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\begin{center}
\vskip 1cm{\LARGE\bf On $s$-Fibonomials
}
\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}
For a given natural number $s$, we study $s$-Fibonacci sequences $F_{sn}$
and the corresponding $s$-Fibonomial coefficients $\binom{n}{p}_{F_{s}}=%
\frac{F_{sn}F_{s\left( n-1\right) }\cdots F_{s\left( n-p+1\right) }}{%
F_{s}F_{2s}\cdots F_{ps}}$. We obtain the $Z$ transform of products of
powers of $s$-Fibonacci sequences. Since the $s$-Fibonomials are involved in
this $Z$ transform, we obtain from it some new results involving products of
sequences of the type $F_{sn+m}^{k}$ together with $s$-Fibonomials.
\end{abstract}


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\section{\label{Sec1}Introduction}

We use $\mathbb{N}$ for the natural numbers and $\mathbb{N}^{\prime }$ for $%
\mathbb{N\cup }\left\{ 0\right\} $. We will be using without further
comments the basic formulas for the Fibonacci sequence $F_{n}$ and Lucas
sequence $L_{n}$ (\seqnum{A000045} and \seqnum{A000032} of 
Sloane's {\it Encyclopedia},
respectively), such as $F_{n}=\frac{1}{\sqrt{5}}\left(
\alpha ^{n}-\beta ^{n}\right) $ and $L_{n}=\alpha ^{n}+\beta ^{n}$, where $%
\alpha =\frac{1}{2}\left( 1+\sqrt{5}\right) $ and $\beta =\frac{1}{2}\left(
1-\sqrt{5}\right) $, together with the well-known relations involving $%
\alpha $ and $\beta $. What we use about Fibonacci identities is contained
in the references \cite{K} and \cite{V}.

Throughout this work, $s$ will denote a natural number.

For a given Fibonacci number $F_{n}$, $n\in \mathbb{N}$, the $s$-\textit{%
Fibonacci factorial} of $F_{n}$, denoted by $\left( F_{n}!\right) _{s}$, is
defined as $\left( F_{n}!\right) _{s}=F_{sn}F_{s\left( n-1\right) }\cdots
F_{s}$. Given $n\in \mathbb{N}^{\prime }$ and $k\in \left\{ 0,1,\ldots
,n\right\} $, the $s$-\textit{Fibonomial coefficient }$\binom{n}{k}_{F_{s}}$
is defined by $\binom{n}{0}_{F_{s}}=\binom{n}{n}_{F_{s}}=1$, and%
\begin{equation*}
\binom{n}{k}_{F_{s}}=\frac{\left( F_{n}!\right) _{s}}{\left( F_{k}!\right)
_{s}\left( F_{n-k}!\right) _{s}},\text{ \ \ }k=1,2,\ldots ,n-1,\text{ \ }
\end{equation*}%
that is%
\begin{equation}
\binom{n}{k}_{F_{s}}=\frac{F_{sn}F_{s\left( n-1\right) }\cdots F_{s\left(
n-k+1\right) }}{F_{s}F_{2s}\cdots F_{ks}}.  \label{1.1}
\end{equation}

(The $1$-Fibonomials are called simply Fibonomials.) It is clear that $%
\binom{n}{k}_{F_{s}}=\binom{n}{n-k}_{F_{s}}$. From the identity $F_{s\left(
n-k\right) +1}F_{sk}+F_{sk-1}F_{s\left( n-k\right) }=F_{sn}$ one can see at
once that%
\begin{equation}
\binom{n}{k}_{F_{s}}=F_{s\left( n-k\right) +1}\binom{n-1}{k-1}%
_{F_{s}}+F_{sk-1}\binom{n-1}{k}_{F_{s}},  \label{1.2}
\end{equation}%
which shows (with a simple induction argument) that $s$-Fibonomials are
integers. (See \cite{Hog}.)

Some examples are the following (as triangular arrays, where the lines
correspond to $n\in \mathbb{N}^{\prime }$, the columns to $k$, $0\leq k\leq
n $, and the array is filled-out with formula (\ref{1.1}) and/or rule (\ref%
{1.2}):

(a) $s=1$ (the Fibonomials: \seqnum{A010048} of 
Sloane's \textit{Encyclopedia}):
\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; 1_{\bigstar } &amp;  &amp; 1_{\flat } &amp;  &amp;  &amp;  &amp;  \\ 
&amp;  &amp;  &amp; 1 &amp;  &amp; 2_{\triangle } &amp;  &amp; 2_{\flat } &amp;  &amp; 1 &amp;  &amp;  &amp;  \\ 
&amp;  &amp; 1 &amp;  &amp; 3_{\maltese } &amp;  &amp; 6_{\flat } &amp;  &amp; 3 &amp;  &amp; 1 &amp;  &amp;  \\ 
&amp; 1 &amp;  &amp; 5 &amp;  &amp; 15_{\flat } &amp;  &amp; 15 &amp;  &amp; 5 &amp;  &amp; 1 &amp;  \\ 
1 &amp;  &amp; 8 &amp;  &amp; 40_{\flat } &amp;  &amp; 60 &amp;  &amp; 40 &amp;  &amp; 8 &amp;  &amp; 1 \\ 
&amp;  &amp;  &amp;  &amp; \cdot &amp;  &amp; \cdot &amp;  &amp; \cdot &amp;  &amp;  &amp;  &amp; 
\end{array}%
\end{equation*}

(b) $s=2$ (\seqnum{A034801} of 
Sloane's \textit{Encyclopedia}):
\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; 3_{\bigstar } &amp;  &amp; 1_{\flat } &amp;  &amp;  &amp;  &amp;  \\ 
&amp;  &amp;  &amp; 1 &amp;  &amp; 8_{\triangle } &amp;  &amp; 8_{\flat } &amp;  &amp; 1 &amp;  &amp;  &amp;  \\ 
&amp;  &amp; 1 &amp;  &amp; 21_{\maltese } &amp;  &amp; 56_{\flat } &amp;  &amp; 21 &amp;  &amp; 1 &amp;  &amp;  \\ 
&amp; 1 &amp;  &amp; 55 &amp;  &amp; 385_{\flat } &amp;  &amp; 385 &amp;  &amp; 55 &amp;  &amp; 1 &amp;  \\ 
1 &amp;  &amp; 144 &amp;  &amp; 2640_{\flat } &amp;  &amp; 6930 &amp;  &amp; 2640 &amp;  &amp; 144 &amp;  &amp; 1 \\ 
&amp;  &amp;  &amp;  &amp; \cdot &amp;  &amp; \cdot &amp;  &amp; \cdot &amp;  &amp;  &amp;  &amp; 
\end{array}%
\end{equation*}

(c) $s=3$ (\seqnum{A034802} of Sloane's
\textit{Encyclopedia}):
\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; 4_{\bigstar } &amp;  &amp; 1_{\flat } &amp;  &amp;  &amp;  &amp;  \\ 
&amp;  &amp;  &amp; 1 &amp;  &amp; 17_{\triangle } &amp;  &amp; 17_{\flat } &amp;  &amp; 1 &amp;  &amp;  &amp;  \\ 
&amp;  &amp; 1 &amp;  &amp; 72_{\maltese } &amp;  &amp; 306_{\flat } &amp;  &amp; 72 &amp;  &amp; 1 &amp;  &amp;  \\ 
&amp; 1 &amp;  &amp; 305 &amp;  &amp; 5490_{\flat } &amp;  &amp; 5490 &amp;  &amp; 305 &amp;  &amp; 1 &amp;  \\ 
1 &amp;  &amp; 1292 &amp;  &amp; 98515_{\flat } &amp;  &amp; 417240 &amp;  &amp; 98515 &amp;  &amp; 1292 &amp;  &amp; 1 \\ 
&amp;  &amp;  &amp;  &amp; \cdot &amp;  &amp; \cdot &amp;  &amp; \cdot &amp;  &amp;  &amp;  &amp; 
\end{array}%
\end{equation*}

We have marked some patterns that can be recognized as general relations
among the $s$-Fibonomials $\binom{n}{1}_{F_{s}}$ (that is, among the
quotients $\frac{F_{sn}}{F_{s}}$, as $s$-sequences ---with fixed $n$) as
follows:

\begin{itemize}
\item (a)%
\begin{equation*}
\dbinom{2^{n}}{1}_{F_{s}}=\frac{F_{2\left( 2^{n-1}s\right) }}{F_{s}}%
=L_{2^{n-1}s}\frac{F_{2^{n-1}s}}{F_{s}}=L_{2^{n-1}s}\dbinom{2^{n-1}}{1}%
_{F_{s}},
\end{equation*}%
and then%
\begin{equation}
\dbinom{2^{n}}{1}_{F_{s}}=L_{s}L_{2s}\cdots L_{2^{n-1}s}.  \label{1.3}
\end{equation}

Thus, for $n=1$ we have the sequence of Lucas numbers $\binom{2}{1}_{F_{s}}=%
\frac{F_{2s}}{F_{s}}=L_{s}$ (corresponding to the mark $\bigstar $ in
previous tables), and for $n=2$ we have the sequence $\binom{4}{1}_{F_{s}}=%
\frac{F_{4s}}{F_{s}}=L_{s}L_{2s}=\left( 3,21,72,\ldots \right) $ 
(\seqnum{A083564} of 
Sloane's \textit{Encyclopedia}; 
corresponding to the
mark $\maltese $ in previous tables).

\item (b)%
\begin{eqnarray*}
\dbinom{3^{n}}{1}_{F_{s}} &amp;=&amp;\frac{F_{3^{n}s}}{F_{s}}=\frac{\left( \alpha
^{3^{n-1}s}\right) ^{3}-\left( \beta ^{3^{n-1}s}\right) ^{3}}{F_{s}}=\left(
\alpha ^{3^{n-1}2s}+\beta ^{3^{n-1}2s}+\left( -1\right) ^{3^{n-1}s}\right) 
\frac{F_{3^{n-1}s}}{F_{s}} \\
&amp;=&amp;\left( L_{3^{n-1}2s}+\left( -1\right) ^{s}\right) \dbinom{3^{n-1}}{1}%
_{F_{s}},
\end{eqnarray*}%
and then%
\begin{equation}
\dbinom{3^{n}}{1}_{F_{s}}=\left( L_{3^{n-1}2s}+\left( -1\right) ^{s}\right)
\left( L_{3^{n-2}2s}+\left( -1\right) ^{s}\right) \cdots \left(
L_{2s}+\left( -1\right) ^{s}\right) .  \label{1.4}
\end{equation}

Thus for $n=1$ we have the sequence
$\binom{3}{1}_{F_{s}}=\frac{F_{3s}}{F_{s}}=L_{2s}+\left( -1\right)
^{s}=\left( 2,8,17,\ldots \right) $ (\seqnum{A047946} of 
Sloane's \textit{Encyclopedia}; corresponding to the mark
$\triangle $ in previous tables).

\item (c)%
\begin{equation*}
\dbinom{2^{m}3^{n}}{1}_{F_{s}}=\frac{F_{2\left( 2^{m-1}3^{n}s\right) }}{F_{s}%
}=L_{2^{m-1}3^{n}s}\frac{F_{2^{m-1}3^{n}s}}{F_{s}}=L_{2^{m-1}3^{n}s}\dbinom{%
2^{m-1}3^{n}}{1}_{F_{s}},
\end{equation*}%
and then%
\begin{equation}
\dbinom{2^{m}3^{n}}{1}_{F_{s}}=L_{2^{m-1}3^{n}s}L_{2^{m-2}3^{n}s}\cdots
L_{3^{n}s}\dbinom{3^{n}}{1}_{F_{s}}.  \label{1.5}
\end{equation}
\end{itemize}

Finally, note that for fixed $s$, the sequence $\left( 1,\bigstar
,\blacktriangle ,\maltese ,\ldots \right) $ corresponds to $\binom{n}{1}%
_{F_{s}}=\frac{F_{ns}}{F_{s}}$, $n\geq 1$. Thus, for $s=1$ we have the
sequence $\frac{F_{n}}{F_{1}}=F_{n}$. For $s=2$ we have the sequence $\frac{%
F_{2n}}{F_{2}}=F_{2n}=\left( 1,3,8,21,\ldots \right) $ (\seqnum{A001906} of 
Sloane's {\it Encyclopedia}), and for $s=3$ we have the
sequence $\frac{F_{3n}}{F_{3}}=\frac{1}{2}F_{3n}=\left( 1,4,17,72,\ldots
\right) $ (\seqnum{A001076} of Sloane's {\it Encyclopedia}).

\begin{remark}
For fixed $s$, the sequence $\binom{n}{2}_{F_{s}}=\frac{F_{sn}F_{s\left(
n-1\right) }}{F_{s}F_{2s}}$, $n\geq 2$, (marked with $\flat $ for each $%
s=1,2,3$ in previous tables) is, in the case of Fibonomials ($s=1$), the
famous golden rectangle sequence $\binom{n}{2}_{F}=F_{n}F_{n-1}=\left(
1,2,6,15,\ldots \right)$ (\seqnum{A001654} of 
Sloane's {\it Encyclopedia}).
The corresponding
sequences for the cases $s=2$ and $s=3$ are also included in \textit{the
mentioned Encyclopedia }(\seqnum{A092521} and \seqnum{A156085}, respectively), but there are
not references to the formulas in which they appear in this work, namely $%
\binom{n}{2}_{F_{2}}=\frac{1}{3}F_{2n}F_{2\left( n-1\right) }=\left(
1,8,56,385,\ldots \right) $ and $\binom{n}{2}_{F_{3}}=\frac{1}{16}%
F_{3n}F_{3\left( n-1\right) }=\left( 1,17,306,5490,\ldots \right) $,
respectively.
\end{remark}

We have found just a few references on $s$-Fibonomials, even though Hoggatt 
\cite{Hog} considered them in 1967. Besides the sequences of $s$-Fibonomials
for $s=1,$ $s=2$ and $s=3$ included in Sloane's
{\it Encyclopedia} 
mentioned above, Gould \cite{Go} studied divisibility
properties of $s$-Fibonomials.

The context in which $s$-Fibonomials are studied in this article is related
to $Z$ transforms (and then to generating functions) of certain Fibonacci
sequences. This story begins with the works of Riordan \cite{R}, Carlitz 
\cite{C} and Horadam \cite{Hor} first, and Shannon \cite{Sh} later. From all
these works we know that the $Z$ transform of the sequence $F_{n}^{k}$ is%
\begin{equation}
\mathcal{Z}\left( F_{n}^{k}\right) =z\frac{\sum_{i=0}^{k}\sum_{j=0}^{i}%
\left( -1\right) ^{\frac{j\left( j+1\right) }{2}}\dbinom{k+1}{j}%
_{F}F_{i-j}^{k}z^{k-i}}{\sum_{i=0}^{k+1}\left( -1\right) ^{\frac{i\left(
i+1\right) }{2}}\dbinom{k+1}{i}_{F}z^{k+1-i}},  \label{1.6}
\end{equation}

(See also \cite{S-T1} and \cite{S-T2}.) In a recent work \cite{P}, we proved
the following formula for the $Z$ transform of the sequence $%
F_{n+m_{1}}^{k_{1}}F_{n+m_{2}}^{k_{2}}$:%
\begin{equation}
\mathcal{Z}\left( F_{n+m_{1}}^{k_{1}}F_{n+m_{2}}^{k_{2}}\right) =z\frac{%
\sum_{i=0}^{k_{1}+k_{2}}\sum_{j=0}^{i}\left( -1\right) ^{\frac{j\left(
j+1\right) }{2}}\dbinom{k_{1}+k_{2}+1}{j}%
_{F}F_{m_{1}+i-j}^{k_{1}}F_{m_{2}+i-j}^{k_{2}}z^{k_{1}+k_{2}-i}}{%
\sum_{i=0}^{k_{1}+k_{2}+1}\left( -1\right) ^{\frac{i\left( i+1\right) }{2}}%
\dbinom{k_{1}+k_{2}+1}{i}_{F}z^{k_{1}+k_{2}+1-i}},  \label{1.7}
\end{equation}%
where $m_{1},m_{2}\in \mathbb{Z}$ and $k_{1},k_{2}\in \mathbb{N}^{\prime }$
are given. So (\ref{1.6}) became a particular case of (\ref{1.7}). After our
work \cite{P}, we were able to find that (\ref{1.7}) is in fact a particular
case of a more general result: it turns out that for given $m_{1},m_{2}\in 
\mathbb{Z},s\in \mathbb{N}$ and $k_{1},k_{2},t_{1},t_{2}\in \mathbb{N}%
^{\prime }$, the $Z$ transform of the sequence $%
F_{t_{1}sn+m_{1}}^{k_{1}}F_{t_{2}sn+m_{2}}^{k_{2}}$ is%
\begin{eqnarray}
&amp;&amp;\mathcal{Z}\left( F_{t_{1}sn+m_{1}}^{k_{1}}F_{t_{2}sn+m_{2}}^{k_{2}}\right)
\label{1.8} \\
&amp;=&amp;z\frac{\sum_{i=0}^{t_{1}k_{1}+t_{2}k_{2}}\sum_{j=0}^{i}\left( -1\right) ^{%
\frac{\left( sj+2(s+1)\right) \left( j+1\right) }{2}}\dbinom{%
t_{1}k_{1}+t_{2}k_{2}+1}{j}_{F_{s}}F_{m_{1}+t_{1}s\left( i-j\right)
}^{k_{1}}F_{m_{2}+t_{2}s\left( i-j\right)
}^{k_{2}}z^{t_{1}k_{1}+t_{2}k_{2}-i}}{\sum_{i=0}^{t_{1}k_{1}+t_{2}k_{2}+1}%
\left( -1\right) ^{\frac{\left( si+2(s+1)\right) \left( i+1\right) }{2}}%
\dbinom{t_{1}k_{1}+t_{2}k_{2}+1}{i}_{F_{s}}z^{t_{1}k_{1}+t_{2}k_{2}+1-i}}, 
\notag
\end{eqnarray}%
which means that for $z\in \mathbb{C}$ outside the disk $\overline{D}%
=\left\{ z\in \mathbb{C}:\left\vert z\right\vert \leq \alpha ^{s\left(
t_{1}k_{1}+t_{2}k_{2}\right) }\right\} $, the right-hand side of this
formula equals to%
\begin{equation}
\sum_{n=0}^{\infty }\frac{F_{t_{1}sn+m_{1}}^{k_{1}}F_{t_{2}sn+m_{2}}^{k_{2}}%
}{z^{n}}.  \notag
\end{equation}

We have to mention that Fibonomials can be seen as \emph{combinatorial
objects} (see \cite{B}), and it is natural to expect that $s$-Fibonomials
could enclose more general combinatorial interpretations. Formula (\ref{1.8}%
) is the main result in this work. But we will see that some interesting
results are obtained as consequences of (\ref{1.8}), so its proof is not our
final goal. In section \ref{Sec2} we introduce some basic facts about $Z$
transform that will be used in the rest of the work, together with some
preliminary results to be used mainly in section \ref{Sec3}. Is in section %
\ref{Sec3} where we prove (\ref{1.8}). In section \ref{Sec4} we establish
some corollaries of (\ref{1.8}), and finally in section \ref{Sec5} we do
some comments on the representation of $s$-Fibonomials as linear
combinations of certain \textquotedblleft homogeneous
terms\textquotedblright .

\section{\label{Sec2}Preliminaries}

We will be working with the so called \textquotedblleft $Z$
transform\textquotedblright , of which we recall some basic facts in this
section. (For more details see \cite{G} and \cite{Vi}.) The $Z$ transform
maps complex sequences $\left( a_{0},a_{1},a_{2},\ldots \right) $ into
complex (holomorphic) functions $A:U\subset \mathbb{C\rightarrow C}$ given
by the Laurent series $A\left( z\right) =\sum_{n=0}^{\infty }a_{n}z^{-n}$.
This function $A$ is defined outside the closure $\overline{D}$ of the disk $%
D$ of convergence of the Taylor series $\sum_{n=0}^{\infty }a_{n}z^{n}$. We
will also denote the $Z$ transform of the sequence $\left( a_{n}\right)
_{n=0}^{\infty }$ by $\mathcal{Z}\left( a_{n}\right) $. Some properties of
the $Z$ transform which we will be using throughout this work are the
following: (avoiding the details of regions of convergence)

(a) $\mathcal{Z}$ is linear and injective.

(b) \textit{Advance-shifting property.} For $k\in \mathbb{N}$ we have%
\begin{equation}
\mathcal{Z}\left( a_{n+k}\right) =z^{k}\left( \mathcal{Z}\left( a_{n}\right)
-a_{0}-\frac{a_{1}}{z}-\cdots -\frac{a_{k-1}}{z^{k-1}}\right) .  \label{2.1}
\end{equation}

Here $a_{n+k}$ is the sequence $a_{n+k}=\left( a_{k},a_{k+1},\ldots \right) $%
.

(c) \textit{Multiplication by the sequence }$\lambda ^{n}$\textit{.} If $%
\mathcal{Z}\left( a_{n}\right) =A\left( z\right) $, then%
\begin{equation}
\mathcal{Z}\left( \lambda ^{n}a_{n}\right) =A\left( \frac{z}{\lambda }%
\right) .  \label{2.2}
\end{equation}

(d) \textit{Convolution theorem.} If $a_{n}$ and $b_{n}$ are two given
sequences, then%
\begin{equation}
\mathcal{Z}\left( a_{n}\ast b_{n}\right) =\mathcal{Z}\left( a_{n}\right) 
\mathcal{Z}\left( b_{n}\right) ,  \label{2.3}
\end{equation}%
where $a_{n}\ast b_{n}=\sum_{t=0}^{n}a_{t}b_{n-t}$ is the convolution of the
sequences $a_{n}$ and $b_{n}$.

If $A\left( z\right) =\mathcal{Z}\left( a_{n}\right) $, we also write $a_{n}=%
\mathcal{Z}^{-1}\left( A\left( z\right) \right) $, and we say that the
sequence $a_{n}$ is the \textit{inverse }$Z$\textit{\ transform} of $A\left(
z\right) $. Clearly $\mathcal{Z}^{-1}\,$is also linear and injective.

Observe that according to (\ref{2.2}), if $\mathcal{Z}\left( a_{n}\right)
=A\left( z\right) $ then%
\begin{equation}
\mathcal{Z}\left( \left( -1\right) ^{n}a_{n}\right) =A\left( -z\right) ,
\label{2.4}
\end{equation}%
and%
\begin{equation}
\mathcal{Z}\left( L_{sn+m}a_{n}\right) =\alpha ^{m}A\left( \frac{z}{\alpha
^{s}}\right) +\beta ^{m}A\left( \frac{z}{\beta ^{s}}\right) .  \label{2.5}
\end{equation}

For given $\lambda \in \mathbb{C}$, $\lambda \neq 0$, the $Z$ transform of
the sequence $\lambda ^{n}$ is plainly%
\begin{equation}
\mathcal{Z}\left( \lambda ^{n}\right) =\sum_{n=0}^{\infty }\frac{\lambda ^{n}%
}{z^{n}}=\frac{z}{z-\lambda },  \label{2.6}
\end{equation}%
(defined for $\left\vert z\right\vert &gt;\left\vert \lambda \right\vert $).
This is an important formula in this article since Fibonacci sequences are
combinations of sequences of the form $\lambda ^{n}$. In particular we have
that the $Z$ transform of the constant sequence $1$ is%
\begin{equation}
\mathcal{Z}\left( 1\right) =\frac{z}{z-1}.  \label{2.7}
\end{equation}

(Observe that if $t_{1}=t_{2}=0$, formula (\ref{1.8}) says that $\mathcal{Z}%
\left( F_{m_{1}}^{k_{1}}F_{m_{2}}^{k_{2}}\right) =z\frac{\left( -1\right)
^{s+1}F_{m_{1}}^{k_{1}}F_{m_{2}}^{k_{2}}}{\left( -1\right) ^{s+1}z+\left(
-1\right) ^{s}}$, which is essentially (\ref{2.7}).)

For given $m\in \mathbb{Z}$ one has%
\begin{equation}
\mathcal{Z}\left( F_{sn+m}\right) =\frac{z\left( F_{m}z+\left( -1\right)
^{m}F_{s-m}\right) }{z^{2}-L_{s}z+\left( -1\right) ^{s}},  \label{2.8}
\end{equation}%
and from this expression one can see that%
\begin{equation}
F_{s}F_{sn+m}=F_{m}F_{s\left( n+1\right) }+\left( -1\right)
^{m}F_{s-m}F_{sn}.  \label{2.9}
\end{equation}

(See \cite{P}.)

