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	\begin{center}
		\vskip 1cm{\LARGE\bf Common Values of Generalized Fibonacci \\
			\vskip .1in 
			and Leonardo Sequences
		}
		\vskip 1cm
		\large
		Bibhu Prasad Tripathy and Bijan Kumar Patel \\
		Department of Mathematics\\
		School of Applied Sciences \\ 
		KIIT University Bhubaneswar \\ 
		Odisha 751024 \\ 
		India \\
		
		
		\href{mailto: bptbibhu@gmail.com}{\tt bptbibhu@gmail.com} \\
		\href{mailto: bijan.patelfma@kiit.ac.in}{\tt bijan.patelfma@kiit.ac.in}
	\end{center}
	
	\vskip .2in
	
	\begin{abstract}
		For an integer $k \geq 2$, let $F_{n}^{(k)}$ be the $k$-generalized Fibonacci sequence that  starts with $0, \dots ,0,1,1$ ($k$ terms) and each term afterwards is the sum of $k$ preceding terms. In this paper, we find all the $k$-generalized Fibonacci numbers that are Leonardo numbers. More explicitly, we solve the Diophantine equation $F_{n}^{(k)} = \Le_{m}$ in positive integers $n, k, m$ with $k \geq 2$.
	\end{abstract}
	\section{Introduction}
	The Fibonacci and Lucas sequence are two fascinating topics in integer sequences. The Leonardo sequence $(\Le_{m})_{m \geq 0}$ is an integer sequence that is related to the Fibonacci and Lucas sequences. Leonardo numbers are discussed by Catarino and Borges \cite{Catarino}. It is the sequence \seqnum{A001595} in the OEIS satisfying the recurrence relation
	\begin{equation}\label{eq-1.1}
		\Le_{m} = \Le_{m-1}+\Le_{m-2} + 1
	\end{equation}
	for  $ m \geq 2 $ with the initial terms  $\Le_{0}= 1$ and $\Le_{1}= 1$.  The first few terms of $(\Le_{m})_{m \geq 0}$ are
	\[
	1, 1, 3, 5, 9, 15, 25, 41, 67, 109, 177, 287, 465, 753, 1219, 1973,  \ldots .
	\]
	
	In the recent past, many aspects of Leonardo sequence have been studied such as hybrid Leonardo numbers \cite{Alp}, incomplete Leonardo numbers \cite{Catarino1}, Leonardo Pisano polynomials, hybrinomials \cite{Kuruz} and $q$-Leonardo hybrid numbers \cite{Ozimamoglu}.
	
	The Fibonacci sequence $(F_n)_{n\geq 0}$ is the binary recurrence sequence  given by 
	\[
	F_{n+2} = F_{n+1} + F_n~~ {\rm for} ~n\geq 0
	\]
	with the initial terms $F_0 = 0$ and $F_1 = 1$. 
	
	Let $k \geq 2$ be an integer. One of numerous generalizations of the Fibonacci sequence, called the $k$-generalized Fibonacci sequence $(F_{n}^{(k)})_{n \geq -(k-2)}$ is given by the recurrence
	
	\begin{equation}\label{eq-1.2}
		F_{n}^{(k)} = F_{n-1}^{(k)} + F_{n-2}^{(k)} + \dots + F_{n-k}^{(k)} = \sum_{i=1}^{k} F_{n-i}^{(k)} ~~\text{for all}~n \geq 2,  
	\end{equation}
	with the initial conditions $F_{-(k-2)}^{(k)} = F_{-(k-3)}^{(k)} = \dots = F_{0}^{(k)} = 0$ and $F_{1}^{(k)} = 1$. Here, $F_{n}^{(k)}$ denotes the $n$th $k$-generalized Fibonacci number. 
	
	Note that for $k = 2$, we have $F_{n}^{(2)} = F_n$, the $n$th Fibonacci number. 
	For $k = 3$, we have $F_{n}^{(3)} = T_n$, the $n$th Tribonacci number. They are followed by the Tetranacci numbers for $k = 4$, and so on.
	
	A Leonardo number is called $k$-Fibonacci Leonardo number if it is a $k$-generalized Fibonacci number. The aim of this paper is to determine all the $k$-Fibonacci Leonardo numbers.
	
	Finding the intersection of two recurrent sequences of positive integers is a topic that has been extensively studied in number theory. Currently, several researchers have been interested in finding the intersection of the $k$-generalized Fibonacci sequence with other number sequences. For instance, one can go through \cite{Bravo3, Bravo1, Bravo5, Gueye, Rihane, Rihane1}.
	
