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\begin{center}
\vskip 1cm{\LARGE\bf 
Powers of Two as Sums of\\
\vskip .1in
Three Lucas Numbers
}
\vskip 1cm
\large
Bilizimb\'ey\'e~Edjeou\\
Universit\'e Gaston Berger de Saint-Louis\\
BP: 234, Saint-Louis\\
Senegal\\
\href{mailto:olivieredjeou@gmail.com}{\tt olivieredjeou@gmail.com} \\
\ \\
Amadou~Tall\\
Universit\'e Cheikh Anta Diop de Dakar\\ 
BP $5005$, Dakar Fann\\
Senegal\\
\href{mailto:amadou7.tall@ucad.edu.sn}{\tt amadou7.tall@ucad.edu.sn}\\
\ \\
Mohamed Ben Fraj Ben~Maaouia\\
Universit\'e Gaston Berger de Saint-Louis\\
BP $234$, Saint-Louis\\
Senegal\\
\href{mailto:maaouiaalg@hotmail.com}{\tt maaouiaalg@hotmail.com}\\
\end{center}

\vskip .2 in

\begin{abstract}  
In this paper, we find all positive integer solutions of the Diophantine
equation $L_k +L_l + L_t = 2^d$ in non-negative integers $k,l,t$,
and $d$, where $(L_n)_{n\geq0}$ is the Lucas sequence. The tools used
to solve our main theorem are linear forms in logarithms, properties
of continued fractions, and a version of the Baker-Davenport reduction
method in Diophantine approximation.
\end{abstract}

\section{Introduction}

The Lucas sequence $(L_k)_{k\geq0}$ is a linear recurrence  given by $L_0 = 2, L_1 = 1$ and 
$$L_{k+2} = L_{k+1} + L_k, \quad \hbox{for $k\geq 0$}.$$
It satisfies the same recurrence
as the Fibonacci sequence $(F_k)_{k\geq0}$ given by $F_0 = 0, F_1 = 1$ and 
$$F_{k+2} = F_{k+1} + F_k, \quad \hbox{for $k\geq 2$},$$ 
whose numbers are found everywhere in nature.
The Fibonacci numbers are famous
for possessing many wonderful and amazing properties.

In 2014, Bravo and Luca \cite{2}  studied   the Diophantine equation
$$L_k + L_l  =  2^t$$
in positive integers $k,l$ and $t$.
Similar equations involving Fibonacci and Padovan sequences are solved in \cite{5,7}.
E. Bravo and J. Bravo  \cite{4} found  also all powers of $2$  which are sums of  three Fibonacci numbers.  Specifically, they proved the following theorems.
\begin{theorem}
\label{thm1} 
The only solutions $(k,l,t)$ of the Diophantine equation $L_k + L_l  = 2^t$ in positive integers $k,l, t $ and with $k \geq l$ are 
$${(0,0,2);(1,1,1);(3,3,3);(2,1,2);(4,1,3);(7,2,5)}.$$
\end{theorem}

\begin{theorem}
\label{thm2} 
All solutions $(k,l,t,d)$ of the Diophantine equation
$$F_k + F_l + F_t = 2^d$$
in non-negative integers $k,l, t $, with $k \geq l \geq t $ and $d$ are
$${(3,1,1,2);(3,2,2,2);(3,2,1,2);(4,4,3,3);(5,3,1,3);(5,3,2,3);(6,5,4,4)};$$ 
$${(7,3,1,4); (7,3,2,4); (8,6,4,5); (10,6,1,6); (10,6,2,6); (11,9,5,7); (13,8,3,8); (16,9,4,10)}.$$
\end{theorem}

In this paper, we prove an extension of Theorem \ref{thm1} when the two Lucas numbers are replaced by three Lucas numbers and determine all the solutions of the Diophantine equation
$$ L_k + L_l+ L_t  =  2^d$$
in non-negative integers $k,l,t$ and $d$. We prove the following result.
 \begin{theorem}
\label{thm3}
All solutions $(k,l,t,d)$ of the Diophantine equation
 \begin{equation}
\label{eq2}
 L_k + L_l + L_t  =  2^d
\end{equation}  
in non-negative integers $k\geq l\geq t$ and $d$, are 
$$(1, 1, 0, 2),(2, 2, 0, 3),(3, 0, 0, 3),(3, 2, 1, 3),(4, 4, 0, 4),(5, 2, 0, 4),(5, 3, 1, 4),(6, 4, 4, 5),$$
  $$(6, 5, 2, 5),(7, 1, 0, 5),(10, 2, 0, 7),(10, 3, 1, 7),(17, 13, 3, 12).$$
 \end{theorem}
Our method of proof is similiar to the method described in \cite{2, 4}.

