right) }\left( x\right) &amp;=&amp;L_{s\left( 2m+1-r\right) }\left( x\right)
F_{s}\left( x\right) ,
\end{eqnarray*}%
and use (\ref{A1}) to write%
\begin{equation*}
F_{s\left( r+1\right) }\left( x\right) F_{s\left( 2m+1-r\right) }\left(
x\right) -F_{sr}\left( x\right) F_{s\left( 2m+2-r\right) }\left( x\right)
=\left( -1\right) ^{sr}F_{s}\left( x\right) F_{s\left( 2m+1-2r\right)
}\left( x\right) .
\end{equation*}

Then, by using again (\ref{A2}) we have 
\begin{eqnarray*}
&amp;&amp;LHS\left( s,r,m,x\right) -LHS\left( s,r+1,m,x\right) \\
&amp;=&amp;-\left( -1\right) ^{s\left( r+1\right) }F_{sr}\left( x\right) F_{s}\left(
x\right) L_{sr}\left( x\right) +\left( -1\right) ^{sr}F_{s\left(
2m+1-r\right) }\left( x\right) F_{s}\left( x\right) L_{s\left( 2m+1-r\right)
}\left( x\right) \\
&amp;&amp;-L_{s\left( 2m+1\right) }\left( x\right) \left( -1\right) ^{sr}F_{s}\left(
x\right) F_{s\left( 2m+1-2r\right) }\left( x\right) \\
&amp;=&amp;\left( -1\right) ^{sr}F_{s}\left( x\right) \left( \left( -1\right)
^{s+1}F_{2sr}\left( x\right) +F_{2s\left( 2m+1-r\right) }-L_{s\left(
2m+1\right) }\left( x\right) F_{s\left( 2m+1-2r\right) }\left( x\right)
\right) \\
&amp;=&amp;0,
\end{eqnarray*}%
as wanted.

The proof of (\ref{2.10.1}) is similar.

\textit{Proof of (\ref{2.6.1}).}\textbf{\ }We want to prove the identity%
\begin{eqnarray*}
&amp;&amp;\left( -1\right) ^{s\left( r+1\right) }H_{s\left( k-1\right) +l}\left(
x\right) F_{sr}\left( x\right) +H_{s\left( 2m+k\right) +l}\left( x\right)
F_{s\left( 2m+1-r\right) }\left( x\right) \\
&amp;=&amp;F_{s\left( 2m+1\right) }\left( x\right) H_{s\left( 2m-r+k\right)
+l}\left( x\right) .
\end{eqnarray*}

By using that $H_{n}\left( x\right) =H_{0}\left( x\right) F_{n-1}\left(
x\right) +H_{1}\left( x\right) F_{n}\left( x\right) $, we can write (\ref%
{2.6.1}) as the two following identities%
\begin{eqnarray*}
&amp;&amp;\left( -1\right) ^{s\left( r+1\right) }F_{s\left( k-1\right) +l-1}\left(
x\right) F_{sr}\left( x\right) +F_{s\left( 2m+k\right) +l-1}\left( x\right)
F_{s\left( 2m+1-r\right) }\left( x\right) \\
&amp;=&amp;F_{s\left( 2m+1\right) }\left( x\right) F_{s\left( 2m-r+k\right)
+l-1}\left( x\right) , \\
&amp;&amp;\left( -1\right) ^{s\left( r+1\right) }F_{s\left( k-1\right) +l}\left(
x\right) F_{sr}\left( x\right) +F_{s\left( 2m+k\right) +l}\left( x\right)
F_{s\left( 2m+1-r\right) }\left( x\right) \\
&amp;=&amp;F_{s\left( 2m+1\right) }\left( x\right) F_{s\left( 2m-r+k\right)
+l}\left( x\right) ,
\end{eqnarray*}%
and we notice that these are true by (\ref{A1}).

The proof of (\ref{2.9.1}) is similar.

\section{Acknowledgments}

I wish to thank the anonymous referee for his/her careful reading of the
first version of this article, and his/her valuable comments and
suggestions, which certainly contributed to improve the final version
presented here.

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\noindent 2010 {\it Mathematics Subject Classification}:
Primary 11Y55; Secondary 11B39.

\noindent \emph{Keywords: } 
Fibonacci polynomials, $s$-polygibonomials, addition formulas.

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\noindent (Concerned with sequences
\seqnum{A000032},
\seqnum{A000045},
\seqnum{A010048}, and
\seqnum{A034801}.)

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\vspace*{+.1in}
\noindent
Received  March 3 2011;
revised version received  July 13 2011.
Published in {\it Journal of Integer Sequences}, September 5 2011.

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\noindent
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