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\begin{center}
\vskip 1cm{\LARGE\bf 
Cassini-Like Formula for Generalized \\
\vskip .1in
Hyper-Fibonacci Numbers
}
\vskip 1cm
\large
Lyes Ait-Amrane\footnote{Corresponding author.}\\
Ecole nationale Sup\'{e}rieure d'Informatique (ESI)\\
BP 68M, Oued Smar, 16309\\
El Harrach, Algiers\\
Algeria\\
\href{mailto:lyesait@gmail.com}{\tt lyesait@gmail.com }\\
\ \\
Djilali Behloul\\
Faculty of Computer Science\\
University of Sciences and Technology Houari Boumediene (USTHB) \\
BP 32, El Alia, 16111\\
Bab Ezzouar, Algiers\\
Algeria\\
\href{mailto:dbehloul@yahoo.fr}{\tt dbehloul@yahoo.fr}\\
\end{center}

\vskip .2 in

\begin{abstract}
In this paper, we give some properties of generalized hyper-Fibonacci numbers in order to obtain a Cassini-like formula for them.
\end{abstract}

\section{Introduction}
\label{intro}
 In this paper, we consider the generalized Fibonacci sequence $(w_n)_{n\ge0}$ defined by
\begin{displaymath}
\begin{cases}
w_0=a, w_1=b; \\
w_{n+2}=\alpha w_{n+1}+\beta w_{n}, & (n\geq0)
\end{cases}
\end{displaymath}
where $a, b, \alpha$, and $\beta$ are integers. Several authors \cite{cer,hor,kes,tag} studied properties of these numbers. If $\alpha^2+4\beta\ne0$, then we have
\begin{equation}\label{eq23}
w_n=as_1^n+(b-as_1)\left(\frac{s_1^n-s_2^n}{s_1-s_2}\right),\qquad n\ge0,
\end{equation}
where $s_1=(\alpha+\sqrt{\alpha+4\beta})/2$ and $s_2=(\alpha-\sqrt{\alpha+4\beta})/2$ are the roots of $x^2-\alpha x-\beta=0$. If $A=b-as_2$ and $B=b-as_1$, then the identity \eqref{eq23} is equivalent to the well-known Binet-like formula
$$w_n=\frac{As_1^n-Bs_2^n}{s_1-s_2},\qquad n\ge0.$$
If $\beta\ne0$ and $\alpha^2+4\beta\ne0$, then the generalized
Fibonacci numbers for negative subscripts are defined, using Identity
\eqref{eq23}, by $w_{-n}=a(s_1^{-n}+s_2^{-n})-(-\beta)^{-n}w_n$
for $n\ge0$. If $\beta\ne0$ and  $\alpha^2+4\beta=0$, then
$\alpha=2t$ and $\beta=-t^2$, where $t$ is the double root of
$x^2-\alpha x-\beta=0$. Thus, $w_n=((2b\alpha^{-1}-a)n+a)t^{n}$ and
$w_{-n}=((a-2b\alpha^{-1})n+a)t^{-n}$, for $n\ge0$.


We introduce the generalized hyper-Fibonacci numbers associated with the sequence $(w_n)_{n\geq0}$ as follows:
\begin{equation}\label{eq06}
w_n^{(r+1)}=\sum_{k=0}^n \alpha^{n-k}w_k^{(r)},\quad w_n^{(0)}=w_n,\quad w_0^{(r)}=a,\quad w_1^{(r)}=\alpha ar+b,
\end{equation}
where $r$ is a nonnegative integer.

The aim of this paper is to extend the well-known Cassini formula \cite{gkp, mil, wer}
\begin{equation}\label{eq03}
F_{n}F_{n+2}-F_{n+1}^2=(-1)^{n+1}
\end{equation}
to the generalized hyper-Fibonacci numbers \eqref{eq06}, where $(F_n)_n$ denotes the classical Fibonacci sequence.
The identity \eqref{eq03} can be written in a determinant form as
\begin{equation}\label{eq04}
\begin{vmatrix}
   F_n & F_{n+1} \\
   F_{n+1} & F_{n+2}
\end{vmatrix}=(-1)^{n+1}.
\end{equation}
In \cite{sta}, Stakhov generalized the Cassini formula \eqref{eq03} to the $p$-Fibonacci numbers and developed a new coding theory based on the
$Q_p$-matrices. By analogy, one can use the compagnon matrices given bellow in the formula \eqref{eq01} instead of the $Q_p$-matrices.
Halici \cite{hal} established the Cassini formula for the Fibonacci quaternions.
Martinjak and Urbiha \cite{mar} extended the Cassini formula \eqref{eq04} to the hyper-Fibonacci numbers defined in \cite{dil,cao} as
\begin{equation*}
F_n^{(r+1)}=\sum_{k=0}^n F_k^{(r)},\quad F_n^{(0)}=F_n,\quad F_0^{(r)}=0,\quad F_1^{(r)}=1,
\end{equation*}
where $r$ is a nonnegative integer. The number $F_n^{(r)}$ is called the $n$th hyper-Fibonacci number of the $r$th generation. The hyper-Fibonacci numbers satisfy
 many interesting number-theoretical and combinatorial properties, e.g., \cite{cao}. Martinjak and Urbiha \cite{mar} defined the matrix
 \begin{equation*}
A_{r,n}=\begin{pmatrix}
   F_n^{(r)} & F_{n+1}^{(r)} & \cdots & F_{n+r+1}^{(r)} \\
   F_{n+1}^{(r)} & F_{n+2}^{(r)} & \cdots & F_{n+r+2}^{(r)}\\
     \vdots & \vdots & \ddots & \vdots\\
    F_{n+r+1}^{(r)} & F_{n+r+2}^{(r)} & \cdots & F_{n+2r+2}^{(r)}
\end{pmatrix}
\end{equation*}
and proved that
\begin{equation}\label{eq24}
 \det(A_{r,n})=(-1)^{n+\lfloor (r+3)/2\rfloor},
 \end{equation}
  where $n$ and $r$ are nonnegative integers. It is clear that for $r = 0$, we obtain \eqref{eq04}. Recently, in \cite{ab1}, the authors generalized
the identity \eqref{eq24} to the $(a,b)$-hyper-Fibonacci numbers.