\section{Preliminaries}

Before proceeding further, we recall the Binet formula for the Lucas numbers $(L_k)_{k\geq0}$, namely
$$L_k = \alpha^k + \beta^k, \quad\hbox{for}\quad  k \geq 0,$$      
where 
$$\alpha = \frac{1 + \sqrt{5}}{2}\quad\hbox{and} \quad \beta = \frac{1 - \sqrt{5}}{2}$$
are the roots of the characteristic equation $x^2 - x - 1 = 0$. In particular, the inequality 
  \begin{equation}
\label{eq3}
\alpha^{k - 1}  \leq  L_k \leq 2\alpha^k
\end{equation}
holds for all $k \geq 0$.

To prove Theorem \ref{thm3}, using a result on linear forms in two logarithms., we require some notation.
 Let $\delta$ be an algebraic number of degree $d$  with minimal polynomial
$$a_0x^d + a_1 x^{d-1} + \cdots + a_d = a_0\prod_{i = 1}^d(X - \delta^{(i)} )$$
where the $a_i$ are relatively prime integers with $a_0 &gt; 0$ and the $\delta^{(i)}$ denotes the conjugates of $\delta$. Then
$$h(\delta) = \frac{1}{d} (\log a_0 + \sum_{i = 1}^d\log(\max\lbrace\vert\delta^{(i)}\vert,1\rbrace))$$
is called the logarithmic height of $\delta$. In particular, if $\delta = p/q$ is a rational number with
$\gcd(p,q) = 1$ and $q &gt; 0$, then 
$$h(\delta) = \log \max \lbrace{\vert p\vert,q}\rbrace.$$

The following properties of the logarithmic height, will be used in the next section. Let $\delta$, $\nu$ be algebraic numbers and $r\in \mathbb{Z}$. Then
\begin{itemize}
 \item $h(\delta \pm \nu) \leq h(\delta) + h(\nu) + \log2$,
\item  $h(\delta\nu^{\pm 1}) \leq h(\delta) + h(\nu)$,
\item $h(\delta^r ) = \vert r \vert h(\delta)$.
\end{itemize}
Using the above notation, we restate  Laurent, Mignotte, and Nesterenko's result \cite[Cor.\ 1]{6}.
\begin{theorem}
\label{thm4} 
Let $\delta_1 ,\delta_2$ be two non-zero algebraic numbers, and let $\log\delta_1$ and $\log\delta_2$ be any
determinations of their logarithms. Set
$$D = [\mathbb{Q}(\delta_1 ,\delta_2 ) :\mathbb{Q} ]/[\mathbb{R}(\delta_1 ,\delta_2 ) : \mathbb{R}]$$
 and
$$\Gamma := b_2\log\delta_2 - b_1\log\delta_1,$$
where $b_1$ and $b_2$ are positive integers. Further, let $A_1, A_2 &gt;  1$ be real numbers  such that
$$\log{A_i} \geq \max\lbrace h(\delta_i) , \frac{\vert h(\delta_i)\vert} {D}, \frac{1}{D}\rbrace,\quad i = 1,2.$$
Then, assuming that $\delta_1$ and $\delta_2$ are multiplicatively independent, we have
$$\log{ \vert \Gamma \vert} &gt; -30.9\cdot D^4(\max{\lbrace \log{b'}, \frac{21}{D},\frac{1}{2} \rbrace)^2} \log{A_1} \log{A_2}, $$
where 
$$b' = \frac{b_1}{D\log{A_2}} + \frac{b_2}{D\log{A_1}}.$$
\end{theorem}

We  also need the following general lower bound for linear forms in logarithms due to Matveev \cite{8}.
\begin{theorem}
\label{thm5}
 Assume that $\delta_1 ,\ldots, \delta_t$ are positive real algebraic numbers in a real algebraic number field $\mathbb{K}$ of degree $D$. Let $b_1 ,\cdots,b_n$ be rational integers, and 
$$\Lambda := \delta_1^{b_1} \cdots \delta_t^{b_t} - 1$$
be not zero. Then
$$\vert \Lambda \vert &gt; \exp{(-1.4\cdot 30^{t+3}\cdot t^{4.5}\cdot D^2(1 + \log D)(1 +  \log B)}A_1 \cdots A_t),$$
where
$$B \geq \max{\lbrace \vert b_1\vert,\ldots,\vert b_t\vert\rbrace},$$
and
$$A_i \geq \max{\lbrace Dh(\delta_i ),\vert\log\delta_i \vert,0.16\rbrace}, \quad \hbox{for all} \quad i = 1,\cdots,t.$$
\end{theorem}