	Motivated by the above literature, we study the Diophantine equation 
	\begin{equation}\label{eq-1.3}
		F_{n}^{(k)} = \Le_{m}.
	\end{equation}
	In particular, our main result is the following.
	\begin{theorem}\label{thm1}
		All the solutions of the Diophantine equation \eqref{eq-1.3} in positive integers with $k \geq 2$ are given by 
		$$(n, k, m) \in \{ (1,k,0), (2,k,0), (1,k,1), (2,k,1), (4, 2, 2), (5, 2, 3), (6, 4, 5)\}. $$ Thus, the only $k$-Fibonacci Leonardo numbers are $1$, $3$, $5$, and $15$. 
	\end{theorem}
	
	\section{Auxiliary results}
	Our proof of Theorem \ref{thm1} is mainly based on linear forms in logarithms of algebraic
	numbers and a reduction algorithm originally introduced by Baker and Davenport \cite{Baker} (and improved by Dujella and Peth\"{o} \cite{Dujella}). Here, we use a variant due to de Weger \cite{Weger}, but first, recall some basic notation from algebraic number theory.
	
	\subsection{Linear forms in logarithms}
	Let $\gamma$ be an algebraic number of degree $d$ with minimal primitive polynomial 
	\[
	f(X):= a_0 X^d+a_1 X^{d-1}+ \cdots +a_d = a_0 \prod_{i=1}^{d}(X- \gamma^{(i)}) \in \mathbb{Z}[X],
	\]
	where the $a_i$ are relatively prime integers, $a_0 >0$, and the $\gamma^{(i)}$ are conjugates of $\gamma$. Then
	\begin{equation}\label{eq-2.4}
		h(\gamma)=\frac{1}{d} \Biggl( \log a_0+\sum_{i=1}^{d}\log \bigl(\max\{|\gamma^{(i)}|,1\} \bigr) \Biggr) 
	\end{equation}
	is called the \emph{logarithmic height} of $\gamma$. 
	
	With the established notation, Matveev (see  \cite{Matveev} or  \cite[Theorem 9.4]{Bugeaud}) proved the following result.
	
	\begin{theorem}\label{thm2}
		Assume that $\eta_1, \ldots, \eta_t$ are positive real algebraic numbers in a real algebraic number field $\mathbb{K}$ of degree $D$, $b_1, \ldots, b_t$ are rational integers, and 
		\[
		\Lambda :=\eta_1^{b_1}\cdots\eta_t^{b_t}-1,
		\]
		is not zero. Then
		\[
		|\Lambda| \geq \exp \bigl(-1.4\cdot 30^{t+3}\cdot t^{4.5}\cdot D^2(1+\log D)(1+\log B)A_1\cdots A_t \bigr),
		\]
		where
		\[
		B\geq \max\{|b_1|,\ldots,|b_t|\},
		\]
		and
		\[
		A_i\geq \max\{Dh(\eta_i),|\log \eta_i|, 0.16\}, ~ \text{for all} ~ i=1,\ldots,t.
		\]
	\end{theorem}
	\subsection{The de Weger reduction algorithm}
	Here we present a variant of the reduction method of Baker and Davenport \cite{Baker}
	(and improved by Dujella and Peth\"{o} \cite{Dujella}) due to de Weger \cite{Weger}.
	
	Let $ \vartheta_{1}, \vartheta_{2}, \delta \in \mathbb{R}$ be given and let $x_{1}, x_{2} \in \mathbb{Z}$ be unknowns. Let 
	\begin{equation} \label{eq-2.5}
		\Lambda = \delta + x_{1}\vartheta_{1} + x_{2}\vartheta_{2}. 
	\end{equation}
	Set $X = \max \{|x_{1},|x_{2}|\}$. Let $X_{0}, Y$ be positive. Assume that 
	\begin{equation} \label{eq-2.6}
		|\Lambda| < c \exp{(-\rho Y)}
	\end{equation}
	and 
	\begin{equation} \label{eq-2.7}
		Y \leq X \leq X_{0},
	\end{equation}
	where $c, \rho$ be positive constants.
	\noindent
	When $\delta = 0$ in \eqref{eq-2.5}, we get 
	\[
	\Lambda = x_{1}\vartheta_{1} + x_{2}\vartheta_{2}.
	\]
	Put $\vartheta = - \vartheta_{1}/\vartheta_{2}$. We assume that 
	$x_{1}$ and $x_{2}$ are coprime. Let the continued fraction expansion of $\vartheta$ be given by 
	\[
	[a_{0}, a_{1}, a_{2}, \dots],
	\]
	and let the $k$th convergent of $\vartheta$ be $p_{k}/q_{k}$ for $k=0,1,2,\dots$. We may assume without loss of generality that $|\vartheta_{1}| < |\vartheta_{2}|$ and $x_{1} > 0$. We have the following results.
	