If $(b_n)_{n\geq0}$ and $(c_n)_{n\geq0}$ are two sequences satisfying the relation
$a_{n+2}=\alpha a_{n+1}+\beta a_{n}$, then we have the identity \cite{vol}
\begin{equation}\label{eq02}
b_nc_{n-1}-b_{n-1}c_n=(-\beta)^{n-1}(b_1c_0-b_0c_1).
\end{equation}
If we take $b_n=F_{n+2}$ and $c_n=F_{n+1}$, then the identity \eqref{eq02} reduces to \eqref{eq03}. For
 $b_n=w_{n+2}$ and $c_n=w_{n+1}$, the identity \eqref{eq02} reduces to
\begin{equation}\label{eq08}
w_{n}w_{n+2}-w_{n+1}^2=(-\beta)^{n}(\beta a^2+\alpha ab-b^2).
\end{equation}
Cassini formula \eqref{eq08} can also be expressed as a determinant in the following way
\begin{equation*}
\begin{vmatrix}
   w_n & w_{n+1} \\
w_{n+1} & w_{n+2}
\end{vmatrix}=(-\beta)^{n-1}(\beta b^2-\alpha\beta ab-\beta^2a^2).
\end{equation*}
 For $n, r\in\mathbb{Z}$ such that $n\ge0$ and $r\ge0$,
let us define the $(r+2)\times (r+2)$ matrix
 \begin{equation*}
W_{r,n}=\begin{pmatrix}
   w_n^{(r)} & w_{n+1}^{(r)} & \cdots & w_{n+r+1}^{(r)} \\
   w_{n+1}^{(r)} & w_{n+2}^{(r)} & \cdots & w_{n+r+2}^{(r)}\\
     \vdots & \vdots & \ddots & \vdots\\
    w_{n+r+1}^{(r)} & w_{n+r+2}^{(r)} & \cdots & w_{n+2r+2}^{(r)}
\end{pmatrix}.
\end{equation*}


In Section \ref{sec2}, we establish some combinatorial properties involving the generalized hyper-Fibonacci sequence $(w_{n}^{(r)})_{n\geq0}$. In Section \ref{sec3}, we evaluate the determinant of the matrix $W_{r,n}$, which gives the identity \eqref{eq24} for $a=0$ and $b=\alpha=\beta=1$. For $a=2$ and $b=\alpha=\beta=1$, we deduce the determinant of the matrix
\begin{equation*}
\begin{pmatrix}
   L_n^{(r)} & L_{n+1}^{(r)} & \cdots & L_{n+r+1}^{(r)} \\
   L_{n+1}^{(r)} & L_{n+2}^{(r)} & \cdots & L_{n+r+2}^{(r)}\\
     \vdots & \vdots & \ddots & \vdots\\
    L_{n+r+1}^{(r)} & L_{n+r+2}^{(r)} & \cdots & L_{n+2r+2}^{(r)}
\end{pmatrix}
\end{equation*}
involving the hyper-Lucas numbers given by \cite{dil,cao}
\begin{equation*}
L_n^{(r+1)}=\sum_{k=0}^n L_k^{(r)},\quad L_n^{(0)}=L_n,\quad L_0^{(r)}=2,\quad L_1^{(r)}=2r+1.
\end{equation*}

We let ${n \choose k}$ denote the binomial coefficient which is defined for a nonnegative integer $n$ and an integer $k$ by
\begin{displaymath}
{n \choose k} = \begin{cases}
\displaystyle\frac{n!}{k!(n-k)!}, & \text{if $0\leq k \leq n$;}\\
0, & \text{otherwise,}
\end{cases}
\end{displaymath}
and for a negative integer $n$ and an integer $k$ by
\begin{displaymath}
{n \choose k} = \begin{cases}
\displaystyle(-1)^k {-n+k-1 \choose k}, & \text{if $k \geq 0$;}\\[.2in]
\displaystyle(-1)^{n-k}{-k-1 \choose n-k}, & \text{if $k \leq n$;}\\
0, & \text{otherwise.}
\end{cases}
\end{displaymath}

In Sections \ref{sec2} and \ref{sec3} we assume that $\beta\ne0$.
\section{Properties of the sequences $(w_{n}^{(r)})_{n\geq0}$}\label{sec2}
 In this section, we give some properties satisfied by the generalized hyper-Fibonacci sequence $(w_{n}^{(r)})_{n\geq0}$ given by the formula \eqref{eq06}.
\begin{lemma}\label{lem01}
Let $n\geq0$ be an integer. Then
\begin{equation}\label{eq11}
w_n^{(1)}=\frac{1}{\beta}w_{n+2}-\frac{\alpha^{n+1}}{\beta}b.
\end{equation}
\end{lemma}
\begin{proof}
We prove the lemma by induction on $n\ge0$. For $n=0$, the identity \eqref{eq11} is trivially checked. Now assume that \eqref{eq11} holds for an integer $n\geq0$. Then
\begin{align*}
w_{n+1}^{(1)}  & = \sum_{k=0}^{n+1}\alpha^{n+1-k}w_k\\
               & = \alpha\sum_{k=0}^{n}\alpha^{n-k}w_k+w_{n+1}\\
               & = \alpha w_n^{(1)}+w_{n+1}\\
               & = \alpha\left(\frac{1}{\beta}w_{n+2}-\frac{\alpha^{n+1}}{\beta}b\right)+w_{n+1}\\
               & = \frac{1}{\beta}w_{n+3}-\frac{\alpha^{n+2}}{\beta}b.
\end{align*}
We conclude that \eqref{eq11} holds for all $n\ge0$.
\end{proof}