Finally,  we present a version of the reduction method  based on the Baker-Davenport Lemma \cite{1}, from Dujella and Peth\H{o}  \cite{3}. This will be one of the key tools used to reduce the upper bounds on the variables of the equation \eqref{eq2}.
\begin{lemma}
\label{lem1}
 Let $N$ be a positive integer, let $p/q$ be a convergent  of the irrational number $\gamma$ such that $q &gt; 6N$, and let ${A,B,\mu}$ be real numbers with ${A &gt; 0}$ and
${B &gt; 1}$. Define 
$$\xi := \Vert \mu q \Vert- N\Vert\gamma q\Vert,$$
where $\Vert \cdot \Vert$ denotes the distance to  the nearest integer.
If $\xi &gt; 0$, then there is no solution to the inequality
$$0 &lt; u\gamma - v + \mu &lt; AB^{-w},$$
in positive integers $u$, $v$, and $w$, with
$u \leq N$ and $w \geq \frac{\log{(Aq/\xi)}}{\log{B}}$.
\end{lemma}

\section{The Proof of Theorem~\ref{thm3}}
First of all, observe that if $k = l = t$, then  equation \eqref{eq2} becomes $3L_k = 2^d$. Since $3\nmid 2$, then equation \eqref{eq2} has no solution.  Subsequently, we assume that either $k &gt; l$ or $l &gt; t$.

 If $k \leq 250$, then a brute force search using 
 {\tt Sagemath} in the range  $0 \leq t \leq l \leq k \leq 250$
produces the solutions 
$$(1, 1, 0, 2),(2, 2, 0, 3),(3, 0, 0, 3),(3, 2, 1, 3),(4, 4, 0, 4),(5, 2, 0, 4),(5, 3, 1, 4),(6, 4, 4, 5),(6, 5, 2, 5),$$
$$(7, 1, 0, 5),(10, 2, 0, 7),(10, 3, 1, 7),(17, 13, 3, 12).$$
Thus, for the remainder of the paper, we assume that $k &gt; 250$. Let us now establish a relation between $k$ and $d$.

 Combining \eqref{eq2} with the right inequality of \eqref{eq3}, one gets that
$$2^d \leq 2\alpha^{k} + 2\alpha^{l} + 2\alpha^{t} &lt;  6\alpha^{k} &lt; 6\cdot 2^{k} &lt; 2^{k+3},$$
which leads to $d \leq k+2$. 

\subsection{Bounding $k-l$ and $k-t$ in terms of $k$}
 We rewrite \eqref{eq2} as
$$\alpha^{k} - 2^d  =  -\beta^{k} - L_l - L_t.$$
Now taking absolute values, we obtain
$$\vert\alpha^k - 2^d \vert  \leq  \vert\beta\vert^k + L_l + L_t &lt;  \frac{1}{2} + 2\alpha^l + 2\alpha^t.$$
Dividing both sides of the above expression by $\alpha^k$ and taking into account that $k \geq l \geq t$, we get
$$\vert1 - 2^d\alpha^{-k} \vert  &lt;  \frac{1}{2}\alpha^{-k} + 2\alpha^{-k+l} + 2\alpha^{-k+t} &lt; 5\alpha^{-k+l}.$$
Thus
\begin{equation}
\label{eq6}
 \vert1 - 2^d\alpha^{-k} \vert  &lt;  \frac{5}{\alpha^{k-l}}.
\end{equation}
We apply Theorem \ref{thm4} to 
$$\Gamma := d \log\alpha - k\log2.$$
Therefore the estimate \eqref{eq6} can be rewritten as
\begin{equation}
\label{eq7}
\vert1 - e^\Gamma \vert  &lt;  \frac{5}{\alpha^{k-l}}.
\end{equation}

