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\begin{document}




\title{On multiple sums of products of Lucas numbers}


\author{Jaroslav Seibert, Pavel Trojovsk\'y \\   
University Hradec Kr\'alov\'e \\
Department of Mathematics  \\
Rokitansk\'eho 62, 500 03 Hradec Kr\'alov\'e, Czech Republic \\ 
pavel.trojovsky@uhk.cz
}

\date{}

\maketitle

\begin{abstract}
This paper studies some sums of products of the Lucas numbers, which are a generalization of the sums of the Lucas numbers from \cite{Va11}. These sums are related to the denominator of the generating function of the $k$-th powers of the Fibonacci numbers. We considered \cite{Se9} a special case for an even positive integer $k$ and we generalize this result to an arbitrary positive integer $k$ in this paper. These sums are expressed as the sum of the binomial and Fibonomial coefficients. We see Theorem 2 and its proof, which is based on special inverse formulas, as the main result of paper.\\
\noindent {\bf AMS Subj. Classification:} 11B39, 05A15, 05A10.\\
\noindent {\bf Key Words and Phrases:} Generating function, Riordan's theorem, Generalized Fibonacci numbers, Fibonomial coefficients.
\end{abstract}

\newcommand{\fbinom}[2]{ { #1 \brack #2 } }
\newcommand{\gbinom}[2]{ \left\{ #1 \atop #2 \right\} }



\section{Introduction}
\medskip

Generating functions are very helpful in finding of relations for sequences of integers. Some authors found miscellaneous 
identities for the Fibonacci numbers $F_n$, defined by recurrence relation $F_{n+2}=F_n+F_{n+1}$, with $F_0=0$, $F_1=1$, and the Lucas numbers $L_n$, defined by the same recurrence but with the initial conditions $L_0=2$, $L_1=1$, by manipulation with their generating functions. Our approach is rather different in this paper.  

In 1718 DeMoivre found the generating function of the Fibonacci numbers $F_n$ and used it for deriving the closed form 
$F_n=\frac{1}{\sqrt5}(\alpha^n-\beta^n)$, with $\alpha=\frac12(1+\sqrt5)$ and $\beta=\frac12(1-\sqrt5)$ (similarly the formula $L_n=\alpha^n+\beta^n$ holds for the Lucas numbers). In 1957 S.~W.~Golomb \cite{Go2} found the generating function for the square of $F_n$ and this result started the effort to find a recurrence or a closed form for the generating function $f_k(x) = \sum_{n=0}^{\infty} F_n^k x^n$ of the $k$-th powers of the Fibonacci numbers. Riordan \cite{Ri6} found a general recurrence for $f_k(x)$. Carlitz \cite{Ca1},  Horadam \cite{Ho4} and Mansour \cite{Ma0} presented some generalizations of Riordan's results and found similar recurrences for the generating functions of powers of any \hbox{second--order} recurrence sequences. 

Horadam gave some closed forms for the numerator and the denominator of this generating function. From his results follows, for example   
\begin{equation}
f_k(x) = \frac{
\sum_{i=0}^{k}
  \sum_{j=0}^i
    (-1)^{\frac{j(j+1)}2}\fbinom{k+1}{j}\; F_{i-j}^k \; x^i
}
{
\sum_{i=0}^{k+1} (-1)^{\frac{i(i+1)}2} \fbinom{k+1}{i}\; x^i}~, \label{eq: R1}   %\tag{1}
\end{equation}
where $\fbinom{n}{k}$ are the Fibonomial coefficients defined for any nonnegative integers $n$ and $k$ by
$$
 \fbinom{n}{k}=\prod_{i=0}^{k-1} \frac{F_{n-i}}{F_{i+1}} =  \frac{F_n F_{n-1}\cdots F_{n-k+1}}{F_1 F_2\cdots F_{k}}~,
$$
with $\fbinom{n}{0}=1$ and $\fbinom{n}{k}=0$ for  $n<k$.

Using Carlitz' method, Shannon \cite{Sh10} obtained some special results for the numerator and the denominator in the expression of the generating function $f_k(x)$. For example, he used the $q$--analog of the terminating binomial theorem (firstly published by Rothe \cite{Ro8}, but from Gauss's posthumous papers it is known that he had found it around 1808, see \cite{Ka5}) and obtained the relation
$$
 \prod_{i=0}^{k} (1-q^i\,x) = \sum_{i=0}^{k+1} (-1)^i q^{\frac{i}2(i-1)}\gbinom{k+1}{i}\,x^i~. 
$$
Q--binomial coefficients are defined $\gbinom{k+1}{i}=\frac{(q^{k+1}-1)(q^k-1)\cdots(q^{k-i+2}-1)}{(q-1)(q^2-1)\cdots(q^i-1)}$ for $i\geq 1$ and any complex numbers $q$, $x$ and any positive integer $k$, where $\gbinom{k+1}{0}=1$. 
Replacing $q$ by $\beta/\alpha$ and $x$ by $\alpha^k\,x$ he got
\begin{equation}
 \prod_{i=0}^{k} (1-\alpha^{k-i}\beta^i\,x) = \sum_{i=0}^{k+1} (-1)^{\frac{i}2(i+1)}\fbinom{k+1}{i}\,x^i~. 
     \label{eq: R2} %\tag{2}
\end{equation}

 We paid attention \cite{Se9} to a generalization of a type of the \hbox{well--known} formulas for the Fibonacci and Lucas numbers, see \cite{Va11} pp.~179--183, for example 
$$
 \sum_{i = 0}^n (-1)^i L_{n - 2i} = 2F_{n+1}~.
$$
In this paper we concentrate on the sums
\begin{equation}
\sum_{i_{n}=0}^{\lfloor\frac{k-1}2\rfloor}
      \sum_{i_{n-1}=i_n+1}^{\lfloor\frac{k-1}2\rfloor}
      \cdots
       \sum_{i_{n-2}=i_{n-1}+1}^{\lfloor\frac{k-1}2\rfloor}
       (-1)^{i_1+i_2+\cdots+i_n}
        \prod_{j=1}^{n} L_{k-2\,i_j}~,\label{eq: R3} %\tag{3}
\end{equation}
where $k$ is an arbitrary positive integer. The special case of \thetag{3} for an odd $k$ was solved up in \cite{Se9}. Here we use analogous method to find formulas for an even integer $k$.