The following proposition expresses a generalized hyper-Fibonacci number of any generation $r\ge1$ in terms of the $0$th generation.
\begin{proposition}
Let $r\geq1$ be an integer. Then
\begin{equation}\label{eq09}
w_n^{(r)}=\frac{1}{\beta^r}w_{n+2r}-\alpha^{n+1}\sum_{l=0}^{r-1}{n+r-l-1 \choose r-l-1}\frac{w_{2l+1}}{\beta^{l+1}}, \quad n\geq0.
\end{equation}
\end{proposition}
\begin{proof}
We prove the proposition by induction on $r\ge1$. We deduce from Lemma \ref{lem01} that \eqref{eq09}
holds for $r=1$. Now assume that \eqref{eq09} holds for an integer $r\geq1$. Then
\begin{align*}
w_{n}^{(r+1)} & = \sum_{k=0}^{n}\alpha^{n-k}w_k^{(r)}\\
              & = \sum_{k=0}^{n}\alpha^{n-k}\left(\frac{1}{\beta^r}w_{k+2r}-\alpha^{k+1}\sum_{l=0}^{r-1}{k+r-l-1 \choose r-l-1} \frac{w_{2l+1}}{\beta^{l+1}}\right)\\
              & = \frac{1}{\beta^r}\sum_{k=0}^{n}\alpha^{n-k}w_{k+2r}-\sum_{k=0}^{n}\sum_{l=0}^{r-1}{k+r-l-1 \choose r-l-1 } \alpha^{n+1}\frac{w_{2l+1}}{\beta^{l+1}}\\
              & = \frac{1}{\beta^r}\sum_{l=0}^{n+2r}\alpha^{n+2r-l}w_{l}-\frac{\alpha^{n+1}}{\beta^r}\sum_{l=0}^{2r-1}\alpha^{2r-1-l}w_{l}
               -\alpha^{n+1}\sum_{l=0}^{r-1}\frac{w_{2l+1}}{\beta^{l+1}}\sum_{k=0}^{n}{k+r-l-1 \choose r-l-1}\\
              & = \frac{1}{\beta^r}w_{n+2r}^{(1)}-\frac{\alpha^{n+1}}{\beta^r}w_{2r-1}^{(1)}-\alpha^{n+1}\sum_{l=0}^{r-1}{n+r-l \choose r-l} \frac{w_{2l+1}}{\beta^{l+1}}\\
              & = \frac{1}{\beta^{r+1}}w_{n+2r+2}-\frac{\alpha^{n+1}}{\beta^{r+1}}w_{2r+1}-\alpha^{n+1}\sum_{l=0}^{r-1}{n+r-l \choose r-l} \frac{w_{2l+1}}{\beta^{l+1}}\\
              & = \frac{1}{\beta^{r+1}}w_{n+2r+2}-\alpha^{n+1}\sum_{l=0}^{r}{n+r-l \choose r-l}\frac{w_{2l+1}}{\beta^{l+1}}.
\end{align*}
We deduce that \eqref{eq09} holds for all $r\geq1$.
\end{proof}

To go further, we need to define the generalized hyper-Fibonacci numbers of negative subscripts. To do this, we first study the particular case $a=0$ and $b=1$ of the sequence $(w_n)_n\geq0$. Thus, let $(u_n)_n$ denote this particular sequence. The definition given by \eqref{eq06} reduces to
\begin{equation}\label{eq10}
u_n^{(r+1)}=\sum_{k=0}^n \alpha^{n-k}u_k^{(r)},\quad u_n^{(0)}=u_n,\quad u_0^{(r)}=0,\quad u_1^{(r)}=1.
\end{equation}

\begin{proposition}\label{prop01}
Let $r\geq0$ be an integer. Then
\begin{equation}\label{eq13}
u_n^{(r+1)}=\frac{1}{\beta}u_{n+2}^{(r)}-{n+r+1 \choose r}\frac{\alpha^{n+1}}{\beta}, \quad n\geq0.
\end{equation}
\end{proposition}
\begin{proof}
We prove the proposition by induction on $r\ge0$. We deduce from Lemma \ref{lem01} that \eqref{eq13}
holds for $r=0$. Now assume that \eqref{eq13} holds for an integer $r\geq0$. Then
\begin{align*}
u_{n}^{(r+2)} & = \displaystyle\sum_{k=0}^{n}\alpha^{n-k}u_k^{(r+1)}\\
              & = \sum_{k=0}^{n}\alpha^{n-k}\left(\frac{1}{\beta}u_{k+2}^{(r)}-{k+r+1 \choose r}\frac{\alpha^{k+1}}{\beta}\right)\\
              & = \frac{1}{\beta}\sum_{k=0}^{n}\alpha^{n-k}u_{k+2}^{(r)}-\frac{1}{\beta}\sum_{k=0}^{n}{k+r+1 \choose r}\alpha^{n+1}\\
              & = \frac{1}{\beta}\sum_{l=2}^{n+2}\alpha^{n+2-l}u_{l}^{(r)}-\frac{1}{\beta}\sum_{l=2}^{n+2}{l+r-1 \choose r}\alpha^{n+1}\\
              & = \frac{1}{\beta}\left(\sum_{l=0}^{n+2}\alpha^{n+2-l}u_{l}^{(r)}-\alpha^{n+1}\right)-\frac{1}{\beta}\left(\sum_{l=1}^{n+2}{l+r-1 \choose r} \alpha^{n+1}-\alpha^{n+1}\right)\\
              & = \frac{1}{\beta}u_{n+2}^{(r+1)}-\frac{1}{\beta}\sum_{l=1}^{n+2}{l+r-1 \choose r}\alpha^{n+1}\\
              & = \frac{1}{\beta}u_{n+2}^{(r+1)}-{n+r+2 \choose r+1}\frac{\alpha^{n+1}}{\beta}.
\end{align*}
We deduce that \eqref{eq13} holds for all $r\geq0$.
\end{proof}

The following corollary allows us to extend the sequence $(u_n)_{n\geq0}$ to negative subscripts.
\begin{corollary}\label{cor03}
Let $r\geq0$ and $n\ge0$ be integers. Then
\begin{equation}\label{eq19}
u_{n+2}^{(r+1)}=\alpha u_{n+1}^{(r+1)}+\beta u_n^{(r+1)}+{n+r+1 \choose r}\alpha^{n+1}.
\end{equation}
\end{corollary}
\begin{proof}
According to \eqref{eq10}, we have
\begin{equation}\label{eq21}
u_{n+1}^{(r)}=u_{n+1}^{(r+1)}-\alpha u_n^{(r+1)},
\end{equation}
where $n\geq0$ and $r\geq0$. From \eqref{eq13} and \eqref{eq21} we get
$$u_n^{(r+1)}=\frac{1}{\beta}u_{n+2}^{(r+1)}-\frac{\alpha}{\beta} u_{n+1}^{(r+1)}-{n+r+1 \choose r}\frac{\alpha^{n+1}}{\beta}.$$
We deduce that
$$u_{n+2}^{(r+1)}=\alpha u_{n+1}^{(r+1)}+\beta u_n^{(r+1)}+{n+r+1 \choose r}\alpha^{n+1}.$$
\end{proof}