	\begin{lemma} \cite[Lemma 3.1]{Weger} \label{lem 2.2}
		If \eqref{eq-2.6} and \eqref{eq-2.7} hold for $x_{1}$, $x_{2}$ with $X \geq 1$ and $\delta = 0$, then $(-x_{2}, x_{1}) = (p_{k},q_{k})$ for an index $k$ that satisfies 
		\[
		k \leq -1 + \frac{\log(1+X_{0}\sqrt{5})}{\log\left( \frac{1+\sqrt{5}}{2}\right)} := Y_{0}.
		\]
	\end{lemma}
	\begin{lemma} \cite[Lemma 3.2]{Weger} \label{lem 2.3}
		Let 
		\[
		A = \displaystyle\max_{0 \leq k \leq Y_{0}} a_{k+1}.
		\]
		If \eqref{eq-2.6} and \eqref{eq-2.7} hold for $x_{1}$, $x_{2}$ with $X \geq 1$ and $\delta = 0$, then 
		\[
		Y < \frac{1}{\rho} \log \left( \frac{c(A+2)}{|\vartheta_{2}|} \right) + \frac{1}{\rho} \log X  < \frac{1}{\rho} \log \left( \frac{c(A+2)X_{0}}{|\vartheta_{2}|} \right).
		\]
		When $\delta \neq 0$ in \eqref{eq-2.5}, put $\vartheta = - \vartheta_{1}/\vartheta_{2}$ and $\psi = \delta / \vartheta_{2}$. Then we have \[
		\frac{\Lambda}{\vartheta_{2}} = \psi - x_{1}\vartheta + x_{2}.
		\]
	\end{lemma}
	Let $p/q$ be a convergent of $\vartheta$ with $q > X_{0}$. For a real number $x,$ we let $
	\|x\| = \min \{|x-n|: n\in \mathbb{Z} \}$ be the distance from $x$ to the nearest integer. We have the following result.
	\begin{lemma} \cite[Lemma 3.3]{Weger} \label{lem-2.4}
		Suppose that 
		\[
		\|q \psi \| > \frac{2X_{0}}{q}.
		\]
		Then the solutions of \eqref{eq-2.6} and \eqref{eq-2.7} satisfy 
		\[
		Y < \frac{1}{\rho} \log \left( \frac{q^{2}c}{|\vartheta_{2}|X_{0}} \right).
		\]
	\end{lemma}
	
	\subsection{Properties of the Leonardo sequence}
	
	The characteristic equation of $(\Le_{m})_{m \geq 0}$ is $x^{3}-2 x^{2} -1 = 0$, which has roots $\alpha  = \frac{1+\sqrt{5}}{2}$ and $\beta = \frac{-1}{\alpha}$ (see \cite{Alp1}). The Binet formula for $\Le_{m}$ is 
	\begin{equation}\label{eq-2.8}
		\Le_{m} = 2 \left(\frac{\alpha^{m+1} - \beta^{m+1}}{\alpha - \beta} \right) - 1 = \frac{\alpha(2\alpha^{m} -1) - \beta(2\beta^{m} - 1)}{\alpha - \beta} \quad \text{for all } m \geq 0.
	\end{equation}
	\begin{lemma}\label{lem 2.5}
		The inequality
		\begin{equation}\label{eq-2.9}
			\alpha^{m} \leq \Le_{m} \leq \alpha^{m+1},
		\end{equation}
		holds for all positive integers $m \geq 2$.
	\end{lemma}
	\begin{proof}
		This can be easily proved by the method of induction on $m$.
	\end{proof}
	\begin{lemma}\cite[Lemma 2.1]{Catarino} \label{lem 2.6}
		For all $m \geq 0$, the $m$-th  Leonardo number $\Le_{m}$ is an odd number.
	\end{lemma}
	
	\subsection{Properties of the $k$-generalized Fibonacci sequence}
	In this subsection, we recall some facts and properties of the $k$-generalized Fibonacci sequence which will be used later. The characteristic polynomial of the $k$-generalized Fibonacci sequence is
	\[
	\Psi_{k}(x) = x^k - x^{k-1} - \dots - x - 1.
	\]
	The polynomial $\Psi_{k}(x)$ is irreducible over $\mathbb{Q}[x]$ and has just one root outside the unit circle. It is real and positive, so it satisfies $\varphi(k) > 1$. The other roots are strictly inside the unit circle. Throughout this paper, $\varphi := \varphi(k)$ denotes that single root, which is located between $2(1-2^{-k})$ and $2$ (see \cite{Wolfram}). To simplify the notation, in general, we omit the dependence of $k$ on $\varphi$.
	