The algebraic number field containing $2 ,\alpha$ is $\mathbb{Q}(\sqrt{5})$, so we can take $D := 2$. By using \eqref{eq2} and the Binet formula for the Lucas sequence, we have
\begin{equation}
\label{eq8}
  \alpha^k = L_k - \beta^k  &lt;  L_k + 1 \leq L_k + L_l + L_t = 2^d.
\end{equation}
Consequently, 
$1 &lt; 2^d\alpha^{-k}$ and so $\Gamma &gt; 0$.  Using the fact that $\log(1 + x) \leq x$ for all $x \in \mathbb{R}^+$, together with \eqref{eq7}, gives
\begin{equation}
\label{eq9}
 0 &lt; \Gamma  &lt;  \frac{5}{\alpha^{k-l}},
 \end{equation}
Hence, 
 \begin{equation}
\label{eq10}
 \log\Gamma  &lt;  \log5 - (k-l)\log\alpha.
 \end{equation}
Note further that $h(\alpha) = \log\alpha/2$ and $h(2) = \log2$. Thus, we can choose 
$$\log A_1 := \log\alpha \quad \hbox{and} \quad \log A_2 := \log2.$$
Finally, recall that $d \leq k+2 $, and so
$$b' = \frac{k}{2\log2} + \frac{d}{2\log\alpha} &lt; 4k.$$
Since $\alpha$ and $2$ are multiplicatively independent, we have, by Theorem \ref{thm4}, that
$$\log{\Gamma}  \geq  {-30.9\cdot 2^4}\cdot (\max \lbrace\log{(4k)},21/2,1/2\rbrace)^2\cdot \log\alpha\cdot \log2.$$
Thus
\begin{equation}
\label{eq11}
 \log{\Gamma}  &gt;  {-174}\cdot (\max\lbrace\log{(4k)},21/2,1/2\rbrace)^2.
\end{equation}  
Combining \eqref{eq10} and \eqref{eq11}, we obtain
\begin{equation}
\label{eq12}
(k-l)\log\alpha  &lt;  180\cdot (\max\lbrace\log(4k),21/2\rbrace)^2.
\end{equation}

Let us now establish a second linear form in logarithms. To this end, we rewrite equation \eqref{eq2} as follows
$$\alpha^k(1 + \alpha^{(l-k)} ) - 2^d  =  -\beta^k- \beta^l - L_t.$$
Taking absolute values in the above relation and using the fact that $\beta = (1 - \sqrt{5})/2$ we get
$$\vert \alpha^k(1 + \alpha^{(l-k)}) - 2^d\vert  = \vert-\beta^k- \beta^l - L_t\vert &lt; 2+ 2\alpha^t$$
for all $k &gt; 250$ and $l \geq t \geq 0$. 
Dividing both sides of the above inequality by the first term of
the left-hand side, we obtain
\begin{equation}
\label{eq15}
\vert 1 - 2^d\alpha^{-k}(1 + \alpha^{(l-k)})^{-1}\vert  &lt;    \frac{2}{\alpha^{k}(1 + \alpha^{(l-k)})} + \frac{2}{\alpha^{k-t}(1 + \alpha^{(l-k)})} &lt; \frac{4}{\alpha^{k-t}}.
\end{equation}

We are now ready to apply Matveev's result given in Theorem \ref{thm5}. To do this, we take the parameters
$n := 3$ and
$$\delta_1 := 2,\quad  \delta_2 := \alpha, \quad \delta_3 := (1 + \alpha^{(l-k)}).$$
We take $b_1 := d$, $b_2 := -k$ and $b_3 := -1$. As before, $\mathbb{K}  := \mathbb{Q}(\sqrt{5})$ contains $ \delta_1 ,\delta_2 ,\delta_3$ and has
$D := [\mathbb{K} : \mathbb{Q}] = 2$. To see why the left-hand side of \eqref{eq15} is not zero, note that otherwise, we
would get the relation
\begin{equation} 
\label{eq16}
  \alpha^k + \alpha^l  =  2^d.
  \end{equation}
Conjugating the above relation in $\mathbb{Q}({\sqrt{5}})$, we get
 \begin{equation}
\label{eq17}
\beta^k + \beta^l = 2^d.
\end{equation}
Further, combining \eqref{eq16} and \eqref{eq17}, we obtain
$$\alpha^k&lt;\alpha^k + \alpha^l = \vert \beta^k + \beta^l\vert&lt;2 .$$
This is impossible because $k&gt;250$.
Thus, 
$$1 - 2^d\alpha^{-k}(1 + \alpha^{(l-k)})^{-1}$$
is not zero.

In this application of Theorem \ref{thm5}, we take $A_1 := 2\log2$ and $A_2 := \log\alpha$. Since $t \leq k+2$,
it follows that we can take $B := k+2$. Let us now estimate $h(\delta_3 )$. We begin by observing that
$$\delta_3  =  (1 + \alpha^{(l-k)}) &lt; 2\quad \hbox{and} \quad \delta_3^{-1} &lt; 1.$$ 
So that
$$0 &lt;\log\delta_3  &lt;  1.$$
 Next, notice that
$$ h(\delta_3 ) \leq  (k-l)\log\alpha + \log2.$$
Hence, we can take
$$A_3 := 2 +(k-l)\log\alpha &gt; \max\lbrace{2h(\delta_3 ),\vert\log\delta_3\vert,0.16}\rbrace.$$
Now Theorem \ref{thm5} implies that a lower bound on the left-hand side of \eqref{eq15} is
$$\log\vert\Lambda\vert &gt; -1.4\cdot 30^6\cdot 3^{4.5}\cdot 2^2(1 + \log2)(1 + \log(k+2))\cdot 2\log2\cdot 2\log\alpha\cdot (2 +(k-l)\log\alpha).$$
 So, inequality \eqref{eq15} yields
 \begin{equation}
\label{eq20}
 k-t  &lt;  2.8\cdot 10^{12}\log(k+2)\cdot (2 + (k-l)\log\alpha), 
 \end{equation}
where we used the inequality $1 + \log(k+2) &lt; 2\log(k+2)$, which holds because $k &gt; 250$.