Now we give a linear relation between the sequences $(w_n)_{n\geq0}$ and $(u_n)_{n\geq0}$.
\begin{lemma}\label{lem04}
Let $n\ge0$ be an integer. Then
\begin{equation}\label{eq14}
w_{n+1}=bu_{n+1}+\beta au_{n}.
\end{equation}
\end{lemma}
\begin{proof}
We prove the lemma by induction on $n\ge0$. For $n=0, 1$ and $2$, the identity \eqref{eq14} is trivially checked. Now assume that \eqref{eq14} holds up to 
some order $n\ge1$. Then
\begin{align*}
w_{n+2} & = \alpha w_{n+1}+\beta w_{n}\\
        & = \alpha\left(bu_{n+1}+\beta au_{n}\right)+\beta\left(bu_{n}+\beta au_{n-1}\right)\\
        & = b\left(\alpha u_{n+1}+\beta u_{n}\right)+\beta a\left(\alpha u_{n}+\beta u_{n-1}\right)\\
        & = bu_{n+2}+\beta au_{n+1}.
\end{align*}
We conclude that \eqref{eq14} holds for all $n\ge0$.
\end{proof}

In the following lemma, we give a relation between the sequences
$(w_{n}^{(1)})_{n\geq0}$ and $(u_{n}^{(1)})_{n\geq0}$.
\begin{lemma}\label{lem05}
Let $n\ge0$ be an integer. Then
\begin{equation}\label{eq15}
w_{n}^{(1)}=bu_{n}^{(1)}+\beta au_{n-1}^{(1)}+a\alpha^n.
\end{equation}
\end{lemma}
\begin{proof}
We prove the lemma by induction on $n\ge0$. Since $u_{0}^{(1)}=u_{-1}^{(1)}=0$, the identity \eqref{eq15} holds for $n=0$. Now assume that \eqref{eq15} holds for an integer $n\ge0$. Then
\begin{align*}
w_{n+1}^{(1)} & = \sum_{k=0}^{n+1}\alpha^{n+1-k}w_{k}\\
              & = w_{n+1}+\alpha\sum_{k=0}^{n}\alpha^{n-k}w_{k}\\
              & = w_{n+1}+\alpha w_{n}^{(1)}\\
              & = bu_{n+1}+\beta au_{n}+\alpha(bu_{n}^{(1)}+\beta au_{n-1}^{(1)}+a\alpha^n)\\
              & = b(u_{n+1}+\alpha u_{n}^{(1)})+\beta a(u_{n}+\alpha u_{n-1}^{(1)})+a\alpha^{n+1} \\
              & = bu_{n+1}^{(1)}+\beta au_{n}^{(1)}+a\alpha^{n+1}.
\end{align*}
We conclude that \eqref{eq15} holds for all $n\ge0$.
\end{proof}

Now we derive a relation between the sequences $(w_{n}^{(r)})_{n\geq0}$ and $(u_{n}^{(r)})_{n\geq0}$, for any generation $r\geq1$.
\begin{proposition}
Let $r\ge1$ be an integer. Then
\begin{equation}\label{eq16}
w_{n}^{(r)}=bu_{n}^{(r)}+\beta au_{n-1}^{(r)}+a{n+r-1 \choose r-1}\alpha^n,\qquad n\ge0.
\end{equation}
\end{proposition}
\begin{proof}
We prove the proposition by induction on $n\ge0$. We deduce from Lemma \ref{lem05} that Identity \eqref{eq16} holds for $r=1$. Now assume that \eqref{eq16} holds for an integer $r\ge1$. Then
\begin{align*}
w_{n}^{(r+1)} & = \sum_{k=0}^{n}\alpha^{n-k}w_{k}\\
              & = \sum_{k=0}^{n}\alpha^{n-k}\left(bu_{k}^{(r)}+\beta au_{k-1}^{(r)}+a{k+r-1 \choose r-1}\alpha^k\right)\\
              & = \sum_{k=0}^{n}\alpha^{n-k}u_{k}^{(r)}+\beta a\sum_{k=0}^{n}\alpha^{n-k}u_{k-1}^{(r)}+a\alpha^n\sum_{k=0}^{n}{k+r-1 \choose r-1}\\
              & = bu_{n}^{(r+1)}+\beta a\sum_{l=0}^{n-1}\alpha^{(n-1)-l}u_{l}^{(r)}+{n+r \choose r}a\alpha^n\\
              & = bu_{n}^{(r+1)}+\beta au_{n-1}^{(r+1)}+{n+r \choose r}a\alpha^n.
\end{align*}
We conclude that \eqref{eq16} holds for all $n\ge0$.
\end{proof}
\begin{remark}\label{rem01}
Using Identity \eqref{eq19}, we extend, for $r\geq 1$ and $\alpha\ne0$, the sequence $(u_n)_{n\geq0}$ to negative subscripts as follows:
 $$u_{-n}^{(r)}=\frac{1}{\beta}u_{-n+2}^{(r)}-\frac{\alpha}{\beta} u_{-n+1}^{(r)}-{-n+r \choose r-1}\frac{\alpha^{-n+1}}{\beta},\quad n\ge0.$$
 It is easy to see that $$u_{-n}^{(r)}=0,\quad \text{for $0\leq n\leq r$ \quad and} \quad u_{-r-1}^{(r)}=\frac{1}{\beta}\left(\frac{-1}{\alpha}\right)^r\ne0.$$
Now, using Identity \eqref{eq16}, we define, for $r\ge1$ and $\alpha\ne0$, the generalized hyper-Fibonacci numbers for negative subscripts by
\begin{equation*}
w_{-n}^{(r)}=bu_{-n}^{(r)}+\beta au_{-n+1}^{(r)}+a{-n+r-1 \choose r-1}\alpha^{-n},\qquad n\ge0.
\end{equation*}
We deduce that
\begin{equation*}
w_{-n}^{(r)}=0\qquad \text{for $1\leq n\leq r$}.
\end{equation*}
 If $\alpha=0$, we have respectively from \eqref{eq06} and \eqref{eq10} that $w_n^{(r)}=w_n$ and $u_n^{(r)}=u_n$ for $n\ge0$. Thus, $$w_{-n}^{(r)}=w_{-n} \quad \text{and} \quad u_{-n}^{(r)}=u_{-n},\quad n\ge0.$$
\end{remark}