	Dresden and Du \cite{Dresden} gave the following simplified Binet-like formula for $F_{n}^{(k)}$:
	\begin{equation}\label{eq-2.10}
		F_{n}^{(k)} = \displaystyle\sum_{i=1}^{k} f_{k} (\varphi_{i})\varphi_{i}^{n-1} = \sum_{i=1}^{k} \frac{\varphi_{i}-1}{2+(k+1)(\varphi_{i}-2)} \varphi_{i}^{n-1} ,
	\end{equation}
	where  $\varphi := \varphi_{1}, \varphi_{2},\dots , \varphi_{k}$  are the roots of the characteristic polynomial $\Psi_{k}(x)$. It was also proved in \cite{Dresden} that the contribution of the roots that 
	lie inside the unit circle to the formula \eqref{eq-2.10} is very small, namely that the approximation 
	\begin{equation}\label{eq-2.11}
		\left| F_{n}^{(k)} -  f_{k} (\varphi) \varphi^{n-1} \right| < \frac{1}{2} \quad \text{holds  for   all} \quad n \geq 2 - k.
	\end{equation} 
	Furthermore, it was shown by Bravo and Luca in \cite{ Bravo} that the inequality
	\begin{equation}\label{eq-2.12}
		\varphi ^ {n-2} \leq F_{n} ^ {(k)} \leq \varphi ^ {n-1} \text{ holds for all }  n \geq 1 ~{\rm and}~ k \geq 2.
	\end{equation}
	The first direct observation is that the first $k + 1$ non-zero terms in $F_{n}^{(k)}$ are powers of $2$, namely
	\[
	F_{1}^{(k)}=1, F_{2}^{(k)}=1, F_{3}^{(k)}=2, F_{4}^{(k)}=4, \dots, F_{k+1}^{(k)} = 2^{k-1},
	\]
	while the next term in the above sequence is $F_{k+2}^{(k)} = 2^k -1$. Thus, we have that
	\[
	F_{n}^{(k)} = 2^{n-2}~\text{holds for all}~2 \leq n \leq k+1.
	\]
	We also observe that the recursion \eqref{eq-1.2} implies the three-term recursion
	\[
	F_{n}^{(k)} = 2F_{n-1}^{(k)} - F_{n-k-1}^{(k)}~\text{for all}~n \geq 3,
	\]
	which shows that the $k$-generalized Fibonacci sequence grows at a rate less than $2^{n-2}$. In fact, the inequality $F_{n}^{(k)} < 2^{n-2}$ holds for all $n \geq k + 2$ (see \cite[ Lemma 2]{Bravo}).
	
	The following result was proved by Bravo and Luca \cite{Bravo}.
	\begin{lemma}\label{lem2.8}
		Let $k \geq 2$, $\varphi$ be the dominant root of   $(F_{n}^{(k)})_{n \geq -(k-2)}$, and consider the function defined in \eqref{eq-2.10}. Then the inequalities
		\[
		\frac{1}{2} < f_{k}(\varphi) < \frac{3}{4} \  \text{and} \  \left|f_{k}(\varphi^{(i)})  \right| < 1 \ \text{ hold for all} \ \ 2 \leq i \leq k.
		\]	
	\end{lemma}
	\noindent
	In addition, they proved that the logarithmic height of $f$ satisfies 
	\begin{equation}\label{eq-2.13}
		h(f_{k}(\varphi)) < \log(k+1) + \log4 \quad \text{for all} \quad k \geq 2.
	\end{equation}
	We finish this subsection with the following estimate due to Bravo, G\'omez, and Luca \cite{Bravo1}, which will be used later.
	\begin{lemma}\label{lem2.9}
		Let  $k \geq 2$ and suppose that $n < 2^{k/2}$. Then
		\[
		F_{n}^{(k)} = 2^{n-2}(1 + \xi) \quad \text{where} \quad  |\xi| < \frac{1}{2^{k/2}}.
		\]
	\end{lemma}
	
	\section{Proof of Theorem \ref{thm1}}

\begin{proof}

First, note that $F_{1}^{(k)} = F_{2}^{(k)} = 1 =
\Le_{0}=\Le_{1}$. Therefore, we may assume that $n \geq 3$. For $3 \leq
n \leq k+1$, we have that $F_{n}^{(k)} = 2^{n-2}$, but $\Le_{m}$ is an
odd number for $m \geq 0$. Thus, there is no solution of \eqref{eq-1.3}
in this range. From now, we assume that $n \geq k+2$ and $k \geq 2$.

	Combining the inequalities \eqref{eq-2.9} and \eqref{eq-2.12} together with equation \eqref{eq-1.3}, we have \[
	\varphi^{n-2} \leq \alpha^{m+1} \quad \text{and} \quad \alpha^{m}
	\leq \varphi^{n-1}.  \] Then, we deduce that \[ (n-2)\left(
	\frac{\log \varphi}{\log \alpha}  \right) -1 \leq m \leq (n-1)
	\left( \frac{\log \varphi}{\log \alpha}  \right).  \] Using
	the fact $2(1-2^{-k}) < \varphi(k) < 2$ for all $k \geq 2$,
	it follows that \begin{equation}\label{eq-3.14}
		0.8n - 2.6 < m < 1.5n -1.5.
	\end{equation}