Now using \eqref{eq12} in the right-most term of  inequality \eqref{eq20} and performing the respective calculations, we obtain
\begin{equation}
\label{eq21}
k-t  &lt;  5.1\cdot 10^{14} \log(k+2)(\max\lbrace{\log(4k),21/2}\rbrace)^2.
\end{equation}

\subsection{ Bounding $k$}

Finally, we consider a third linear form in logarithms. We now rewrite equation \eqref{eq2} as follows
$$\alpha^k(1 + \alpha^{(l-k)} + \alpha^{(t-k)} ) - 2^d  =  -\beta^k- \beta^l -\beta^t.$$
Taking absolute values in the above relation and using the fact that $\beta = (1 - \sqrt{5})/2,$ we get
$$\vert \alpha^k(1 + \alpha^{(l-k)} + \alpha^{(t-k)}) - 2^d \vert  = \vert-\beta^k- \beta^l -\beta^t\vert &lt; 3$$
for all $k &gt; 250$ and $l \geq t \geq 0$.
Dividing both sides of the above inequality by the first term of
the left-hand side, we obtain
\begin{equation}
\label{eq24}
\vert 1 - 2^d\alpha^{-k}(1 + \alpha^{(l-k)} + \alpha^{(t-k)})^{-1}\vert  &lt;    \frac{3}{\alpha^{k}(1 + \alpha^{(l-k)} + \alpha^{(t-k)})}  &lt; \frac{3}{\alpha^{k}}.
\end{equation}
We apply Theorem \ref{thm5} to $$\Lambda = 1 - 2^d\alpha^{-k}(1 + \alpha^{(l-k)} + \alpha^{(t-k)})^{-1},$$ with the parameters
$n := 3$, $\delta_1 := 2$, $\delta_2 := \alpha$, $\delta_3 := (1 + \alpha^{(l-k)} + \alpha^{(t-k)})$,
 $b_1 := d$, $b_2 := -k$ and $b_3 := -1$, $\mathbb{K}  := \mathbb{Q}(\sqrt{5})$ contains $ \delta_1 ,\delta_2 , \delta_3$,
$D := [\mathbb{K} : \mathbb{Q}] = 2$. To see why the left-hand side of \eqref{eq24} is not zero, note that otherwise, we
would get the relation
\begin{equation}
\label{eq25}
  \alpha^k + \alpha^l + \alpha^t  =  2^d.
  \end{equation}
  Conjugating the above relation in $\mathbb{Q}({\sqrt{5}})$, we get
  \begin{equation}
\label{eq26}
   \beta^k + \beta^l + \beta^t = 2^d.
   \end{equation} 
Furthermore, combining \eqref{eq25} and \eqref{eq26}, we obtain
     $$\alpha^k&lt;\alpha^k + \alpha^l + \alpha^t = \vert \beta^k + \beta^l+\beta^t\vert&lt;3.$$
This is impossible because $k&gt;250$. Thus, 
   $$1 - 2^d\alpha^{-k}(1 + \alpha^{(l-k)} + \alpha^{(t-k)})^{-1}$$
    is not zero.
  We now apply Theorem \ref{thm5} with $A_1 := 2\log2$, $A_2 := \log\alpha$. Since $d \leq k+2$;
it follows that we can take $B := k+2$. Let us now estimate $h(\delta_3 )$. We begin by observing that
$$\gamma_3  =  (1 + \alpha^{(l-k)} + \alpha^{(t-k)}) &lt; 3$$
 and 
$$0 &lt;\log\gamma_3  &lt;  \log3.$$ 
Next, notice that
 $$h(\delta_3 ) \leq  (k-l)\log\alpha + (k-t)\log\alpha + 2\log2 \leq 2(k-t)\log\alpha + 2\log2.$$
Hence, we can take
$$A_3 := 4 +2(k-t)\log\alpha &gt; \max\lbrace{2h(\delta_3 ),\vert\log \delta_3\vert,0.16}\rbrace.$$
Now, from Theorem \ref{thm5} we have 
$$\log\vert\Lambda\vert &gt; -1.4\cdot 30^6\cdot 3^{4.5}\cdot 2^2(1 + \log2)(1 + \log(k+2))\cdot 2\log2\cdot 2\log\alpha\cdot (4 +2(k-t)\log\alpha).$$
So, inequality \eqref{eq24} gives
\begin{equation}
\label{eq29}
 k  &lt;  10^{13}\log(k+2)\cdot (2 + (k-t)\log\alpha),  
 \end{equation}
where we used the inequality $1 + \log(k+2) &lt; 2\log(k+2)$, which holds because $k &gt; 250$.