Now we begin with a list of preliminary results that will be used in
sections \ref{Sec3} and \ref{Sec4}.

\begin{proposition}
\label{Prop2.1}\textit{For given }$k\in \mathbb{N}^{\prime }$\textit{\ we
have }%
\begin{equation}
\left( -1\right) ^{s+1}\dprod\limits_{j=0}^{k}\left( z-\alpha ^{sj}\beta
^{s\left( k-j\right) }\right) =\sum_{i=0}^{k+1}\left( -1\right) ^{\frac{%
\left( si+2(s+1)\right) \left( i+1\right) }{2}}\dbinom{k+1}{i}%
_{F_{s}}z^{k+1-i}.  \label{2.10}
\end{equation}
\end{proposition}

\begin{proof}
We proceed by induction on $k$. For $k=0$ the result is clearly true (both
sides are equal to $\left( -1\right) ^{s+1}\left( z-1\right) $). Let us
suppose the formula is true for a given $k$. Then%
\begin{eqnarray*}
&amp;&amp;\left( -1\right) ^{s+1}\dprod\limits_{j=0}^{k+1}\left( z-\alpha ^{sj}\beta
^{s\left( k+1-j\right) }\right) \\
&amp;=&amp;\left( -1\right) ^{s+1}\left( z-\alpha ^{s\left( k+1\right) }\right)
\beta ^{s\left( k+1\right) }\dprod\limits_{j=0}^{k}\left( \frac{z}{\beta ^{s}%
}-\alpha ^{sj}\beta ^{s\left( k-j\right) }\right) \\
&amp;=&amp;\left( -1\right) ^{s+1}\left( z-\alpha ^{s\left( k+1\right) }\right)
\beta ^{s\left( k+1\right) }\left( -1\right) ^{s+1}\sum_{i=0}^{k+1}\left(
-1\right) ^{\frac{\left( si+2(s+1)\right) \left( i+1\right) }{2}}\dbinom{k+1%
}{i}_{F_{s}}\left( \frac{z}{\beta ^{s}}\right) ^{k+1-i} \\
&amp;=&amp;\left( z-\alpha ^{s\left( k+1\right) }\right) \sum_{i=0}^{k+1}\left(
-1\right) ^{\frac{\left( si+2(s+1)\right) \left( i+1\right) }{2}}\dbinom{k+1%
}{i}_{F_{s}}\beta ^{si}z^{k+1-i} \\
&amp;=&amp;\sum_{i=0}^{k+1}\left( -1\right) ^{\frac{\left( si+2(s+1)\right) \left(
i+1\right) }{2}}\dbinom{k+1}{i}_{F_{s}}\beta ^{si}z^{k+2-i}-\alpha ^{s\left(
k+1\right) }\sum_{i=1}^{k+2}\left( -1\right) ^{\frac{\left( s\left(
i-1\right) +2(s+1)\right) i}{2}}\dbinom{k+1}{i-1}_{F_{s}}\beta ^{s\left(
i-1\right) }z^{k+2-i} \\
&amp;=&amp;\sum_{i=0}^{k+2}\left( -1\right) ^{\frac{\left( si+2(s+1)\right) \left(
i+1\right) }{2}}\dbinom{k+2}{i}_{F_{s}}\frac{1}{F_{s\left( k+2\right) }}%
\left( \beta ^{si}F_{s\left( k+2-i\right) }+\left( -1\right) ^{-s\left(
i+1\right) }\alpha ^{s\left( k+1\right) }\beta ^{s\left( i-1\right)
}F_{si}\right) z^{k+2-i} \\
&amp;=&amp;\sum_{i=0}^{k+2}\left( -1\right) ^{\frac{\left( si+2(s+1)\right) \left(
i+1\right) }{2}}\dbinom{k+2}{i}_{F_{s}}z^{k+2-i}
\end{eqnarray*}%
as wanted. Here we used that%
\begin{equation*}
\beta ^{si}F_{s\left( k+2-i\right) }+\left( -1\right) ^{-s\left( i+1\right)
}\alpha ^{s\left( k+1\right) }\beta ^{s\left( i-1\right) }F_{si}=F_{s\left(
k+2\right) },
\end{equation*}%
which can be proved easily by writing the $F$'s in terms of $\alpha $ and $%
\beta $.
\end{proof}

We will denote the $\left( k+1\right) $-th degree polynomial $\left(
-1\right) ^{s+1}\prod_{j=0}^{k}\left( z-\alpha ^{sj}\beta ^{s\left(
k-j\right) }\right) $ as $D_{s,k+1}\left( z\right) $. By combining pairs of
adequate linear factors, it is possible to express $D_{s,k+1}\left( z\right) 
$ as product of \textquotedblleft Lucas factors\textquotedblright\
(quadratic factors in which the coefficient of the linear term is a Lucas
number, like in the denominator of (\ref{2.8}); see also \cite{St}). We have
two cases:

(a) If $k$ is even, $k=2p$ say, then%
\begin{equation}
D_{s,2p+1}\left( z\right) =\left( -1\right) ^{s+1}\left( z-\left( -1\right)
^{sp}\right) \dprod\limits_{j=0}^{p-1}\left( z^{2}-\left( -1\right)
^{sj}L_{2s\left( p-j\right) }z+1\right) .  \label{2.11}
\end{equation}

(b) If $k$ is odd, $k=2p-1$ say, then%
\begin{equation}
D_{s,2p}\left( z\right) =\left( -1\right)
^{s+1}\dprod\limits_{j=0}^{p-1}\left( z^{2}-\left( -1\right) ^{sj}L_{s\left(
2p-1-2j\right) }z+\left( -1\right) ^{s}\right) .  \label{2.12}
\end{equation}

Indeed, we have%
\begin{eqnarray*}
D_{s,2p+1}\left( z\right) &amp;=&amp;\left( -1\right)
^{s+1}\dprod\limits_{j=0}^{2p}\left( z-\alpha ^{sj}\beta ^{s\left(
2p-j\right) }\right) \\
&amp;=&amp;\left( -1\right) ^{s+1}\left( z-\left( -1\right) ^{sp}\right)
\dprod\limits_{j=0}^{p-1}\left( z-\alpha ^{sj}\beta ^{s\left( 2p-j\right)
}\right) \dprod\limits_{j=p+1}^{2p}\left( z-\alpha ^{sj}\beta ^{s\left(
2p-j\right) }\right) \\
&amp;=&amp;\left( -1\right) ^{s+1}\left( z-\left( -1\right) ^{sp}\right)
\dprod\limits_{j=0}^{p-1}\left( z-\alpha ^{sj}\beta ^{s\left( 2p-j\right)
}\right) \dprod\limits_{j=0}^{p-1}\left( z-\alpha ^{s\left( 2p-j\right)
}\beta ^{sj}\right) \\
&amp;=&amp;\left( -1\right) ^{s+1}\left( z-\left( -1\right) ^{sp}\right)
\dprod\limits_{j=0}^{p-1}\left( z^{2}-\left( -1\right) ^{sj}L_{2s\left(
p-j\right) }z+1\right) ,
\end{eqnarray*}%
which proves (\ref{2.11}). Similarly we have%
\begin{eqnarray*}
D_{s,2p}\left( z\right) &amp;=&amp;\left( -1\right)
^{s+1}\dprod\limits_{j=0}^{2p-1}\left( z-\alpha ^{sj}\beta ^{s\left(
2p-1-j\right) }\right) \\
&amp;=&amp;\left( -1\right) ^{s+1}\dprod\limits_{j=0}^{p-1}\left( z-\alpha
^{sj}\beta ^{s\left( 2p-1-j\right) }\right) \dprod\limits_{j=p}^{2p-1}\left(
z-\alpha ^{sj}\beta ^{s\left( 2p-1-j\right) }\right) \\
&amp;=&amp;\left( -1\right) ^{s+1}\dprod\limits_{j=0}^{p-1}\left( z-\alpha
^{sj}\beta ^{s\left( 2p-1-j\right) }\right) \dprod\limits_{j=0}^{p-1}\left(
z-\alpha ^{s\left( 2p-1-j\right) }\beta ^{sj}\right) \\
&amp;=&amp;\left( -1\right) ^{s+1}\dprod\limits_{j=0}^{p-1}\left( z^{2}-\left(
-1\right) ^{sj}L_{s\left( 2p-1-2j\right) }z+\left( -1\right) ^{s}\right) ,
\end{eqnarray*}%
which proves (\ref{2.12}).

Before we establish the next proposition, we would like to see how (\ref%
{2.10}) and (\ref{2.11}) have some interesting identities (involving $s$%
-Fibonomials) hidden. First observe that we can write (\ref{2.10}) as
follows:%
\begin{equation}
\dprod\limits_{j=0}^{k}\left( z-\alpha ^{sj}\beta ^{s\left( k-j\right)
}\right) =\sum_{i=0}^{k+1}\left( -1\right) ^{\frac{i\left( s\left(
i+1\right) +2(s+1)\right) }{2}}\dbinom{k+1}{i}_{F_{s}}z^{k+1-i}.
\label{2.13}
\end{equation}

Since%
\begin{equation*}
\alpha ^{sj}\beta ^{s\left( k-j\right) }=\left( -1\right) ^{s\left(
k+j\right) }\alpha ^{s\left( 2j-k\right) }=\frac{\left( -1\right) ^{s\left(
k+j\right) }}{2}\left( L_{s\left( 2j-k\right) }+\sqrt{5}F_{s\left(
2j-k\right) }\right) ,
\end{equation*}%
we see that the numbers $a_{s,j,k}=\frac{\left( -1\right) ^{s\left(
k+j\right) }}{2}\left( L_{s\left( 2j-k\right) }+\sqrt{5}F_{s\left(
2j-k\right) }\right) $, $j,k\in \mathbb{Z}$, form an abelian multiplicative
group $a_{s,j_{1},k_{1}}a_{s,j_{2},k_{2}}=a_{s,j_{1}+j_{2},k_{1}+k_{2}}$,
(with identity $a_{s,0,0}$ and inverses $a_{s,j,k}^{-1}=a_{s,-j,-k}$). Thus,
(\ref{2.13}) together with Vi\`{e}te's formulas tell us that%
\begin{eqnarray}
&amp;&amp;\left( -1\right) ^{\frac{r\left( s\left( r+1\right) +2(s+1)\right) }{2}%
-ksr+r}2\dbinom{k+1}{r}_{F_{s}}  \label{2.14} \\
&amp;=&amp;\sum_{0\leq j_{1}&lt;j_{2}&lt;\cdots &lt;j_{r}\leq k}\left( -1\right) ^{s\left(
j_{1}+j_{2}+\cdots +j_{r}\right) }\left( L_{2s\left( j_{1}+j_{2}+\cdots
+j_{r}\right) -ksr}+\sqrt{5}F_{2s\left( j_{1}+j_{2}+\cdots +j_{r}\right)
-ksr}\right) ,  \notag
\end{eqnarray}%
where $r\in \left\{ 1,2,\ldots ,k+1\right\} $ is given. Moreover, since the
left-hand side of (\ref{2.14}) is an integer, we have%
\begin{equation}
\sum_{0\leq j_{1}&lt;j_{2}&lt;\cdots &lt;j_{r}\leq k}\left( -1\right) ^{s\left(
j_{1}+j_{2}+\cdots +j_{r}\right) }L_{2s\left( j_{1}+j_{2}+\cdots
+j_{r}\right) -ksr}=\left( -1\right) ^{\frac{r\left( s\left( r+1\right)
+2(s+1)\right) }{2}-ksr+r}2\dbinom{k+1}{r}_{F_{s}},  \label{2.15}
\end{equation}%
and%
\begin{equation}
\sum_{0\leq j_{1}&lt;j_{2}&lt;\cdots &lt;j_{r}\leq k}\left( -1\right) ^{s\left(
j_{1}+j_{2}+\cdots +j_{r}\right) }F_{2s\left( j_{1}+j_{2}+\cdots
+j_{r}\right) -ksr}=0.  \label{2.16}
\end{equation}

In particular we have the following identities, corresponding to (\ref{2.15}%
) and (\ref{2.16}) with $r=1:$%
\begin{equation}
\sum_{j=0}^{k}\left( -1\right) ^{sj}L_{s\left( 2j-k\right) }=\frac{\left(
-1\right) ^{ks}}{F_{s}}2F_{s\left( k+1\right) }\ \ \ \ ,\ \ \ \
\sum_{j=0}^{k}\left( -1\right) ^{sj}F_{s\left( 2j-k\right) }=0,  \label{2.17}
\end{equation}%
and with $r=2:$

\begin{equation}
\sum_{j=1}^{k}\sum_{i=0}^{j-1}\left( -1\right) ^{s\left( i+j\right)
}L_{2s\left( i+j-k\right) }=\frac{2\left( -1\right) ^{s}}{F_{s}F_{2s}}%
F_{sk}F_{s\left( k+1\right) }\text{ \ \ \ \ , \ \ \ \ }\sum_{j=1}^{k}%
\sum_{i=0}^{j-1}\left( -1\right) ^{s\left( i+j\right) }F_{2s\left(
i+j-k\right) }=0.  \label{2.18}
\end{equation}

On the other hand, observe that according to (\ref{2.10}) and (\ref{2.11})
we have%
\begin{equation*}
\sum_{i=0}^{2p+1}\left( -1\right) ^{\frac{\left( si+2(s+1)\right) \left(
i+1\right) }{2}}\dbinom{2p+1}{i}_{F_{s}}z^{2p+1-i}=\left( -1\right)
^{s+1}\left( z-\left( -1\right) ^{sp}\right) \dprod\limits_{j=0}^{p-1}\left(
z^{2}-\left( -1\right) ^{sj}L_{2s\left( p-j\right) }z+1\right) ,
\end{equation*}%
which can be written as follows:%
\begin{eqnarray}
&amp;&amp;\sum_{i=0}^{p}\left( \left( -1\right) ^{\frac{\left( si+2(s+1)\right)
\left( i+1\right) }{2}}z^{2p+1-i}+\left( -1\right) ^{\frac{\left( s\left(
2p+1-i\right) +2(s+1)\right) \left( 2p+2-i\right) }{2}}z^{i}\right) \dbinom{%
2p+1}{i}_{F_{s}}  \label{2.19} \\
&amp;=&amp;\left( -1\right) ^{s+1}\left( z-\left( -1\right) ^{sp}\right)
\dprod\limits_{j=0}^{p-1}\left( z^{2}-\left( -1\right) ^{sj}L_{2s\left(
p-j\right) }z+1\right) .  \notag
\end{eqnarray}

If $s$ is even, $s=2\sigma $ say ($\sigma \in \mathbb{N}$), and we set $z=-1$
we get from (\ref{2.19}) that%
\begin{equation*}
\sum_{i=0}^{p}\left( \left( -1\right) ^{\sigma i\left( i+1\right) }+\left(
-1\right) ^{\sigma i\left( i-1\right) }\right) \dbinom{2p+1}{i}_{F_{2\sigma
}}=2\dprod\limits_{j=0}^{p-1}\left( 2+L_{4\sigma \left( p-j\right) }\right) .
\end{equation*}

That is, we have the identity%
\begin{equation}
\sum_{i=0}^{p}\dbinom{2p+1}{i}_{F_{2\sigma
}}=\dprod\limits_{j=1}^{p}L_{2\sigma j}^{2}.  \label{2.20}
\end{equation}

Similarly, if $s$ is odd, $s=2\sigma -1$ say ($\sigma \in \mathbb{N}$),
formula (\ref{2.19}) can be written as follows:%
\begin{eqnarray}
&amp;&amp;\sum_{i=0}^{p}\left( \left( -1\right) ^{\frac{i\left( i+1\right) }{2}%
}z^{2p+1-i}+\left( -1\right) ^{\frac{\left( 2p+1-i\right) \left(
2p+2-i\right) }{2}}z^{i}\right) \dbinom{2p+1}{i}_{F_{2\sigma -1}}
\label{2.21} \\
&amp;=&amp;\left( z-\left( -1\right) ^{p}\right) \dprod\limits_{j=0}^{p-1}\left(
z^{2}-\left( -1\right) ^{j}L_{2\left( 2\sigma -1\right) \left( p-j\right)
}z+1\right) .  \notag
\end{eqnarray}

If in (\ref{2.21}) $p$ is replaced by $2p-1$ and we set $z=1$, then we get%
\begin{equation*}
\sum_{i=0}^{2p-1}\left( \left( -1\right) ^{\frac{i\left( i+1\right) }{2}%
}+\left( -1\right) ^{\frac{i\left( i+1\right) }{2}}\right) \dbinom{2\left(
2p-1\right) +1}{i}_{F_{2\sigma -1}}=2\dprod\limits_{j=0}^{2p-2}\left(
2-\left( -1\right) ^{j}L_{2\left( 2\sigma -1\right) \left( 2p-1-j\right)
}\right) ,
\end{equation*}%
from where we obtain the identity%
\begin{equation}
\sum_{i=0}^{2p-1}\left( -1\right) ^{\frac{i\left( i+1\right) }{2}}\dbinom{%
4p-1}{i}_{F_{2\sigma -1}}=\left( -1\right)
^{p}\dprod\limits_{j=1}^{2p-1}L_{\left( 2\sigma -1\right) j}^{2}.
\label{2.22}
\end{equation}

If now in (\ref{2.21}) $p\,$\ is replaced by $2p$ and we set $z=-1$, then we
get%
\begin{equation*}
\sum_{i=0}^{2p}2\left( -1\right) ^{\frac{i\left( i-1\right) }{2}+1}\dbinom{%
4p+1}{i}_{F_{2\sigma -1}}=-2\dprod\limits_{j=0}^{2p-1}\left( 2+\left(
-1\right) ^{j}L_{2\left( 2\sigma -1\right) \left( 2p-j\right) }\right) ,
\end{equation*}%
from where we obtain the identity%
\begin{equation}
\sum_{i=0}^{2p}\left( -1\right) ^{\frac{i\left( i-1\right) }{2}}\dbinom{4p+1%
}{i}_{F_{2\sigma -1}}=\left( -1\right) ^{p}\dprod\limits_{j=1}^{2p}L_{\left(
2\sigma -1\right) j}^{2}.  \label{2.23}
\end{equation}

\begin{proposition}
\label{Prop2.2}\textit{Let }$t,k\in \mathbb{N}^{\prime }$\textit{, }$m\in 
\mathbb{Z}$\textit{\ be given. Then}

\textit{(a)}%
\begin{eqnarray}
&amp;&amp;\frac{\alpha ^{sk}}{\sum_{i=0}^{t+1}\left( -1\right) ^{\frac{\left(
si+2(s+1)\right) \left( i+1\right) }{2}}\dbinom{t+1}{i}_{F_{s}}\alpha
^{si}z^{t+1-i}}+\frac{\beta ^{sk}}{\sum_{i=0}^{t+1}\left( -1\right) ^{\frac{%
\left( si+2(s+1)\right) \left( i+1\right) }{2}}\dbinom{t+1}{i}_{F_{s}}\beta
^{si}z^{t+1-i}}  \notag \\
&amp;=&amp;\frac{L_{sk}z+\left( -1\right) ^{sk+1}L_{s\left( t-k+1\right) }}{%
\sum_{i=0}^{t+2}\left( -1\right) ^{\frac{\left( si+2(s+1)\right) \left(
i+1\right) }{2}}\dbinom{t+2}{i}_{F_{s}}z^{t+2-i}}.  \label{2.24}
\end{eqnarray}

\textit{(b)}%
\begin{eqnarray}
&amp;&amp;\frac{\alpha ^{m+sk}}{\sum_{i=0}^{t+1}\left( -1\right) ^{\frac{\left(
si+2(s+1)\right) \left( i+1\right) }{2}}\dbinom{t+1}{i}_{F_{s}}\alpha
^{si}z^{t+1-i}}-\frac{\beta ^{m+sk}}{\sum_{i=0}^{t+1}\left( -1\right) ^{%
\frac{\left( si+2(s+1)\right) \left( i+1\right) }{2}}\dbinom{t+1}{i}%
_{F_{s}}\beta ^{si}z^{t+1-i}}  \notag \\
&amp;=&amp;\frac{\sqrt{5}\left( F_{sk+m}z+\left( -1\right) ^{sk+m}F_{s\left(
t-k+1\right) -m}\right) }{\sum_{i=0}^{t+2}\left( -1\right) ^{\frac{\left(
si+2(s+1)\right) \left( i+1\right) }{2}}\dbinom{t+2}{i}_{F_{s}}z^{t+2-i}}.
\label{2.25}
\end{eqnarray}
\end{proposition}

\begin{proof}
Observe that, according to proposition \ref{Prop2.1}, we have that%
\begin{eqnarray*}
\sum_{i=0}^{t+1}\left( -1\right) ^{\frac{\left( si+2(s+1)\right) \left(
i+1\right) }{2}}\dbinom{t+1}{i}_{F_{s}}\alpha ^{si}z^{t+1-i} &amp;=&amp;\alpha
^{s\left( t+1\right) }\sum_{i=0}^{t+1}\left( -1\right) ^{\frac{\left(
si+2(s+1)\right) \left( i+1\right) }{2}}\dbinom{t+1}{i}_{F_{s}}\left( \frac{z%
}{\alpha ^{s}}\right) ^{t+1-i} \\
&amp;=&amp;\left( -1\right) ^{s+1}\alpha ^{s\left( t+1\right)
}\dprod\limits_{j=0}^{t}\left( \frac{z}{\alpha ^{s}}-\alpha ^{sj}\beta
^{s\left( t-j\right) }\right) \\
&amp;=&amp;\left( -1\right) ^{s+1}\dprod\limits_{j=0}^{t}\left( z-\alpha ^{s\left(
j+1\right) }\beta ^{s\left( t-j\right) }\right) ,
\end{eqnarray*}%
and similarly%
\begin{eqnarray*}
\sum_{i=0}^{t+1}\left( -1\right) ^{\frac{\left( si+2(s+1)\right) \left(
i+1\right) }{2}}\dbinom{t+1}{i}_{F_{s}}\beta ^{si}z^{t+1-i} &amp;=&amp;\left(
-1\right) ^{s+1}\beta ^{s\left( t+1\right) }\dprod\limits_{j=0}^{t}\left( 
\frac{z}{\beta ^{s}}-\alpha ^{sj}\beta ^{s\left( t-j\right) }\right) \\
&amp;=&amp;\left( -1\right) ^{s+1}\dprod\limits_{j=0}^{t}\left( z-\alpha ^{sj}\beta
^{s\left( t-j+1\right) }\right) .
\end{eqnarray*}