	\subsection{An inequality for $n$ and $m$ in terms of $k$}
	By using \eqref{eq-1.3}, \eqref{eq-2.8}, \eqref{eq-2.11} and
	taking absolute value, we obtain \begin{equation}\label{eq-3.15}
		\left| f_{k}(\varphi) \varphi^{n-1} -
		\frac{2 \alpha^{m+1}}{\sqrt{5}} \right| <
		\frac{1}{2} + \frac{2|\beta|^{m+1}}{\sqrt{5}}+
		\frac{|-\alpha|}{\sqrt{5}}+\frac{|\beta|}{\sqrt{5}}.
	\end{equation} Dividing both sides of the above inequality
	by $\frac{2 \alpha^{m+1}}{\sqrt{5}}$, we conclude that
	\begin{equation}\label{eq-3.16}
		\left| \left(\frac{\sqrt{5}f_{k}(\varphi)}{2}\right)
		\varphi^{n-1}  \alpha^{-(m+1)} - 1 \right| <   4
		\alpha^{-m}.
	\end{equation} Let \begin{equation}\label{eq-3.17}
		\Lambda_{1} :=
		\left(\frac{\sqrt{5}f_{k}(\varphi)}{2}\right)
		\varphi^{n-1}  \alpha^{-(m+1)} - 1,
	\end{equation} and inequality \eqref{eq-3.16} becomes
	\begin{equation}\label{eq-3.18}
		|\Lambda_{1}| < 4 \alpha^{-m}.
	\end{equation} Before applying Theorem \ref{thm2},
	we need to prove that $\Lambda_{1} \neq 0$. Assume
	that $\Lambda_{1}= 0$, then we get \[ f_{k}(\varphi)
	= \frac{2}{\sqrt{5}}\varphi^{-(n-1)}\alpha^{(m+1)}, \]
	and so $f_{k}(\varphi)$ is an algebraic integer, which is
	impossible. Thus, $\Lambda_{1} \neq 0$. Therefore, we apply
	Theorem \ref{thm2} to get a lower bound for  $\Lambda_{1}$
	given by \eqref{eq-3.17} with the parameters: \[ \eta_{1} :=
	\frac{\sqrt{5}f_{k}(\varphi)}{2}, \quad \eta_{2} := \varphi ,
	\quad \eta_{3} := \alpha, \] and \[ b_{1}:= 1, \quad b_{2}:=
	n-1, \quad b_{3}:= -(m+1).  \] Note that $\eta_{1}, \eta_{2},
	\eta_{3}$ are positive real numbers and belong to the field
	$\mathbb{K} := \mathbb{Q}(\varphi,\sqrt{5})$. So we can take $D
	:= [\mathbb{K}:\mathbb{Q}] \leq 2k$. Since $h(\eta_{2}) = (\log
	\varphi) / k < (\log 2)/k$ and $h(\eta_{3}) = (\log \alpha) / 2$,
	we choose \[ \max\{2kh(\eta_{2}),|\log \eta_{2}|,0.16\} = 2\log
	2 := A_{2} \] and \[ \max\{2kh(\eta_{3}),|\log \eta_{3}|,0.16\} =
	k\log \alpha := A_{3}.	\] By using the estimate \eqref{eq-2.13}
	and the properties of logarithmic height, we get that for all
	$k \geq 2$ \begin{align*}
		h(\eta_{1}) &\leq h(f_{k}(\varphi))+
		h\left(\frac{\sqrt{5}}{2}\right)\\ &< \log(k+1) + \log
		4 + \log \left( 2 \sqrt{5} \right) \\ &< 5.8 \log k.
	\end{align*} Thus, we obtain \[ \max\{2kh(\eta_{1}),|\log
	\eta_{1}|,0.16\} = 11.6 k \log k := A_{1}.  \] In addition, by
	\eqref{eq-3.14} we take $B:= 1.5n$. Then by Theorem \ref{thm2},
	we have \begin{equation}\label{eq-3.19}
		\hspace{-0.20cm} |\Lambda_{1}| > \exp \left( -1.432
		\times 10^{11} (2k)^{2} (1+ \log2k) (1 +\log 1.5n)
		(11.6 k \log k) (2\log2) (k \log \alpha) \right).
	\end{equation} Comparing  \eqref{eq-3.18} and \eqref{eq-3.19},
	taking logarithms and then performing the respective calculations,
	we get that \[ m \log \alpha - \log 4 < 4.43\times 10^{12} k^{4}
	\log k (1 + \log 2k)(1+ \log 1.5n).  \] Taking into consideration
	the facts $1+\log 2k < 3.5 \log k$ for all $k \geq 2$ and $1+ \log
	1.5n < 2.1 \log n$ for all $n \geq 4$, we conclude that \[ m <
	6.77\times 10^{13} k^{4} \log^{2} k \log n.  \] Using \eqref{eq-3.14}, the last inequality becomes \begin{equation}\label{eq-3.20}
		\frac{n}{\log n} < 8.47 \times 10^{13} k^{4} \log^{2} k.
	\end{equation} Since the function $x \mapsto x/ \log x$
	is increasing for all $x > e$, it is easy to check that
	\begin{equation}\label{eq-3.21}
		\frac{x}{\log x} < S \implies x < 2S\log S \quad
		\text{whenever} \quad S \geq 3.
	\end{equation} Then, taking $x:= n$ and $S := 8.47 \times 10^{13}
	k^{4} \log^{2} k$, inequality \eqref{eq-3.21} together with
	$32.07 +4\log k + 2 \log \log k < 49.3 \log k$ for all $k \geq
	2$, yields \begin{align*}
		n &< 2 \left( 8.47 \times 10^{13} k^{4} \log^{2} k
		\right ) \log \left( 8.47 \times 10^{13} k^{4} \log^{2}
		k \right) \\ &< (1.69 \times 10^{14} k^{4} \log^{2}
		k )(32.07+4\log k + 2 \log \log k ) \\ &< 8.35 \times
		10^{15} k^{4} \log^{3} k.
	\end{align*} By combining the above results, we obtain the
	following lemma.  \begin{lemma}\label{lem3.1}
		If $(n, k, m)$ is a solution in integers of equation
		\eqref{eq-1.3} with $k \geq 2$ and $n \geq k+2$, then
		the inequalities \begin{equation}\label{eq-3.22}
			0.6m < n < 8.35 \times 10^{15} k^{4} \log^{3} k
		\end{equation} hold.
	\end{lemma}