Throughout the paper we adopt the conventions that the sum and the product over an empty set is 0 and 1, respectively,  
$\lfloor x \rfloor$ represents the greatest integer less than or equal to $x$, the relation $f(x)\sim g(x)$ means
that $f(x)$ is asymptotic to $g(x)$ and Iverson's notation (see, e.~g.,~\cite{Gr3}) that 

\begin{displaymath}
  [P(k)]=
      \left\{\begin{array}{c} 
        1,\quad \mbox{ if statement } P(k) \mbox { is true\,;} \\ 
        0,\quad \mbox{ if statement } P(k) \mbox { is false\,.} \\ 
       \end{array}\right.
\end{displaymath}





\bigskip
\section{The main results}
\medskip


\begin{definition}
Let $k$ be any positive integer.  We define the sequence $\{S_n(k)\}_{n=0}^{\infty}$ in the following way
$$
 S_0(k)=1~, \quad
 S_1(k) = \sum_{i_1=0}^{\lfloor\frac{k-1}2\rfloor} (-1)^{i_1} L_{k-2\,i_1}
$$
and 
\begin{equation}
S_n(k)=\sum_{i_{n}=0}^{\lfloor\frac{k-1}2\rfloor}
      \sum_{i_{n-1}=i_n+1}^{\lfloor\frac{k-1}2\rfloor}
      \cdots
       \sum_{i_1=i_2+1}^{\lfloor\frac{k-1}2\rfloor}
       (-1)^{i_1+i_2+\cdots+i_n}
        \prod_{j=1}^{n} L_{k-2\,i_j}~,\label{eq: R4} %\tag{4}
\end{equation}
for any integer $n>1$. 
\end{definition}


Let us denote 
$$
 \Theta(i,k,n) =  \binom{\lfloor\frac{k+1}2\rfloor-n+i}{i}+\binom{\lfloor\frac{k+1}2\rfloor-n+i-1}{i-1}
$$
for any positive integers $i$, $k$ and any nonnegative integer $n$.




\begin{theorem}%1
Let $n$ be any nonnegative integer and let $k$ be any positive integer. 
Then
\begin{equation}
  S_{n}(k) = 
   \sum_{i=0}^{\lfloor\frac{n}2\rfloor} (-1)^{\lfloor\frac{n}2\rfloor - i} \, \Theta(i,k,n) \, \fbinom{k+1}{n-2i} 
   \label{eq: R5} %\tag{5}
\end{equation}
if $k$  is odd and
\begin{equation}
S_{n}(k) =
\sum_{i=0}^{\lfloor\frac{n}2\rfloor} 
\sum_{j=0}^{n-2i} 
(-1)^{i + n(\frac{k}2+1) + \frac{j}2(j+k+1)} \, \Theta(i,k,n) \, \fbinom{k+1}{j}\label{eq: R6} %\tag{6}
\end{equation}
if $k$ is even.
\end{theorem}



\begin{corollary} Let $n$ be any nonnegative integer and let $k$ be any positive integer. Then the asymptotic formula 
\begin{equation}
  S_{n}(k) \sim  \sum_{i=0}^{\lfloor\frac{n}2\rfloor} (-1)^{\lfloor\frac{n}2\rfloor + i\,k} \, \Theta(i,k,n)\,
 \fbinom{k+1}{n-2i}\label{eq: R7} %\tag{7}
\end{equation}
holds as $k\to\infty$.
\end{corollary}


\begin{theorem}
Let $m$ be any integer and let $k$ be any even positive integer. Then
$$
 \sum_{j=0}^{m} (-1)^{\frac{j}2(j+k+1)} \fbinom{k+1}{j} = 
     (-1)^{\frac{m}2(m+k+1)}\frac1{F_{\frac{k}2+1}} \sum_{i=0}^{\lfloor\frac{m}2\rfloor} (-1)^i 
    \fbinom{k+2}{m-2i}\,F_{\frac{k+2}2-m+2i}~.
$$
\end{theorem}


\begin{corollary}
Let $n$ be any nonnegative integer and let $k$ be any even positive integer. 
Then
\begin{equation}
 S_{n}(k) = \frac{(-1)^{\lfloor\frac{n}2\rfloor}}{F_{\frac{k}2+1}} 
     \sum_{i=0}^{\lfloor\frac{n}2\rfloor}\sum_{j=i}^{\lfloor\frac{n}2\rfloor} \, (-1)^{i+j}\, 
     \Theta(i,k,n) \,\fbinom{k+2}{n-j} \, F_{\frac{k+2}2-n+2j}~.\label{eq: R8} %\tag{8}
\end{equation}
\end{corollary}


\begin{theorem}%Theorem~3
Let $m$ be  any integer. Then
\begin{align*}
 \sum_{j=0}^{m}  (-1)^{\frac{j}2(j+k+1)} \fbinom{k+1}{j}   &=    
\frac{(-1)^{\frac{m}2(m+k+1)}}{F_{\frac{k}2+1}F_{k+3}F_{k+4}} 
 \sum_{i=0}^{\lfloor\frac{m}4\rfloor} \fbinom{k+4}{m-4i} \times \\
& \times\left(F_{\frac{k}2+1-(m-4i)}\,L_{\frac{k}2+2-(m-4i)}\,F_{k+3}-F_{m-4i}\,F_{m-4i-1}\right)~.
\end{align*}
\end{theorem}


\begin{corollary} Let $n$ be any nonnegative integer. Then 
\begin{equation}
  \sum_{i=0}^{\lfloor\frac{n}2\rfloor} (-1)^{i} 
  \left(
   \binom{\frac{k+1}2-n+i}{i}+\binom{\frac{k-1}2-n+i}{i-1} 
  \right)\, \fbinom{k+1}{n-2i} = 0 \label{eq: R9} %\tag{9}
 \end{equation}
if $k$ is an odd positive integer, $k<2n-1$, and
\begin{equation}
 \sum_{i=0}^{\lfloor\frac{n}2\rfloor} 
 \sum_{j=0}^{n-2i} 
 (-1)^{i+\frac{j}2(j+k+1)} 
 \left(
   \binom{\frac{k}{2}-n+i}{i}+\binom{\frac{k-2}{2}-n+i}{i-1} 
 \right)\, \fbinom{k+1}{j} = 0   \label{eq: R10} %\tag{10}
\end{equation}
if $k$ is an even, $k<2n$.
\end{corollary}

\begin{corollary}
Let $k$ be any even positive integer. Then
\begin{align*}
 \sum_{i=0}^{\frac{k-2}2} (-1)^{i}\, L_{k-2i} &= F_{k+1} - (-1)^{\frac{k}2}~,                                \\
 \sum_{i_2=0}^{\frac{k-2}2}\sum_{i_1=i_2+1}^{\frac{k-2}2} (-1)^{i_1+i_2+1}\, L_{k-2i_1}\,L_{k-2i_2} &= 
 \frac{k-2}2 + (-1)^{\frac{k}2}F_{k+1} + F_k\,F_{k+1}                               
\end{align*}
and
\begin{align*}
\sum_{i_3=0}^{\frac{k-2}2}\sum_{i_2=i_3+1}^{\frac{k-2}2}&\sum_{i_1=i_2+1}^{\frac{k-2}2} 
 (-1)^{i_1+i_2+i_3}\, L_{k-2i_1}\,L_{k-2i_2}\,L_{k-2i_3}                                                    \\
&=\frac{k-4}2 \left((-1)^{\frac{k}2} - F_{k+1}\right) + F_k\,F_{k+1}\left((-1)^{\frac{k}2} - \frac{1}2 F_{k-1} \right) ~.
\end{align*}
\end{corollary}