Then%
\begin{eqnarray*}
&amp;&amp;\frac{\alpha ^{sk}}{\sum_{i=0}^{t+1}\left( -1\right) ^{\frac{\left(
si+2(s+1)\right) \left( i+1\right) }{2}}\dbinom{t+1}{i}_{F_{s}}\alpha
^{si}z^{t+1-i}}+\frac{\beta ^{sk}}{\sum_{i=0}^{t+1}\left( -1\right) ^{\frac{%
\left( si+2(s+1)\right) \left( i+1\right) }{2}}\dbinom{t+1}{i}_{F_{s}}\beta
^{si}z^{t+1-i}} \\
&amp;=&amp;\frac{\alpha ^{sk}}{\left( -1\right) ^{s+1}\prod_{j=0}^{t}\left( z-\alpha
^{s\left( j+1\right) }\beta ^{s\left( t-j\right) }\right) }+\frac{\beta ^{sk}%
}{\left( -1\right) ^{s+1}\prod_{j=0}^{t}\left( z-\alpha ^{sj}\beta ^{s\left(
t-\left( j-1\right) \right) }\right) } \\
&amp;=&amp;\frac{\alpha ^{sk}}{\left( -1\right) ^{s+1}\prod_{j=1}^{t+1}\left(
z-\alpha ^{sj}\beta ^{s\left( t-\left( j-1\right) \right) }\right) }+\frac{%
\beta ^{sk}}{\left( -1\right) ^{s+1}\prod_{j=0}^{t}\left( z-\alpha
^{sj}\beta ^{s\left( t-\left( j-1\right) \right) }\right) } \\
&amp;=&amp;\frac{\alpha ^{sk}\left( z-\beta ^{s\left( t+1\right) }\right) +\beta
^{sk}\left( z-\alpha ^{s\left( t+1\right) }\right) }{\left( -1\right)
^{s+1}\prod_{j=0}^{t+1}\left( z-\alpha ^{sj}\beta ^{s\left( t+1-j\right)
}\right) } \\
&amp;=&amp;\frac{L_{sk}z+\left( -1\right) ^{sk+1}L_{s\left( t-k+1\right) }}{%
\sum_{i=0}^{t+2}\left( -1\right) ^{\frac{\left( si+2(s+1)\right) \left(
i+1\right) }{2}}\dbinom{t+2}{i}_{F_{s}}z^{t+2-i}},
\end{eqnarray*}%
which proves (a). The proof of (b) is similar.
\end{proof}

\begin{lemma}
\label{Lem2.3}\textit{For given }$i,t\in \mathbb{N}^{\prime }$\textit{, the
following identity holds}%
\begin{equation}
\left( -1\right) ^{si}F_{s\left( t+2-i\right) }F_{s\left( t+1-i\right)
}+L_{s\left( t+1\right) }F_{s\left( t+2-i\right) }F_{si}+\left( -1\right)
^{s\left( t+i\right) }F_{si}F_{s\left( i-1\right) }=F_{s\left( t+2\right)
}F_{s\left( t+1\right) }.  \label{2.26}
\end{equation}
\end{lemma}

\begin{proof}
Leaving the details for the reader, the proof is as follows: first, use $%
\alpha $'s and $\beta $'s to prove that $F_{s\left( t+1-i\right) }\left(
-1\right) ^{si}+L_{s\left( t+1\right) }F_{si}=F_{s\left( t+1+i\right) }$.
Then write the left-hand side of (\ref{2.26}) as $F_{s\left( t+2-i\right)
}F_{s\left( t+1+i\right) }+\left( -1\right) ^{s\left( t+i\right)
}F_{si}F_{s\left( i-1\right) }$. Finally use the standard identity $%
F_{m}F_{n}-F_{m+k}F_{n-k}=\left( -1\right) ^{n-k}F_{m+k-n}F_{k}$ to obtain (%
\ref{2.26}).
\end{proof}

\begin{proposition}
\label{Prop2.4}\textit{Let }$t\in \mathbb{N}^{\prime }$\textit{\ be given.
Then}%
\begin{eqnarray}
&amp;&amp;\sum_{i=0}^{t+2}\left( -1\right) ^{\frac{\left( si+2(s+1)\right) \left(
i+1\right) }{2}}\dbinom{t+2}{i}_{F_{s}}z^{t+2-i}  \label{2.27} \\
&amp;=&amp;\left( z^{2}-L_{s\left( t+1\right) }z+\left( -1\right) ^{s\left(
t+1\right) }\right) \sum_{i=0}^{t}\left( -1\right) ^{\frac{\left(
si+2(s+1)\right) \left( i+1\right) }{2}}\dbinom{t}{i}_{F_{s}}\left(
-1\right) ^{si}z^{t-i}.  \notag
\end{eqnarray}
\end{proposition}

\begin{proof}
We have%
\begin{eqnarray*}
&amp;&amp;\left( z^{2}-L_{s\left( t+1\right) }z+\left( -1\right) ^{s\left(
t+1\right) }\right) \sum_{i=0}^{t}\left( -1\right) ^{\frac{\left(
si+2(s+1)\right) \left( i+1\right) }{2}}\dbinom{t}{i}_{F_{s}}\left(
-1\right) ^{si}z^{t-i} \\
&amp;=&amp;\sum_{i=0}^{t}\left( -1\right) ^{\frac{\left( si+2(s+1)\right) \left(
i+1\right) }{2}}\dbinom{t}{i}_{F_{s}}\left( -1\right)
^{si}z^{t+2-i}-L_{s\left( t+1\right) }\sum_{i=1}^{t+1}\left( -1\right) ^{%
\frac{\left( si-s+2(s+1)\right) i}{2}}\dbinom{t}{i-1}_{F_{s}}\left(
-1\right) ^{s\left( i-1\right) }z^{t+2-i} \\
&amp;&amp;+\left( -1\right) ^{s\left( t+1\right) }\sum_{i=2}^{t+2}\left( -1\right) ^{%
\frac{\left( si-2s+2(s+1)\right) \left( i-1\right) }{2}}\dbinom{t}{i-2}%
_{F_{s}}\left( -1\right) ^{si}z^{t+2-i} \\
&amp;=&amp;\sum_{i=0}^{t+2}\left( -1\right) ^{\frac{\left( si+2(s+1)\right) \left(
i+1\right) }{2}}\dbinom{t+2}{i}_{F_{s}}\frac{1}{F_{s\left( t+2\right)
}F_{s\left( t+1\right) }}\left( 
\begin{array}{c}
\left( -1\right) ^{si}F_{s\left( t+2-i\right) }F_{s\left( t+1-i\right) } \\ 
+L_{s\left( t+1\right) }F_{s\left( t+2-i\right) }F_{si} \\ 
+\left( -1\right) ^{s\left( t+i\right) }F_{si}F_{s\left( i-1\right) }%
\end{array}%
\right) z^{t+2-i} \\
&amp;=&amp;\sum_{i=0}^{t+2}\left( -1\right) ^{\frac{\left( si+2(s+1)\right) \left(
i+1\right) }{2}}\dbinom{t+2}{i}_{F_{s}}z^{t+2-i},
\end{eqnarray*}%
as wanted. In the last step we used lemma \ref{Lem2.3}.
\end{proof}

\begin{proposition}
\label{Prop2.5}\textit{For given} $i,t\in \mathbb{N}^{\prime }$ \textit{the
following identity holds}%
\begin{equation}
\sum\limits_{j=0}^{i}\left( -1\right) ^{\frac{\left( s\left( i-j\right)
+2(s+1)\right) \left( i-j+1\right) }{2}}\dbinom{t+1}{i-j}_{F_{s}}F_{stj+m}=%
\left( -1\right) ^{\frac{is}{2}\left( i-1\right) +i+s+m}\dbinom{t}{i}%
_{F_{s}}F_{is-m}.  \label{2.28}
\end{equation}
\end{proposition}

\begin{proof}
We proceed by induction on $t$. For $t=0$ we need to check that%
\begin{equation*}
\sum\limits_{j=0}^{i}\left( -1\right) ^{\frac{\left( s\left( i-j\right)
+2(s+1)\right) \left( i-j+1\right) }{2}}\dbinom{1}{i-j}_{F_{s}}F_{m}=\left(
-1\right) ^{\frac{is}{2}\left( i-1\right) +i+s+m}\dbinom{0}{i}%
_{F_{s}}F_{is-m}.
\end{equation*}

If $i=0$ we have a trivial equality (both sides are equal to $\left(
-1\right) ^{s+1}F_{m}$). If $i\geq 1$ we have%
\begin{equation*}
\sum\limits_{j=0}^{i}\left( -1\right) ^{\frac{\left( s\left( i-j\right)
+2(s+1)\right) \left( i-j+1\right) }{2}}\dbinom{1}{i-j}_{F_{s}}F_{m}=\left(
-1\right) ^{s+1}F_{m}+\left( -1\right) ^{s}F_{m}=0,
\end{equation*}%
as expected. Suppose now that (\ref{2.28}) is valid for a given $t$. We have
(by using (\ref{1.2}))%
\begin{eqnarray*}
&amp;&amp;\sum\limits_{j=0}^{i}\left( -1\right) ^{\frac{\left( s\left( i-j\right)
+2(s+1)\right) \left( i-j+1\right) }{2}}\dbinom{t+2}{i-j}_{F_{s}}F_{s\left(
t+1\right) j+m} \\
&amp;=&amp;\sum\limits_{j=0}^{i}\left( -1\right) ^{\frac{\left( s\left( i-j\right)
+2(s+1)\right) \left( i-j+1\right) }{2}}F_{s\left( t+1\right) j+m}\left(
F_{s\left( t+2-i+j\right) +1}\dbinom{t+1}{i-j-1}_{F_{s}}+F_{s\left(
i-j\right) -1}\dbinom{t+1}{i-j}_{F_{s}}\right) \\
&amp;=&amp;\sum\limits_{j=1}^{i+1}\left( -1\right) ^{\frac{\left( s\left(
i-j+1\right) +2(s+1)\right) \left( i-j+2\right) }{2}}F_{s\left( t+1\right)
\left( j-1\right) +m}F_{s\left( t+1-i+j\right) +1}\dbinom{t+1}{i-j}_{F_{s}}
\\
&amp;&amp;+\sum\limits_{j=0}^{i}\left( -1\right) ^{\frac{\left( s\left( i-j\right)
+2(s+1)\right) \left( i-j+1\right) }{2}}F_{s\left( t+1\right) j+m}F_{s\left(
i-j\right) -1}\dbinom{t+1}{i-j}_{F_{s}} \\
&amp;=&amp;\sum\limits_{j=0}^{i}\left( -1\right) ^{\frac{\left( s\left( i-j\right)
+2(s+1)\right) \left( i-j+1\right) }{2}}\dbinom{t+1}{i-j}_{F_{s}}\left( 
\begin{array}{c}
\left( -1\right) ^{s\left( i-j\right) +1}F_{s\left( t+1\right) \left(
j-1\right) +m}F_{s\left( t+1-i+j\right) +1} \\ 
+F_{s\left( t+1\right) j+m}F_{s\left( i-j\right) -1}%
\end{array}%
\right) \\
&amp;&amp;-\left( -1\right) ^{\frac{\left( s\left( i+1\right) +2(s+1)\right) \left(
i+2\right) }{2}}F_{s\left( t+1\right) \left( -1\right) +m}F_{s\left(
t+1-i\right) +1}\dbinom{t+1}{i}_{F_{s}}.
\end{eqnarray*}

Next we will use the identity%
\begin{eqnarray}
&amp;&amp;\left( -1\right) ^{s\left( i-j\right) +1}F_{s\left( t+1\right) \left(
j-1\right) +m}F_{s\left( t+1-i+j\right) +1}+F_{s\left( t+1\right)
j+m}F_{s\left( i-j\right) -1}  \label{2.29} \\
&amp;=&amp;\frac{F_{s\left( t+1\right) }}{F_{st}}\left( F_{s\left( i-1\right)
-1}F_{stj+m}+\left( -1\right) ^{s\left( i-1\right) -1}F_{s\left(
t-i+1\right) +1}F_{st\left( j-1\right) +m}\right) ,  \notag
\end{eqnarray}%
which can be proved in two steps (we leave the details for the reader):
first use $\alpha $'s and $\beta $'s to prove that%
\begin{equation*}
\left( -1\right) ^{s\left( i-j\right) +1}F_{s\left( t+1\right) \left(
j-1\right) +m}F_{s\left( t+1-i+j\right) +1}+F_{s\left( t+1\right)
j+m}F_{s\left( i-j\right) -1}=F_{s\left( t+1\right) }F_{st\left( j-1\right)
+m+s\left( i-1\right) -1},
\end{equation*}%
and then use (\ref{2.9}) to write%
\begin{equation*}
F_{st\left( j-1\right) +m+s\left( i-1\right) -1}=\frac{1}{F_{st}}\left(
F_{s\left( i-1\right) -1}F_{stj+m}+\left( -1\right) ^{s\left( i-1\right)
-1}F_{s\left( t-i+1\right) +1}F_{st\left( j-1\right) +m}\right) ,
\end{equation*}%
obtaining in this way (\ref{2.29}).

Thus, by using (\ref{2.29}) we can write%
\begin{eqnarray*}
&amp;&amp;\sum\limits_{j=0}^{i}\left( -1\right) ^{\frac{\left( sj+2(s+1)\right)
\left( j+1\right) }{2}}\dbinom{t+2}{j}_{F_{s}}F_{s\left( t+1\right) \left(
i-j\right) +m} \\
&amp;=&amp;\frac{F_{s\left( t+1\right) }}{F_{st}}\sum\limits_{j=0}^{i}\left(
-1\right) ^{\frac{\left( s\left( i-j\right) +2(s+1)\right) \left(
i-j+1\right) }{2}}\dbinom{t+1}{i-j}_{F_{s}}\left( 
\begin{array}{c}
F_{s\left( i-1\right) -1}F_{stj+m} \\ 
+\left( -1\right) ^{s\left( i-1\right) -1}F_{s\left( t-i+1\right)
+1}F_{st\left( j-1\right) +m}%
\end{array}%
\right)  \\
&amp;&amp;-\left( -1\right) ^{\frac{\left( s\left( i+1\right) +2(s+1)\right) \left(
i+2\right) }{2}}F_{s\left( t+1\right) \left( -1\right) +m}F_{s\left(
t+1-i\right) +1}\dbinom{t+1}{i},
\end{eqnarray*}%
and by using the induction hypothesis (together with some simplifications),
we get%
\begin{eqnarray*}
&amp;&amp;\sum\limits_{j=0}^{i}\left( -1\right) ^{\frac{\left( sj+2(s+1)\right)
\left( j+1\right) }{2}}\dbinom{t+2}{j}_{F_{s}}F_{s\left( t+1\right) \left(
i-j\right) +m} \\
&amp;=&amp;\frac{F_{s\left( t+1\right) }}{F_{st}}F_{s\left( i-1\right)
-1}\sum\limits_{j=0}^{i}\left( -1\right) ^{\frac{\left( s\left( i-j\right)
+2(s+1)\right) \left( i-j+1\right) }{2}}F_{stj+m}\dbinom{t+1}{i-j}_{F_{s}} \\
&amp;&amp;+\frac{F_{s\left( t+1\right) }}{F_{st}}F_{s\left( t-i+1\right) +1}\left(
-1\right) ^{s\left( i-1\right) -1}\sum\limits_{j=0}^{i-1}\left( -1\right) ^{%
\frac{\left( s\left( i-j-1\right) +2(s+1)\right) \left( i-j\right) }{2}%
}F_{stj+m}\dbinom{t+1}{i-1-j}_{F_{s}} \\
&amp;&amp;-\left( -1\right) ^{\frac{\left( s\left( i+1\right) +2(s+1)\right) \left(
i+2\right) }{2}}F_{s\left( t+1\right) \left( -1\right) +m}F_{s\left(
t+1-i\right) +1}\dbinom{t+1}{i} \\
&amp;&amp;+\frac{F_{s\left( t+1\right) }}{F_{st}}F_{s\left( t-i+1\right) +1}\left(
-1\right) ^{s\left( i-1\right) -1}\left( -1\right) ^{\frac{\left(
si+2(s+1)\right) \left( i+1\right) }{2}}F_{-st+m}\dbinom{t+1}{i}_{F_{s}} \\
&amp;=&amp;\frac{F_{s\left( t+1\right) }}{F_{st}}F_{s\left( i-1\right) -1}\left(
-1\right) ^{\frac{is}{2}\left( i-1\right) +i+s+m}\dbinom{t}{i}%
_{F_{s}}F_{is-m} \\
&amp;&amp;+\frac{F_{s\left( t+1\right) }}{F_{st}}F_{s\left( t-i+1\right) +1}\left(
-1\right) ^{s\left( i-1\right) -1}\left( -1\right) ^{\frac{\left( i-1\right)
s}{2}i+i-1+\left( 2-i\right) s+m}\dbinom{t}{i-1}_{F_{s}}F_{\left( i-1\right)
s-m} \\
&amp;&amp;+\dbinom{t+1}{i}\left( 
\begin{array}{c}
\frac{F_{s\left( t+1\right) }}{F_{st}}F_{s\left( t-i+1\right) +1}\left(
-1\right) ^{s\left( i-1\right) -1}\left( -1\right) ^{\frac{\left( si+2\left(
s+1\right) \right) \left( i+1\right) }{2}}F_{-st+m} \\ 
-\left( -1\right) ^{\frac{\left( s\left( i+1\right) +2\left( s+1\right)
\right) \left( i+2\right) }{2}}F_{s\left( t+1\right) \left( -1\right)
+m}F_{s\left( t+1-i\right) +1}%
\end{array}%
\right)  \\
&amp;=&amp;\left( -1\right) ^{\frac{is}{2}\left( i-1\right) +i+s+m}\dbinom{t+1}{i}%
_{F_{s}}\frac{1}{F_{st}}\left( F_{is-m}F_{s\left( i-1\right) -1}F_{s\left(
t-i+1\right) }+F_{is}F_{\left( i-1\right) s-m}F_{s\left( t-i+1\right)
+1}\right)  \\
&amp;&amp;+\dbinom{t+1}{i}\left( -1\right) ^{\frac{is}{2}\left( i-1\right) +i+s+m}%
\frac{\left( -1\right) ^{si+m}}{F_{st}}F_{s\left( t-i+1\right) +1}\left( 
\begin{array}{c}
\left( -1\right) ^{s}F_{s\left( t+1\right) }F_{-st+m} \\ 
-\left( -1\right) ^{si+m}F_{st}F_{s\left( t+1\right) \left( -1\right) +m}%
\end{array}%
\right)  \\
&amp;=&amp;\left( -1\right) ^{\frac{is}{2}\left( i-1\right) +i+s+m}\dbinom{t+1}{i}%
_{F_{s}}F_{is-m},
\end{eqnarray*}%
as wanted. In the last step we used the identity%
\begin{eqnarray*}
&amp;&amp;F_{is-m}F_{s\left( i-1\right) -1}F_{s\left( t-i+1\right) }+F_{is}F_{\left(
i-1\right) s-m}F_{s\left( t-i+1\right) +1} \\
&amp;&amp;+\left( -1\right) ^{si+m}F_{s\left( t-i+1\right) +1}\left( \left(
-1\right) ^{s}F_{s\left( t+1\right) }F_{-st+m}-F_{st}F_{-s\left( t+1\right)
+m}\right)  \\
&amp;=&amp;F_{st}F_{is-m},
\end{eqnarray*}%
which proof is an easy exercise left to the reader.
\end{proof}

The proof of the following proposition is similar to the proof of
proposition \ref{Prop2.5}, with some changes in signs and, of course, some $%
F $'s substituted by $L$'s. We leave it for the reader.

\begin{proposition}
\label{Prop2.6}\textit{For given} $i,t\in \mathbb{N}^{\prime }$ \textit{the
following identity holds}\textbf{\ }%
\begin{equation}
\sum\limits_{j=0}^{i}\left( -1\right) ^{\frac{\left( s\left( i-j\right)
+2\left( s+1\right) \right) \left( i-j+1\right) }{2}}\dbinom{t+1}{i-j}%
_{F_{s}}L_{stj+m}=\left( -1\right) ^{\left( i+1\right) \left( \frac{is}{2}%
+1\right) -is+m+s}\dbinom{t}{i}_{F_{s}}L_{is-m}.  \label{2.30}
\end{equation}
\end{proposition}

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

We begin this section by noting that we can write explicitly the sequence $%
F_{sn+m_{1}}^{k_{1}}F_{sn+m_{2}}^{k_{2}}$ (where $m_{1},m_{2}\in \mathbb{Z}$
and $k_{1},k_{2}\in \mathbb{N}^{\prime }$ are given) as follows:%
\begin{eqnarray*}
&amp;&amp;F_{sn+m_{1}}^{k_{1}}F_{sn+m_{2}}^{k_{2}} \\
&amp;=&amp;\left( \frac{\alpha ^{sn+m_{1}}-\beta ^{sn+m_{1}}}{\sqrt{5}}\right)
^{k_{1}}\left( \frac{\alpha ^{sn+m_{2}}-\beta ^{sn+m_{2}}}{\sqrt{5}}\right)
^{k_{2}} \\
&amp;=&amp;5^{-\frac{k_{1}+k_{2}}{2}}\sum_{i=0}^{k_{1}}\dbinom{k_{1}}{i}\left(
\alpha ^{sn+m_{1}}\right) ^{i}\left( -\beta ^{sn+m_{1}}\right)
^{k_{1}-i}\sum_{j=0}^{k_{2}}\dbinom{k_{2}}{j}\left( \alpha
^{sn+m_{2}}\right) ^{j}\left( -\beta ^{sn+m_{2}}\right) ^{k_{2}-j} \\
&amp;=&amp;5^{-\frac{k_{1}+k_{2}}{2}}\beta
^{m_{1}k_{1}+m_{2}k_{2}}\sum_{j=0}^{k_{1}+k_{2}}\sum_{i=0}^{k_{1}}\left(
-1\right) ^{k_{1}+k_{2}-j}\dbinom{k_{1}}{i}\dbinom{k_{2}}{j-i}\left( \frac{%
\alpha }{\beta }\right) ^{\left( m_{1}-m_{2}\right) i+m_{2}j}\left( \alpha
^{sj}\beta ^{s\left( k_{1}+k_{2}-j\right) }\right) ^{n}.
\end{eqnarray*}

Then the $Z$ transform of $F_{sn+m_{1}}^{k_{1}}F_{sn+m_{2}}^{k_{2}}$ is%
\begin{eqnarray}
&amp;&amp;\mathcal{Z}\left( F_{sn+m_{1}}^{k_{1}}F_{sn+m_{2}}^{k_{2}}\right)
\label{3.1} \\
&amp;=&amp;5^{-\frac{k_{1}+k_{2}}{2}}\beta
^{m_{1}k_{1}+m_{2}k_{2}}\sum_{j=0}^{k_{1}+k_{2}}\sum_{i=0}^{k_{1}}\left(
-1\right) ^{k_{1}+k_{2}-j}\dbinom{k_{1}}{i}\dbinom{k_{2}}{j-i}\left( \frac{%
\alpha }{\beta }\right) ^{\left( m_{1}-m_{2}\right) i+m_{2}j}\frac{z}{%
z-\alpha ^{sj}\beta ^{s\left( k_{1}+k_{2}-j\right) }}.  \notag
\end{eqnarray}

Our first main result says that this expression can be written in a special
form.

\begin{theorem}
\label{Th3.1}\textit{Let }$m_{1},m_{2}\in \mathbb{Z}$\textit{\ and }$%
k_{1},k_{2}\in \mathbb{N}^{\prime }$\textit{\ be given. The }$Z$\textit{\
transform of the sequence }$F_{sn+m_{1}}^{k_{1}}F_{sn+m_{2}}^{k_{2}}$\textit{%
\ is }%
\begin{equation}
\mathcal{Z}\left( F_{sn+m_{1}}^{k_{1}}F_{sn+m_{2}}^{k_{2}}\right) =z\frac{%
\sum_{i=0}^{k_{1}+k_{2}}\sum_{j=0}^{i}\left( -1\right) ^{\frac{\left(
si+2(s+1)\right) \left( i+1\right) }{2}}\dbinom{k_{1}+k_{2}+1}{j}%
_{F_{s}}F_{m_{1}+s\left( i-j\right) }^{k_{1}}F_{m_{2}+s\left( i-j\right)
}^{k_{2}}z^{k_{1}+k_{2}-i}}{\sum_{i=0}^{k_{1}+k_{2}+1}\left( -1\right) ^{%
\frac{\left( si+2(s+1)\right) \left( i+1\right) }{2}}\dbinom{k_{1}+k_{2}+1}{i%
}_{F_{s}}z^{k_{1}+k_{2}+1-i}}.  \label{3.2}
\end{equation}
\end{theorem}

\begin{proof}
We have to show that%
\begin{eqnarray}
&amp;&amp;5^{-\frac{k_{1}+k_{2}}{2}}\beta
^{m_{1}k_{1}+m_{2}k_{2}}\sum_{j=0}^{k_{1}+k_{2}}\sum_{i=0}^{k_{1}}\dbinom{%
k_{1}}{i}\dbinom{k_{2}}{j-i}\left( -1\right) ^{k_{1}+k_{2}-j}\left( \frac{%
\alpha }{\beta }\right) ^{\left( m_{1}-m_{2}\right) i+m_{2}j}\frac{z}{%
z-\alpha ^{sj}\beta ^{s\left( k_{1}+k_{2}-j\right) }}  \notag \\
&amp;=&amp;z\frac{\sum_{i=0}^{k_{1}+k_{2}}\sum_{j=0}^{i}\left( -1\right) ^{\frac{%
\left( sj+2(s+1)\right) \left( j+1\right) }{2}}\dbinom{k_{1}+k_{2}+1}{j}%
_{F_{s}}F_{m_{1}+s\left( i-j\right) }^{k_{1}}F_{m_{2}+s\left( i-j\right)
}^{k_{2}}z^{k_{1}+k_{2}-i}}{\sum_{i=0}^{k_{1}+k_{2}+1}\left( -1\right) ^{%
\frac{\left( si+2(s+1)\right) \left( i+1\right) }{2}}\dbinom{k_{1}+k_{2}+1}{i%
}_{F_{s}}z^{k_{1}+k_{2}+1-i}}.  \label{3.3}
\end{eqnarray}