	\subsection{The case of small $k$} Suppose now that $k \in
	[2, 230]$. In order to apply Lemma \ref{lem-2.4}, we let
	\begin{equation}\label{eq-3.23}
		\Gamma_{1} := (n-1)\log \varphi -(m+1) \log \alpha +
		\log \left(\frac{\sqrt{5}f_{k}(\varphi)}{2}\right).
	\end{equation} Then $e^{\Gamma_{1}} - 1 := \Lambda_{1}$, where
	$\Lambda_{1}$ is defined by \eqref{eq-3.17}. Therefore, \eqref{eq-3.18} can be written as \begin{equation}\label{eq-3.24}
		|e^{\Gamma_{1}} -1 | < 4 \alpha^{-m}.
	\end{equation} Note that $\Gamma_{1} \neq 0.$ Since $\Lambda_{1}
	\neq 0$, we distinguish the following cases. If $\Gamma_{1} >  0$,
	then $e^{\Gamma_{1}} -1 > 0$. Using the fact $x \leq e^{x} -1 $
	for all $x \in \mathbb{R}$ and the inequality \eqref{eq-3.24}, we
	obtain \[ 0 < \Gamma_{1} < 4 \alpha^{-m}.  \] If, on the contrary,
	$\Gamma_{1} < 0$, then $4 \alpha^{-m} < 1/2$ holds for all $m \geq
	5$. Thus, from \eqref{eq-3.24}, we have $|e^{\Gamma_{1}} -1 |
	< 1/2$ and therefore $e^{|\Gamma_{1}|} < 2$. Since $\Gamma_{1}
	< 0$, we have \[ 0 < |\Gamma_{1}| \leq e^{|\Gamma_{1}|} - 1 =
	e^{|\Gamma_{1}|}|e^{\Gamma_{1}} - 1 | < 8 \alpha^{-m}.	\]
	In both cases, we have \begin{equation}\label{eq-3.25}
		0 < |\Gamma_{1}| <  8 \alpha^{-m}  < 8 \exp (- 0.48
		\times m).
	\end{equation} Let \[ c := 8, \quad \rho := 0.48 , \quad \psi :=
	\frac{\log \left(\frac{\sqrt{5}f_{k}(\varphi)}{2}\right)}{\log
	\varphi}, \] and \[ \vartheta:= \frac{\log \alpha}{\log
	\varphi}, \quad \vartheta_{1}:= - \log \alpha,
	\quad \vartheta_{2}:= \log \varphi, \quad \delta:= \log
	\left(\frac{\sqrt{5}f_{k}(\varphi)}{2}\right).	\] For each $k
	\in [2, 230]$, we find a good approximation of $\varphi$ and a
	convergent $p_{l}/q_{l}$ of the continued fraction of $\vartheta$
	such that $q_{l} > X_{0}$, where $X_{0} = \left\lfloor 8.35
	\times 10^{15} k^{4} \log^{3} k \right \rfloor$, which is an upper
	bound of $\max \{ n-1, m \}$ from Lemma \ref{lem3.1}. After doing
	this, we use Lemma \ref{lem-2.4} on inequality \eqref{eq-3.25}. A
	computer search with \textit{Mathematica} revealed that if $k \in
	[2,230]$, then the maximum value of  $\left\lfloor \frac{1}{\rho}
	\log \left( q^{2} c / |\vartheta_{2}| X_{0}\right) \right \rfloor$
	is 318, which is an upper bound on $m$ according to Lemma \ref{lem-2.4}. Hence, we deduce that the possible solutions $(n, m, k)$
	of the equation \eqref{eq-1.3} for which $k \in [2,230]$ have
	$m \leq 318$, therefore we use inequalities \eqref{eq-3.14}
	to obtain $n \leq 401$.