Now using \eqref{eq20} in the rightmost term of the above inequality \eqref{eq29} and performing the respective calculations, we obtain
\begin{equation}
\label{eq30}
k  &lt;  2.6\cdot 10^{27} (\log(k+2))^2(\max\lbrace{\log(4k),21/2}\rbrace)^2.
\end{equation}
If $\max\lbrace{\log(4k),21/2}\rbrace = 21/2$, it then follows from \eqref{eq30} that 
 $$k &lt; 287\cdot 10^{27}(\log(k+2))^2,$$        
 giving
 $$k &lt; 15\cdot 10^{32}.$$
If on the other hand we have that $\max\lbrace{\log(4k),21/2}\rbrace = \log(4k)$, then inequality \eqref{eq30} gives that
  $$k &lt; 2.6\cdot 10^{27}(\log(k+2))^2(\log(4k))^2,$$
   and so 
  $$k &lt; 12\cdot 10^{34}.$$
 In any case, we have that
 $$k &lt; 12\cdot 10^{34}$$
always holds. We summarize what we have so far in the following lemma.   
 \begin{lemma}
\label{lem2}
 If $(k,l,t,d)$ is a solution in positive integers of equation \eqref{eq2} with $k \geq l \geq t$ and
$k &gt; 250$, then inequalities
 $$d \leq k+2 \quad\hbox{and}\quad k &lt; 12\cdot 10^{34}$$
 hold.
\end{lemma}

\section{The final computations}
In this section, we will reduce the upper bound on $k$.  Firstly, we determine  a suitable upper bound on $k-l,  k-t$, and later
we use Lemma \ref{lem1} to conclude that $k$ must be  smaller than $250$.

Turning back to inequality \eqref{eq9}, we obtain
$$0 &lt; d\log2 - k\log \alpha &lt; \frac{5}{\alpha^{k-l}}.$$
Dividing across by $\log\alpha$, we get
\begin{equation}
\label{eq31}
0  &lt;  d\gamma - k &lt; \frac{11}{\alpha^{k-l}},
\end{equation}
where 
 $$\gamma := \frac{\log2}{\log\alpha}.$$
 
 Let $[a_0 ,a_1 ,a_2 ,a_3 ,a_4,a_5,a_6,a_7,\ldots] = [1,2,3,1,2,3,2,4\ldots]$ be the continued fraction expansion of $\gamma$, and let
denote $p_n /q_n$ its $n$th convergent. Recall also that $d &lt; 12\cdot10^{34}$ by Lemma \ref{lem2}.
A quick inspection using {\tt Sagemath} reveals that
\begin{align*}
&amp; 37527245802242661673724926130723830 = q_{73} &lt; 12\cdot10^{34} &lt;  \\
&amp; q_{74} = 175184858909722330004986691804684639.
\end{align*}
Furthermore, $a_N := \max\lbrace{a_i; i = 0,1,\ldots  ,44}\rbrace = a_{17} = 134$. So, from the known properties
of continued fractions, we obtain that
\begin{equation}
\label{eq32}
\vert d\gamma - k\vert  &gt;  \frac{1}{(a_N + 2)d}.
\end{equation}
Comparing estimates \eqref{eq31} and \eqref{eq32}, we get right away that
$$\alpha^{k-l}  &lt;  11\cdot 136\cdot d &lt; 18\cdot 10^{37},$$
leading to $k-l &lt; 184$.