We will proceed by induction on $k_{1}$ and/or $k_{2}$. Observe that the
case $k_{1}=k_{2}=0$ is (\ref{2.7}). Let us consider the case $k_{1}=k_{2}=1$
and then we construct the induction argument on $k_{1}$ (note that (\ref{3.2}%
) is symmetric with respect to $k_{1}$ and $k_{2}$). For the case $%
k_{1}=k_{2}=1$ we use that $L_{2s}+\left( -1\right) ^{s}=\frac{F_{3s}}{F_{s}}
$ to write%
\begin{eqnarray*}
&amp;&amp;5^{-\frac{1+1}{2}}\beta ^{m_{1}+m_{2}}\sum_{i=0}^{1}\sum_{j=0}^{1}\dbinom{1%
}{i}\dbinom{1}{j}\left( -1\right) ^{i+j}\left( \frac{\alpha }{\beta }\right)
^{m_{1}i+m_{2}j}\frac{z}{z-\alpha ^{s\left( i+j\right) }\beta ^{s\left(
2-i-j\right) }} \\
&amp;&amp; \\
&amp;=&amp;5^{-1}z\left( \frac{\left( \alpha ^{m_{1}+m_{2}}+\beta
^{m_{1}+m_{2}}\right) z-\beta ^{m_{1}+m_{2}}\alpha ^{2s}-\alpha
^{m_{1}+m_{2}}\beta ^{2s}}{z^{2}-L_{2s}z+1}-\frac{\alpha ^{m_{1}}\beta
^{m_{2}}+\alpha ^{m_{2}}\beta ^{m_{1}}}{z-\left( -1\right) ^{s}}\right) \\
&amp;&amp; \\
&amp;=&amp;5^{-1}\frac{z}{z^{3}-\left( L_{2s}+\left( -1\right) ^{s}\right)
z^{2}+\left( 1+\left( -1\right) ^{s}L_{2s}\right) z-\left( -1\right) ^{s}}%
\times \\
&amp;&amp;\times \left( \!\!\!%
\begin{array}{c}
\left( \alpha ^{m_{1}+m_{2}}+\beta ^{m_{1}+m_{2}}-\alpha ^{m_{1}}\beta
^{m_{2}}-\alpha ^{m_{2}}\beta ^{m_{1}}\right) z^{2} \\ 
\\ 
-\left( \beta ^{m_{1}+m_{2}}\alpha ^{2s}\!+\!\alpha ^{m_{1}+m_{2}}\beta
^{2s}\!-\!\left( \alpha ^{m_{1}}\beta ^{m_{2}}\!+\!\alpha ^{m_{2}}\beta
^{m_{1}}\right) \left( \alpha ^{2s}\!+\!\beta ^{2s}\right) \!+\!\left(
-1\right) ^{s}\left( \alpha ^{m_{1}+m_{2}}\!+\!\beta ^{m_{1}+m_{2}}\right)
\right) z \\ 
\\ 
+\left( -1\right) ^{s}\beta ^{m_{1}+m_{2}}\alpha ^{2s}+\left( -1\right)
^{s}\alpha ^{m_{1}+m_{2}}\beta ^{2s}-\alpha ^{m_{1}}\beta ^{m_{2}}-\alpha
^{m_{2}}\beta ^{m_{1}}%
\end{array}%
\!\!\!\right) \\
&amp;&amp; \\
&amp;=&amp;\frac{z}{\left( -1\right) ^{s+1}z^{3}+\left( -1\right) ^{s}\frac{F_{3s}}{%
F_{s}}z^{2}-\frac{F_{3s}}{F_{s}}z+1}\times \\
&amp;&amp;\times \left( 
\begin{array}{c}
\left( -1\right) ^{s+1}F_{m_{1}}F_{m_{2}}z^{2}+\left( -1\right) ^{s+1}\left(
F_{m_{1}+s}F_{m_{2}+s}-\frac{F_{3s}}{F_{s}}F_{m_{1}}F_{m_{2}}\right) z \\ 
\\ 
+\left( -1\right) ^{s+1}F_{m_{1}+2s}F_{m_{2}+2s}+\left( -1\right) ^{s}\frac{%
F_{3s}}{F_{s}}F_{m_{1}+s}F_{m_{2}+s}-\frac{F_{3s}}{F_{s}}F_{m_{1}}F_{m_{2}}%
\end{array}%
\right) \\
&amp;&amp; \\
&amp;=&amp;z\frac{\sum_{i=0}^{2}\sum_{j=0}^{i}\left( -1\right) ^{\frac{\left(
sj+2(s+1)\right) \left( j+1\right) }{2}}\dbinom{3}{j}_{F_{s}}F_{m_{1}+s%
\left( i-j\right) }F_{m_{2}+s\left( i-j\right) }z^{2-i}}{\sum_{i=0}^{3}%
\left( -1\right) ^{\frac{\left( si+2(s+1)\right) \left( i+1\right) }{2}}%
\dbinom{3}{i}_{F_{s}}z^{3-i}},
\end{eqnarray*}%
which is (\ref{3.3}) with $k_{1}=k_{2}=1$. Suppose now that (\ref{3.3}) is
true for a given $k_{1}$. We will show that it is also true for $k_{1}+1$.
We have%
\begin{eqnarray*}
\mathcal{Z}\left( F_{sn+m_{1}}^{k_{1}+1}F_{sn+m_{2}}^{k_{2}}\right) &amp;=&amp;5^{-%
\frac{k_{1}+k_{2}+1}{2}}\beta ^{m_{1}\left( k_{1}+1\right)
+m_{2}k_{2}}\sum_{j=0}^{k_{1}+k_{2}+1}\sum_{i=0}^{k_{1}+1}\left( -1\right)
^{k_{1}+k_{2}+1-j}\dbinom{k_{1}+1}{i}\dbinom{k_{2}}{j-i}\times \\
&amp;&amp;\times \left( \frac{\alpha }{\beta }\right) ^{\left( m_{1}-m_{2}\right)
i+m_{2}j}\frac{z}{z-\alpha ^{sj}\beta ^{s\left( k_{1}+k_{2}+1-j\right) }}.
\end{eqnarray*}

If we use that $\binom{k_{1}+1}{i}=\binom{k_{1}}{i}+\binom{k_{1}}{i-1}$,
separate in the corresponding two terms, and shift the indices of the second
term, we get%
\begin{eqnarray}
&amp;&amp;\mathcal{Z}\left( F_{sn+m_{1}}^{k_{1}+1}F_{sn+m_{2}}^{k_{2}}\right)
\label{3.4} \\
&amp;=&amp;\!\!-5^{-\frac{k_{1}+k_{2}+1}{2}}\beta ^{m_{1}k_{1}+m_{2}k_{2}}\beta
^{m_{1}}\!\!\sum_{j=0}^{k_{1}+k_{2}}\!\sum_{i=0}^{k_{1}}\!\left( -1\right)
^{k_{1}+1+k_{2}-j}\!\dbinom{k_{1}}{i}\!\dbinom{k_{2}}{j-i}\!\!\left( \frac{%
\alpha }{\beta }\right) ^{\left( m_{1}-m_{2}\right) i+m_{2}j}\!\!\frac{\frac{%
z}{\beta ^{s}}}{\frac{z}{\beta ^{s}}-\alpha ^{sj}\beta ^{s\left(
k_{1}+k_{2}-j\right) }}  \notag \\
&amp;&amp;+5^{-\frac{k_{1}+k_{2}+1}{2}}\beta ^{m_{1}k_{1}+m_{2}k_{2}}\alpha
^{m_{1}}\!\!\sum_{j=0}^{k_{1}+k_{2}}\!\sum_{i=0}^{k_{1}}\!\left( -1\right)
^{k_{1}+k_{2}-j}\!\dbinom{k_{1}}{i}\!\dbinom{k_{2}}{j-i}\!\!\left( \frac{%
\alpha }{\beta }\right) ^{\left( m_{1}-m_{2}\right) i+m_{2}j}\!\!\frac{\frac{%
z}{\alpha ^{s}}}{\frac{z}{\alpha ^{s}}-\alpha ^{sj}\beta ^{s\left(
k_{1}+k_{2}-j\right) }}.  \notag
\end{eqnarray}

By using the induction hypothesis and proposition \ref{Prop2.2} (b), we can
write (\ref{3.4}) as%
\begin{eqnarray*}
&amp;&amp;\mathcal{Z\!}\left( F_{sn+m_{1}}^{k_{1}+1}F_{sn+m_{2}}^{k_{2}}\right) \!\!
\\
&amp;=&amp;\frac{\alpha ^{m_{1}}}{5^{\frac{1}{2}}}\frac{\frac{z}{\alpha ^{s}}%
\sum_{i=0}^{k_{1}+k_{2}}\sum_{j=0}^{i}\left( -1\right) ^{\frac{\left(
sj+2(s+1)\right) \left( j+1\right) }{2}}\dbinom{k_{1}+k_{2}+1}{j}%
_{F_{s}}F_{m_{1}+s\left( i-j\right) }^{k_{1}}F_{m_{2}+s\left( i-j\right)
}^{k_{2}}\left( \frac{z}{\alpha ^{s}}\right) ^{k_{1}+k_{2}-i}}{%
\sum_{i=0}^{k_{1}+k_{2}+1}\left( -1\right) ^{\frac{\left( si+2(s+1)\right)
\left( i+1\right) }{2}}\dbinom{k_{1}+k_{2}+1}{i}_{F_{s}}\left( \frac{z}{%
\alpha ^{s}}\right) ^{k_{1}+k_{2}+1-i}} \\
&amp;&amp;-\frac{\beta ^{m_{1}}}{5^{\frac{1}{2}}}\frac{\frac{z}{\beta ^{s}}%
\sum_{i=0}^{k_{1}+k_{2}}\sum_{j=0}^{i}\left( -1\right) ^{\frac{\left(
sj+2(s+1)\right) \left( j+1\right) }{2}}\dbinom{k_{1}+k_{2}+1}{j}%
_{F_{s}}F_{m_{1}+s\left( i-j\right) }^{k_{1}}F_{m_{2}+s\left( i-j\right)
}^{k_{2}}\left( \frac{z}{\beta ^{s}}\right) ^{k_{1}+k_{2}-i}}{%
\sum_{i=0}^{k_{1}+k_{2}+1}\left( -1\right) ^{\frac{\left( si+2(s+1)\right)
\left( i+1\right) }{2}}\dbinom{k_{1}+k_{2}+1}{i}_{F_{s}}\left( \frac{z}{%
\beta ^{s}}\right) ^{k_{1}+k_{2}+1-i}} \\
&amp;&amp; \\
&amp;=&amp;\frac{z}{5^{\frac{1}{2}}}\sum_{i=0}^{k_{1}+k_{2}}\sum_{j=0}^{i}\left(
-1\right) ^{\frac{\left( sj+2(s+1)\right) \left( j+1\right) }{2}}\dbinom{%
k_{1}+k_{2}+1}{j}_{F_{s}}F_{m_{1}+s\left( i-j\right)
}^{k_{1}}F_{m_{2}+s\left( i-j\right) }^{k_{2}}\times \\
&amp;&amp;\times \left( 
\begin{array}{c}
\dfrac{\alpha ^{m_{1}+si}}{\sum_{i=0}^{k_{1}+k_{2}+1}\left( -1\right) ^{%
\frac{\left( si+2(s+1)\right) \left( i+1\right) }{2}}\dbinom{k_{1}+k_{2}+1}{i%
}_{F_{s}}\alpha ^{si}z^{k_{1}+k_{2}+1-i}} \\ 
\\ 
-\text{ }\dfrac{\beta ^{m_{1}+si}}{\sum_{i=0}^{k_{1}+k_{2}+1}\left(
-1\right) ^{\frac{\left( si+2(s+1)\right) \left( i+1\right) }{2}}\dbinom{%
k_{1}+k_{2}+1}{i}_{F_{s}}\beta ^{si}z^{k_{1}+k_{2}+1-i}}%
\end{array}%
\right) z^{k_{1}+k_{2}-i} \\
&amp;&amp; \\
&amp;=&amp;\frac{z\sum_{i=0}^{k_{1}+k_{2}}\sum_{j=0}^{i}\left( -1\right) ^{\frac{%
\left( sj+2(s+1)\right) \left( j+1\right) }{2}}\dbinom{k_{1}+k_{2}+1}{j}%
_{F_{s}}F_{m_{1}+s\left( i-j\right) }^{k_{1}}F_{m_{2}+s\left( i-j\right)
}^{k_{2}}}{\sum_{i=0}^{k_{1}+k_{2}+2}\left( -1\right) ^{\frac{\left(
si+2(s+1)\right) \left( i+1\right) }{2}}\dbinom{k_{1}+k_{2}+2}{i}%
_{F_{s}}z^{k_{1}+k_{2}+2-i}}\times \\
&amp;&amp;\times \left( F_{m_{1}+si}z+\left( -1\right) ^{si+m_{1}}F_{s\left(
k_{1}+k_{2}-i+1\right) -m_{1}}\right) z^{k_{1}+k_{2}-i} \\
&amp;&amp; \\
&amp;=&amp;\frac{z}{\sum_{i=0}^{k_{1}+k_{2}+2}\left( -1\right) ^{\frac{\left(
si+2(s+1)\right) \left( i+1\right) }{2}}\dbinom{k_{1}+k_{2}+2}{i}%
_{F_{s}}z^{k_{1}+k_{2}+2-i}}\times \\
&amp;&amp;\times \left( 
\begin{array}{c}
\sum\limits_{i=0}^{k_{1}+k_{2}}\sum\limits_{j=0}^{i}\left( -1\right) ^{\frac{%
\left( sj+2(s+1)\right) \left( j+1\right) }{2}}\dbinom{k_{1}+k_{2}+1}{j}%
_{F_{s}}\times \\ 
\times F_{m_{1}+s\left( i-j\right) }^{k_{1}}F_{m_{2}+s\left( i-j\right)
}^{k_{2}}F_{m_{1}+si}z^{k_{1}+k_{2}+1-i} \\ 
+ \\ 
\sum\limits_{i=0}^{k_{1}+k_{2}}\sum\limits_{j=0}^{i}\left( -1\right) ^{\frac{%
\left( sj+2(s+1)\right) \left( j+1\right) }{2}}\dbinom{k_{1}+k_{2}+1}{j}%
_{F_{s}}\times \\ 
\times F_{m_{1}+s\left( i-j\right) }^{k_{1}}F_{m_{2}+s\left( i-j\right)
}^{k_{2}}\left( -1\right) ^{si+m_{1}}F_{s\left( k_{1}+k_{2}-i+1\right)
-m_{1}}z^{k_{1}+k_{2}-i}%
\end{array}%
\right)
\end{eqnarray*}

Some further simplifications give us%
\begin{eqnarray*}
&amp;&amp;\mathcal{Z\!}\left( F_{sn+m_{1}}^{k_{1}+1}F_{sn+m_{2}}^{k_{2}}\right) \!\!
\\
&amp;=&amp;\frac{z}{\sum_{i=0}^{k_{1}+k_{2}+2}\left( -1\right) ^{\frac{\left(
si+2(s+1)\right) \left( i+1\right) }{2}}\dbinom{k_{1}+k_{2}+2}{i}%
_{F_{s}}z^{k_{1}+k_{2}+2-i}}\times  \\
&amp;&amp;\times \left( 
\begin{array}{c}
\sum\limits_{i=0}^{k_{1}+k_{2}}\sum\limits_{j=0}^{i}\left( -1\right) ^{\frac{%
\left( sj+2(s+1)\right) \left( j+1\right) }{2}}\dbinom{k_{1}+k_{2}+1}{j}%
_{F_{s}}\times  \\ 
\times F_{m_{1}+s\left( i-j\right) }^{k_{1}}F_{m_{2}+s\left( i-j\right)
}^{k_{2}}F_{m_{1}+si}z^{k_{1}+k_{2}+1-i} \\ 
+ \\ 
\sum\limits_{i=1}^{k_{1}+k_{2}+1}\sum\limits_{j=1}^{i}\left( -1\right) ^{%
\frac{\left( sj-s+2(s+1)\right) j}{2}}\dbinom{k_{1}+k_{2}+1}{j-1}%
_{F_{s}}\times  \\ 
\times F_{m_{1}+s\left( i-j\right) }^{k_{1}}F_{m_{2}+s\left( i-j\right)
}^{k_{2}}\left( -1\right) ^{si-s+m_{1}}F_{s\left( k_{1}+k_{2}-i+2\right)
-m_{1}}z^{k_{1}+k_{2}+1-i}%
\end{array}%
\right)  \\
&amp;&amp; \\
&amp;=&amp;\frac{z}{\sum_{i=0}^{k_{1}+k_{2}+2}\left( -1\right) ^{\frac{\left(
si+2(s+1)\right) \left( i+1\right) }{2}}\dbinom{k_{1}+k_{2}+2}{i}%
_{F_{s}}z^{k_{1}+k_{2}+2-i}}\times  \\
&amp;&amp;\times \sum_{i=0}^{k_{1}+k_{2}+1}\sum_{j=0}^{i}\left( 
\begin{array}{c}
\left( -1\right) ^{\frac{\left( sj+2(s+1)\right) \left( j+1\right) }{2}%
}F_{m_{1}+s\left( i-j\right) }^{k_{1}}F_{m_{2}+s\left( i-j\right) }^{k_{2}}%
\dbinom{k_{1}+k_{2}+2}{j}_{F_{s}}\frac{1}{F_{s\left( k_{1}+k_{2}+2\right) }}%
\times  \\ 
\times \left( F_{s\left( k_{1}+k_{2}+2-j\right) }F_{m_{1}+si}+\left(
-1\right) ^{m_{1}+1+s\left( i-j\right) }F_{sj}F_{s\left(
k_{1}+k_{2}-i+2\right) -m_{1}}\right) 
\end{array}%
\right) z^{k_{1}+k_{2}-i+1} \\
&amp;&amp; \\
&amp;=&amp;z\frac{\sum_{i=0}^{k_{1}+k_{2}+1}\sum_{j=0}^{i}\left( -1\right) ^{\frac{%
\left( sj+2(s+1)\right) \left( j+1\right) }{2}}F_{m_{1}+s\left( i-j\right)
}^{k_{1}+1}F_{m_{2}+s\left( i-j\right) }^{k_{2}}\dbinom{k_{1}+k_{2}+2}{j}%
_{F_{s}}z^{k_{1}+k_{2}-i+1}}{\sum_{i=0}^{k_{1}+k_{2}+2}\left( -1\right) ^{%
\frac{\left( si+2(s+1)\right) \left( i+1\right) }{2}}\dbinom{k_{1}+k_{2}+2}{i%
}_{F_{s}}z^{k_{1}+k_{2}+2-i}},
\end{eqnarray*}%
as wanted. We used that%
\begin{equation*}
\sum_{j=0}^{k_{1}+k_{2}+1}\left( -1\right) ^{\frac{\left( sj+2(s+1)\right)
\left( j+1\right) }{2}}\dbinom{k_{1}+k_{2}+1}{j}_{F_{s}}F_{m_{1}+s\left(
k_{1}+k_{2}+1-j\right) }^{k_{1}}F_{m_{2}+s\left( k_{1}+k_{2}+1-j\right)
}^{k_{2}}=0,
\end{equation*}%
(which comes from the induction hypothesis). We also used in the last step
that%
\begin{equation*}
F_{s\left( k_{1}+k_{2}+2-j\right) }F_{m_{1}+si}+\left( -1\right)
^{m_{1}+1+s\left( i-j\right) }F_{sj}F_{s\left( k_{1}+k_{2}-i+2\right)
-m_{1}}=F_{s\left( k_{1}+k_{2}+2\right) }F_{m_{1}+s\left( i-j\right) },
\end{equation*}%
which is a direct consequence of the known identity $%
F_{m}F_{n}-F_{m+k}F_{n-k}=\left( -1\right) ^{n-k}F_{m+k-n}F_{k}$.
\end{proof}

Our second important result of this section is the following theorem.

\begin{theorem}
\label{Th3.2}\textit{Let }$m_{1},m_{2}\in \mathbb{Z}$\textit{\ and }$%
t_{1},t_{2}\in \mathbb{N}^{\prime }$\textit{\ be given. The }$Z$\textit{\
transform of the sequence }$%
F_{t_{1}sn+m_{1}}^{k_{1}}F_{t_{2}sn+m_{2}}^{k_{2}}$\textit{\ is given by}%
\begin{eqnarray}
&amp;&amp;\mathcal{Z}\left( F_{t_{1}sn+m_{1}}F_{t_{2}sn+m_{2}}\right)  \label{3.5} \\
&amp;=&amp;z\frac{\sum_{i=0}^{t_{1}+t_{2}}\sum_{j=0}^{i}\left( -1\right) ^{\frac{%
\left( sj+2(s+1)\right) \left( j+1\right) }{2}}\dbinom{t_{1}+t_{2}+1}{j}%
_{F_{s}}F_{m_{1}+t_{1}s\left( i-j\right) }F_{m_{2}+t_{2}s\left( i-j\right)
}z^{t_{1}+t_{2}-i}}{\sum_{i=0}^{t_{1}+t_{2}+1}\left( -1\right) ^{\frac{%
\left( si+2(s+1)\right) \left( i+1\right) }{2}}\dbinom{t_{1}+t_{2}+1}{i}%
_{F_{s}}z^{t_{1}+t_{2}+1-i}}.  \notag
\end{eqnarray}
\end{theorem}

\begin{proof}
We will proceed by induction on the parameters $t_{1}$ and/or $t_{2}$.
Observe that case $t_{1}=t_{2}=0$ is trivial (it is essentially (\ref{2.7}))
and in the case $t_{1}=t_{2}=1$ the result is true by theorem \ref{Th3.1}.
Suppose now the result is true for a given $t_{1}$ together with all its
values $\leq t_{1}$, and let us prove it is also true for $t_{1}+1$. (As in
the proof of \ref{Th3.1}, we have a symmetry property that allows us to
proceed in this way.) We will use that%
\begin{equation*}
F_{\left( t_{1}+1\right) sn+m_{1}}=F_{t_{1}sn+m_{1}}L_{sn}-\left( -1\right)
^{sn}F_{\left( t_{1}-1\right) sn+m_{1}},
\end{equation*}%
(easy to prove). Then we have that%
\begin{eqnarray}
\mathcal{Z}\left( F_{\left( t_{1}+1\right) sn+m_{1}}F_{t_{2}sn+m_{2}}\right)
&amp;=&amp;\mathcal{Z}\left( \left( F_{t_{1}sn+m_{1}}L_{sn}-\left( -1\right)
^{sn}F_{\left( t_{1}-1\right) sn+m_{1}}\right) F_{t_{2}sn+m_{2}}\right)
\label{3.6} \\
&amp;=&amp;\mathcal{Z}\left( L_{sn}F_{t_{1}sn+m_{1}}F_{t_{2}sn+m_{2}}\right) -%
\mathcal{Z}\left( \left( -1\right) ^{sn}F_{\left( t_{1}-1\right)
sn+m_{1}}F_{t_{2}sn+m_{2}}\right) .  \notag
\end{eqnarray}

Now we use (\ref{2.2}) and (\ref{2.5}), together with the induction
hypothesis to write%
\begin{eqnarray*}
&amp;&amp;\mathcal{Z}\left( F_{\left( t_{1}+1\right)
sn+m_{1}}F_{t_{2}sn+m_{2}}\right) \\
&amp;=&amp;\frac{z}{\alpha ^{s}}\frac{\sum_{i=0}^{t_{1}+t_{2}}\sum_{j=0}^{i}\left(
-1\right) ^{\frac{\left( sj+2(s+1)\right) \left( j+1\right) }{2}}\dbinom{%
t_{1}+t_{2}+1}{j}_{F_{s}}F_{m_{1}+t_{1}s\left( i-j\right)
}F_{m_{2}+t_{2}s\left( i-j\right) }\left( \frac{z}{\alpha ^{s}}\right)
^{t_{1}+t_{2}-i}}{\sum_{i=0}^{t_{1}+t_{2}+1}\left( -1\right) ^{\frac{\left(
si+2(s+1)\right) \left( i+1\right) }{2}}\dbinom{t_{1}+t_{2}+1}{i}%
_{F_{s}}\left( \frac{z}{\alpha ^{s}}\right) ^{t_{1}+t_{2}+1-i}} \\
&amp;&amp;+\frac{z}{\beta ^{s}}\frac{\sum_{i=0}^{t_{1}+t_{2}}\sum_{j=0}^{i}\left(
-1\right) ^{\frac{\left( sj+2(s+1)\right) \left( j+1\right) }{2}}\dbinom{%
t_{1}+t_{2}+1}{j}_{F_{s}}F_{m_{1}+t_{1}s\left( i-j\right)
}F_{m_{2}+t_{2}s\left( i-j\right) }\left( \frac{z}{\beta ^{s}}\right)
^{t_{1}+t_{2}-i}}{\sum_{i=0}^{t_{1}+t_{2}+1}\left( -1\right) ^{\frac{\left(
si+2(s+1)\right) \left( i+1\right) }{2}}\dbinom{t_{1}+t_{2}+1}{i}%
_{F_{s}}\left( \frac{z}{\beta ^{s}}\right) ^{t_{1}+t_{2}+1-i}} \\
&amp;&amp;-\frac{z}{\left( -1\right) ^{s}}\frac{\sum_{i=0}^{t_{1}+t_{2}-1}%
\sum_{j=0}^{i}\left( -1\right) ^{\frac{\left( sj+2(s+1)\right) \left(
j+1\right) }{2}}\dbinom{t_{1}+t_{2}}{j}_{F_{s}}F_{m_{1}+\left(
t_{1}-1\right) s\left( i-j\right) }F_{m_{2}+t_{2}s\left( i-j\right) }\left( 
\frac{z}{\left( -1\right) ^{s}}\right) ^{t_{1}-1+t_{2}-i}}{%
\sum_{i=0}^{t_{1}+t_{2}}\left( -1\right) ^{\frac{\left( si+2(s+1)\right)
\left( i+1\right) }{2}}\dbinom{t_{1}+t_{2}}{i}_{F_{s}}\left( \frac{z}{\left(
-1\right) ^{s}}\right) ^{t_{1}+t_{2}-i}} \\
&amp;=&amp;z\sum_{i=0}^{t_{1}+t_{2}}\sum_{j=0}^{i}\left( -1\right) ^{\frac{\left(
sj+2(s+1)\right) \left( j+1\right) }{2}}\dbinom{t_{1}+t_{2}+1}{j}%
_{F_{s}}F_{m_{1}+t_{1}s\left( i-j\right) }F_{m_{2}+t_{2}s\left( i-j\right)
}\times \\
&amp;&amp;\times \left( 
\begin{array}{c}
\dfrac{\alpha ^{si}}{\sum_{i=0}^{t_{1}+t_{2}+1}\left( -1\right) ^{\frac{%
\left( si+2(s+1)\right) \left( i+1\right) }{2}}\dbinom{t_{1}+t_{2}+1}{i}%
_{F_{s}}\alpha ^{si}z^{t_{1}+t_{2}+1-i}} \\ 
\\ 
+\text{ }\dfrac{\beta ^{si}}{\sum_{i=0}^{t_{1}+t_{2}+1}\left( -1\right) ^{%
\frac{\left( si+2(s+1)\right) \left( i+1\right) }{2}}\dbinom{t_{1}+t_{2}+1}{i%
}_{F_{s}}\beta ^{si}z^{t_{1}+t_{2}+1-i}}%
\end{array}%
\right) z^{t_{1}+t_{2}-i} \\
&amp;&amp;-z\frac{\sum_{i=0}^{t_{1}+t_{2}-1}\sum_{j=0}^{i}\left( -1\right) ^{\frac{%
\left( sj+2(s+1)\right) \left( j+1\right) }{2}}\dbinom{t_{1}+t_{2}}{j}%
_{F_{s}}F_{m_{1}+\left( t_{1}-1\right) s\left( i-j\right)
}F_{m_{2}+t_{2}s\left( i-j\right) }\left( -1\right) ^{si}z^{t_{1}-1+t_{2}-i}%
}{\sum_{i=0}^{t_{1}+t_{2}}\left( -1\right) ^{\frac{\left( si+2(s+1)\right)
\left( i+1\right) }{2}}\dbinom{t_{1}+t_{2}}{i}_{F_{s}}\left( -1\right)
^{si}z^{t_{1}+t_{2}-i}}.
\end{eqnarray*}