\bigskip
\section{The preliminary results}
\medskip

\begin{lemma}%1
Let $k$ be any positive integer. Then $S_n(k) = 0$ for each positive integer $n>\left\lfloor\frac{k+1}2\right\rfloor$.
\end{lemma}
\begin{proof}%[\bf Proof of Lemma~7] 
After rewriting relation \thetag{4} from Definition 1 into the form
$$
S_n(k) = 
\sum_
{\substack{i_1, i_2,\dots, i_n \\ 0\leq i_n< i_{n-1}< \cdots < i_1 \leq \left\lfloor\frac{k-1}2\right\rfloor}}
     (-1)^{i_1+i_2+\cdots+i_n}
        \prod_{j=1}^{n} L_{k-2\,i_j}
$$
the assertion easily follows from the condition
$$
 0\leq i_n< i_{n-1}< \cdots < i_1 \leq \left\lfloor\frac{k-1}2\right\rfloor
$$
which does not hold for any values $i_1, i_2,\dots, i_n$ if $\left\lfloor\frac{k-1}2\right\rfloor<n-1$.
\end{proof}


\begin{lemma}%2
Let $k$ be any even positive integer and let $n$ be any positive integer. Then
\begin{alignat*}{2}
  &(i)& \qquad \sum_{i=0}^{n}
    \binom{\frac{k}2-2i}{n-i} S_{2i}(k) &= 0 \text{\quad for \quad} n\geq \frac{k}2+1  \\
   &(ii)& \qquad
  \sum_{i=0}^{n}
   \binom{\frac{k}2 - (2i+1)}{n-i} S_{2i+1}(k) &= 0 \text{\quad for \quad} n\geq \frac{k}2~.
\end{alignat*}
\end{lemma}
\begin{proof}%[\bf Proof of Lemma~8] 
We show the proof of $(i)$. Case $(ii)$ can be proved by an analogous way.
Each positive integer $n\geq \frac{k}2+1$ can be written in the form $n=\frac{k}2+l$, where $l$ is any positive integer. 
We will present that just one of factors in the product $\binom{\frac{k}2-2i}{n-i} S_{2i}(k)$ is equal to zero. Concretely, the first one equals zero for $i\leq\lfloor\frac{k}4\rfloor$ and the second one equals zero for 
$i>\lfloor\frac{k}4\rfloor$. For the sum in $(i)$ the following holds: 
$$
\sum_{i=0}^{\frac{k}2+l}
   \binom{\frac{k}2-2i}{\frac{k}2+l-i} S_{2i}(k) = Q_1(k,l) + Q_2(k,l)~,
$$
where
$$
 Q_1(k,l) = \sum_{i=0}^{\lfloor\frac{k}4\rfloor}
   \binom{\frac{k}2-2i}{\frac{k}2+l-i} S_{2i}(k)
$$
and
$$
 Q_2(k,l) = \sum_{i=\lfloor\frac{k}4\rfloor+1}^{\frac{k}2+l}
   \binom{\frac{k}2-2i}{\frac{k}2+l-i} S_{2i}(k)                
 =        \sum_{p=1}^{\frac{k}2-\lfloor\frac{k}4\rfloor+l}
   \binom{\frac{k}2-2\lfloor\frac{k}4\rfloor-2p}{\frac{k}2-\lfloor\frac{k}4\rfloor+l-p} 
   S_{2\lfloor\frac{k}4\rfloor+ 2p} (k)~.
$$
It is obvious that $\binom{\frac{k}2-2i}{\frac{k}2+l-i}=0$ if $i\leq\lfloor\frac{k}4\rfloor$ 
and therefore $Q_1(k,l)=0$ for any $k$ and $l$. Since the equality 
$S_{2\lfloor\frac{k}4\rfloor+2p}(k)=0$ is implied by Lemma~9 for any nonnegative integer $p$,
it follows that $Q_2(k,l)=0$.
\end{proof}



\begin{lemma}%3
Let $n$ be any positive integer and let $q$ be any integer. Then the following inverse formula holds:
$$
 a_n = \sum_{i=0}^{\lfloor \frac{n}2 \rfloor} (-1)^n \binom{q-n+2i}{i} b_{n-2i}
$$
if and only if 
\begin{equation}
 b_n = \sum_{i=0}^{\lfloor \frac{n}2 \rfloor} 
  (-1)^{n+i} \left( \binom{q-n+i}{i} + \binom{q-n+i-1}{i-1} \right)\, a_{n-2i}~.\label{eq: R11}%\tag{11}
\end{equation}
\end{lemma}
\begin{proof}%[\bf Proof of Lemma~9] 
Riordan \cite{Ri7} (p.~243) gave the following inverse formula:
$$
 a_n = \sum_{i=0}^{n} \binom{q-2i}{n-i} b_i
$$
if and only if 
$$
 b_n = \sum_{i=0}^{n} (-1)^{n+i} \left( \binom{q-n-i}{n-i} + \binom{q-n-i-1}{n-i-1} \right)\, a_i~.
$$
To get Lemma~11 from this formula first we substitute  $\{a_{n}\}$ by $\{a_{2n}\}$, $\{b_{i}\}$ by $\{b_{2i}\}$, $n$ by $\frac{n}{2}$ and $i$ by $\frac{n}{2}-i$ and then $\{a_{n}\}$ by $\{a_{2n+1}\}$, $\{b_{i}\}$ by $\{-b_{2i+1}\}$, $n$ by $\frac{n-1}{2}$, $i$ by $\frac{n-1}{2}-i$ and $q$ by $q-1$. This leads to the proved formula.
\end{proof}


\begin{lemma}%4
Let $n$, $k$, $l$ be any positive integers, $l<n<k$. Let $c_i$, $i=1,2,\dots, n$, be any real numbers, $c_n\not=0$.  Then 
$$
(i)\quad   \lim_{k\to\infty}\fbinom{k}{l} {\fbinom{k}{n}}^{-1} = 0~, \qquad
(ii)\quad  \sum_{i=l}^{n} c_i\,\fbinom{k}{i} \sim c_n\, \fbinom{k}{n} \quad \text{as}\quad k\to\infty~.     
$$
\end{lemma}
\begin{proof}%[\bf Proof of Lemma~10] 
Relation $(i)$ follows from the definition of the Fibonomial coefficients and the obvious fact that $\lim_{k\to\infty} F_k = \infty$. Thus,
\begin{align*}
\lim_{k\to\infty} \fbinom{k}{l} {\fbinom{k}{n}}^{-1} &=
\lim_{k\to\infty} \frac{F_k F_{k-1}\cdots F_{k-l+1}}{F_1F_2\cdots F_l} \cdot \frac{F_1F_2\cdots F_n}{F_k F_{k-1}\cdots F_{k-n+1}}=\\
&=\frac{F_1F_2\cdots F_n}{F_1 F_{2}\cdots F_l}\,
\lim_{k\to\infty} \frac{F_k F_{k-1}\cdots F_{k-l+1}}{F_k F_{k-1}\cdots F_{k-n+1}}=\\
&= \prod_{i=l+1}^{n} F_i\, \cdot \,
\lim_{k\to\infty} \frac1{F_{k-l}\cdots F_{k-n+1}}=0~.
\end{align*}
 Asymptotic formula $(ii)$ is implied by $(i)$.
\end{proof}