	Finally, we used \textit{Mathematica} to compare $F_{n}^{(k)}$
	and $\Le_{m}$ for the range $4 \leq  n \leq 401$ and $2 \leq  m
	\leq 318$, with $m < n/0.6$ and checked that the solutions of
	equation \eqref{eq-1.3} are \[ F_{4}^{(2)} =3= \Le_{2}, \quad
	F_{5}^{(2)} = 5= \Le_{3}, \quad \text{and} \quad F_{6}^{(4)} =
	15= \Le_{5}.  \]


	\subsection{The case of large $k$} From now on, we assume that
	$k > 230$. Here, it follows from Lemma \ref{lem3.1} that \[
	0.6m < n < 8.35 \times 10^{15} k^{4} \log^{3} k < 2^{k/2}.
	\] Using \eqref{eq-2.8} and Lemma \ref{lem2.9}, we can write
	\eqref{eq-1.3} as \begin{equation}\label{eq-3.26}
		2^{n-2} - \frac{2\alpha^{m+1}}{\sqrt{5}} = 2^{n-2} \xi -
		\frac{2\beta^{m+1}}{\sqrt{5}} - \frac{\alpha}{\sqrt{5}}
		+ \frac{\beta}{\sqrt{5}}.
	\end{equation} Taking absolute values on both sides
	of \eqref{eq-3.26}, we have that \[ \left| 2^{n-2} -
	\frac{2\alpha^{m+1}}{\sqrt{5}} \right| < \frac{2^{n-2}}{2^{k/2}}
	+ \frac{5}{\sqrt{5}}.  \] Dividing both sides of the above
	inequality  by $2^{n-2}$ and taking into account that $1/2^{n-2}
	< 1/2^{k/2}$ for $n \geq k+2$, we obtain \begin{equation}
	\label{eq-3.27}
		\left|1 - 2^{-n} \frac{8}{\sqrt{5}} \alpha^{m+1} \right|
		< \frac{3.23}{2^{k/2}}.
	\end{equation} Applying Theorem \ref{thm2} for the left-hand side,
	we set \[ \Lambda_{2} := 2^{-n} \frac{8}{\sqrt{5}} \alpha^{m+1}
	- 1.  \] Note that $\Lambda_{2} \neq 0$. Indeed, if  $\Lambda_{2}
	= 0$, then $\alpha^{2(m+1)}$ is a rational number, which is not
	possible for all positive integers $m$. Therefore $\Lambda_{2}
	\neq 0$. We take $t := 3$, \[ \eta_{1}:= 2, \quad \eta_{2}:=
	\frac{8}{\sqrt{5}}, \quad \eta_{3}:= \alpha, \] and \[
	b_{1}:= -n, \quad b_{2}:= 1, \quad b_{3}:= m+1.  \] Note that
	$\mathbb{K} := \mathbb{Q}(\alpha)$ contains $\eta_{1}, \eta_{2},
	\eta_{3}$ and has $D:= 2$. Since $m < 1.5n$, we deduce that $B:=
	\max\{|b_{1}|,|b_{2}|,|b_{3}|\}= 1.5n$. The logarithmic heights
	for $\eta_{1}$, $\eta_{2}$, and $\eta_{3}$ are calculated as
	follows: \[ h(\eta_{1}) = \log2, \quad h(\eta_{2}) = \log
	\left(8\sqrt{5}\right) \quad \text{and} \quad h(\eta_{3})
	= \frac{\log \alpha}{2}.  \] Thus, we can take \[ A_{1} :=
	2 \log2, \quad A_{2} := 5.8 \quad \text{and} \quad A_{3} :=
	\log \alpha.  \] As before, by applying Theorem \ref{thm2},
	we have \begin{equation}\label{eq-3.28}
		|\Lambda_{2}| > \exp \left( -8.27 \times 10^{12} \log
		n \right),
	\end{equation} where $1 + \log 1.5n < 2.1 \log n $ holds for
	all $n \geq 4$. Comparing \eqref{eq-3.27} and \eqref{eq-3.28},
	we obtain \[ k < 2.39 \times 10^{13} \log n.  \] By Lemma
	\ref{lem3.1} and using the fact $36.66 +4 \log k + 3 \log \log
	k < 11.7 \log k$ for all $k > 230$, we get \begin{align*}
		k &< 2.39 \times 10^{13} \log \left( 8.35 \times
		10^{15} k^{4} \log^{2} k \right )  \\ &< ( 2.39 \times
		10^{13})(36.66 +4 \log k + 3 \log \log k ) \\ &< 2.8
		\times 10^{14} \log k.
	\end{align*} Solving the above inequality by using the relation
	\eqref{eq-3.21} gives \[ k < 1.87 \times 10^{16}.  \] Again from
	Lemma \ref{lem3.1}, we obtain \begin{equation}\label{eq-3.29}