Next we use proposition \ref{Prop2.2} (a) to obtain that%
\begin{eqnarray*}
&amp;&amp;\mathcal{Z}\left( F_{\left( t_{1}+1\right)
sn+m_{1}}F_{t_{2}sn+m_{2}}\right) \\
&amp;=&amp;\frac{z}{\sum_{i=0}^{t_{1}+t_{2}+2}\left( -1\right) ^{\frac{\left(
si+2(s+1)\right) \left( i+1\right) }{2}}\dbinom{t_{1}+t_{2}+2}{i}%
_{F_{s}}z^{t_{1}+t_{2}+2-i}}\times \\
&amp;&amp;\times \sum_{i=0}^{t_{1}+t_{2}}\sum_{j=0}^{i}\left( -1\right) ^{\frac{%
\left( sj+2(s+1)\right) \left( j+1\right) }{2}}\dbinom{t_{1}+t_{2}+1}{j}%
_{F_{s}}F_{m_{1}+t_{1}s\left( i-j\right) }F_{m_{2}+t_{2}s\left( i-j\right)
}\times \\
&amp;&amp;\times \left( L_{si}z+\left( -1\right) ^{si+1}L_{s\left(
t_{1}+t_{2}-i+1\right) }\right) z^{t_{1}+t_{2}-i} \\
&amp;&amp;-z\frac{\sum_{i=0}^{t_{1}+t_{2}-1}\sum_{j=0}^{i}\left( -1\right) ^{\frac{%
\left( sj+2(s+1)\right) \left( j+1\right) }{2}}\dbinom{t_{1}+t_{2}}{j}%
_{F_{s}}F_{m_{1}+\left( t_{1}-1\right) s\left( i-j\right)
}F_{m_{2}+t_{2}s\left( i-j\right) }\left( -1\right) ^{si}z^{t_{1}-1+t_{2}-i}%
}{\sum_{i=0}^{t_{1}+t_{2}}\left( -1\right) ^{\frac{\left( si+2(s+1)\right)
\left( i+1\right) }{2}}\dbinom{t_{1}+t_{2}}{i}_{F_{s}}\left( -1\right)
^{si}z^{t_{1}+t_{2}-i}},
\end{eqnarray*}%
and then, by proposition \ref{Prop2.4} we have%
\begin{eqnarray}
&amp;&amp;\mathcal{Z}\left( F_{\left( t_{1}+1\right)
sn+m_{1}}F_{t_{2}sn+m_{2}}\right)  \label{3.7} \\
&amp;=&amp;\frac{z}{\sum_{i=0}^{t_{1}+t_{2}+2}\left( -1\right) ^{\frac{\left(
si+2(s+1)\right) \left( i+1\right) }{2}}\dbinom{t_{1}+t_{2}+2}{i}%
_{F_{s}}z^{t_{1}+t_{2}+2-i}}\times  \notag \\
&amp;&amp;\times \left( 
\begin{array}{c}
\sum\limits_{i=0}^{t_{1}+t_{2}}\sum\limits_{j=0}^{i}\left( -1\right) ^{\frac{%
\left( sj+2(s+1)\right) \left( j+1\right) }{2}}\dbinom{t_{1}+t_{2}+1}{j}%
_{F_{s}}F_{m_{1}+t_{1}s\left( i-j\right) }F_{m_{2}+t_{2}s\left( i-j\right)
}\times \\ 
\times \left( L_{si}z+\left( -1\right) ^{si+1}L_{s\left(
t_{1}+t_{2}-i+1\right) }\right) z^{t_{1}+t_{2}-i} \\ 
\\ 
-\sum\limits_{i=0}^{t_{1}+t_{2}-1}\sum\limits_{j=0}^{i}\left( -1\right) ^{%
\frac{\left( sj+2(s+1)\right) \left( j+1\right) }{2}}\dbinom{t_{1}+t_{2}}{j}%
_{F_{s}}F_{m_{1}+\left( t_{1}-1\right) s\left( i-j\right)
}F_{m_{2}+t_{2}s\left( i-j\right) }\times \\ 
\times \left( -1\right) ^{si}z^{t_{1}-1+t_{2}-i}\left( z^{2}-L_{s\left(
t_{1}+t_{2}+1\right) }z+\left( -1\right) ^{s\left( t_{1}+t_{2}+1\right)
}\right)%
\end{array}%
\right) .  \notag
\end{eqnarray}

We have now the expected denominator (of (\ref{3.5}) with $t_{1}$ replaced
by $t_{1}+1$). Let us work with the numerator of (\ref{3.7}), $z\left(
A-B\right) $ say, where%
\begin{eqnarray*}
A &amp;=&amp;\sum_{i=0}^{t_{1}+t_{2}}\sum_{j=0}^{i}\left( -1\right) ^{\frac{\left(
sj+2(s+1)\right) \left( j+1\right) }{2}}\dbinom{t_{1}+t_{2}+1}{j}%
_{F_{s}}F_{m_{1}+t_{1}s\left( i-j\right) }F_{m_{2}+t_{2}s\left( i-j\right)
}\times  \\
&amp;&amp;\times \left( L_{si}z+\left( -1\right) ^{si+1}L_{s\left(
t_{1}+t_{2}-i+1\right) }\right) z^{t_{1}+t_{2}-i},
\end{eqnarray*}%
and%
\begin{eqnarray*}
B &amp;=&amp;\sum_{i=0}^{t_{1}+t_{2}-1}\sum_{j=0}^{i}\left( -1\right) ^{\frac{\left(
sj+2(s+1)\right) \left( j+1\right) }{2}}\dbinom{t_{1}+t_{2}}{j}%
_{F_{s}}F_{m_{1}+\left( t_{1}-1\right) s\left( i-j\right)
}F_{m_{2}+t_{2}s\left( i-j\right) }\times  \\
&amp;&amp;\times \left( -1\right) ^{si}z^{t_{1}-1+t_{2}-i}\left( z^{2}-L_{s\left(
t_{1}+t_{2}+1\right) }z+\left( -1\right) ^{s\left( t_{1}+t_{2}+1\right)
}\right) .
\end{eqnarray*}

We have that%
\begin{eqnarray*}
A &amp;=&amp;\sum_{i=0}^{t_{1}+t_{2}}\sum_{j=0}^{i}\left( -1\right) ^{\frac{\left(
sj+2(s+1)\right) \left( j+1\right) }{2}}\dbinom{t_{1}+t_{2}+1}{j}%
_{F_{s}}F_{m_{1}+t_{1}s\left( i-j\right) }F_{m_{2}+t_{2}s\left( i-j\right)
}L_{si}z^{t_{1}+t_{2}+1-i} \\
&amp;&amp;+\sum_{i=1}^{t_{1}+t_{2}+1}\sum_{j=1}^{i}\left( -1\right) ^{\frac{\left(
sj-s+2(s+1)\right) j}{2}}\dbinom{t_{1}+t_{2}+1}{j-1}_{F_{s}}\times \\
&amp;&amp;\times F_{m_{1}+t_{1}s\left( i-j\right) }F_{m_{2}+t_{2}s\left( i-j\right)
}\left( -1\right) ^{s\left( i-1\right) +1}L_{s\left( t_{1}+t_{2}-i+2\right)
}z^{t_{1}+t_{2}+1-i} \\
&amp;=&amp;\sum_{i=0}^{t_{1}+t_{2}+1}\sum_{j=0}^{i}\left( -1\right) ^{\frac{\left(
sj+2(s+1)\right) \left( j+1\right) }{2}}\dbinom{t_{1}+t_{2}+2}{j}_{F_{s}}%
\frac{F_{m_{2}+t_{2}s\left( i-j\right) }F_{m_{1}+t_{1}s\left( i-j\right) }}{%
F_{s\left( t_{1}+t_{2}+2\right) }}\times \\
&amp;&amp;\times \left( F_{s\left( t_{1}+t_{2}+2-j\right) }L_{si}+\left( -1\right)
^{s\left( i-j\right) }F_{sj}L_{s\left( t_{1}+t_{2}-i+2\right) }\right)
z^{t_{1}+1+t_{2}-i}.
\end{eqnarray*}

But we have the identity%
\begin{equation*}
F_{s\left( t_{1}+t_{2}+2-j\right) }L_{si}+\left( -1\right) ^{s\left(
i-j\right) }F_{sj}L_{s\left( t_{1}+t_{2}-i+2\right) }=L_{s\left( i-j\right)
}F_{s\left( t_{1}+t_{2}+2\right) },
\end{equation*}%
thus%
\begin{equation*}
A=\sum_{i=0}^{t_{1}+t_{2}+1}\sum_{j=0}^{i}\left( -1\right) ^{\frac{\left(
sj+2(s+1)\right) \left( j+1\right) }{2}}\dbinom{t_{1}+t_{2}+2}{j}%
_{F_{s}}F_{m_{2}+t_{2}s\left( i-j\right) }F_{m_{1}+t_{1}s\left( i-j\right)
}L_{s\left( i-j\right) }z^{t_{1}+1+t_{2}-i}.
\end{equation*}

Now let us work with $B:$%
\begin{eqnarray*}
&amp;&amp;B=\sum_{i=0}^{t_{1}+t_{2}-1}\sum_{j=0}^{i}\left( -1\right) ^{\frac{\left(
sj+2(s+1)\right) \left( j+1\right) }{2}}\dbinom{t_{1}+t_{2}}{j}%
_{F_{s}}F_{m_{1}+\left( t_{1}-1\right) s\left( i-j\right)
}F_{m_{2}+t_{2}s\left( i-j\right) }\left( -1\right) ^{si}z^{t_{1}+1+t_{2}-i}
\\
&amp;&amp;-\sum_{i=1}^{t_{1}+t_{2}}\sum_{j=1}^{i}\left( -1\right) ^{\frac{\left(
sj-s+2(s+1)\right) j}{2}}\dbinom{t_{1}+t_{2}}{j-1}_{F_{s}}\left( 
\begin{array}{c}
F_{m_{1}+\left( t_{1}-1\right) s\left( i-j\right) }F_{m_{2}+t_{2}s\left(
i-j\right) }\times \\ 
\times L_{s\left( t_{1}+t_{2}+1\right) }\left( -1\right) ^{s\left(
i-1\right) }z^{t_{1}+t_{2}+1-i}%
\end{array}%
\right) \\
&amp;&amp;+\sum_{i=2}^{t_{1}+t_{2}+1}\sum_{j=2}^{i}\left( -1\right) ^{\frac{\left(
sj-2s+2(s+1)\right) \left( j-1\right) }{2}}\dbinom{t_{1}+t_{2}}{j-2}%
_{F_{s}}\left( 
\begin{array}{c}
F_{m_{1}+\left( t_{1}-1\right) s\left( i-j\right) }F_{m_{2}+t_{2}s\left(
i-j\right) }\times \\ 
\times \left( -1\right) ^{si}z^{t_{1}+1+t_{2}-i}\left( -1\right) ^{s\left(
t_{1}+t_{2}+1\right) }%
\end{array}%
\right) \\
&amp;=&amp;\sum_{i=0}^{t_{1}+t_{2}+1}\sum_{j=0}^{i}\left( -1\right) ^{\frac{\left(
sj+2(s+1)\right) \left( j+1\right) }{2}}\dbinom{t_{1}+t_{2}+2}{j}_{F_{s}}%
\frac{F_{m_{2}+t_{2}s\left( i-j\right) }F_{m_{1}+\left( t_{1}-1\right)
s\left( i-j\right) }}{F_{s\left( t_{1}+t_{2}+2\right) }F_{s\left(
t_{1}+t_{2}+1\right) }}\times \\
&amp;&amp;\times \left( -1\right) ^{s\left( i-j\right) }\left( 
\begin{array}{c}
\left( -1\right) ^{sj}F_{s\left( t_{1}+t_{2}+1-j\right) }F_{s\left(
t_{1}+t_{2}+2-j\right) } \\ 
+F_{sj}F_{s\left( t_{1}+t_{2}+2-j\right) }L_{s\left( t_{1}+t_{2}+1\right) }
\\ 
+\left( -1\right) ^{s\left( t_{1}+t_{2}+j\right) }F_{sj}F_{s\left(
j-1\right) }%
\end{array}%
\right) z^{t_{1}+1+t_{2}-i} \\
&amp;=&amp;\sum_{i=0}^{t_{1}+t_{2}+1}\sum_{j=0}^{i}\left( -1\right) ^{\frac{\left(
sj+2(s+1)\right) \left( j+1\right) }{2}}\dbinom{t_{1}+t_{2}+2}{j}%
_{F_{s}}\left( -1\right) ^{s\left( i-j\right) }F_{m_{2}+t_{2}s\left(
i-j\right) }F_{m_{1}+\left( t_{1}-1\right) s\left( i-j\right)
}z^{t_{1}+1+t_{2}-i}.
\end{eqnarray*}

(We used lemma \ref{Lem2.3} in the last step.)

Thus, the numerator of (\ref{3.7}) is%
\begin{eqnarray}
z\left( A-B\right)  &amp;=&amp;z\sum_{i=0}^{t_{1}+t_{2}+1}\sum_{j=0}^{i}\left(
-1\right) ^{\frac{\left( sj+2(s+1)\right) \left( j+1\right) }{2}}\dbinom{%
t_{1}+t_{2}+2}{j}_{F_{s}}F_{m_{2}+t_{2}s\left( i-j\right) }\times 
\label{3.8} \\
&amp;&amp;\times \left( F_{m_{1}+t_{1}s\left( i-j\right) }L_{s\left( i-j\right)
}-\left( -1\right) ^{s\left( i-j\right) }F_{m_{1}+\left( t_{1}-1\right)
s\left( i-j\right) }\right) z^{t_{1}+1+t_{2}-i}.  \notag
\end{eqnarray}

Finally, with a simple calculation with $\alpha $'s and $\beta $'s (left to
the reader), we see that%
\begin{equation*}
F_{m_{1}+t_{1}s\left( i-j\right) }L_{s\left( i-j\right) }-\left( -1\right)
^{s\left( i-j\right) }F_{m_{1}+\left( t_{1}-1\right) s\left( i-j\right)
}=F_{m_{1}+\left( t_{1}+1\right) s\left( i-j\right) },
\end{equation*}%
so (\ref{3.8}) is%
\begin{equation*}
z\left( A-B\right) =z\sum_{i=0}^{t_{1}+t_{2}+1}\sum_{j=0}^{i}\left(
-1\right) ^{\frac{\left( sj+2(s+1)\right) \left( j+1\right) }{2}}\dbinom{%
t_{1}+t_{2}+2}{j}_{F_{s}}F_{m_{1}+\left( t_{1}+1\right) s\left( i-j\right)
}F_{m_{2}+t_{2}s\left( i-j\right) }z^{t_{1}+1+t_{2}-i},
\end{equation*}%
which is the expected numerator (of (\ref{3.5}) with $t_{1}$ replaced by $%
t_{1}+1$). This ends our induction argument.
\end{proof}

The natural generalization of (\ref{3.2}) (theorem \ref{Th3.1}) is clearly
as follows:%
\begin{eqnarray}
&amp;&amp;\mathcal{Z}\left( F_{sn+m_{1}}^{k_{1}}\cdots F_{sn+m_{l}}^{k_{l}}\right)
\label{3.9} \\
&amp;=&amp;z\frac{\sum_{i=0}^{k_{1}+\cdots +k_{l}}\sum_{j=0}^{i}\left( -1\right) ^{%
\frac{\left( sj+2(s+1)\right) \left( j+1\right) }{2}}\dbinom{k_{1}+\cdots
+k_{l}+1}{j}_{F_{s}}F_{m_{1}+s\left( i-j\right) }^{k_{1}}\cdots
F_{m_{l}+s\left( i-j\right) }^{k_{l}}z^{k_{1}+\cdots +k_{l}-i}}{%
\sum_{i=0}^{k_{1}+\cdots +k_{l}+1}\left( -1\right) ^{\frac{\left(
si+2(s+1)\right) \left( i+1\right) }{2}}\dbinom{k_{1}+\cdots +k_{l}+1}{i}%
_{F_{s}}z^{k_{1}+\cdots +k_{l}+1-i}},  \notag
\end{eqnarray}%
(which can be proved with the same sort of arguments used in the proof of (%
\ref{3.2})), and the natural generalization of (\ref{3.5}) (theorem \ref%
{Th3.2}) is%
\begin{eqnarray}
&amp;&amp;\mathcal{Z}\left( F_{t_{1}sn+m_{1}}\cdots F_{t_{l}sn+m_{l}}\right)
\label{3.10} \\
&amp;=&amp;z\frac{\sum_{i=0}^{t_{1}+\cdots +t_{l}}\sum_{j=0}^{i}\left( -1\right) ^{%
\frac{\left( sj+2(s+1)\right) \left( j+1\right) }{2}}\dbinom{t_{1}+\cdots
+t_{l}+1}{j}_{F_{s}}F_{m_{1}+t_{1}s\left( i-j\right) }\cdots
F_{m_{l}+t_{l}s\left( i-j\right) }z^{t_{1}+\cdots +t_{l}-i}}{%
\sum_{i=0}^{t_{1}+\cdots +t_{l}+1}\left( -1\right) ^{\frac{\left(
si+2(s+1)\right) \left( i+1\right) }{2}}\dbinom{t_{1}+\cdots +t_{l}+1}{i}%
_{F_{s}}z^{t_{1}+\cdots +t_{l}+1-i}}.  \notag
\end{eqnarray}

But then, from (\ref{3.10}) we see that%
\begin{eqnarray}
&amp;&amp;\mathcal{Z}\left( F_{t_{1}sn+m_{1}}^{k_{1}}\cdots
F_{t_{l}sn+m_{l}}^{k_{l}}\right)  \label{3.11} \\
&amp;=&amp;z\frac{\sum\limits_{i=0}^{k_{1}t_{1}+\cdots
+k_{l}t_{l}}\!\sum\limits_{j=0}^{i}\left( -1\right) ^{\frac{\left(
sj+2(s+1)\right) \left( j+1\right) }{2}}\!\dbinom{k_{1}t_{1}+\cdots
+k_{l}t_{l}+1}{j}_{F_{s}}\!F_{m_{1}+t_{1}s\left( i-j\right)
}^{k_{1}}\!\cdots \!F_{m_{l}+t_{l}s\left( i-j\right)
}^{k_{l}}z^{k_{1}t_{1}+\cdots +k_{l}t_{l}-i}}{\sum\limits_{i=0}^{k_{1}t_{1}+%
\cdots +k_{l}t_{l}+1}\left( -1\right) ^{\frac{\left( si+2(s+1)\right) \left(
i+1\right) }{2}}\dbinom{k_{1}t_{1}+\cdots +k_{l}t_{l}+1}{i}%
_{F_{s}}z^{k_{1}t_{1}+\cdots +k_{l}t_{l}+1-i}}.  \notag
\end{eqnarray}

That is, the generalization (\ref{3.9}) is indeed a particular case of (\ref%
{3.11}). This is the result we wanted to prove in this section.

\section{\label{Sec4}Some corollaries}

Our starting point in this section is formula (\ref{3.11}). We want to
present here some consequences of it.

\begin{corollary}
\label{Cor4.1}\textit{For }$p\in \mathbb{N}^{\prime }$\textit{\ given, the }$%
Z$\textit{\ transform of the sequence }$\binom{n}{p}_{F_{s}}$\textit{\ is}%
\begin{equation}
\mathcal{Z}\left( \dbinom{n}{p}_{F_{s}}\right) =\frac{\left( -1\right)
^{s+1}z}{D_{s,p+1}\left( z\right) }.  \label{4.1}
\end{equation}
\end{corollary}

\begin{proof}
We use (\ref{3.11}) to write%
\begin{eqnarray*}
\mathcal{Z}\left( \dbinom{n}{p}_{F_{s}}\right) &amp;=&amp;\frac{1}{\left(
F_{p}!\right) _{s}}\mathcal{Z}\left( F_{s\left( n-p+1\right) }F_{s\left(
n-p+2\right) }\cdots F_{sn}\right) \\
&amp;=&amp;\frac{1}{\left( F_{p}!\right) _{s}}z\frac{\sum_{i=0}^{p}\sum_{j=0}^{i}%
\left( -1\right) ^{\frac{\left( sj+2(s+1)\right) \left( j+1\right) }{2}}%
\dbinom{p+1}{j}_{F_{s}}F_{s\left( 1-p+i-j\right) }F_{s\left( 2-p+i-j\right)
}\cdots F_{s\left( i-j\right) }z^{p-i}}{\sum_{i=0}^{p+1}\left( -1\right) ^{%
\frac{\left( si+2(s+1)\right) \left( i+1\right) }{2}}\dbinom{p+1}{i}%
_{F_{s}}z^{p+1-i}}.
\end{eqnarray*}

But the product $F_{s\left( 1-p+i-j\right) }F_{s\left( 2-p+i-j\right)
}\cdots F_{s\left( i-j\right) }$ is different from zero if and only if $i=p$
and $j=0$. In such a case that product is $\left( F_{p}!\right) _{s}$ and
the numerator reduces to $\left( -1\right) ^{s+1}\left( F_{p}!\right) _{s}$.
Thus (\ref{4.1}) follows.
\end{proof}

(Formula (\ref{4.1}) corresponding to $s=1$ was proved by a different method
in \cite{St}.)