\begin{lemma}%5
Let $\{a_n\}$, $\{b_n\}$ be any sequences of real numbers, with $b_{-1}=0$, and let $h$ be any integer. 
Then for an arbitrary positive integer $n$
\begin{equation}
  a_n = b_n - (-1)^h\, b_{n-1} \label{eq: R12}%\tag{12}
\end{equation}
if and only if
\begin{equation}
 b_n = \sum_{i=0}^{n} (-1)^{h(n+i)} \, a_i~.\label{eq: R13}%\tag{13}
\end{equation}
\end{lemma}
\begin{proof}%[\bf Proof of Lemma~11] 
Let us show that identity \thetag{12} implies identity \thetag{13}. We have
\begin{align*}
 \sum_{i=0}^{n} (-1)^{h(n+i)} \, a_i & = \sum_{i=0}^{n} (-1)^{h(n+i)} \, (b_i - (-1)^h\,b_{i-1}) \\ 
   &= \sum_{i=0}^{n} (-1)^{h(n+i)} \, b_i - \sum_{i=1}^{n}(-1)^{h(n-1+i)}\,b_{i-1} - (-1)^{h(n-1)}\,b_{-1} \\
   &= \sum_{i=0}^{n} (-1)^{h(n+i)} \, b_i - \sum_{j=0}^{n-1}(-1)^{h(n+j)}\,b_j = b_n~.
\end{align*}
Thus, this part of the assertion is true and similarly we can prove the reversed implication.
\end{proof}


\begin{lemma}%6
Let $k$ be any even positive integer and let $a$ be any positive integer. Then
$$
 \fbinom{k+1}{a} + (-1)^{\frac{k}2+a}\, \fbinom{k+1}{a-1} = \frac{F_{\frac{k}2+1-a}}{F_{\frac{k}2 + 1}}\,  
\fbinom{k+2}{a}. 
$$
\end{lemma}
\begin{proof}%[\bf Proof of Lemma~12] 
Using the definition of the Fibonomial coefficients we get the relation
$$
  F_{\frac{k}2-a+1}\,F_{k+2} = F_{\frac{k}2+1} \left( F_{k-a+2} + (-1)^{\frac{k}2+a}\,F_a \right)~,
$$
which can be written in the form 
$$
 F_{\frac{k}2-a+1}\,L_{\frac{k}2+1} = F_{k-a+2} + (-1)^{\frac{k}2+a}\,F_a
$$
as $F_{2n}=F_n\,L_n$ (\cite{Va11}, p.~176). We get the previous relation by setting \linebreak $l=\frac{k}2-a+1$ and $n=\frac{k}2+1$ into the identity $($\cite{Va11}, p.~177$)$
\begin{equation}
 F_{l+n} = F_l\,L_n + (-1)^{n+1} F_{l-n}~,\label{eq: R14}%\tag{14}
\end{equation}
which holds for any integers $l$, $n$. The assertion follows at once.
\end{proof}

The following form of $\Theta(i,k,n)$ is more effective for the computation of the sums $S_n(k)$:

\begin{lemma}%7
Let $i$, $n$ be any integers and let $k$ be any even positive integer. Then
\begin{displaymath}
 \Theta(i,k,n) = 
  \left\{\begin{array}{cc} 
    0, & i<0~;        \\
    1, & i=0~;        \\
    \frac{k - 2(n - 2i)}{2i}\,  \prod_{j=1}^{i-1} \frac{k - 2(n + j - i)}{2(i - j)}~, & i>0~.
\end{array}
 \right.
\end{displaymath}
\end{lemma}
\begin{proof}%[\bf Proof of Lemma~13] 
The cases for $i\leq 0$ are clear. For $i>0$ we can write:
\begin{multline*}
 \Theta(i,k,n) =  \binom{\frac{k}2-n+i}{i}+\binom{\frac{k}2-n+i-1}{i-1} \\
               =     \frac{k-2(n-2i)}{2i}\; \binom{\frac{k}{2}-n+i-1}{i-1} = 
           \frac{k-2(n-2i)}{2i}\; \prod_{j=1}^{i-1} \frac{\frac{k}2 - n + i - j}{i - j}
\end{multline*}
and the proof is over.
\end{proof}

\bigskip
\section{Additional properties of the inner sum}
\medskip

Now we will investigate properties of the inner sum involved in \thetag{6}. Let us denote
\begin{equation}
  \sigma_k(m) = \sigma(m) := \sum_{j=0}^{k-m} (-1)^{\frac{j}2(j+k+1)} \fbinom{k+1}{j}~,\label{eq: R15}%\tag{15}
\end{equation}
where $k$ is any even positive integer and $m$ is any integer.


\begin{lemma}%8
Let $k$ be any even positive integer and let $m$ be any integer. Then

$(i)$
$$
  \sigma(m) = 0~, \text{ for }\quad m\leq -1 \quad \text{or} \quad m\geq k+1~, 
$$

$(ii)$
$$
 \sigma(k-m) = \sigma(m)~,
$$

$(iii)$
\begin{alignat*}{2}
  \sigma(0) &= 1~, \quad & \sigma(1)=&\, 1 + (-1)^{\frac{k-2}2}F_{k+1}~,            \\
  \sigma(2) &= 1 -  L_{\frac{k+2}2} F_{k+1}F_{\frac{k-2}2}~, \quad 
    & \sigma(3)=&\, 1 - \frac12(-1)^{\frac{k}2}F_{k+1} \left(2 - F_k\,F_{\frac{k-4}2} L_{\frac{k+2}2}\right)~.
\end{alignat*}
\end{lemma}
\begin{proof}%[\bf Proof of Lemma~14] 
$(i)$ First we prove the case for $m=-1$:
\begin{align*}
 \sigma(-1)  &=  \sum_{j=0}^{k+1} (-1)^{\frac{j}2(j+k+1)} \fbinom{k+1}{j}              \\
           &=  \sum_{j=0}^{\frac{k}2} (-1)^{\frac{j}2(j+k+1)} \fbinom{k+1}{j} +
               \sum_{j=\frac{k}2+1}^{k+1} (-1)^{\frac{j}2(j+k+1)} \fbinom{k+1}{j}    \\
            &= \sum_{j=0}^{\frac{k}2} (-1)^{\frac{j}2(j+k+1)} \fbinom{k+1}{j} +
               \sum_{i=0}^{\frac{k}2} (-1)^{\frac{k+1-i}2(2k+2-i)}  \fbinom{k+1}{k+1-i}  \\
            &= \sum_{j=0}^{\frac{k}2} (-1)^{\frac{j}2(j+k+1)} \fbinom{k+1}{j} +
               \sum_{i=0}^{\frac{k}2} (-1)^{-1} (-1)^{\frac{i}2(i+k+1)} \fbinom{k+1}{i} = 0~.
\end{align*}
For $m\geq k+1$ the assertion is obvious, according to defining formula \thetag{15}. The case for $m<-1$ follows from  $\sigma(-1)=0$ and $\fbinom{k+1}{i}=0$, for $i>k+1$, with respect to the definition of the Fibonomial coefficients.