Let us now go back to \eqref{eq15} and determine an improved
upper bound on $k-t$. Put
\begin{equation}
\label{eq34}
\omega_1  :=  d\log2 - k\log\alpha - \log(1 + \alpha^{-(k-l)}).
\end{equation}
 Therefore, \eqref{eq15} implies that
 \begin{equation}
\label{eq35}
\vert1 - e^{\omega_1}\vert  &lt;  \frac{4}{\alpha^{k-t}}.
\end{equation}
Note that $\omega_1 \neq 0$, by using \eqref{eq2} and the Binet formula for the Lucas sequence, we have
$$\alpha^k + \alpha^l = L_t -\beta^k -\beta^l &lt; L_k+L_l+L_t=2^d.$$
Therefore, 
$$1 &lt; 2^d\alpha^{-k}(1 + \alpha^{(l-k)})^{-1}$$
and so $\omega_1&gt;0$.
Thus
\begin{equation}
\label{eq36}
0 &lt; \omega_1 \leq  e^{\omega_1} -  1  &lt;  \frac{4}{\alpha^{k-t}}.
\end{equation}
Replacing $\omega_1$ in the above inequality by its formula \eqref{eq34} and dividing both sides of the resulting
inequality by $\log\alpha$, we get
 \begin{equation}
\label{eq37}
 0  &lt;  d\left(\frac{\log2}{\log\alpha}\right) - k - \frac{\log(1 + \alpha^{-(k-l)})}{\log\alpha} &lt; \frac{9}{\alpha^{k-t}}.
 \end{equation}
 We now put
$$\gamma := \frac{\log2}{\log\alpha}, \quad \mu := - \frac{\log(1 + \alpha^{-(k-l)})}{\log\alpha}, \quad A := 9 \quad\hbox{and}\quad B := \alpha.$$
Clearly $\gamma$ is an irrational number. We also put $N := 12\cdot 10^{34}$ , which is an upper bound on $d$ by
Lemma \ref{lem2}. We therefore apply Lemma \ref{lem1} to inequality \eqref{eq37} for all choices $k-l \in \lbrace1,\ldots,184\rbrace$
except when $k-l = 1,3$ and get that
$$k-t &lt; \frac{\log(Aq/\xi)}{\log B},$$
where $q &gt; 6N$ is a denominator of a convergent of the continued fraction of $\gamma$ such that
$\xi = \Vert \mu q\Vert - N\Vert \gamma q\Vert &gt; 0$. Indeed, using 
{\tt Sagemath}, we have that 
$$q = q_{75} = 1439006117080021301713618460568200942.$$ 
We find that if $(k,l,t,d)$ is
a possible solution of the equation \eqref{eq2} with $\omega_1 &gt; 0$ and $k-l \in \lbrace1,\ldots,184\rbrace$
except when $k-l = 1,3$, then 
$$k-t &lt; 178.$$ 

Let us now treat the cases where  $k-l = 1$ and $3$. The discussion of these cases will be different from the previous ones, because when applying Lemma \ref{lem1} to the expression \eqref{eq37}, the corresponding parameter $\mu$ appearing in Lemma \ref{lem1} is

\begin{displaymath}
\frac{\log(1 + \alpha^{-(k-l)})}{\log\alpha} = \begin{cases}
-1, &amp; \text{if  $k-l=1$;} \\
 1-\frac{\log2}{\log\alpha}, &amp; \text{if $k-l=3$.}
 \end{cases}
\end{displaymath}
In both cases, the parameters $\gamma$ and $\mu$ are linearly dependent, which yields that the corresponding value of $\xi$ from Lemma \ref{lem1} is always negative and therefore the reduction method is not useful for reducing the bound on $k-t$ in these instances. One can see that if $k-l = 1, 3,$ then the resulting inequality from \eqref{eq37} has the shape
$$0 &lt; \vert a \gamma - b \vert &lt; \frac{9}{\alpha^{k-t}},$$
with $\gamma$ being an irrational number and $a,b \in \mathbb{Z}$. So, one can appeal to the known properties of the convergents of the continued fractions to obtain a nontrivial lower bound for
$$\vert a \gamma - b \vert.$$
When $k-l= 1$, from \eqref{eq37}, we get that

\begin{equation}
\label{eq38}
0  &lt;  d\gamma - (k+1)  &lt; \frac{9}{\alpha^{k-t}}.
\end{equation}

Let $[a_0 ,a_1 ,a_2 ,a_3 ,a_4,a_5,a_6,a_7,...] = [1,2,3,1,2,3,2,4\ldots]$ be the continued fraction expansion of $\gamma$, and let
denote $p_n /q_n$ its $n$th convergent. Recall also that $d &lt; 12\cdot10^{34}$ by Lemma \ref{eq7}.
Furthermore, $a_N := \max\lbrace{a_i: i = 0,1,\ldots,44}\rbrace = a_{17} = 134$. So, from the known properties
of continued fractions, we obtain that
\begin{equation}
\label{eq39}
\vert d\gamma - (k+1)\vert  &gt;  \frac{1}{(a_N + 2)d}.
\end{equation}
Comparing estimates \eqref{eq38} and \eqref{eq39}, we get right away that
$$\alpha^{k-t}  &lt;  9\cdot 136\cdot d &lt; 15\cdot 10^{37},$$
leading to $k-t &lt; 184$.