Observe that according to the advance-shifting property (\ref{2.1}) and (\ref%
{4.1}), if $0\leq p_{0}\leq p$, then%
\begin{equation}
\mathcal{Z}\left( \dbinom{n+p_{0}}{p}_{F_{s}}\right) =z^{p_{0}}\frac{\left(
-1\right) ^{s+1}z}{\sum_{i=0}^{p+1}\left( -1\right) ^{\frac{\left(
si+2(s+1)\right) \left( i+1\right) }{2}}\dbinom{p+1}{i}_{F_{s}}z^{p+1-i}}.
\label{4.2}
\end{equation}

By using that $G_{sn+m}=G_{m}F_{sn-1}+G_{m+1}F_{sn}$, we can see that
formula (\ref{3.11}) is valid for Gibonacci sequences $G_{n}$ replacing the
Fibonacci ones. In fact (it suffices to check the case $%
G_{st_{1}n+m_{1}}^{k_{1}}G_{st_{2}n+m_{2}}^{k_{2}}$), we have 
\begin{eqnarray*}
&amp;&amp;\mathcal{Z}\left(
G_{st_{1}n+m_{1}}^{k_{1}}G_{st_{2}n+m_{2}}^{k_{2}}\right)  \\
&amp;=&amp;\sum_{l_{1}=0}^{k_{1}}\sum_{l_{2}=0}^{k_{2}}\dbinom{k_{1}}{l_{1}}\dbinom{%
k_{2}}{l_{2}}%
G_{m_{1}}^{l_{1}}G_{m_{1}+1}^{k_{1}-l_{1}}G_{m_{2}}^{l_{2}}G_{m_{2}+1}^{k_{2}-l_{2}}%
\mathcal{Z}\left(
F_{st_{1}n-1}^{l_{1}}F_{st_{1}n}^{k_{1}-l_{1}}F_{st_{2}n-1}^{l_{2}}F_{st_{2}n}^{k_{2}-l_{2}}\right) 
\\
&amp;=&amp;\sum_{l_{1}=0}^{k_{1}}\sum_{l_{2}=0}^{k_{2}}\dbinom{k_{1}}{l_{1}}\dbinom{%
k_{2}}{l_{2}}%
G_{m_{1}}^{l_{1}}G_{m_{1}+1}^{k_{1}-l_{1}}G_{m_{2}}^{l_{2}}G_{m_{2}+1}^{k_{2}-l_{2}}z\times 
\\
&amp;&amp;\times \!\!\frac{\sum\limits_{i=0}^{t_{1}k_{1}+t_{2}k_{2}}\!\sum%
\limits_{j=0}^{i}\!\left( -1\right) ^{\frac{\left( sj+2(s+1)\right) \left(
j+1\right) }{2}}\!\!\dbinom{t_{1}k_{1}\!+\!t_{2}k_{2}\!+\!1}{j}%
_{F_{s}}\!\!\!F_{st_{1}\left( i-j\right) -1}^{l_{1}}F_{st_{1}\left(
i-j\right) }^{k_{1}-l_{1}}F_{st_{2}\left( i-j\right)
-1}^{l_{2}}F_{st_{2}\left( i-j\right)
}^{k_{2}-l_{2}}z^{t_{1}k_{1}+t_{2}k_{2}-i}}{\sum%
\limits_{i=0}^{t_{1}k_{1}+t_{2}k_{2}+1}\left( -1\right) ^{\frac{\left(
si+2(s+1)\right) \left( i+1\right) }{2}}\dbinom{t_{1}k_{1}+t_{2}k_{2}+1}{i}%
_{F_{s}}z^{t_{1}k_{1}+t_{2}k_{2}+1-i}} \\
&amp;=&amp;\frac{z}{\sum\limits_{i=0}^{t_{1}k_{1}+t_{2}k_{2}+1}\left( -1\right) ^{%
\frac{\left( si+2(s+1)\right) \left( i+1\right) }{2}}\dbinom{%
t_{1}k_{1}+t_{2}k_{2}+1}{i}_{F_{s}}z^{t_{1}k_{1}+t_{2}k_{2}+1-i}}\times  \\
&amp;&amp;\times \left( \!%
\begin{array}{c}
\sum\limits_{i=0}^{t_{1}k_{1}+t_{2}k_{2}}\sum\limits_{j=0}^{i}\left(
-1\right) ^{\frac{\left( sj+2(s+1)\right) \left( j+1\right) }{2}}\dbinom{%
t_{1}k_{1}+t_{2}k_{2}+1}{j}_{F_{s}}\times  \\ 
\times \!\!\sum\limits_{l_{1}=0}^{k_{1}}\sum\limits_{l_{2}=0}^{k_{2}}\!%
\dbinom{k_{1}}{l_{1}}\!\dbinom{k_{2}}{l_{2}}%
G_{m_{1}}^{l_{1}}G_{m_{1}+1}^{k_{1}-l_{1}}G_{m_{2}}^{l_{2}}G_{m_{2}+1}^{k_{2}-l_{2}}F_{st_{1}\left( i-j\right) -1}^{l_{1}}F_{st_{1}\left( i-j\right) }^{k_{1}-l_{1}}F_{st_{2}\left( i-j\right) -1}^{l_{2}}F_{st_{2}\left( i-j\right) }^{k_{2}-l_{2}}z^{t_{1}k_{1}+t_{2}k_{2}-i}%
\end{array}%
\!\right)  \\
&amp;&amp; \\
&amp;=&amp;z\frac{\sum\limits_{i=0}^{t_{1}k_{1}+t_{2}k_{2}}\sum\limits_{j=0}^{i}%
\left( -1\right) ^{\frac{\left( sj+2(s+1)\right) \left( j+1\right) }{2}}%
\dbinom{t_{1}k_{1}+t_{2}k_{2}+1}{j}_{F_{s}}G_{st_{1}\left( i-j\right)
+m_{1}}^{k_{1}}G_{st_{2}\left( i-j\right)
+m_{2}}^{k_{2}}z^{t_{1}k_{1}+t_{2}k_{2}-i}}{\sum%
\limits_{i=0}^{t_{1}k_{1}+t_{2}k_{2}+1}\left( -1\right) ^{\frac{\left(
si+2(s+1)\right) \left( i+1\right) }{2}}\dbinom{t_{1}k_{1}+t_{2}k_{2}+1}{j}%
_{F_{s}}z^{t_{1}k_{1}+t_{2}k_{2}+1-i}},
\end{eqnarray*}%
as expected. In general we have

\begin{eqnarray}
&amp;&amp;\mathcal{Z}\left( G_{st_{1}n+m_{1}}^{k_{1}}\cdots
G_{st_{l}n+m_{l}}^{k_{l}}\right)  \label{4.3} \\
&amp;=&amp;\!z\frac{\sum\limits_{i=0}^{t_{1}k_{1}+\cdots
+t_{l}k_{l}}\!\sum\limits_{j=0}^{i}\!\left( -1\right) ^{\frac{\left(
sj+2(s+1)\right) \left( j+1\right) }{2}}\!\!\dbinom{t_{1}k_{1}+\cdots
+t_{l}k_{l}+1}{j}_{F_{s}}\!\!G_{m_{1}+st_{1}\left( i-j\right)
}^{k_{1}}\cdots G_{m_{l}+st_{l}\left( i-j\right)
}^{k_{l}}z^{t_{1}k_{1}+\cdots +t_{l}k_{l}-i}}{\sum\limits_{i=0}^{t_{1}k_{1}+%
\cdots +t_{l}k_{l}+1}\left( -1\right) ^{\frac{\left( si+2(s+1)\right) \left(
i+1\right) }{2}}\dbinom{t_{1}k_{1}+\cdots +t_{l}k_{l}+1}{i}%
_{F_{s}}z^{t_{1}k_{1}+\cdots +t_{l}k_{l}+1-i}}.  \notag
\end{eqnarray}

\begin{corollary}
\label{Cor4.2}\textit{Let }$m_{1},\ldots ,m_{l}\in \mathbb{Z}$\textit{\ and }%
$t_{1},\ldots ,t_{l},k_{1},\ldots ,k_{l}\in \mathbb{N}^{\prime }$\textit{\
be given. The sequence }$\prod\limits_{i=1}^{l}G_{t_{i}sn+m_{i}}^{k_{i}}$%
\textit{\ can be expressed as a linear combination of }$s$-\textit{%
Fibonomials }$\binom{n+t_{1}k_{1}+\cdots +t_{l}k_{l}-i}{t_{1}k_{1}+\cdots
+t_{l}k_{l}}_{F_{s}}$\textit{, }$i=0,1,\ldots ,t_{1}k_{1}+\cdots +t_{l}k_{l}$%
, \textit{according to}%
\begin{eqnarray}
G_{t_{1}sn+m_{1}}^{k_{1}}\cdots G_{t_{l}sn+m_{l}}^{k_{l}} &amp;=&amp;\left(
-1\right) ^{s+1}\sum_{i=0}^{t_{1}k_{1}+\cdots
+t_{l}k_{l}}\sum_{j=0}^{i}\left( -1\right) ^{\frac{\left( sj+2(s+1)\right)
\left( j+1\right) }{2}}\dbinom{t_{1}k_{1}+\cdots +t_{l}k_{l}+1}{j}%
_{F_{s}}\times  \notag \\
&amp;&amp;\times G_{m_{1}+t_{1}s\left( i-j\right) }^{k_{1}}\cdots
G_{m_{l}+t_{l}s\left( i-j\right) }^{k_{l}}\dbinom{n+t_{1}k_{1}+\cdots
+t_{l}k_{l}-i}{t_{1}k_{1}+\cdots +t_{l}k_{l}}_{F_{s}}.  \label{4.4}
\end{eqnarray}
\end{corollary}

\begin{proof}
This follows directly from (\ref{4.3}) and (\ref{4.2}).
\end{proof}

Some examples of (\ref{4.4}) are the following (after simplifications on the
coefficients of the $s$-Fibonomials sequences of the right-hand side):

\begin{equation}
F_{2sn+m}=F_{m}\dbinom{n+2}{2}_{F_{s}}+\left( -1\right) ^{m}F_{s-m}L_{s}%
\dbinom{n+1}{2}_{F_{s}}+\left( -1\right) ^{s+m+1}F_{2s-m}\dbinom{n}{2}%
_{F_{s}}.  \label{4.5a}
\end{equation}

\begin{equation}
F_{sn}^{3}=F_{s}^{3}\left( \dbinom{n+2}{3}_{F_{s}}+2\left( -1\right)
^{s}L_{s}\dbinom{n+1}{3}_{F_{s}}+\left( -1\right) ^{s}\dbinom{n}{3}%
_{F_{s}}\right) .  \label{4.5}
\end{equation}

\begin{equation}
F_{2sn}^{2}=F_{2s}^{2}\left( \dbinom{n+3}{4}_{F_{s}}+\dbinom{n}{4}%
_{F_{s}}+\left( -1\right) ^{s+1}\frac{L_{3s}}{L_{s}}\left( \dbinom{n+2}{4}%
_{F_{s}}+\dbinom{n+1}{4}_{F_{s}}\right) \right) .  \label{4.6}
\end{equation}

\begin{equation}
L_{sn}^{2}=4\dbinom{n+2}{2}_{F_{s}}-\left( 3L_{s}^{2}+4\left( -1\right)
^{s+1}\right) \dbinom{n+1}{2}_{F_{s}}+\left( -1\right) ^{s}L_{s}^{2}\dbinom{n%
}{2}_{F_{s}}.  \label{4.7}
\end{equation}

\begin{equation}
F_{sn}^{4}=F_{s}^{4}\left( \dbinom{n+3}{4}_{F_{s}}+\dbinom{n}{4}%
_{F_{s}}+\left( 3\left( -1\right) ^{s}\frac{F_{3s}}{F_{s}}+2\right) \left( 
\dbinom{n+2}{4}_{F_{s}}+\dbinom{n+1}{4}_{F_{s}}\right) \right) .
\label{4.7b}
\end{equation}

\begin{equation}
L_{2sn}F_{sn}=F_{s}\left( L_{2s}\dbinom{n+2}{3}_{F_{s}}-2L_{s}\dbinom{n+1}{3}%
_{F_{s}}+\left( -1\right) ^{s}L_{2s}\dbinom{n}{3}_{F_{s}}\right) .
\label{4.8}
\end{equation}%
\begin{equation}
L_{2sn}F_{sn}^{2}=F_{s}^{2}\left( L_{2s}\left( \dbinom{n+3}{4}_{F_{s}}+%
\dbinom{n}{4}_{F_{s}}\right) +\left( -1\right) ^{s}\left(
5F_{s}F_{3s}-2\right) \left( \dbinom{n+2}{4}_{F_{s}}+\dbinom{n+1}{4}%
_{F_{s}}\right) \right) .  \label{4.9}
\end{equation}

From (\ref{4.4}) we see that%
\begin{equation}
G_{tsn+m}=\left( -1\right) ^{s+1}\sum_{i=0}^{t}\sum_{j=0}^{i}\left(
-1\right) ^{\frac{\left( sj+2(s+1)\right) \left( j+1\right) }{2}}\dbinom{t+1%
}{j}_{F_{s}}G_{ts\left( i-j\right) +m}\dbinom{n+t-i}{t}_{F_{s}}.
\label{4.10}
\end{equation}

But when $G=F$ or $G=L$, this formula can be written in a simpler form.

\begin{corollary}
\label{Cor4.3}\textit{Let }$t\in \mathbb{N}^{\prime }$\textit{\ be given.
The following identities hold}

(a)%
\begin{equation}
F_{tsn+m}=\sum\limits_{i=0}^{t}\left( -1\right) ^{\frac{is\left( i-1\right) 
}{2}+i+m+1}\dbinom{t}{i}_{F_{s}}F_{is-m}\dbinom{n+t-i}{t}_{F_{s}}.
\label{4.11}
\end{equation}

(b) 
\begin{equation}
L_{tsn+m}=\sum\limits_{i=0}^{t}\left( -1\right) ^{\frac{is(i-1)}{2}+i+m}%
\dbinom{t}{i}_{F_{s}}L_{is-m}\dbinom{n+t-i}{t}_{F_{s}}.  \label{4.11b}
\end{equation}
\end{corollary}

\begin{proof}
We have that%
\begin{eqnarray*}
F_{tsn+m} &amp;=&amp;\left( -1\right) ^{s+1}\sum_{i=0}^{t}\sum_{j=0}^{i}\left(
-1\right) ^{\frac{\left( sj+2\left( s+1\right) \right) \left( j+1\right) }{2}%
}\dbinom{t+1}{j}_{F_{s}}F_{ts\left( i-j\right) +m}\dbinom{n+t-i}{t}_{F_{s}}
\\
&amp;=&amp;\left( -1\right) ^{s+1}\sum_{i=0}^{t}\sum_{j=0}^{i}\left( -1\right) ^{%
\frac{\left( s\left( i-j\right) +2\left( s+1\right) \right) \left(
i-j+1\right) }{2}}\dbinom{t+1}{i-j}_{F_{s}}F_{tsj+m}\dbinom{n+t-i}{t}%
_{F_{s}}.
\end{eqnarray*}

According to proposition \ref{Prop2.5} we can write%
\begin{eqnarray*}
F_{tsn+m} &amp;=&amp;\left( -1\right) ^{s+1}\sum_{i=0}^{t}\left( -1\right) ^{\frac{is%
}{2}\left( i-1\right) +i+s+m}\dbinom{t}{i}_{F_{s}}F_{is-m}\dbinom{n+t-i}{t}%
_{F_{s}} \\
&amp;=&amp;\sum\limits_{i=0}^{t}\left( -1\right) ^{\frac{is}{2}\left( i-1\right)
+i+m+1}\dbinom{t}{i}_{F_{s}}F_{is-m}\dbinom{n+t-i}{t}_{F_{s}},
\end{eqnarray*}%
which proves (\ref{4.11}).

The proof of (\ref{4.11b}) is similar (using proposition \ref{Prop2.6}).
\end{proof}

In the following corollary we consider sequences involving \textquotedblleft 
$s$-Gibonomials\textquotedblright\ $\binom{n}{p}_{G_{s}}=\frac{G_{sn}\cdots
G_{s\left( n-p+1\right) }}{G_{s}\cdots G_{sp}}$.

\begin{corollary}
\label{Cor4.4}\textit{Let }$t_{1},\ldots ,t_{l}\in \mathbb{N}$ \textit{\ and
\ }$r_{1}\ldots ,r_{l},p_{1}\ldots ,p_{l}\in \mathbb{N}^{\prime }$\textit{\
be given. Then the }$Z$\textit{\ transform of the sequence }$\binom{n}{p_{1}}%
_{G_{st_{1}}}^{r_{1}}\cdots \binom{n}{p_{k}}_{G_{st_{k}}}^{r_{k}}$\textit{\
is given by}%
\begin{eqnarray}
&amp;&amp;\mathcal{Z}\left( \dbinom{n}{p_{1}}_{G_{st_{1}}}^{r_{1}}\cdots \dbinom{n}{%
p_{k}}_{G_{st_{k}}}^{r_{k}}\right)  \label{4.12} \\
&amp;=&amp;\frac{z}{\sum\limits_{i=0}^{t_{1}r_{1}p_{1}+\cdots
+t_{k}r_{k}p_{k}+1}\left( -1\right) ^{\frac{\left( si+2(s+1)\right) \left(
i+1\right) }{2}}\dbinom{t_{1}r_{1}p_{1}+\cdots +t_{k}r_{k}p_{k}+1}{i}%
_{F_{s}}z^{t_{1}r_{1}p_{1}+\cdots +t_{k}r_{k}p_{k}+1-i}}\times  \notag \\
&amp;&amp;\times \sum\limits_{i=0}^{t_{1}r_{1}p_{1}+\cdots
+t_{k}r_{k}p_{k}}\sum\limits_{j=0}^{i}\left( -1\right) ^{\frac{\left(
sj+2(s+1)\right) \left( j+1\right) }{2}}\dbinom{t_{1}r_{1}p_{1}+\cdots
+t_{k}r_{k}p_{k}+1}{j}_{F_{s}}\times  \notag \\
&amp;&amp;\times \dbinom{i-j}{p_{1}}_{G_{st_{1}}}^{r_{1}}\cdots \dbinom{i-j}{p_{k}}%
_{G_{st_{k}}}^{r_{k}}z^{t_{1}r_{1}p_{1}+\cdots +t_{k}r_{k}p_{k}-i}.  \notag
\end{eqnarray}
\end{corollary}

\begin{proof}
First we write%
\begin{eqnarray*}
&amp;&amp;\mathcal{Z}\left( \dbinom{n}{p_{1}}_{G_{st_{1}}}^{r_{1}}\cdots \dbinom{n}{%
p_{k}}_{G_{st_{k}}}^{r_{k}}\right) \\
&amp;=&amp;\frac{1}{G_{st_{1}}^{r_{1}}\cdots G_{st_{1}p_{1}}^{r_{1}}\cdots
G_{st_{k}}^{r_{k}}\cdots G_{st_{k}p_{k}}^{r_{k}}}\mathcal{Z}\left(
G_{st_{1}n}^{r_{1}}\cdots G_{st_{1}\left( n-p_{1}+1\right) }^{r_{1}}\cdots
G_{st_{k}n}^{r_{k}}\cdots G_{st_{k}\left( n-p_{k}+1\right) }^{r_{k}}\right) ,
\end{eqnarray*}%
and then we use (\ref{4.3}) to get%
\begin{eqnarray*}
&amp;&amp;\mathcal{Z}\left( \dbinom{n}{p_{1}}_{G_{st_{1}}}^{r_{1}}\cdots \dbinom{n}{%
p_{k}}_{G_{st_{k}}}^{r_{k}}\right) \\
&amp;=&amp;\frac{1}{G_{st_{1}}^{r_{1}}\cdots G_{st_{1}p_{1}}^{r_{1}}\cdots
G_{st_{k}}^{r_{k}}\cdots G_{st_{k}p_{k}}^{r_{k}}}\times \\
&amp;&amp;\times \frac{z}{\sum\limits_{i=0}^{t_{1}r_{1}p_{1}+\cdots
+t_{k}r_{k}p_{k}+1}\left( -1\right) ^{\frac{\left( si+2(s+1)\right) \left(
i+1\right) }{2}}\dbinom{t_{1}r_{1}p_{1}+\cdots +t_{k}r_{k}p_{k}+1}{i}%
_{F_{s}}z^{t_{1}r_{1}p_{1}+\cdots +t_{k}r_{k}p_{k}+1-i}}\times \\
&amp;&amp;\times \sum\limits_{i=0}^{t_{1}r_{1}p_{1}+\cdots
+t_{k}r_{k}p_{k}}\sum\limits_{j=0}^{i}\left( -1\right) ^{\frac{\left(
sj+2(s+1)\right) \left( j+1\right) }{2}}\dbinom{t_{1}r_{1}p_{1}+\cdots
+t_{k}r_{k}p_{k}+1}{j}_{F_{s}}\times \\
&amp;&amp;\times G_{st_{1}\left( i-j\right) }^{r_{1}}\cdots G_{st_{1}\left(
i-j-p_{1}+1\right) }^{r_{1}}\cdots G_{st_{k}\left( i-j\right)
}^{r_{k}}\cdots G_{st_{k}\left( i-j-p_{k}+1\right)
}^{r_{k}}z^{t_{1}r_{1}p_{1}+\cdots +t_{k}r_{k}p_{k}-i},
\end{eqnarray*}%
which implies the desired formula (\ref{4.12}).
\end{proof}

\begin{corollary}
\label{Cor4.5}\textit{Let }$t_{1},\ldots ,t_{l}\in \mathbb{N}$ \textit{\ and
\ }$r_{1}\ldots ,r_{l},p_{1}\ldots ,p_{l}\in \mathbb{N}^{\prime }$\textit{\
be given. The sequence }$\prod\limits_{i=1}^{k}\binom{n}{p_{i}}%
_{G_{st_{i}}}^{r_{i}}$\textit{\ can be expressed as a linear combination of }%
$s$-\textit{Fibonomials }$\binom{n+t_{1}r_{1}p_{1}+\cdots +t_{k}r_{k}p_{k}-i%
}{t_{1}r_{1}p_{1}+\cdots +t_{k}r_{k}p_{k}}_{F_{s}}$\textit{, }$i=0,1,\ldots
,t_{1}r_{1}p_{1}+\cdots +t_{k}r_{k}p_{k}$, \textit{according to}%
\begin{eqnarray}
\prod\limits_{i=1}^{k}\binom{n}{p_{i}}_{G_{st_{i}}}^{r_{i}} &amp;=&amp;\left(
-1\right) ^{s+1}\sum\limits_{i=0}^{t_{1}r_{1}p_{1}+\cdots
+t_{k}r_{k}p_{k}}\sum\limits_{j=0}^{i}\left( -1\right) ^{\frac{\left(
sj+2(s+1)\right) \left( j+1\right) }{2}}\dbinom{t_{1}r_{1}p_{1}+\cdots
+t_{k}r_{k}p_{k}+1}{j}_{F_{s}}\times   \notag \\
&amp;&amp;\times \dbinom{i-j}{p_{1}}_{G_{st_{1}}}^{r_{1}}\cdots \dbinom{i-j}{p_{k}}%
_{G_{st_{k}}}^{r_{k}}\dbinom{n+t_{1}r_{1}p_{1}+\cdots +t_{k}r_{k}p_{k}-i}{%
t_{1}r_{1}p_{1}+\cdots +t_{k}r_{k}p_{k}}_{F_{s}}.  \label{4.13}
\end{eqnarray}
\end{corollary}

\begin{proof}
This comes directly from (\ref{4.12}) and (\ref{4.2}).
\end{proof}

Some examples of (\ref{4.13}) are the following (after simplifications on
the coefficients of the $s$-Fibonomials sequences of the right-hand side):

\begin{equation}
\dbinom{n}{2}_{F_{2s}}=\dbinom{n+2}{4}_{F_{s}}+\left( -1\right) ^{s+1}L_{2s}%
\dbinom{n+1}{4}_{F_{s}}+\dbinom{n}{4}_{F_{s}}.  \label{4.14}
\end{equation}

\begin{equation}
\dbinom{n}{2}_{F_{s}}\dbinom{n}{3}_{F_{s}}=\frac{F_{3s}}{F_{s}}\dbinom{n+2}{5%
}_{F_{s}}+\frac{L_{s}F_{3s}}{F_{s}}\dbinom{n+1}{5}_{F_{s}}+\dbinom{n}{5}%
_{F_{s}}.  \label{4.15}
\end{equation}

\begin{equation}
\dbinom{n}{2}_{F_{s}}^{2}=\dbinom{n+2}{4}_{F_{s}}+\left( -1\right)
^{s}L_{s}^{2}\dbinom{n+1}{4}_{F_{s}}+\dbinom{n}{4}_{F_{s}}.  \label{4.16}
\end{equation}

\begin{equation}
\dbinom{n}{2}_{L_{s}}=\frac{2\left( -1\right) ^{s}}{L_{2s}}\dbinom{n+2}{2}%
_{F_{s}}+2\left( -1\right) ^{s+1}\dbinom{n+1}{2}_{F_{s}}+\dbinom{n}{2}%
_{F_{s}}.  \label{4.17}
\end{equation}

\begin{equation}
\dbinom{n}{2}_{L_{s}}\dbinom{n}{2}_{F_{s}}=\dbinom{n+2}{4}_{F_{s}}+\left(
-1\right) ^{s+1}L_{2s}\dbinom{n+1}{4}_{F_{s}}+\dbinom{n}{4}_{F_{s}}.
\label{4.18}
\end{equation}

\begin{equation}
\dbinom{n}{2}_{L_{s}}\dbinom{n}{3}_{F_{s}}=\frac{L_{3s}}{L_{s}}\dbinom{n+2}{5%
}_{F_{s}}-L_{3s}\dbinom{n+1}{5}_{F_{s}}+\dbinom{n}{5}_{F_{s}}.  \label{4.19}
\end{equation}

\begin{eqnarray}
\dbinom{n}{2}_{L_{s}}\dbinom{n}{2}_{F_{s}}^{2} &amp;=&amp;\dbinom{n+4}{6}%
_{F_{s}}+L_{2s}\dbinom{n+3}{6}_{F_{s}}+\left( -1\right) ^{s}L_{2s}\left(
L_{2s}^{2}-1\right) \dbinom{n+2}{6}_{F_{s}}  \notag \\
&amp;&amp;+L_{2s}\dbinom{n+1}{6}_{F_{s}}+\dbinom{n}{6}_{F_{s}}.  \label{4.19b}
\end{eqnarray}

\begin{corollary}
\label{Cor4.6}\textit{(a) Let }$m_{1},\ldots ,m_{l}\in \mathbb{Z}$\textit{\
and }$t_{1},\ldots ,t_{l},k_{1}\ldots ,k_{l}\in \mathbb{N}^{\prime }$\textit{%
\ be given. For }$n\geq k_{1}+\cdots +k_{l}+1$\textit{\ we have that}%
\begin{equation}
\sum_{j=0}^{t_{1}k_{1}+\cdots +t_{l}k_{l}+1}\left( -1\right) ^{\frac{\left(
sj+2(s+1)\right) \left( j+1\right) }{2}}\dbinom{t_{1}k_{1}+\cdots
+t_{l}k_{l}+1}{j}_{F_{s}}G_{m_{1}+st_{1}\left( n-j\right) }^{k_{1}}\cdots
G_{m_{l}+t_{l}s\left( n-j\right) }^{k_{l}}=0.  \label{4.20}
\end{equation}

\textit{(b) Let }$t_{1},\ldots ,t_{l}\in \mathbb{N}$ \textit{and }$%
r_{1}\ldots ,r_{l},p_{1}\ldots ,p_{l}\in \mathbb{N}^{\prime }$\textit{\ be
given. For }$n\geq t_{1}r_{1}p_{1}+\cdots +t_{k}r_{k}p_{k}+1$\textit{\ we
have that}%
\begin{equation}
\sum_{j=0}^{t_{1}r_{1}p_{1}+\cdots +t_{k}r_{k}p_{k}+1}\left( -1\right) ^{%
\frac{\left( sj+2(s+1)\right) \left( j+1\right) }{2}}\dbinom{%
t_{1}r_{1}p_{1}+\cdots +t_{k}r_{k}p_{k}+1}{j}_{F_{s}}\dbinom{n-j}{p_{1}}%
_{G_{st_{1}}}^{r_{1}}\cdots \dbinom{n-j}{p_{k}}_{G_{st_{k}}}^{r_{k}}=0.
\label{4.21}
\end{equation}
\end{corollary}

\begin{proof}
These results are direct consequences of (the numerators in) formulas (\ref%
{4.3}) and (\ref{4.12}).
\end{proof}

\begin{corollary}
\label{Cor4.8}\textit{Let} $p\in \mathbb{N}^{\prime }$\textit{\ be given.
The following identities hold}