\medskip
$(ii)$ We can write successively
\begin{align*}
 \sigma(k-m) &= \sum_{j=0}^{m} (-1)^{\frac{j}2(j+k+1)} \fbinom{k+1}{j} = 
                \sum_{i=k-m+1}^{k+1} (-1)^{\frac{k+1-i}2(2k+2-i)} \fbinom{k+1}{k+1-i} \\
            &= \sum_{i=k-m+1}^{k+1} (-1)^1 (-1)^{\frac{i}2(i+k+1)} \fbinom{k+1}{i}   \\
            &= \sum_{i=0}^{k+1} (-1)^1 (-1)^{\frac{i}2(i+k+1)} \fbinom{k+1}{i} -
               \sum_{i=0}^{k-m} (-1)^1 (-1)^{\frac{i}2(i+k+1)} \fbinom{k+1}{i} \\
            &= -\sigma(-1) + \sum_{i=0}^{k-m} (-1)^{\frac{i}2(i+k+1)} \fbinom{k+1}{i} = \sigma(m)~.
\end{align*}


\medskip
$(iii)$ Identities for $\sigma(0)$ and $\sigma(1)$ are directly implied by $\sigma(-1)=0$.  Using case $(ii)$ and identity \thetag{14} we have 
\begin{align*}
 \sigma(2) &= \sum_{j=0}^{k-2} (-1)^{\frac{j}2(j+k+1)} \fbinom{k+1}{j}  = 1 + (-1)^{\frac{k-2}2}F_{k+1} - F_{k+1}F_{k} \\
           &= 1 - F_{k+1}\left( F_k + (-1)^{\frac{k}2} \right)  =  1- F_{k+1}\,L_{\frac{k}2+1}\,F_{\frac{k}2-1}~,
\end{align*}
\begin{align*}
 \sigma(3) &= \sigma(2) - \frac12 (-1)^{\frac{k-2}2} F_{k+1}F_{k}F_{k-1}    \\
&=  1 - F_{k+1}F_{k} - (-1)^{\frac{k}2}F_{k+1} + \frac12 (-1)^{\frac{k}2} F_{k+1}F_{k}F_{k-1} \\
&=  1 - \frac12(-1)^{\frac{k}2} F_{k+1} \left( 2 - F_k\left( F_{k-1}-2(-1)^{\frac{k}2} \right)\right) \\
&=  1 - \frac12(-1)^{\frac{k}2} F_{k+1} \left( 2 - F_k\,F_{\frac{k}2-2}\,L_{\frac{k}2+1} \right)~. 
\end{align*}
This finishes the proof.
\end{proof}

\medskip
The sum $\sigma (m)$ can be simplified by the following lemma.

\begin{lemma}%9
Let $k$ be any even positive integer and let $m$ be any integer. Then
$$
  \sigma(m) - \sigma(m-2) = (-1)^{\frac{m}2(m+k+1)} \fbinom{k+2}{m} \frac{F_{\frac{k}2+1-m}}{F_{\frac{k}2+1}}~.
$$
\end{lemma}
\begin{proof}%[\bf Proof of Lemma~15] 
For $m<2$ the assertion follows from the definition of the Fibonomial coefficients and Lemma~16.
For $m\geq 2$ we have, with respect to Lemma~16,
\begin{align*}
 \sigma(m)-\sigma(m-2) &= \sigma(k-m)-\sigma(k-m+2) \\
                       &= \sum_{j=0}^{m} (-1)^{\frac{j}2(j+k+1)} \fbinom{k+1}{j} - 
                          \sum_{j=0}^{m-2} (-1)^{\frac{j}2(j+k+1)} \fbinom{k+1}{j}  \\
                       &= (-1)^{\frac{m}2(m+k+1)} \fbinom{k+1}{m} + 
                          (-1)^{\frac{m-1}2((m-1)+k+1)} \fbinom{k+1}{m-1}  \\
                       &= (-1)^{\frac{m}2(m+k+1)} 
                               \left( \fbinom{k+1}{m} + (-1)^{\frac{k}2+m}\fbinom{k+1}{m-1} \right)~,
\end{align*}
which, by Lemma~14, implies the assertion.
\end{proof}

\begin{lemma}%10 
Let $k$ be any even positive integer and let $m$ be any integer. Then
\begin{equation}
 \sigma(m) - \sigma(m-4) = (-1)^{\frac{m}2(m+k+1)}\fbinom{k+4}{m} \frac{F_{\frac{k}2+2-m}} {F_{\frac{k}2+1}F_{k+3}F_{k+4}}
\;\;\omega (m,k)~,\label{eq: R16}%\tag{16}
\end{equation}
where
$$
\omega (m,k) = F_{\frac{k}2+1-m}\,L_{\frac{k}2+2-m}\,F_{k+3}-F_{m}\,F_{m-1}~.
$$
\end{lemma}
\begin{proof}%[\bf Proof of Lemma~16]
With respect to Lemma~17 we have for any integer $m$
\begin{align*}
\sigma(m&) - \sigma(m-4) = (\sigma(m)-\sigma(m-2))+(\sigma(m-2)-\sigma(m-4)) =  \\
 &=   (-1)^{\frac{m}2(m+k+1)} \frac1{F_{\frac{k}2+1}}
\left(
  F_{\frac{k}2+1-m}\fbinom{k+2}{m} - F_{\frac{k}2+3-m}\fbinom{k+2}{m-2}
\right)~.
\end{align*}
The bracket term can be rewritten as 
\begin{align*}
  F_{\frac{k}2+1-m} & \fbinom{k+2}{m} - F_{\frac{k}2+3-m}\fbinom{k+2}{m-2} =  \\
       &=  \fbinom{k+4}{m} \frac1{F_{k+3}F_{k+4}}
\left(
  F_{\frac{k}2+1-m}\,F_{k+3-m}\,F_{k+4-m} - F_{\frac{k}2+3-m}\,F_m\,F_{m-1}~.
\right)   
\end{align*}
The identity
$$
  F_{k+3-m}\,F_{k+4-m} = F_{k+4-2m}\,F_{k+3}+F_m\,F_{m-1}
$$
follows from the identity (\cite{Va11}, p. 177)
$$
  F_{n+h}\,F_{n+l} - F_{n}\,F_{n+h+l} = (-1)^n F_h\,F_l~,
$$
with any integers $h$, $n$, $l$. Hence, we obtain
\begin{align*}
  &F_{\frac{k}2+1-m} \fbinom{k+2}{m} - F_{\frac{k}2+3-m}\fbinom{k+2}{m-2}  \\
 &=  \fbinom{k{+}4}{m} \frac1{F_{k+3}F_{k+4}}
\left(
  F_{\frac{k+2}2{-}m}\,\bigg(F_{k+4-2m}\,F_{k+3} {-} F_m\,F_{m-1}\bigg) - F_{\frac{k+6}2-m}\,F_m\,F_{m-1}
\right)   \\
&=  \fbinom{k{+}4}{m} \frac1{F_{k+3}F_{k+4}}
\left(
  F_{\frac{k}2+1-m}\,F_{k+4-2m}\,F_{k+3} - \left(F_{\frac{k}2+3-m} - F_{\frac{k}2+1-m}\right) F_m\,F_{m-1}
\right)   \\
&=  \fbinom{k{+}4}{m} \frac{F_{\frac{k}2+2-m}}{F_{k+3}F_{k+4}}
\left(
  F_{\frac{k}2+1-m}\,L_{\frac{k}2+2-m}\,F_{k+3} - F_{m}\,F_{m-1}
\right) 
\end{align*}
and the assertion follows.
\end{proof}