		n < 5.38 \times 10^{85} \quad \text{and} \quad m <
		8.97 \times 10^{85}.
	\end{equation} Let \begin{equation}\label{eq-3.30}
		\Gamma_{2} :=  (m+1) \log \alpha -n \log 2 + \log
		\left(\frac{8}{\sqrt{5}}\right).
	\end{equation} Using a similar method to show the inequality
	\eqref{eq-3.25}, one can see that \begin{equation}\label{eq-3.31}
		0 < |\Gamma_{2}| <  \frac{6.46}{2^{k/2}}  < 6.46 \times
		\exp (- 0.34 \times k)
	\end{equation} holds for all $k > 230$. The inequality
	\eqref{eq-3.29} implies that we can take $X_{0}:= 8.97 \times
	10^{85}$. Further, we choose \[ c := 6.46, \quad \rho := 0.34 ,
	\quad \psi := -\frac{\log \left(\frac{8}{\sqrt{5}}\right)}{\log
	2}, \] and \[ \vartheta:= \frac{\log \alpha}{\log 2}, \quad
	\vartheta_{1}:=  \log \alpha, \quad \vartheta_{2}:= - \log 2,
	\quad \delta:= \log \left(\frac{8}{\sqrt{5}}\right).  \] Using
	Lemma \ref{lem 2.2} with  $c := 6.46$, $\rho := 0.34$ and $X_{0}
	:= 8.97 \times 10^{85}$, we get $Y_{0} := 411.954\dots$. Let \[
	[a_{0}, a_{1}, a_{2},\dots] := [0, 1, 2, 3, 1, 2, 3, 2, 4, 2,
	1, 2, 11, 2, 1, 11, 1, 1, 134, 2, 2, 2, 1, 4, 1, 1, 3, 1,\dots]
	\] be the continued fraction expansion of $\log \alpha / \log
	2$. With the help of \textit{Mathematica}, we find that \[
	\displaystyle\max_{0 \leq k \leq Y_{0}} a_{k+1} =  880 :=A.  \]
	Then by Lemma \ref{lem 2.3}, we have \[ k < \frac{1}{0.34} \cdot
	\log \left (\frac{6.46 \cdot 882 \cdot 8.97 \cdot 10^{85}}{\log
	2} \right ) < 609.  \] With the above upper bound on $k$ and by
	Lemma \ref{lem3.1}, we have \begin{equation}\label{eq-3.32}
		n <  3.1\times 10^{29} \quad \text{and} \quad m < 5.2
		\times 10^{29}.
	\end{equation} We apply again Lemma \ref{lem 2.3} with $X_{0}:=
	5.2 \times 10^{29}$. Hence by Lemma \ref{lem 2.2}, we obtain
	$Y_{0} := 142.863\dots$ and $A := 134$. According to Lemma
	\ref{lem 2.3}, it becomes \[ k < \frac{1}{0.34} \cdot \log
	\left (\frac{6.46 \cdot 136 \cdot 5.2 \cdot 10^{29}}{\log 2}
	\right ) < 223, \] which contradicts our assumption that $k >
	230$. Hence, we have shown that there are no solutions $(n, k,
	m)$ to equation \eqref{eq-1.3} with $k > 230$. This completes
	the proof of Theorem \ref{thm1}.
\end{proof}

	\section{Acknowledgments} The authors would like to thank the
	anonymous referee for their useful comments and suggestions
	that improved the exposition of the paper.



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	\bigskip \hrule \bigskip \noindent	 2020 {\it Mathematics
	Subject Classification}: Primary 11B39; Secondary 11J86, 11R52.

	\noindent{\it Keywords:} $k$-generalized Fibonacci number,
	Leonardo number, linear form in logarithms, reduction method.

	\bigskip \hrule \bigskip

\noindent (Concerned with sequences
\seqnum{A000032},
\seqnum{A000045}, and
\seqnum{A001595}.)

\bigskip
\hrule
\bigskip

\vspace*{+.1in}
\noindent
Received January 31 2023;
revised versions received  February 8 2023; May 17 2023; May 26 2023.
Published in {\it Journal of Integer Sequences}, June 12 2023.

\bigskip
\hrule
\bigskip

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

                                                                                