The following theorem is the key assertion behind the computation of the generalized Cassini formula.
\begin{theorem}\label{thm02} Assume that $\alpha\ne0$ and let $r\ge0$ be an integer. Then
\begin{equation}\label{eq18}
w_{n+r+2}^{(r)}=\sum_{k=0}^{r+1}(-1)^{r+1-k}\alpha^{r-k}\left(\alpha^2{r+1 \choose k-1}-\beta {r \choose k}\right)w_{n+k}^{(r)},\qquad n\ge-r.
\end{equation}
\end{theorem}
\begin{proof}
Let us proceed by induction on $r$. For $r=0$, the identity \eqref{eq18} holds by definition of the sequence $(w_n)_n$. Now assume that \eqref{eq18} holds for an integer $r\ge0$. Since $n+1\geq n\ge-r$, we have
\begin{equation}\label{eq20}
w_{n+r+3}^{(r)}=\sum_{k=0}^{r+1}(-1)^{r+1-k}\alpha^{r-k}\left(\alpha^2{r+1 \choose k-1}-\beta {r \choose k}\right)w_{n+k+1}^{(r)}.
\end{equation}
Since $n+r+3\geq n+r+2\ge0$, we have $w_{n+r+3}^{(r)}=w_{n+r+3}^{(r+1)}-\alpha w_{n+r+2}^{(r+1)}$. For $k=0, 1 \ldots, r+1$, we have
$w_{n+k+1}^{(r)}=w_{n+k+1}^{(r+1)}-\alpha w_{n+k}^{(r+1)}$ because
\begin{itemize}
  \item If $n+k+1<0$ then we obtain $0=0-0$.
  \item If $n+k+1\ge0$ and $n+k<0$ then $n+k=-1$, we obtain $a=a-0$.
  \item If $n+k\ge0$ then $n+k+1>0$ and we obtain $w_{n+k+1}^{(r)}=w_{n+k+1}^{(r+1)}-\alpha w_{n+k}^{(r+1)}$.
\end{itemize}
Thus, from \eqref{eq20} we get
$$w_{n+r+3}^{(r+1)}-\alpha w_{n+r+2}^{(r+1)}=\sum_{k=0}^{r+1}(-1)^{r+1-k}\alpha^{r-k}\left(\alpha^2{r+1 \choose k-1}-\beta {r \choose k }\right)(w_{n+k+1}^{(r+1)}-\alpha w_{n+k}^{(r+1)}).$$
We deduce that
\begin{align*}
w_{n+r+3}^{(r+1)} & = \alpha w_{n+r+2}^{(r+1)}+\sum_{k=0}^{r+1}(-1)^{r+1-k}\alpha^{r-k}\left(\alpha^2{r+1 \choose k-1}-\beta {r \choose k}
                    \right)w_{n+k+1}^{(r+1)}\\
                  & \quad +\sum_{k=0}^{r+1}(-1)^{r-k}\alpha^{r+1-k}\left(\alpha^2{r+1 \choose k-1}-\beta {r \choose k}\right) w_{n+k}^{(r+1)}\\
                  & = (\alpha r+2\alpha) w_{n+r+2}^{(r+1)}+\sum_{k=0}^{r}(-1)^{r+1-k}\alpha^{r-k}\left(\alpha^2{r+1 \choose k-1}-\beta
                   {r \choose k}\right)w_{n+k+1}^{(r+1)}\\
                  & \quad +(-1)^{r+1}\alpha^{r+1}\beta w_n^{(r+1)}+\sum_{k=1}^{r+1}(-1)^{r-k}\alpha^{r+1-k}\left(\alpha^2{r+1 \choose k-1}-\beta {r \choose k} \right) w_{n+k}^{(r+1)}\\
                  & = (\alpha r+2\alpha) w_{n+r+2}^{(r+1)}+\sum_{l=1}^{r+1}(-1)^{r-l}\alpha^{r+1-l}\left(\alpha^2{r+1 \choose l-2}-\beta
                   {r \choose l-1}\right)w_{n+l}^{(r+1)}\\
                  & \quad +(-1)^{r+1}\alpha^{r+1}\beta w_n^{(r+1)}+\sum_{k=1}^{r+1}(-1)^{r-k}\alpha^{r+1-k}\left(\alpha^2{r+1 \choose k-1}-\beta {r \choose k} \right) w_{n+k}^{(r+1)}
\end{align*}
\begin{align*}
                  & = \sum_{k=0}^{r+2}(-1)^{r-k}\alpha^{r+1-k}\left(\alpha^2{r+1 \choose k-2}+\alpha^2{r+1 \choose k-1}-\beta {r \choose k-1}-\beta
                   {r \choose k}\right) w_{n+k}^{(r+1)}\\
                  & = \sum_{k=0}^{r+2}(-1)^{r-k}\alpha^{r+1-k}\left(\alpha^2{r+2 \choose k-1}-\beta {r+1 \choose k}\right) w_{n+k}^{(r+1)}.
\end{align*}
We conclude that \eqref{eq18} holds for all $r\ge0$.
\end{proof}
\section{A generalization of the Cassini formula}\label{sec3}
Cassini formula \eqref{eq08} can be expressed as a $2\times 2$ determinant in the following way
\begin{equation*}
\begin{vmatrix}
   w_n & w_{n+1} \\
w_{n+1} & w_{n+2}
\end{vmatrix}=(-\beta)^{n-1}(\beta b^2-\alpha\beta ab-\beta^2a^2).
\end{equation*}
 For $n, r\in\mathbb{Z}$ such that $n\ge0$ and $r\ge0$,
let us define the $(r+2)\times (r+2)$ matrix
 \begin{equation*}
W_{r,n}=\begin{pmatrix}
   w_n^{(r)} & w_{n+1}^{(r)} & \cdots & w_{n+r+1}^{(r)} \\