\begin{lemma}%11 
Let $m\geq 5$ be any integer and let $k$ be any positive even integer in one of the following forms
$$
 (i)\ k =  m - 4 + [2\nmid m]~, \quad (ii) \ k =  2(m-3)~, \quad (iii) \ k =  2(m-1)~.
$$
Then $\omega(m,k)$ can be factored into a product of the Fibonacci or Lucas numbers. 
\end{lemma}
\begin{proof}%[\bf Proof of Lemma~18]
Condition $(i)$, with respect to the identities (\cite{Va11}, pp.~176--177) $F_{-n}=(-1)^{n+1}F_n$, $L_{-n}=(-1)^{n}L_n$ and 
$F_{2n}=F_n\,L_n$, leads to the relation
\begin{align*}
 \omega(m,m-3) &= F_{-\frac{m+1}2}\,F_m\,L_{-\frac{m-1}2} - F_{m}\,F_{m-1} 
               = F_{m}\,L_{\frac{m-1}2} (F_{\frac{m+1}2} - F_{\frac{m-1}2}  )  \\
               &= F_{m}\,F_{\frac{m-3}2}\,L_{\frac{m-1}2}
\end{align*}
if $m$ is odd and to the relation 
\begin{align*}
 \omega(m,m-4) &= F_{\frac{m+2}2}\,F_{m-1}\,L_{\frac{m}2} - F_{m}\,F_{m-1} 
                = F_{m-1}\,L_{\frac{m}2} ( F_{\frac{m+2}2} - F_{\frac{m}2}) \\
                &= F_{m-1}\,F_{\frac{m-2}2}\,L_{\frac{m}2}
\end{align*}
if $m$ is even. 

Using the identity $F_{n+1}^2+F_n^2=F_{2n+1}$ (\cite{Va11}, p. 177), we have from condition $(ii)$
\begin{align*}
 \omega(m,2(m-3)) &= F_{2m-3} - F_m\,F_{m-1} = F_{m-2}^2 + F_{m-1}^2 - F_m\,F_{m-1}          \\
                  &= F_{m-2}^2 - F_{m-1}\,(F_m - F_{m-1})  =  F_{m-2}^2 - F_{m-1}\,F_{m-2} \\
                  &= F_{m-2}(F_{m-2} - F_{m-1}) = - F_{m-2}\,F_{m-3}~.
\end{align*}
Condition $(iii)$ gives $\omega(m,2(m-1))= - F_m\,F_{m-1}$.
\end{proof}

\begin{rem}%Remark~1
{\rm
The right--hand side of \thetag{16} can not be factored in a product of the Fibonacci or Lucas numbers for arbitrary values of $k$ and $m$. The trivial factorization can be done for $m=0$ and $m=1$. Table~1 lists the values of $m$ and $k$, $2\leq m\leq 10$, $2\leq k\leq 170$, for which $\omega(m,k)$ can be factored into a product of the Fibonacci or Lucas numbers. These values were found by computer. The computer search for $10 \leq m\leq 100$ showed that $\omega(m,k)$ can be factored into a product of the Fibonacci or Lucas numbers only at values of $m$, $k$ satisfying conditions from Lemma~20.

}
\bigskip

\newpage

\begin{center}
 {\rm Table~1. The values for which $\omega(m,k)$ is factorizable. 

\vspace{5mm}

\begin{tabular}{|c||ccccc|}
\hline
$m$ &  &  &    $k$  & &  \\
\hline
\hline
2 & 2 & 6 &      &  &    \\
3 & 2 & 4 & 6    &  &    \\
4 & 2 & 4 & 6 & 8  &     \\
5 & 2 & 4 & 8 & 10 &     \\
6 & 2 & 6 & 10 &  &      \\
7 & 4 & 6 & 8 & 12 &     \\
8 & 4 & 6 & 8 & 10 & 14  \\
9 & 2 & 6 & 10 & 12 & 16 \\
10 & 2 & 6 & 14 & 18 &   \\
\hline
\end{tabular}
}

\end{center}


\end{rem}



\medskip



\bigskip
\section{The proofs of the main results}
\medskip

\begin{proof}[\bf Proof of Theorem~2]%Theorem~1

First we prove identity \thetag{5}. We showed \cite{Se9} that for any positive odd integer $k$ and any positive integer 
$n$
\begin{equation}
S_{2n-1}(k) = \sum_{i=1}^{n} (-1)^{i+1} 
 \left(
   \binom{\frac{k+3}{2}-n-i}{n-i}+\binom{\frac{k+1}{2}-n-i}{n-i-1} 
 \right)\, \fbinom{k+1}{2i-1} \label{eq: R17}%\tag{17}
\end{equation}
and
\begin{equation}
S_{2(n-1)}(k) = 
\sum_{i=1}^{n} (-1)^{i+1} 
 \left(
   \binom{\frac{k+5}{2}-n-i}{n-i}+\binom{\frac{k+3}{2}-n-i}{n-i-1} 
 \right)\, \fbinom{k+1}{2(i-1)}~. \label{eq: R18}%\tag{18}
\end{equation}
Relation \thetag{5} can be obtained from \thetag{17} and \thetag{18}. 
Replacing $n$ by $n+1$ and $i$ by $n+1-i$ we have for any nonnegative integer $n$
$$
S_{2n+1}(k) = \sum_{i=0}^{n} (-1)^{n-i} 
 \left(
   \binom{\frac{k+1}{2}{-}(2n{+}1){+}i}{i}+\binom{\frac{k-1}{2}{-}(2n{+}1){+}i}{i-1} 
 \right)\, \fbinom{k+1}{2n{+}1{-}2i}
$$
and
$$
S_{2n}(k) = 
\sum_{i=0}^{n} (-1)^{n-i} 
 \left(
   \binom{\frac{k+1}{2}-2n+i}{i}+\binom{\frac{k-1}{2}-2n+i}{i-1} 
 \right)\, \fbinom{k+1}{2n - 2i}~,
$$
which can be joined into the proved identity. 

We begin the proof of relation \thetag{6} by defining the polynomial 
\begin{equation}
 P_{k}(x) = \sum_{i=0}^{k} p_i(k)\, x^i=\prod_{j=0}^{\frac{k}2-1} \left(1-(-1)^j L_{k-2j}\,x + x^2\right)
\label{eq: R19}%\tag{19}
\end{equation}
for an even nonnegative integer $k$. By direct multiplication of the factors in \thetag{19} we get the identities 
\begin{equation}
 p_{2i+1}(k) = - \sum_{j=0}^{i} \binom{\frac{k}{2}-(2j+1)}{i-j} \; S_{2j+1}(k)~,
\label{eq: R20}%\tag{20}
\end{equation}
for $i=0,1,2,\dots,\frac{k-2}{2}$, and
\begin{equation}
 p_{2i}(k) = \sum_{j=0}^{i} \binom{\frac{k}{2}-2j}{i-j} \; S_{2j}(k)~,\label{eq: R21}%\tag{21}
\end{equation}
for $i=0,1,2,\dots,\frac{k}{2}$. By shifting indexes of summation it is possible to join \thetag{20} and \thetag{21} into 
the relation 
\begin{equation}
 p_n(k) = \sum_{i=0}^{\lfloor\frac{n}2\rfloor} (-1)^{n} \binom{\frac{k}{2}-n+2i}{i} \; S_{n-2i}(k)~,
\label{eq: R22}%\tag{22}
\end{equation}
for $n=0,1,2,\dots,k$. This identity can be extended to any positive integer $n$ with respect to Lemma~9,
as $p_{n}(k)=0$ for $n<0$ or $n>k$.