\textit{(a)}%
\begin{equation}
\dbinom{n+1}{p+2}_{F_{s}}=\frac{1}{F_{s\left( p+2\right) }}F_{s\left(
p+2\right) n}\ast \left( -1\right) ^{s\left( n+p\right) }\dbinom{n}{p}%
_{F_{s}}.  \label{4.23}
\end{equation}

\textit{(b)}%
\begin{equation}
\dbinom{n+2}{p+4}_{F_{s}}=\frac{1}{F_{s\left( p+4\right) }F_{s\left(
p+2\right) }}F_{s\left( p+4\right) n}\ast \left( -1\right) ^{s\left(
n+1\right) }F_{s\left( p+2\right) n}\ast \dbinom{n}{p}_{F_{s}}.  \label{4.24}
\end{equation}
\end{corollary}

\begin{proof}
(a) First observe that%
\begin{eqnarray*}
&amp;&amp;D_{s,p+3}\left( z\right) \\
&amp;=&amp;\dprod\limits_{j=0}^{p+2}\left( z-\alpha ^{sj}\beta ^{s\left(
p+2-j\right) }\right) =\dprod\limits_{j=-1}^{p+1}\left( z-\alpha ^{s\left(
j+1\right) }\beta ^{s\left( p+1-j\right) }\right)
=\dprod\limits_{j=-1}^{p+1}\left( z-\left( -1\right) ^{s}\alpha ^{sj}\beta
^{s\left( p-j\right) }\right) \\
&amp;=&amp;\left( z-\left( -1\right) ^{s}\alpha ^{-s}\beta ^{s\left( p+1\right)
}\right) \left( z-\left( -1\right) ^{s}\alpha ^{s\left( p+1\right) }\beta
^{-s}\right) \dprod\limits_{j=0}^{p}\left( -1\right) ^{s}\left( \left(
-1\right) ^{s}z-\alpha ^{sj}\beta ^{s\left( p-j\right) }\right) \\
&amp;=&amp;\left( -1\right) ^{s\left( p+1\right) }\left( z^{2}-L_{s\left( p+2\right)
}z+\left( -1\right) ^{sp}\right) D_{s,p+1}\left( \left( -1\right)
^{s}z\right) .
\end{eqnarray*}

Then%
\begin{eqnarray*}
\mathcal{Z}\left( \dbinom{n+1}{p+2}_{F_{s}}\right)  &amp;=&amp;\frac{\left(
-1\right) ^{s+1}z^{2}}{D_{s,p+3}\left( z\right) } \\
&amp;=&amp;\frac{\left( -1\right) ^{s+1}z^{2}}{\left( -1\right) ^{s\left( p+1\right)
}\left( z^{2}-L_{s\left( p+2\right) }z+\left( -1\right) ^{sp}\right)
D_{s,p+1}\left( \left( -1\right) ^{s}z\right) } \\
&amp;=&amp;\left( -1\right) ^{s\left( p+1\right) }\left( -1\right) ^{s}\frac{1}{%
F_{s\left( p+2\right) }}\frac{F_{s\left( p+2\right) }z}{z^{2}-L_{s\left(
p+2\right) }z+\left( -1\right) ^{sp}}\frac{\left( -1\right) ^{s+1}\left(
-1\right) ^{s}z}{D_{s,p+1}\left( \left( -1\right) ^{s}z\right) },
\end{eqnarray*}%
from where (according to (\ref{2.8}) and the convolution theorem)%
\begin{eqnarray*}
\dbinom{n+1}{p+2}_{F_{s}} &amp;=&amp;\left( -1\right) ^{sp}\frac{1}{F_{s\left(
p+2\right) }}F_{s\left( p+2\right) n}\ast \left( -1\right) ^{sn}\dbinom{n}{p}%
_{F_{s}} \\
&amp;=&amp;\frac{1}{F_{s\left( p+2\right) }}F_{s\left( p+2\right) n}\ast \left(
-1\right) ^{s\left( n+p\right) }\dbinom{n}{p}_{F_{s}},
\end{eqnarray*}%
as wanted.

(b) Let us consider the polynomial $D_{s,p+5}\left( z\right) $ and observe
that%
\begin{eqnarray*}
&amp;&amp;D_{s,p+5}\left( z\right) \\
&amp;=&amp;\dprod\limits_{j=0}^{p+4}\left( z-\alpha ^{sj}\beta ^{s\left(
p+4-j\right) }\right) =\dprod\limits_{j=-2}^{p+2}\left( z-\alpha ^{s\left(
j+2\right) }\beta ^{s\left( p+2-j\right) }\right)
=\dprod\limits_{j=-2}^{p+2}\left( z-\alpha ^{sj}\beta ^{s\left( p-j\right)
}\right) \\
&amp;=&amp;\left( z-\alpha ^{-2s}\beta ^{s\left( p+2\right) }\right) \left( z-\alpha
^{s\left( p+2\right) }\beta ^{-2s}\right) \left( z-\alpha ^{-s}\beta
^{s\left( p+1\right) }\right) \left( z-\alpha ^{s\left( p+1\right) }\beta
^{-s}\right) \dprod\limits_{j=0}^{p}\left( z-\alpha ^{sj}\beta ^{s\left(
p-j\right) }\right) \\
&amp;=&amp;\left( z^{2}-L_{s\left( p+4\right) }z+\left( -1\right) ^{sp}\right)
\left( z^{2}-\left( -1\right) ^{s}L_{s\left( p+2\right) }z+\left( -1\right)
^{sp}\right) D_{s,p+1}\left( z\right) .
\end{eqnarray*}

Then%
\begin{eqnarray*}
&amp;&amp;\mathcal{Z}\left( \dbinom{n+2}{p+4}_{F_{s}}\right) \\
&amp;=&amp;\frac{\left( -1\right) ^{s+1}z^{3}}{D_{s,p+5}\left( z\right) } \\
&amp;=&amp;\frac{\left( -1\right) ^{s+1}z^{3}}{\left( z^{2}-L_{s\left( p+4\right)
}z+\left( -1\right) ^{sp}\right) \left( z^{2}-\left( -1\right)
^{s}L_{s\left( p+2\right) }z+\left( -1\right) ^{sp}\right) D_{s,p+1}\left(
z\right) } \\
&amp;=&amp;\frac{1}{F_{s\left( p+4\right) }}\frac{F_{s\left( p+4\right) }z}{%
z^{2}-L_{s\left( p+4\right) }z+\left( -1\right) ^{sp}}\frac{1}{F_{s\left(
p+2\right) }}\left( -1\right) ^{s}\frac{F_{s\left( p+2\right) }\left(
-1\right) ^{s}z}{\left( z^{2}-\left( -1\right) ^{s}L_{s\left( p+2\right)
}z+\left( -1\right) ^{sp}\right) }\frac{\left( -1\right) ^{s+1}z}{%
D_{s,p+1}\left( z\right) },
\end{eqnarray*}%
from where (according to (\ref{2.8}) and the convolution theorem)%
\begin{eqnarray*}
\dbinom{n+2}{p+4}_{F_{s}} &amp;=&amp;\frac{1}{F_{s\left( p+4\right) }}F_{s\left(
p+4\right) n}\ast \frac{1}{F_{s\left( p+2\right) }}\left( -1\right)
^{s}\left( -1\right) ^{ns}F_{s\left( p+2\right) n}\ast \dbinom{n}{p}_{F_{s}}
\\
&amp;=&amp;\frac{1}{F_{s\left( p+4\right) }F_{s\left( p+2\right) }}F_{s\left(
p+4\right) n}\ast \left( -1\right) ^{s\left( n+1\right) }F_{s\left(
p+2\right) n}\ast \dbinom{n}{p}_{F_{s}},
\end{eqnarray*}%
as wanted.
\end{proof}

Some examples are 
\begin{equation}
\dbinom{n+1}{4}_{F_{s}}=\frac{1}{F_{4s}}\sum_{i=0}^{n}\left( -1\right)
^{si}F_{4s\left( n-i\right) }\dbinom{i}{2}_{F_{s}}.  \label{4.25}
\end{equation}%
\begin{equation}
\dbinom{n+2}{6}_{F_{s}}=\frac{1}{F_{6s}F_{4s}}\sum_{i=0}^{n}\sum_{j=0}^{i}%
\left( -1\right) ^{s\left( j+1\right) }F_{4sj}F_{6s\left( i-j\right) }%
\dbinom{n-i}{2}_{F_{s}}.  \label{4.26}
\end{equation}

In the following corollary we will deal with $k$ given sequences $\left(
a_{n}\right) _{0},\left( a_{n}\right) _{1},\cdots ,\left( a_{n}\right) _{k}$
and the convolution of them $\left( a_{n}\right) _{0}\ast \left(
a_{n}\right) _{1}\ast \cdots \ast \left( a_{n}\right) _{k}$, which we will
denote as $\ast _{j=0}^{k}\left( a_{n}\right) _{j}$.

\begin{corollary}
\label{Cor4.9}\textit{Let} $p\in \mathbb{N}$\textit{\ be given. The
following identities hold}

\textit{(a)} 
\begin{equation}
\dbinom{n+p}{2p}_{F_{s}}=\left( -1\right) ^{spn}\ast _{j=0}^{p-1}\frac{%
\left( -1\right) ^{sj\left( n+1\right) }}{F_{2s\left( p-j\right) }}%
F_{2s\left( p-j\right) n}.  \label{4.27}
\end{equation}

\textit{(b)}%
\begin{equation}
\dbinom{n+p-1}{2p-1}_{F_{s}}=\ast _{j=0}^{p-1}\frac{\left( -1\right)
^{sj\left( n+1\right) }}{F_{s\left( 2p-1-2j\right) }}F_{s\left(
2p-1-2j\right) n}.  \label{4.28}
\end{equation}
\end{corollary}

\begin{proof}
(a) According to (\ref{4.1}) and (\ref{2.11}) we have that%
\begin{equation*}
z^{p}\mathcal{Z}\left( \dbinom{n}{2p}_{F_{s}}\right) =\frac{z^{p+1}}{\left(
z-\left( -1\right) ^{sp}\right) \dprod\limits_{j=0}^{p-1}\left( z^{2}-\left(
-1\right) ^{sj}L_{2s\left( p-j\right) }z+1\right) },
\end{equation*}%
or (by using (\ref{4.2}))%
\begin{equation*}
\mathcal{Z}\left( \dbinom{n+p}{2p}_{F_{s}}\right) =\frac{z}{z-\left(
-1\right) ^{sp}}\dprod\limits_{j=0}^{p-1}\frac{z}{z^{2}-\left( -1\right)
^{sj}L_{2s\left( p-j\right) }z+1},
\end{equation*}%
from where (\ref{4.27}) follows (by using (\ref{2.8}) and the convolution
theorem).

(b) According to (\ref{4.1}) and (\ref{2.12}) we have that%
\begin{eqnarray*}
z^{p-1}\mathcal{Z}\left( \dbinom{n}{2p-1}_{F_{s}}\right) &amp;=&amp;\frac{z^{p}}{%
\dprod\limits_{j=0}^{p-1}\left( z^{2}-\left( -1\right) ^{sj}L_{s\left(
2p-1-2j\right) }z+\left( -1\right) ^{s}\right) } \\
&amp;=&amp;\dprod\limits_{j=0}^{p-1}\frac{z}{z^{2}-\left( -1\right) ^{sj}L_{s\left(
2p-1-2j\right) }z+\left( -1\right) ^{s}},
\end{eqnarray*}%
from where (\ref{4.28}) follows (with (\ref{4.2}), (\ref{2.8}) and the
convolution theorem).
\end{proof}

Some examples are%
\begin{equation}
\dbinom{n+1}{2}_{F_{s}}=\frac{1}{F_{2s}}\sum_{j=0}^{n}\left( -1\right)
^{s\left( n-j\right) }F_{2sj}.  \label{4.29}
\end{equation}

\begin{equation}
\dbinom{n+2}{4}_{F_{s}}=\frac{1}{F_{4s}F_{2s}}\sum_{i=0}^{n}\sum_{j=0}^{i}%
\left( -1\right) ^{s\left( j+1\right) }F_{2sj}F_{4s\left( i-j\right) }.
\label{4.30}
\end{equation}

In the following corollary we present decompositions of some sequences of $s$%
-Fibonomials as linear combinations of certain Fibonacci sequences.

\begin{corollary}
\label{Cor4.10}\textit{We have the following decompositions of the }$s$%
\textit{-Fibonomials:}%
\begin{eqnarray}
&amp;&amp;\dbinom{n+p}{2p}_{F_{s}}  \label{4.31} \\
&amp;=&amp;\sum_{j=0}^{p-1}\frac{1}{F_{2s\left( p-j\right) }\dprod_{i=0,i\neq
j}^{p-1}\left( \left( -1\right) ^{sj}L_{2s\left( p-j\right) }-\left(
-1\right) ^{si}L_{2s\left( p-i\right) }\right) }\sum_{t=0}^{n}\left(
-1\right) ^{s\left( p\left( n-t\right) +j\left( t+1\right) \right)
}F_{2s\left( p-j\right) t}.\text{ \ \ }\left( p\geq 1\right)  \notag
\end{eqnarray}%
\begin{eqnarray}
&amp;&amp;\dbinom{n+p-1}{2p-1}_{F_{s}}  \label{4.32} \\
&amp;=&amp;\sum_{j=0}^{p-1}\frac{\left( -1\right) ^{js\left( n+1\right) }}{%
F_{s\left( 2p-1-2j\right) }\dprod_{i=0,i\neq j}^{p-1}\left( \left( -1\right)
^{sj}L_{s\left( 2p-1-2j\right) }-\left( -1\right) ^{si}L_{s\left(
2p-1-2i\right) }\right) }F_{s\left( 2p-1-2j\right) n}.\text{ \ \ \ }\left(
p\geq 1\right)  \notag
\end{eqnarray}%
\begin{eqnarray}
\dbinom{n+p-1}{2p}_{F_{s}} &amp;=&amp;\sum_{t=0}^{n}\sum_{j=0}^{p-1}\frac{\left(
-1\right) ^{s\left( p\left( n-t\right) +jt\right) +1}}{\prod_{i=0,i\neq
j}^{p-1}\left( \left( -1\right) ^{sj}L_{2s\left( p-j\right) }-\left(
-1\right) ^{si}L_{2s\left( p-i\right) }\right) }\times  \label{4.33} \\
&amp;&amp;\times \left( F_{2s\left( p-j\right) t+1}+\frac{F_{2s\left( p-j\right)
-1}-L_{2s\left( p-j\right) }}{F_{2s\left( p-j\right) }}F_{2s\left(
p-j\right) t}\right) .\text{ \ \ \ }\left( p\geq 2\right)  \notag
\end{eqnarray}%
\begin{eqnarray}
\dbinom{n+p-2}{2p-1}_{F_{s}} &amp;=&amp;\sum_{j=0}^{p-1}\frac{\left( -1\right)
^{s\left( jn+1\right) +1}}{\prod_{i=0,i\neq j}^{p-1}\left( \left( -1\right)
^{sj}L_{s\left( 2p-1-2j\right) }-\left( -1\right) ^{si}L_{s\left(
2p-1-2i\right) }\right) }\times  \label{4.34} \\
&amp;&amp;\times \left( F_{s\left( 2p-1-2j\right) n+1}+\frac{F_{s\left(
2p-1-2j\right) -1}-L_{s\left( 2p-1-2j\right) }}{F_{s\left( 2p-1-2j\right) }}%
F_{s\left( 2p-1-2j\right) n}\right) .\text{ \ \ \ }\left( p\geq 2\right) 
\notag
\end{eqnarray}%
\begin{eqnarray}
\dbinom{n+p-2}{2p}_{F_{s}} &amp;=&amp;\sum_{t=0}^{n}\sum_{j=0}^{p-1}\frac{\left(
-1\right) ^{s\left( p\left( n-t\right) +j\left( t+1\right) \right)
+1}L_{2s\left( p-j\right) }}{\prod_{i=0,i\neq j}^{p-1}\left( \left(
-1\right) ^{sj}L_{2s\left( p-j\right) }-\left( -1\right) ^{si}L_{2s\left(
p-i\right) }\right) }\times  \label{4.35} \\
&amp;&amp;\times \left( F_{2s\left( p-j\right) t+1}+\frac{F_{2s\left( p-j\right) -1}-%
\frac{L_{4s\left( p-j\right) }+1}{L_{2s\left( p-j\right) }}}{F_{2s\left(
p-j\right) }}F_{2s\left( p-j\right) t}\right) .\text{ \ \ \ }\left( p\geq
3\right)  \notag
\end{eqnarray}%
\begin{eqnarray}
\dbinom{n+p-3}{2p-1}_{F_{s}} &amp;=&amp;\sum_{j=0}^{p-1}\frac{\left( -1\right)
^{sj\left( n+1\right) +1}L_{s\left( 2p-1-2j\right) }}{\prod_{i=0,i\neq
j}^{p-1}\left( \left( -1\right) ^{sj}L_{s\left( 2p-1-2j\right) }-\left(
-1\right) ^{si}L_{s\left( 2p-1-2i\right) }\right) }\times  \label{4.36} \\
&amp;&amp;\times \left( F_{s\left( 2p-1-2j\right) n+1}+\frac{F_{s\left(
2p-1-2j\right) -1}-\frac{L_{2s\left( 2p-1-2j\right) }+\left( -1\right) ^{s}}{%
L_{s\left( 2p-1-2j\right) }}}{F_{s\left( 2p-1-2j\right) }}F_{s\left(
2p-1-2j\right) n}\right) .\text{ \ \ \ }\left( p\geq 3\right)  \notag
\end{eqnarray}%
\begin{eqnarray}
\dbinom{n+p-3}{2p}_{F_{s}} &amp;=&amp;\sum_{t=0}^{n}\sum_{j=0}^{p-1}\frac{\left(
-1\right) ^{s\left( p\left( n-t\right) +jt\right) +1}\left( L_{4s\left(
p-j\right) }+1\right) }{\prod_{i=0,i\neq j}^{p-1}\left( \left( -1\right)
^{sj}L_{2s\left( p-j\right) }-\left( -1\right) ^{si}L_{2s\left( p-i\right)
}\right) }\times  \label{4.37} \\
&amp;&amp;\times \left( F_{2s\left( p-j\right) t+1}+\frac{F_{2s\left( p-j\right) -1}-%
\frac{L_{6s\left( p-j\right) }+L_{2s\left( p-j\right) }}{L_{4s\left(
p-j\right) }+1}}{F_{2s\left( p-j\right) }}F_{2s\left( p-j\right) t}\right) .%
\text{ \ \ \ }\left( p\geq 4\right)  \notag
\end{eqnarray}%
\begin{eqnarray}
&amp;&amp;\dbinom{n+p-4}{2p-1}_{F_{s}}  \label{4.38} \\
&amp;=&amp;\sum_{j=0}^{p-1}\frac{\left( -1\right) ^{s\left( jn+1\right) +1}\left(
L_{2s\left( 2p-1-2j\right) }+\left( -1\right) ^{s}\right) }{\prod_{i=0,i\neq
j}^{p-1}\left( \left( -1\right) ^{sj}L_{s\left( 2p-1-2j\right) }-\left(
-1\right) ^{si}L_{s\left( 2p-1-2i\right) }\right) }\times  \notag \\
&amp;&amp;\times \left( F_{s\left( 2p-1-2j\right) n+1}+\frac{F_{s\left(
2p-1-2j\right) -1}-\frac{L_{3s\left( 2p-1-2j\right) }+\left( -1\right)
^{s}L_{s\left( 2p-1-2j\right) }}{L_{2s\left( 2p-1-2j\right) }+\left(
-1\right) ^{s}}}{F_{s\left( 2p-1-2j\right) }}F_{s\left( 2p-1-2j\right)
n}\right) .\text{ \ \ \ }\left( p\geq 4\right)  \notag
\end{eqnarray}
\end{corollary}

\begin{proof}
The proof is similar in all the four cases. It depends on an adequate
partial fractions decomposition. We show all the steps of the proof only for
the case (a), and indicate the corresponding decompositions used in the
remaining cases. For the case (a) we use that for $p\geq 1$ one has the
following partial fractions decompositions:%
\begin{eqnarray}
&amp;&amp;\frac{z^{p}}{\dprod\limits_{j=0}^{p-1}\left( z^{2}-\left( -1\right)
^{sj}L_{2s\left( p-j\right) }z+1\right) }  \label{4.39} \\
&amp;=&amp;\sum_{j=0}^{p-1}\frac{1}{\dprod_{i=0,i\neq j}^{p-1}\left( \left(
-1\right) ^{sj}L_{2s\left( p-j\right) }-\left( -1\right) ^{si}L_{2s\left(
p-i\right) }\right) }\frac{z}{z^{2}-\left( -1\right) ^{sj}L_{2s\left(
p-j\right) }z+1},  \notag
\end{eqnarray}%
and%
\begin{eqnarray}
&amp;&amp;\frac{z^{p}}{\dprod\limits_{j=0}^{p-1}\left( z^{2}-\left( -1\right)
^{js}L_{s\left( 2p-1-2j\right) }z+\left( -1\right) ^{s}\right) }
\label{4.40} \\
&amp;=&amp;\sum_{j=0}^{p-1}\frac{1}{\dprod_{i=0,i\neq j}^{p-i}\left( \left(
-1\right) ^{sj}L_{s\left( 2p-1-2j\right) }-\left( -1\right) ^{si}L_{s\left(
2p-1-2i\right) }\right) }\frac{z}{z^{2}-\left( -1\right) ^{js}L_{s\left(
2p-1-2j\right) }z+\left( -1\right) ^{s}}.  \notag
\end{eqnarray}

We begin by noting that with (\ref{4.1}) and (\ref{2.11}) we can write%
\begin{eqnarray*}
z^{p}\mathcal{Z}\left( \dbinom{n}{2p}_{F_{s}}\right) &amp;=&amp;\frac{z^{p+1}}{%
\left( z-\left( -1\right) ^{sp}\right) \dprod\limits_{j=0}^{p-1}\left(
z^{2}-\left( -1\right) ^{sj}L_{2s\left( p-j\right) }z+1\right) } \\
&amp;=&amp;\frac{z}{z-\left( -1\right) ^{sp}}\frac{z^{p}}{\dprod\limits_{j=0}^{p-1}%
\left( z^{2}-\left( -1\right) ^{sj}L_{2s\left( p-j\right) }z+1\right) },
\end{eqnarray*}%
and then, by using (\ref{4.39}) we have%
\begin{eqnarray*}
&amp;&amp;z^{p}\mathcal{Z}\left( \dbinom{n}{2p}_{F_{s}}\right) \\
&amp;=&amp;\frac{z}{z-\left( -1\right) ^{sp}}\sum_{j=0}^{p-1}\frac{1}{%
\dprod_{i=0,i\neq j}^{p-1}\left( \left( -1\right) ^{sj}L_{2s\left(
p-j\right) }-\left( -1\right) ^{si}L_{2s\left( p-i\right) }\right) }\frac{z}{%
z^{2}-\left( -1\right) ^{sj}L_{2s\left( p-j\right) }z+1} \\
&amp;&amp;\frac{z}{z-\left( -1\right) ^{sp}}\sum_{j=0}^{p-1}\frac{\left( -1\right)
^{sj}}{F_{2s\left( p-j\right) }\dprod_{i=0,i\neq j}^{p-1}\left( \left(
-1\right) ^{sj}L_{2s\left( p-j\right) }-\left( -1\right) ^{si}L_{2s\left(
p-i\right) }\right) }\frac{F_{2s\left( p-j\right) }\left( -1\right) ^{sj}z}{%
z^{2}-\left( -1\right) ^{sj}L_{2s\left( p-j\right) }z+1}.
\end{eqnarray*}

Thus, according to (\ref{4.2}), convolution theorem and (\ref{2.8}) we have
that%
\begin{eqnarray*}
&amp;&amp;\dbinom{n+p}{2p}_{F_{s}} \\
&amp;=&amp;\left( -1\right) ^{spn}\ast \sum_{j=0}^{p-1}\frac{\left( -1\right) ^{sj}}{%
F_{2s\left( p-j\right) }\dprod_{i=0,i\neq j}^{p-1}\left( \left( -1\right)
^{sj}L_{2s\left( p-j\right) }-\left( -1\right) ^{si}L_{2s\left( p-i\right)
}\right) }\left( -1\right) ^{sjn}F_{2s\left( p-j\right) n} \\
&amp;=&amp;\sum_{t=0}^{n}\left( -1\right) ^{sp\left( n-t\right) }\sum_{j=0}^{p-1}%
\frac{1}{F_{2s\left( p-j\right) }\dprod_{i=0,i\neq j}^{p-1}\left( \left(
-1\right) ^{sj}L_{2s\left( p-j\right) }-\left( -1\right) ^{si}L_{2s\left(
p-i\right) }\right) }\left( -1\right) ^{sj\left( t+1\right) }F_{2s\left(
p-j\right) t} \\
&amp;=&amp;\sum_{j=0}^{p-1}\frac{1}{F_{2s\left( p-j\right) }\dprod_{i=0,i\neq
j}^{p-1}\left( \left( -1\right) ^{sj}L_{2s\left( p-j\right) }-\left(
-1\right) ^{si}L_{2s\left( p-i\right) }\right) }\left( -1\right) ^{s\left(
p\left( n-t\right) +j\left( t+1\right) \right) }\sum_{t=0}^{n}F_{2s\left(
p-j\right) t},
\end{eqnarray*}%
which proves (\ref{4.31}). Also, by using (\ref{4.1}), (\ref{2.12}) and (\ref%
{4.40}) we have that%
\begin{eqnarray*}
&amp;&amp;z^{p-1}\mathcal{Z}\left( \dbinom{n}{2p-1}_{F_{s}}\right) \\
&amp;=&amp;\frac{z^{p}}{\dprod\limits_{j=0}^{p-1}\left( z^{2}-\left( -1\right)
^{sj}L_{s\left( 2p-1-2j\right) }z+\left( -1\right) ^{s}\right) } \\
&amp;=&amp;\sum_{j=0}^{p-1}\frac{1}{\dprod_{i=0,i\neq j}^{p-i}\left( \left(
-1\right) ^{sj}L_{s\left( 2p-1-2j\right) }-\left( -1\right) ^{si}L_{s\left(
2p-1-2i\right) }\right) }\frac{z}{z^{2}-\left( -1\right) ^{sj}L_{s\left(
2p-1-2j\right) }z+\left( -1\right) ^{s}} \\
&amp;=&amp;\sum_{j=0}^{p-1}\frac{1}{\dprod_{i=0,i\neq j}^{p-i}\left( \left(
-1\right) ^{sj}L_{s\left( 2p-1-2j\right) }-\left( -1\right) ^{si}L_{s\left(
2p-1-2i\right) }\right) }\frac{\left( -1\right) ^{js}}{F_{s\left(
2p-1-2j\right) }}\frac{F_{s\left( 2p-1-2j\right) }\left( -1\right) ^{js}z}{%
z^{2}-\left( -1\right) ^{sj}L_{s\left( 2p-1-2j\right) }z+\left( -1\right)
^{s}}
\end{eqnarray*}%
from where (by using (\ref{4.2}) and (\ref{2.8}))%
\begin{eqnarray*}
&amp;&amp;\dbinom{n+p-1}{2p-1}_{F_{s}} \\
&amp;=&amp;\sum_{j=0}^{p-1}\frac{1}{\dprod_{i=0,i\neq j}^{p-i}\left( \left(
-1\right) ^{sj}L_{s\left( 2p-1-2j\right) }-\left( -1\right) ^{si}L_{s\left(
2p-1-2i\right) }\right) }\frac{\left( -1\right) ^{js}}{F_{s\left(
2p-1-2j\right) }}\left( -1\right) ^{jsn}F_{s\left( 2p-1-2j\right) n} \\
&amp;=&amp;\sum_{j=0}^{p-1}\frac{\left( -1\right) ^{js\left( n+1\right) }}{%
F_{s\left( 2p-1-2j\right) }\dprod_{i=0,i\neq j}^{p-i}\left( \left( -1\right)
^{sj}L_{s\left( 2p-1-2j\right) }-\left( -1\right) ^{si}L_{s\left(
2p-1-2i\right) }\right) }F_{s\left( 2p-1-2j\right) n},
\end{eqnarray*}%
which proves (\ref{4.32}).