   w_{n+1}^{(r)} & w_{n+2}^{(r)} & \cdots & w_{n+r+2}^{(r)}\\
     \vdots & \vdots & \ddots & \vdots\\
    w_{n+r+1}^{(r)} & w_{n+r+2}^{(r)} & \cdots & w_{n+2r+2}^{(r)}
\end{pmatrix}.
\end{equation*}
Our purpose is to evaluate the determinant of the matrix $W_{r,n}$. Note that \begin{equation*}
W_{0,n}=\begin{pmatrix}
w_n & w_{n+1} \\
w_{n+1} & w_{n+2}\end{pmatrix}.
\end{equation*}
From Theorem \ref{thm02}, we can write  $$w_{n+r+2}^{(r)}=\sum_{k=0}^{r+1}q_k w_{n+k}^{(r)},\qquad n\ge-r,$$
 where $$q_k=(-1)^{r+1-k}\alpha^{r-k}\left(\alpha^2{r+1 \choose k-1}-\beta {r \choose k}\right),\quad 0\leq k\leq r+1.$$
Let
\begin{equation}\label{eq01}
V_{r+2}=\begin{pmatrix}
   0 & 1 & 0 & \cdots & 0 & 0 \\
   0 & 0 & 1 & \cdots & 0 & 0 \\
     \vdots & \vdots & \vdots & \ddots & \vdots & \vdots\\
   0 & 0 & 0 & \cdots & 1 & 0\\
   0 & 0 & 0 & \cdots & 0 & 1\\
   q_0 & q_1 & q_2 & \cdots & q_{r} & q_{r+1}
\end{pmatrix}
\end{equation}
denote the companion matrix of the generalized hyper-Fibonacci sequence $\left(w_{n}^{(r)}\right)_{n}$.
We deduce from Theorem \ref{thm02} that the generalized hyper-Fibonacci sequence $(w_n^{(r)})_n$ can be defined by the vector recurrence relation
 \begin{equation}\label{eq31}
\begin{pmatrix}
   w_{n+1}^{(r)} \\
   w_{n+2}^{(r)} \\
     \vdots \\
    w_{n+r+2}^{(r)}
\end{pmatrix}=V_{r+2}\begin{pmatrix}
   w_{n}^{(r)} \\
   w_{n+1}^{(r)} \\
     \vdots \\
    w_{n+r+1}^{(r)}
\end{pmatrix},
\end{equation}
where $n+r\ge0$.
\begin{lemma}\label{lem02}
Let $n\ge0$ and $r\ge0$ be integers. Then
$$W_{r,n}=V_{r+2}^n W_{r,0}.$$
\end{lemma}
\begin{proof}
From the relation \eqref{eq31} we can write $W_{r,n}=V_{r+2} W_{r,n-1}$. It follows that
$$W_{r,n}=V_{r+2} W_{r,n-1}=V_{r+2}^2 W_{r,n-2}=\cdots=V_{r+2}^n W_{r,0}.$$
\end{proof}
\begin{lemma}\label{lem03}
Let $r\ge0$ be an integer. Then
$$\det(V_{r+2})=-\alpha^r\beta.$$
\end{lemma}
\begin{proof}
It is clear that $$\det(V_{r+2})=(-1)^{r+3}q_0=(-1)^{r+3}(-1)^{r+2}\alpha^r\beta=-\alpha^r\beta.$$
\end{proof}
\begin{theorem}\label{thm01}
Let $n\ge0$ and $r\ge0$ be integers. Then
\begin{equation}\label{eq36}\det(W_{r,n})=(-1)^{n+\lfloor (r+3)/2\rfloor}\alpha^{nr+r^2}\beta^{n+r}b^r(b^2-\alpha ab-\beta a^2).
\end{equation}
\end{theorem}
\begin{proof}
For $r=0$, the result follows from Identity \eqref{eq08}. Assume that $r\ge1$. For $\alpha\ne0$, we deduce from \eqref{eq31} that multiplication by $V_{r+2}^{-1}$ decreases by $1$ the subscript of each component, i.e.,
$$V_{r+2}^{-1}W_{r,0}=\begin{pmatrix}
   w_{-1}^{(r)} & w_{0}^{(r)} & \cdots & w_{r}^{(r)} \\
   w_{0}^{(r)} & w_{1}^{(r)} & \cdots & w_{r+1}^{(r)}\\
     \vdots & \vdots & \ddots & \vdots\\
    w_{r}^{(r)} & w_{r+1}^{(r)} & \cdots & w_{2r+1}^{(r)}
\end{pmatrix}.$$
Thus,
$$V_{r+2}^{-r}W_{r,0}=\begin{pmatrix}
   w_{-r}^{(r)} & w_{1-r}^{(r)} & \cdots & w_{1}^{(r)} \\
   w_{1-r}^{(r)} & w_{2-r}^{(r)} & \cdots & w_{2}^{(r)}\\
     \vdots & \vdots & \ddots & \vdots\\
    w_{1}^{(r)} & w_{2}^{(r)} & \cdots & w_{r+2}^{(r)}
\end{pmatrix}.$$
Since $w_{-n}^{(r)}=0$ for $1\leq n\leq r$, then
$$V_{r+2}^{-r}W_{r,0}=\begin{pmatrix}
   0 & 0 & \cdots & 0 & w_{0}^{(r)} & w_{1}^{(r)} \\
    0 & 0 & \cdots  & w_{0}^{(r)} & w_{1}^{(r)} & w_{2}^{(r)}\\
      \vdots & \vdots & \ddots & \vdots & \vdots & \vdots\\
    0 &  w_{0}^{(r)} & \cdots & w_{r-2}^{(r)} & w_{r-1}^{(r)} & w_{r}^{(r)}\\
        w_{0}^{(r)} & w_{1}^{(r)} & \cdots & w_{r-1}^{(r)} & w_{r}^{(r)} & w_{r+1}^{(r)}\\
    w_{1}^{(r)} & w_{2}^{(r)} & \cdots &  w_{r}^{(r)} & w_{r+1}^{(r)} & w_{r+2}^{(r)}
\end{pmatrix}.$$