If $k$ is an even positive integer, the denominator in \thetag{1} is a polynomial of an odd degree $k+1$: 
$$
 D_{k+1}(x)=\sum_{i=0}^{k+1} d_{k+1,i}\, x^i~, 
$$
where integers $d_{k+1,i}=(-1)^{\frac{i(i+1)}2}\fbinom{k+1}{i}$ are terms of the sequence which was named as 
``signed Fibonomial triangle'' in the on-line encyclopedia of integer sequences (maintained by Sloane)
with ID Number A055870, see \cite{WWW12}. Identity \thetag{2} implies
\begin{align*}
 D_{k+1}(x) &= \prod_{j=0}^{k} (1-\alpha^{k-j}\,\beta^j\,x) = (1-(\alpha \beta)^{\frac{k}2}x)
               \prod_{\substack{ j=0 \\ j\not=\frac{k}2}}^{k} (1-\alpha^{k-j}\,\beta^j\,x)  \\
            &=(1-(-1)^{\frac{k}{2}}x) \prod_{j=0}^{\frac{k}2-1} \left(1-(-1)^j\alpha^{k-2j}\,x\right)\, 
                         \left(1-(-1)^j\beta^{k-2j}\,x\right)   \\
            &=(1-(-1)^{\frac{k}{2}}x) \prod_{j=0}^{\frac{k}2-1} 
                  \left(1-(-1)^j (\alpha^{k-2j}+\beta^{k-2j})\,x + (\alpha\beta)^{k-2j}x^2 \right) \\
            &=(1-(-1)^{\frac{k}{2}}x) \prod_{j=0}^{\frac{k}2-1} (1-(-1)^j L_{k-2j}x + x^2)~,
\end{align*} 
according to the relation $\alpha\beta=-1$ and the formula \hbox{$L_{k{-}2j}=\alpha^{k{-}2j}{+}\beta^{k{-}2j}$.} 
Thus, with respect to \thetag{19}, $D_{k+1}(x) = (1-(-1)^{\frac{k}{2}}x) \, P_k(x)$.
By multiplying on the \hbox{right--hand} side and comparing coefficients of $x^i$ we have the following relations between coefficients $d_{k+1,i}$ of $D_{k+1}(x)$ and coefficients $p_i(k)$ of $P_k(x)$ 
\begin{align*}
 d_{k+1,0} &= p_0(k) = 1~,  \\
 d_{k+1,i} &= p_i(k) + (-1)^{\frac{k}{2}+1}\, p_{i-1}(k)~, \ i=1,2,\dots,k~,  \\
 d_{k+1,k+1} &= (-1)^{\frac{k}{2}+1} p_k(k) = (-1)^{\frac{k}{2}+1}~.  \\
\end{align*}
As $p_{n}(k)=0$ for $n<0$ or $n>k$ we can rewrite the previous relations in the recurrence
$$
 p_n(k) + (-1)^{\frac{k}2+1} \, p_{n-1}(k) = d_{k+1,n}~,
$$
which holds for any integer $n$. Using Lemma~13 we have 
\begin{equation}
  p_n(k)= \sum_{j=0}^{n} (-1)^{\frac{k}{2}(n+j)}  d_{k+1,j} \label{eq: R23}%\tag{23}
\end{equation}
for any nonnegative integer $n$.

To complete the proof of \thetag{6} we have to invert identity \thetag{22}. Setting \linebreak $a_n = p_{2n}(k)$, 
$b_n = S_{2n}(k)$ and $q=\frac{k}{2}$ in inverse formula \thetag{11} we obtain
\begin{equation}
 S_{n}(k) = \sum_{i=0}^{\lfloor \frac{n}2\rfloor} (-1)^{n+i} 
 \left(
   \binom{\frac{k}{2}-n+i}{i}+\binom{\frac{k}{2}-n+i-1}{i-1} 
 \right)\, p_{n-2i}(k)~.\label{eq: R24}%\tag{24} 
\end{equation}
From \thetag{23} and \thetag{24} we deduce that
$$
 S_{n}(k) = \sum_{i=0}^{\lfloor \frac{n}2\rfloor}\sum_{j=0}^{n-2i} (-1)^{n+i}(-1)^{\frac{k}2(n+j)} 
 \left(
   \binom{\frac{k}{2}-n+i}{i}+\binom{\frac{k}{2}-n+i-1}{i-1} 
 \right)\, d_{k+1,j}~. 
$$
Putting $d_{k+1,j}=(-1)^{\frac{j}2(j+1)}\fbinom{k+1}{j}$ we obtain \thetag{6} after simplification.
\end{proof}



\begin{proof}[\bf Proof of Corollary~3]%Corollary~1
The assertion is obviously true with respect to \thetag{5} if $k$ is any odd integer.
For even values of $k$ identity \thetag{6} can be written using \thetag{15} as
$$
 S_{n}(k) = \sum_{i=0}^{\lfloor\frac{n}2\rfloor} (-1)^{n+i+n\frac{k}2} \,\sigma (n-2i)\; \Theta(i,k,n)~.
$$
With respect to Lemma~12 for $k\to\infty$
$$
 \sigma (n-2i) \sim (-1)^{\frac{n-2i}2\,(n-2i+k+1)} \fbinom{k+1}{n-2i} =
(-1)^i(-1)^{\frac{n}2\,(n+k+1)} \fbinom{k+1}{n-2i}~.
$$ 
Hence, we obtain
\begin{align*}
 S_{n}(k) &\sim \sum_{i=0}^{\lfloor\frac{n}2\rfloor} (-1)^{n+i+n\frac{k}2}\,(-1)^i (-1)^{\frac{n}2\,(n+k+1)}\; 
                 \Theta(i,k,n)\,\fbinom{k+1}{n-2i} \\
          &= \sum_{i=0}^{\lfloor\frac{n}2\rfloor} (-1)^{\frac{n}2\,(n-1)}\; 
                 \Theta(i,k,n)\,\fbinom{k+1}{n-2i} 
\end{align*}
and the assertion follows from the congruence $\frac{n}2(n{-}1)\equiv \lfloor\frac{n}2\rfloor\pmod 2$.
\end{proof}