For the case (b), use the partial fractions decomposition (valid for $p\geq
2 $):%
\begin{eqnarray*}
&amp;&amp;\frac{z^{p-2}}{\prod_{j=0}^{p-1}\left( z^{2}-\left( -1\right)
^{sj}L_{2s\left( p-j\right) }z+1\right) } \\
&amp;=&amp;-\sum_{j=0}^{p-1}\frac{1}{\prod_{i=0,i\neq j}^{p-1}\left( \left(
-1\right) ^{sj}L_{2s\left( p-j\right) }-\left( -1\right) ^{si}L_{2s\left(
p-i\right) }\right) }\frac{z-\left( -1\right) ^{sj}L_{2s\left( p-j\right) }}{%
z^{2}-\left( -1\right) ^{sj}L_{2s\left( p-j\right) }z+1},
\end{eqnarray*}%
and%
\begin{eqnarray*}
&amp;&amp;\frac{z^{p-2}}{\prod_{j=0}^{p-1}\left( z^{2}-\left( -1\right)
^{sj}L_{s\left( 2p-1-2j\right) }z+\left( -1\right) ^{s}\right) } \\
&amp;=&amp;\sum_{j=0}^{p-1}\frac{\left( -1\right) ^{s+1}}{\prod_{i=0,i\neq
j}^{p-1}\left( \left( -1\right) ^{sj}L_{s\left( 2p-1-2j\right) }-\left(
-1\right) ^{si}L_{s\left( 2p-1-2i\right) }\right) }\frac{z-\left( -1\right)
^{s}L_{s\left( 2p-1-2j\right) }}{z^{2}-\left( -1\right) ^{sj}L_{s\left(
2p-1-2j\right) }z+\left( -1\right) ^{s}}.
\end{eqnarray*}

For the case (c), use the partial fractions decomposition (valid for $p\geq
3 $):%
\begin{eqnarray*}
\frac{z^{p-3}}{\prod_{j=0}^{p-1}\left( z^{2}-\left( -1\right)
^{sj}L_{2s\left( p-j\right) }z+1\right) } &amp;=&amp;\sum_{j=0}^{p-1}\frac{1}{%
\prod_{i=0,i\neq j}^{p-1}\left( \left( -1\right) ^{sj}L_{2s\left( p-j\right)
}-\left( -1\right) ^{si}L_{2s\left( p-i\right) }\right) }\times \\
&amp;&amp;\times \frac{\left( -1\right) ^{sj+1}L_{2s\left( p-j\right) }z+L_{4s\left(
p-j\right) }+1}{z^{2}-\left( -1\right) ^{sj}L_{2s\left( p-j\right) }z+1},
\end{eqnarray*}%
and%
\begin{eqnarray*}
\frac{z^{p-3}}{\prod_{j=0}^{p-1}\left( z^{2}-\left( -1\right)
^{sj}L_{s\left( 2p-1-2j\right) }z+\left( -1\right) ^{s}\right) }
&amp;=&amp;\sum_{j=0}^{p-1}\frac{1}{\prod_{i=0,i\neq j}^{p-1}\left( \left( -1\right)
^{sj}L_{s\left( 2p-1-2j\right) }-\left( -1\right) ^{si}L_{s\left(
2p-1-2i\right) }\right) }\times \\
&amp;&amp;\times \frac{\left( -1\right) ^{sj+1}L_{s\left( 2p-1-2j\right)
}z+L_{2s\left( 2p-1-2j\right) }+\left( -1\right) ^{s}}{z^{2}-\left(
-1\right) ^{sj}L_{s\left( 2p-1-2j\right) }z+\left( -1\right) ^{s}},
\end{eqnarray*}

Finally, For the case (d) use the partial fractions decomposition (valid for 
$p\geq 4$):%
\begin{eqnarray*}
\frac{z^{p-4}}{\prod_{j=0}^{p-1}\left( z^{2}-\left( -1\right)
^{sj}L_{2s\left( p-j\right) }z+1\right) } &amp;=&amp;-\sum_{j=0}^{p-1}\frac{1}{%
\prod_{i=0,i\neq j}^{p-1}\left( \left( -1\right) ^{sj}L_{2s\left( p-j\right)
}-\left( -1\right) ^{si}L_{2s\left( p-i\right) }\right) }\times \\
&amp;&amp;\times \frac{\left( L_{4s\left( p-j\right) }+1\right) z-\left( -1\right)
^{sj}\left( L_{6s\left( p-j\right) }+L_{2s\left( p-j\right) }\right) }{%
z^{2}-\left( -1\right) ^{sj}L_{2s\left( p-j\right) }z+1},
\end{eqnarray*}%
and%
\begin{eqnarray*}
&amp;&amp;\frac{z^{p-4}}{\prod_{j=0}^{p-1}\left( z^{2}-\left( -1\right)
^{sj}L_{s\left( 2p-1-2j\right) }z+\left( -1\right) ^{s}\right) } \\
&amp;=&amp;\left( -1\right) ^{s+1}\sum_{j=0}^{p-1}\frac{1}{\prod_{i=0,i\neq
j}^{p-1}\left( \left( -1\right) ^{sj}L_{s\left( 2p-1-2j\right) }-\left(
-1\right) ^{si}L_{s\left( 2p-1-2i\right) }\right) }\times \\
&amp;&amp;\times \frac{\left( L_{2s\left( 2p-1-2j\right) }+\left( -1\right)
^{s}\right) z-\left( -1\right) ^{sj}\left( L_{3s\left( 2p-1-2j\right)
}+\left( -1\right) ^{s}L_{s\left( 2p-1-2j\right) }\right) }{z^{2}-\left(
-1\right) ^{sj}L_{s\left( 2p-1-2j\right) }z+\left( -1\right) ^{s}}.
\end{eqnarray*}
\end{proof}

Some examples are the following:

\begin{equation}
\dbinom{n+2}{4}_{F_{s}}=\frac{1}{L_{4s}+\left( -1\right) ^{s+1}L_{2s}}%
\sum_{t=0}^{n}\left( \frac{F_{4st}}{F_{4s}}-\left( -1\right) ^{s\left(
t+1\right) }\frac{F_{2st}}{F_{2s}}\right) .  \label{4.41}
\end{equation}

\begin{equation}
\dbinom{n+1}{4}_{F_{s}}=\frac{1}{\left( -1\right) ^{s}L_{2s}-L_{4s}}%
\sum_{t=0}^{n}\left( 
\begin{array}{c}
F_{4st+1}+\frac{F_{4s-1}-L_{4s}}{F_{4s}}F_{4st} \\ 
+\left( -1\right) ^{st+1}\left( F_{2st+1}+\frac{F_{2s-1}-L_{2s}}{F_{2s}}%
F_{2st}\right)%
\end{array}%
\right) .  \label{4.42}
\end{equation}

\begin{equation}
\dbinom{n}{3}_{F_{s}}=\frac{\left( -1\right) ^{s+1}}{L_{3s}-\left( -1\right)
^{s}L_{s}}\left( F_{3sn+1}+\frac{F_{3s-1}-L_{3s}}{F_{3s}}F_{3sn}+\left(
-1\right) ^{sn+1}\left( F_{sn+1}+\frac{F_{s-1}-L_{s}}{F_{s}}F_{sn}\right)
\right) .  \label{4.43}
\end{equation}

\begin{eqnarray}
\dbinom{n}{5}_{F_{s}} &amp;=&amp;\frac{-L_{5s}}{\left( L_{5s}-\left( -1\right)
^{s}L_{3s}\right) \left( L_{5s}-L_{s}\right) }\left( F_{5sn+1}+\frac{%
F_{5s-1}-\frac{L_{10s}+\left( -1\right) ^{s}}{L_{5s}}}{F_{5s}}F_{5sn}\right)
\notag \\
&amp;&amp;+\frac{\left( -1\right) ^{s\left( n+1\right) +1}L_{3s}}{\left( \left(
-1\right) ^{s}L_{3s}-L_{5s}\right) \left( \left( -1\right)
^{s}L_{3s}-L_{s}\right) }\left( F_{3sn+1}+\frac{F_{3s-1}-\frac{L_{6s}+\left(
-1\right) ^{s}}{L_{3s}}}{F_{3s}}F_{3sn}\right)  \notag \\
&amp;&amp;+\frac{-L_{s}}{\left( L_{s}-L_{5s}\right) \left( L_{s}-\left( -1\right)
^{s}L_{3s}\right) }\left( F_{sn+1}+\frac{F_{s-1}-\frac{L_{2s}+\left(
-1\right) ^{s}}{L_{s}}}{F_{s}}F_{sn}\right) .  \label{4.43b}
\end{eqnarray}

\begin{equation}
\dbinom{n+1}{6}_{F_{s}}=\sum_{t=0}^{n}\left( 
\begin{array}{c}
\dfrac{\left( -1\right) ^{s\left( n-t\right) +1}L_{6s}}{\left( L_{6s}-\left(
-1\right) ^{s}L_{4s}\right) \left( L_{6s}-L_{2s}\right) }\left( F_{6st+1}+%
\dfrac{F_{6s-1}-\frac{L_{12s}+1}{L_{6s}}}{F_{6s}}F_{6st}\right) \\ 
+\dfrac{\left( -1\right) ^{s\left( n+1\right) +1}L_{4s}}{\left( \left(
-1\right) ^{s}L_{4s}-L_{6s}\right) \left( \left( -1\right)
^{s}L_{4s}-L_{2s}\right) }\left( F_{4st+1}+\dfrac{F_{4s-1}-\frac{L_{8s}+1}{%
L_{4s}}}{F_{4s}}F_{4st}\right) \\ 
+\dfrac{\left( -1\right) ^{s\left( n-t\right) +1}L_{2s}}{\left(
L_{2s}-L_{6s}\right) \left( L_{2s}-\left( -1\right) ^{s}L_{4s}\right) }%
\left( F_{2st+1}+\dfrac{F_{2s-1}-\frac{L_{4s}+1}{L_{2s}}}{F_{2s}}%
F_{2st}\right)%
\end{array}%
\right) .  \label{4.44}
\end{equation}

\section{\label{Sec5}Final remarks}

We can write (\ref{4.4}) (with $G=F$ and $m_{1}=\cdots =m_{l}=0$) as

\begin{equation}
\left( -1\right) ^{s+1}\sum_{i=0}^{p}\sum_{j=0}^{i}\left( -1\right) ^{\frac{%
\left( sj+2(s+1)\right) \left( j+1\right) }{2}}\dbinom{p+1}{j}%
_{F_{s}}F_{t_{1}s\left( i-j\right) }^{k_{1}}\cdots F_{t_{l}s\left(
i-j\right) }^{k_{l}}\dbinom{n+p-i}{p}_{F_{s}}=F_{t_{1}sn}^{k_{1}}\cdots
F_{t_{l}sn}^{k_{l}},  \label{5.1}
\end{equation}%
where $p=t_{1}k_{1}+\cdots +t_{l}k_{l}$. It is possible to see (\ref{5.1})
as a linear system in the $p$ indeterminates $\binom{n+p-i}{p}_{F_{s}}$, $%
i=1,2,\ldots ,p$, with $m$ equations, where $m$ is the number of terms $%
F_{t_{1}sn}^{k_{1}}\cdots F_{t_{l}sn}^{k_{l}}$ we can form such that $%
t_{1}k_{1}+\cdots +t_{l}k_{l}=p$ (of course we refer to non-trivial terms
and up to natural equivalences).

\begin{conjecture}
\textit{Any }$s$\textit{-Fibonomial }$\binom{n+k}{p}_{F_{s}}$\textit{, }$%
k=0,1,\ldots ,p-1$\textit{, can be written as a linear combination of
homogeneous terms }$F_{t_{1}sn}^{k_{1}}F_{t_{2}sn}^{k_{2}}\cdots
F_{t_{l}sn}^{k_{l}}$ \textit{where }$t_{1}k_{1}+t_{2}k_{2}+\cdots
+t_{l}k_{l}=p$\textit{. }
\end{conjecture}

In the simplest case $p=2$ we have two homogeneous terms, namely $F_{sn}^{2}$
and $F_{2sn}$, and (\ref{5.1}) can be solved for $\binom{n}{2}_{F_{s}}$ and $%
\binom{n+1}{2}_{F_{s}}$ to obtain

\begin{equation}
\dbinom{n}{2}_{F_{s}}=\frac{\left( -1\right) ^{s}}{2}\left( \frac{F_{sn}^{2}%
}{F_{s}^{2}}-\frac{F_{2sn}}{F_{2s}}\right) .  \label{5.2}
\end{equation}

\begin{equation}
\dbinom{n+1}{2}_{F_{s}}=\frac{1}{2}\left( \frac{F_{sn}^{2}}{F_{s}^{2}}+\frac{%
F_{2sn}}{F_{2s}}\right) .  \label{5.3}
\end{equation}

In the case $p=3$ we have three homogeneous terms $F_{3sn}$, $F_{2sn}F_{sn}$
and $F_{sn}^{3}$, and (\ref{5.1}) can be solved for $\binom{n}{3}_{F_{s}}$, $%
\binom{n+1}{3}_{F_{s}}$ and $\binom{n+2}{3}_{F_{s}}$ to obtain

\begin{equation}
\binom{n}{3}_{F_{s}}=\frac{1}{2}\left( -1\right) ^{s+1}\frac{F_{2sn}F_{sn}}{%
F_{2s}F_{s}}+\frac{1}{3}\left( -1\right) ^{s}\frac{F_{3sn}}{F_{3s}}+\frac{1}{%
6}\left( -1\right) ^{s}\frac{F_{sn}^{3}}{F_{s}^{3}}.  \label{5.4}
\end{equation}

\begin{equation}
\binom{n+1}{3}_{F_{s}}=\frac{\left( -1\right) ^{s+1}}{3L_{s}}\frac{F_{3sn}}{%
F_{3s}}+\frac{\left( -1\right) ^{s}}{3L_{s}}\frac{F_{sn}^{3}}{F_{s}^{3}}.
\label{5.5}
\end{equation}

\begin{equation}
\binom{n+2}{3}_{F_{s}}=\frac{1}{2}\frac{F_{2sn}F_{sn}}{F_{2s}F_{s}}+\frac{1}{%
3}\frac{F_{3sn}}{F_{3s}}+\frac{1}{6}\frac{F_{sn}^{3}}{F_{s}^{3}}.
\label{5.6}
\end{equation}

In the case $p=4$ we have 5 homogeneous terms%
\begin{equation}
F_{4sn}\text{ },\text{ }F_{3sn}F_{sn}\text{ },\text{ }F_{sn}^{4}\text{ },%
\text{ }F_{2sn}^{2}\text{ },\text{ }F_{2sn}F_{sn}^{2}.  \label{5.7}
\end{equation}

The corresponding system (\ref{5.1}) can be solved for the $s$-Fibonomials $%
\binom{n-i}{4}_{F_{s}}$, $i=0,1,2,3$, to obtain

\begin{equation}
\dbinom{n}{4}_{F_{s}}=-\frac{1}{4}\frac{F_{4sn}}{F_{4s}}+\frac{L_{3s}}{%
10L_{s}F_{s}^{2}}\frac{F_{3sn}F_{sn}}{F_{3s}F_{s}}+\frac{\left( -1\right)
^{s+1}}{10F_{s}^{2}}\frac{F_{2sn}^{2}}{F_{2s}^{2}}-\frac{1}{4}\frac{%
F_{2sn}F_{sn}^{2}}{F_{s}^{2}F_{2s}}.  \label{5.8}
\end{equation}

\begin{equation}
\dbinom{n+1}{4}_{F_{s}}=\frac{\left( -1\right) ^{s}F_{s}}{4F_{3s}}\frac{%
F_{4sn}}{F_{4s}}+\frac{\left( -1\right) ^{s}}{10F_{s}^{2}}\frac{F_{3sn}F_{sn}%
}{F_{3s}F_{s}}+\frac{\left( -1\right) ^{s+1}}{10F_{s}^{2}}\frac{F_{2sn}^{2}}{%
F_{2s}^{2}}+\frac{\left( -1\right) ^{s+1}F_{s}}{4F_{3s}}\frac{%
F_{2sn}F_{sn}^{2}}{F_{s}^{2}F_{2s}}.  \label{5.9}
\end{equation}

\begin{equation}
\dbinom{n+2}{4}_{F_{s}}=\frac{\left( -1\right) ^{s+1}F_{s}}{4F_{3s}}\frac{%
F_{4sn}}{F_{4s}}+\frac{\left( -1\right) ^{s}}{10F_{s}^{2}}\frac{F_{3sn}F_{sn}%
}{F_{3s}F_{s}}+\frac{\left( -1\right) ^{s+1}}{10F_{s}^{2}}\frac{F_{2sn}^{2}}{%
F_{2s}^{2}}+\frac{\left( -1\right) ^{s}F_{s}}{4F_{3s}}\frac{F_{2sn}F_{sn}^{2}%
}{F_{s}^{2}F_{2s}}.  \label{5.10}
\end{equation}

\begin{equation}
\dbinom{n+3}{4}_{F_{s}}=\frac{1}{4}\frac{F_{4sn}}{F_{4s}}+\frac{L_{3s}}{%
10L_{s}F_{s}^{2}}\frac{F_{3sn}F_{sn}}{F_{3s}F_{s}}+\frac{\left( -1\right)
^{s+1}}{10F_{s}^{2}}\frac{F_{2sn}^{2}}{F_{2s}^{2}}+\frac{1}{4}\frac{%
F_{2sn}F_{sn}^{2}}{F_{s}^{2}F_{2s}}.  \label{5.11}
\end{equation}

However, in this case a linear dependence relation appears, namely:

\begin{equation}
5F_{sn}^{4}+3F_{2sn}^{2}-4F_{3sn}F_{sn}=0,  \label{5.12}
\end{equation}%
so the way of representing the $s$-Fibonomials as linear combinations of
homogeneous terms stated in our conjecture should be not unique.

It remains (for a future work) to have a proof of the conjecture above and
to identify what kind of linear dependencies exist among the homogeneous
terms.

\section{Acknowledgments}

I would like to thank the anonymous referees for their careful reading of
the original version and their valuable observations.

\begin{thebibliography}{99}
\bibitem{B} Arthur T. Benjamin and Sean S. Plott, A combinatorial approach
to Fibonomial coefficients, \textit{Fibonacci Quart.} \textbf{46/47}
(2008/2009), 7--9. See also Errata: \textbf{48} (2010), 276.

\bibitem{C} L. Carlitz, Generating functions for powers of certain sequence
of numbers, \textit{Duke Math. J. }\textbf{29} (1962), 521--537.

\bibitem{G} Urs Graf, \textit{Applied Laplace Transforms and z-Transforms
for Scientists and Engineers. A Computational Approach using a `Mathematica'
Package}, Birkh\"{a}user, 2004.

\bibitem{Go} H. W. Gould, Generalization of Hermite's divisibility theorems
and the Mann-Shanks primality criterion for $s$-Fibonomial arrays, \textit{%
Fibonacci Quart.} \textbf{12} (1974), 157--166.

\bibitem{Hog} V. E. Hoggatt, Jr. Fibonacci numbers and generalized binomial
coefficients, \textit{Fibonacci Quart.} \textbf{5} (1967), 383--400.

\bibitem{Hor} A. F. Horadam, Generating functions for powers of a certain
generalised sequence of numbers, \textit{Duke Math. J.} \textbf{32} (1965),
437--446.

\bibitem{K} Thomas Koshy, \textit{Fibonacci and Lucas Numbers with
Applications}, John Wiley \&amp; Sons, Inc. 2001.

\bibitem{P} C. Pita, More on Fibonomials, \textit{Proceedings of the XIV
Conference on Fibonacci Numbers and Their Applications} (2010), to appear.

\bibitem{R} J. Riordan, Generating functions for powers of Fibonacci
numbers, \textit{Duke Math. J. }\textbf{29} (1962), 5--12.

\bibitem{S-T1} J. Seibert and P. Trojovsk\'{y}, On some identities for the
Fibonomial coefficients, \textit{Math. Slovaca} \textbf{55} (2005), 9--19.

\bibitem{S-T2} J. Seibert and P. Trojovsk\'{y}, On sums of certain products
of Lucas numbers, \textit{Fibonacci Quart.} \textbf{44} (2006), 172--180.

\bibitem{Sh} A. G. Shannon, A method of Carlitz applied to the $k$-th power
generating function for Fibonacci numbers, \textit{Fibonacci Quart.} \textbf{%
12} (1974), 293--299.

\bibitem{St} I. Strazdins, Lucas factors and a Fibonomial generating
function, in G. E. Bergum, A. N. Philippou and A. F. Horadam, eds.,  \textit{%
Applications of Fibonacci Numbers, Vol. 7}, Kluwer Academic Publishers
(1998), 401--404.

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

\bibitem{Vi} Robert Vilch, \textit{Z Transform Theory and Applications}, D.
Reidel Publishing Company, 1987.
\end{thebibliography}


\bigskip
\hrule
\bigskip

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

\noindent \emph{Keywords: } 
$s$-Fibonomial coefficients, Z transform.

\bigskip
\hrule
\bigskip

\noindent (Concerned with sequence
\seqnum{A000032},
\seqnum{A000045},
\seqnum{A001076},
\seqnum{A001654},
\seqnum{A001906},
\seqnum{A010048},
\seqnum{A034801},
\seqnum{A034802},
\seqnum{A047946},
\seqnum{A083564},
\seqnum{A092521}, and
\seqnum{A156085}.)

\bigskip
\hrule
\bigskip

\vspace*{+.1in}
\noindent
Received October 19 2010;
revised version received March 9 2011. 
Published in {\it Journal of Integer Sequences}, March 26 2011.

\bigskip
\hrule
\bigskip

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

\end{document}

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