Thus, \begin{equation}\label{eq34}
\det(W_{r,0})=\det(V_{r+2})^{r}\cdot\Delta,
\end{equation}
 where
$$\Delta=\begin{vmatrix}
   0 & 0 & \cdots & 0 & w_{0}^{(r)} & w_{1}^{(r)} \\
    0 & 0 & \cdots  & w_{0}^{(r)} & w_{1}^{(r)} & w_{2}^{(r)}\\
      \vdots & \vdots & \ddots & \vdots & \vdots & \vdots\\
    0 &  w_{0}^{(r)} & \cdots & w_{r-2}^{(r)} & w_{r-1}^{(r)} & w_{r}^{(r)}\\
        w_{0}^{(r)} & w_{1}^{(r)} & \cdots & w_{r-1}^{(r)} & w_{r}^{(r)} & w_{r+1}^{(r)}\\
    w_{1}^{(r)} & w_{2}^{(r)} & \cdots &  w_{r}^{(r)} & w_{r+1}^{(r)} & w_{r+2}^{(r)}\end{vmatrix}.$$
Let $L_j$ denote the $j$th line of $\Delta$, where $j=1, 2, \ldots, r+2$.
First, we replace $L_{i+1}$ by $L_{i+1}-\alpha L_i$ for $i=r+1, r, \ldots, 1$. Since $w_{i+1}^{(r-1)}=w_{i+1}^{(r)}-\alpha w_{i}^{(r)}$, we get
$$\Delta=\begin{vmatrix}
   0 & 0 & \cdots &  w_{0}^{(r-1)} & w_{1}^{(r-1)}+2\alpha \\
    0 & 0 & \cdots & w_{1}^{(r-1)} & w_{2}^{(r-1)} \\
       \vdots & \vdots& \ddots  & \vdots & \vdots\\
       w_{0}^{(r-1)} & w_{1}^{(r-1)} & \cdots  & w_{r}^{(r-1)} & w_{r+1}^{(r-1)} \\
  w_{1}^{(r-1)} & w_{2}^{(r-1)} & \cdots  & w_{r+1}^{(r-1)} & w_{r+2}^{(r-1)}
\end{vmatrix}.$$
Using the same method again $(r-1)$ times, we obtain
$$\Delta=\begin{vmatrix}
   0 & 0 &  \cdots & 0 &  w_{0} & w_{1}+d_1\\
    0 & 0 &  \cdots &w_{0} & w_{1} & w_{2}+d_2 \\
    \vdots &   \vdots & \ddots& \vdots  & \vdots & \vdots\\
      0 & w_{0} &  \cdots & w_{r-2} & w_{r-1} & w_{r}+d_r \\
       w_{0} & w_{1} &  \cdots  & w_{r-1} & w_{r} & w_{r+1} \\
  w_{1} & w_{2} & \cdots & w_{r} & w_{r+1} & w_{r+2}
\end{vmatrix},$$
where $d_i=a(-1)^{i-1}{r \choose i}\alpha^{i}$ for $1\leq i\leq r$.
Now let $C_j$ denote the $j$th column of this last determinant, where $j=1, \ldots, r+2$. Replacing the column
$C_{i}$ by $C_{i}-(\alpha C_{i-1}+\beta C_{i-2})$ for $i=r+2, r+1,\ldots,3$ and using the fact that $w_i=\alpha w_{i-1}+\beta w_{i-2}$ we get
$$\Delta=\begin{vmatrix}
   0 & 0 & 0 & \cdots & 0 &  w_{0} & w_{1}-\alpha w_0+d_1\\
    0 & 0 & 0 & \cdots &w_{0} & w_{1}-\alpha w_0 & d_2 \\
     0 & 0 & 0 & \cdots & w_{1}-\alpha w_0 & 0 & d_3 \\
    \vdots & \vdots&  \vdots & \ddots & \vdots  & \vdots & \vdots\\
      0 & w_{0} &  w_{1}-\alpha w_0 & \cdots & 0 & 0 & d_r \\
       w_{0} & w_{1} & 0&   \cdots  & 0 & 0 & 0 \\
  w_{1} & w_{2} & 0 & \cdots & 0 & 0 & 0
\end{vmatrix}.$$
Now we permute the column $C_i$ with column $C_{r+3-i}$ for $1\leq i\leq \lfloor(r+2)/2\rfloor$, and obtain
$$\Delta=(-1)^{\lfloor\frac{r+2}{2}\rfloor}\begin{vmatrix}
   w_{1}-\alpha w_0+d_1 & w_{0} & 0 & \cdots & 0 &  0 & 0\\
    d_2 & w_{1}-\alpha w_0  & w_{0} & \cdots & 0 & 0 & 0 \\
     d_3  & 0 &  w_{1}-\alpha w_0 & \cdots & 0 & 0 & 0\\
    \vdots & \vdots&  \vdots & \ddots& \vdots  & \vdots & \vdots\\
      d_r & 0 &  0 & \cdots & w_{1}-\alpha w_0 & w_{0} & 0\\
       0 & 0&  0&  \cdots  & 0 & w_{1} &  w_{0} \\
 0 & 0 & 0 & \cdots & 0 & w_{2} & w_{1}
\end{vmatrix}.$$
We deduce that
\begin{equation}\label{eq35}
\Delta=(-1)^{\lfloor\frac{r+2}{2}\rfloor}\Delta^\prime\begin{vmatrix}
  w_{1} &  w_{0} \\
 w_{2} & w_{1}
\end{vmatrix},
\end{equation}
where
\begin{equation*}
\Delta^\prime=\begin{vmatrix}
   d_1-(\alpha a-b) & a & 0 &  \cdots & 0 & 0 \\
    d_2 & -(\alpha a-b) & a & \cdots & 0 & 0 \\
     d_3 & 0 & -(\alpha a-b) & \cdots & 0 & 0 \\
    \vdots & \vdots& \vdots & \ddots& \vdots & \vdots  \\
      d_{r-1} & 0 & 0 & \cdots & -(\alpha a-b) & a \\
      d_r & 0 & 0 & \cdots & 0 & -(\alpha a-b)
\end{vmatrix}.
\end{equation*}
We distinguish two cases for the calculation of $\Delta^\prime$.