\begin{proof}[\bf Proof of Theorem~4]%Theorem~2
For any even $m$ we have
$$
 \sum_{i=0}^{\frac{m}2} (\sigma(m-2i) - \sigma(m-2(i+1))) = \sigma(m) - \sigma(-2)
$$
and analogously for any odd $m$
$$
 \sum_{i=0}^{\frac{m-1}2} ( \sigma(m-2i) - \sigma(m-2(i+1)) ) = \sigma(m) - \sigma(-1)~.
$$
Thus, using Lemma~16 we obtain for any integer $m$
$$
 \sigma(m) = \sum_{i=0}^{\lfloor\frac{m}2\rfloor} (\sigma(m-2i) - \sigma(m-2(i+1))) 
$$
and with respect to Lemma~17 
\begin{align*}
 \sigma(m) &= \sum_{i=0}^{\lfloor\frac{m}2\rfloor} 
(-1)^{\frac{m-2i}2(m-2i+k+1)}\frac1{F_{\frac{k}2+1}} \fbinom{k+2}{m-2i}\,F_{\frac{k}2+1-(m-2i)}  \\
  &=(-1)^{\frac{m}2(m+k+1)}\frac1{F_{\frac{k}2+1}} \sum_{i=0}^{\lfloor\frac{m}2\rfloor} (-1)^i \fbinom{k+2}{m-2i}\,F_{\frac{k+2}2-m+2i}~.
\end{align*}
\end{proof}


\begin{proof}[\bf Proof of Corollary~5]%Corollary~4 

Applying Theorem~2~and Theorem~4, consecutively, we get
\begin{align*}
 S_{n}(k) &= \sum_{i=0}^{\lfloor\frac{n}2\rfloor} (-1)^{n+i+n\frac{k}2} \,\sigma (n-2i)\; \Theta(i,k,n) \\
          &=\sum_{i=0}^{\lfloor\frac{n}2\rfloor} (-1)^{n+i+n\frac{k}2} \,  \Theta(i,k,n)\, \frac{(-1)^{\frac{n}2(n+k+1)}}{F_{\frac{k}2+1}}
             \, \sum_{j=i}^{\lfloor\frac{n}2\rfloor} (-1)^j \,\fbinom{k+2}{n-2j} \, F_{\frac{k+2}2-n+2j} \\
&=\sum_{i=0}^{\lfloor\frac{n}2\rfloor} (-1)^{\frac{n}2(n-1)+i} \, \frac1{F_{\frac{k}2+1}}
             \, \Theta(i,k,n)\;
               \sum_{j=i}^{\lfloor\frac{n}2\rfloor} (-1)^j \,\fbinom{k+2}{n-2j} \, F_{\frac{k+2}2-n+2j}  \\
&=\sum_{i=0}^{\lfloor\frac{n}2\rfloor} (-1)^{\lfloor\frac{n}2\rfloor+i} \, \Theta(i,k,n) \, \frac1{F_{\frac{k}2+1}}
             \;\sum_{j=i}^{\lfloor\frac{n}2\rfloor} (-1)^j \,\fbinom{k+2}{n-2j} \, F_{\frac{k+2}2-n+2j}                  \\
&= \frac{(-1)^{\lfloor\frac{n}2\rfloor}}{F_{\frac{k}2+1}} 
\sum_{i=0}^{\lfloor\frac{n}2\rfloor}\sum_{j=i}^{\lfloor\frac{n}2\rfloor} \, (-1)^{i+j}\,  
\Theta(i,k,n) \,\fbinom{k+2}{n-2j} \, F_{\frac{k+2}2-n+2j}~.
\end{align*}
\end{proof}


\begin{proof}[\bf Proof of Theorem~6]
Similarly as in the proof of Theorem~4 we obtain for any integer $m$ the relation
$$
 \sum_{i=0}^{\lfloor\frac{m}4\rfloor} (\sigma(m-4i) - \sigma(m-4(i+1))) = 
\sigma(m) -  
   \sigma\left(m-4\left(\left\lfloor\frac{m}4\right\rfloor + 1\right)\right)~.
$$
Thus, using Lemma~16 we obtain
$$
 \sigma(m) = \sum_{i=0}^{\lfloor\frac{m}4\rfloor} (\sigma(m-4i) - \sigma(m-4(i+1)))~. 
$$
With respect to Lemma~18 we have
\begin{align*}
 \sigma(m) &= \sum_{i=0}^{\lfloor\frac{m}4\rfloor} 
(-1)^{\frac{m-4i}2(m-4i+k+1)}
\fbinom{k+4}{m-4i} \frac{F_{\frac{k}2+2-(m-4i)}}{F_{\frac{k}2+1}F_{k+3}F_{k+4}} \\
& \cdot \left(
  F_{\frac{k}2 + 1 - (m-4i)}\,L_{\frac{k}2 + 2 - (m-4i)}\,F_{k+3} - F_{m-4i}\,F_{m-4i-1}
 \right)   \\
  &=\frac{(-1)^{\frac{m}2(m+k+1)}}{F_{\frac{k}2+1}F_{k+3}F_{k+4}} 
 \sum_{i=0}^{\lfloor\frac{m}4\rfloor} \fbinom{k+4}{m-4i}
F_{\frac{k}2+2-(m-4i)} \\
& \cdot \left(
  F_{\frac{k}2+1-(m-4i)}\,L_{\frac{k}2+2-(m-4i)}\,F_{k+3}-F_{m-4i}\,F_{m-4i-1}
 \right) ~.
\end{align*}
\end{proof}




\begin{proof}[\bf Proof of Corollary~7]
Identities \thetag{9} and \thetag{10} can be obtained from identities \thetag{5} and \thetag{6} with respect to 
$S_n(k)=0$ for positive integers $k$, $n>\lfloor \frac{k+1}2\rfloor$ (see Lemma~9).
\end{proof}

\begin{proof}[\bf Proof of Corollary~8]
Each of these three sums follows from identity \thetag{6} after some tedious simplification.
\end{proof}

\bigskip
\section{Concluding remark}
\medskip

It is interesting to compare the effectiveness of formulas \thetag{6} and \thetag{8} in contrast to defining formula \thetag{4} for computation of $S_n(k)$. Therefore, we found the CPU time (in seconds) required for computation of sums 
$S_3(k)$ for some values of $k$ using the system Mathematica on a standard PC. There is the measured time in Table~2.


\bigskip


\newcommand{\mez}{0.3truecm}

\begin{center}
Table~2. CPU time for $S_3(k)$ 

\vspace{5mm}

\hbox{\hskip 0.6truecm
\vbox{
\begin{tabular}{|ccccccccc|}
\hline
          & \qquad &  &  & $k$ & & &   &                         \\
\hline
  \hskip 0.2truecm  100 & \hskip 0.3truecm  200 & \hskip 0.3truecm 300 & \hspace*{\mez}  400 & 
\hskip 0.5truecm 500 & \hskip 0.5truecm 600 & \hskip 0.6truecm  700   & \hskip 0.8truecm  800 \hspace{-0.17truecm} &    \\
\hline
\end{tabular}
}}

\begin{tabular}{|c|cccccccc|}
\hline
\thetag{4} \; & 0.297 & 2.438  &   8.547  &  21.296 &  43.172   & 77.078  &   130.125  & 203.594 \\
\thetag{6} \; & 0     &   0    &    0.047  &  0.094  &  0.172   &  0.297   &    0.484  &   0.719 \\
\thetag{8} \; & 0     &   0    &    0.015  &  0.046  &  0.078   &  0.156   &    0.25   &   0.359 \\
\hline
\end{tabular}

\end{center}



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