$\bullet$ If $\alpha a=b\ne0$, the expansion with respect to the first column gives
\begin{equation*}
\Delta^\prime=\begin{vmatrix}
   d_1 & a & 0 &  \cdots & 0 & 0 \\
    d_2 & 0 & a & \cdots & 0 & 0 \\
     d_3 & 0 & 0 & \cdots & 0 & 0 \\
    \vdots & \vdots& \vdots & \ddots& \vdots & \vdots  \\
      d_{r-1} & 0 & 0 & \cdots & 0 & a \\
      d_r & 0 & 0 & \cdots & 0 & 0\end{vmatrix}=(-1)^{r+1}d_ra^{r-1}=(\alpha a)^r.
\end{equation*}

$\bullet$ If $b-\alpha a\ne 0$, let $L_j$ denote the $j$th line of $\Delta^\prime$, where $j=1,\ldots, r$. Replacing the line $L_i$ by $L_i+\displaystyle\frac{a}{\alpha a-b}L_{i+1}$ for $i=r-1,\ldots,1$ gives
\begin{equation*}
\Delta^\prime=\begin{vmatrix}
  \displaystyle(b-\alpha a)+\sum_{i=1}^r\left(\frac{a}{\alpha a-b}\right)^{i-1}d_i & 0 &   \cdots & 0 & 0 \\
   & b-\alpha a &   \cdots & 0 & 0 \\
    \vdots & \vdots&  \ddots& \vdots & \vdots  \\
      d_{r-1}+\displaystyle\frac{a}{\alpha a-b}d_r & 0 & \cdots & b-\alpha a & 0 \\
      d_r & 0 & \cdots & 0 & b-\alpha a
\end{vmatrix}.
\end{equation*}
We deduce that
\begin{align*}
\Delta^\prime & = (b-\alpha a)^{r-1}\left((b-\alpha a)+\sum_{i=1}^r\left(\frac{a}{\alpha a-b}\right)^{i-1}d_i\right)\\
              & = (b-\alpha a)^{r-1}\left((b-\alpha a)+a\sum_{i=1}^r\left(\frac{a}{b-\alpha a}\right)^{i-1}{r \choose i}\alpha^i\right) \\
              & = (b-\alpha a)^{r-1}\left((b-\alpha a)+(b-\alpha a)\sum_{i=1}^r\left(\frac{\alpha a}{b-\alpha a}\right)^{i}{r \choose i}\right) \\
              & = (b-\alpha a)^{r-1}\left((b-\alpha a)\sum_{i=0}^r\left(\frac{\alpha a}{b-\alpha a}\right)^{i}{r \choose i}\right) \\
              & = (b-\alpha a)^{r}\left(\frac{\alpha a}{b-\alpha a}+1\right)^r\\
              & = (b-\alpha a)^{r}\left(\frac{b}{b-\alpha a}\right)^r\\
              & = b^r,
\end{align*}
which coincides with the case $\alpha a=b\ne0$. Since $\begin{vmatrix}
  w_{1} &  w_{0} \\
 w_{2} & w_{1}
\end{vmatrix}=b^2-\alpha ab-\beta a^2$, we deduce from Identities \eqref{eq34}, \eqref{eq35} and Lemmas \ref{lem02}, \ref{lem03} that
$$\det\left(W_{r,n}\right)=(-1)^{n+\lfloor (r+3)/2\rfloor}\alpha^{nr+r^2}\beta^{n+r}b^r(b^2-\alpha ab-\beta a^2).$$
\end{proof}
\begin{corollary}
Let $n\ge0$ and $r\ge0$ be integers. Then
 \begin{equation*}
\begin{vmatrix}
   F_n^{(r)} & F_{n+1}^{(r)} & \cdots & F_{n+r+1}^{(r)} \\
   F_{n+1}^{(r)} & F_{n+2}^{(r)} & \cdots & F_{n+r+2}^{(r)}\\
     \vdots & \vdots & \ddots & \vdots\\
    F_{n+r+1}^{(r)} & F_{n+r+2}^{(r)} & \cdots & F_{n+2r+2}^{(r)}
\end{vmatrix}=(-1)^{n+\lfloor (r+3)/2\rfloor}.
\end{equation*}
\end{corollary}
\begin{proof}
Follows from Identity \eqref{eq36} for $a=0$ and $b=\alpha=\beta=1$.
\end{proof}
\begin{corollary}
Let $n\ge0$ and $r\ge0$ be integers. Then
 \begin{equation*}
\begin{vmatrix}
   L_n^{(r)} & L_{n+1}^{(r)} & \cdots & L_{n+r+1}^{(r)} \\
   L_{n+1}^{(r)} & L_{n+2}^{(r)} & \cdots & L_{n+r+2}^{(r)}\\
     \vdots & \vdots & \ddots & \vdots\\
    L_{n+r+1}^{(r)} & L_{n+r+2}^{(r)} & \cdots & L_{n+2r+2}^{(r)}
\end{vmatrix}=5(-1)^{n+\lfloor (r+1)/2\rfloor}.
\end{equation*}
\end{corollary}
\begin{proof}
Follows from Identity \eqref{eq36} for $a=2$ and $b=\alpha=\beta=1$.
\end{proof}

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\noindent 2010 {\it Mathematics Subject Classification}: Primary 11B39; Secondary 11B37, 20G20, 15B36.


\noindent \emph{Keywords:} Fibonacci number, generalized Fibonacci number, hyper-Fibonacci number, generalized hyper-Fibonacci number, Cassini formula.

\bigskip
\hrule
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\noindent (Concerned with sequences
\seqnum{A000032},
\seqnum{A000045},
\seqnum{A000071},
\seqnum{A001610},
\seqnum{A001924}, and
\seqnum{A023548}.)


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\vspace*{+.1in}
\noindent
Received  November 27 2021; 
revised versions received  March 8 2022.
Published in {\it Journal of Integer Sequences},  March 10 2022.

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


\end{document}