By the same argument as the one we did before, we get that
$k-t&lt;184$ in the case when $k-l = 3$. This completes the analysis of the cases when $k-l = 1,3$. Consequently,
$k-t&lt; 184$ always holds.

Finally, we shall use \eqref{eq24}  to reduce the upper bound on $k$. Put
\begin{equation}
\label{eq41}
\omega_2  =  d\log2 -k\log\alpha - \log{\varphi(u,v)}
\end{equation}
where $\varphi$ is the function given by the formula $\varphi(u,v) = 1+\alpha^{-u}+\alpha^{-v}$, where $u=k-l, v=k-t$.
Note that $\omega_2 \neq 0$. Thus, we distinguish the following cases. If $\omega_2 &gt; 0$ then, from \eqref{eq24}, we
obtain
$$0 &lt; \omega_2 \leq e^{\omega_2} - 1 &lt; \frac{3}{\alpha^k}.$$
Replacing $\omega_2$ in the above inequality by its formula \eqref{eq41} and dividing both sides of the resulting inequality by $\log\alpha$, we get
 \begin{equation}
\label{eq42}
 0  &lt;  d\left(\frac{\log2}{\log\alpha}\right) - k - \frac{\log(1 + \alpha^{-u} + \alpha^{-v})}{\log\alpha} &lt; \frac{7}{\alpha^k}.
 \end{equation}
 We now put

$$\gamma := \frac{\log2}{\log\alpha},\quad  \mu := - \frac{\log(1 + \alpha^{-u} + \alpha^{-v})}{\log\alpha}, \quad A := 7 \quad\hbox{and}\quad B := \alpha.$$
Clearly $\gamma$ is an irrational number. We also put $N := 12\cdot 10^{34}$ , which is an upper bound on $d$ by Lemma \ref{lem2}. We therefore apply Lemma \ref{lem1} to inequality \eqref{eq42} for all choices of $u \in \lbrace1,\ldots,184\rbrace, v \in \lbrace1,\ldots,184\rbrace$
except when 
\begin{align*}
(u,v) \in \varpi_1 &amp;= \lbrace(4, 2),(13, 1),(17, 4),(17, 5),(18, 7),(19, 4),(19, 5),(22, 2),(23, 1),(23, 2),(24, 1), \\
&amp; (24, 5),(25, 1),(28, 2),(30, 4),(31, 1),(32, 7),(33, 1),(33, 4),(33, 5),(36, 1),(36, 2), \\
&amp; (37, 2),(38, 2),(39, 1),(43, 4),(44, 1),(45, 4),(48, 1),(48, 5),(50, 1),(51, 9),(52, 1), \\
&amp; (52, 2),(53, 4),(54, 2),(55, 2),(56, 2),(57, 2),(59, 2),(61, 1),(62, 4),(64, 1),(66, 1), \\
&amp; (67, 1),(69, 2),(70, 5), (75, 1),(75, 2),(83, 5),(87, 1),(87, 7),(89, 2),(90, 2),(95, 1), \\
&amp; (95, 7),(97, 1),(98, 1),(99, 5),(100, 1),(102, 2),(107, 2),(110, 4),(111, 2),(112, 1), \\
&amp;  (112, 2),(113, 1),(113, 5),(113, 6),(114, 1),(118, 1),(122, 1),(122, 4),(122, 6), \\
&amp; (128, 2),(129, 1),(129, 2), (130, 1),(130, 4),(130, 5),(132, 2),(133, 2),(136, 4), \\
&amp; (138, 2),(139, 1),(139, 2),(141, 1),(141, 2),(142, 4),(147, 1),(148, 1),(158, 7)\rbrace
\end{align*}
 and get that
$$k &lt; \frac{\log(Aq/\xi)}{\log B},$$
where $q &gt; 6N$ is a denominator of a convergent of the continued fraction of $\gamma$ such that
$\xi = \Vert \mu q\Vert - N\Vert \gamma q\Vert &gt; 0$. Indeed, using  
{\tt Sagemath}, we have  that 
$$q = q_{82} = 12054118444825786260212254516320106583249.$$
 We find that if $(k,l,t,d)$ is
a possible solution of the equation \eqref{eq2} with $\omega_2 &gt; 0$ and $ (u,v) \notin \varpi_1$, then $k &lt; 196$. This is
false because our assumption is that $k &gt; 250$.

Let us now work with the cases when $(u,v) \in \varpi_1 $. We cannot study these
cases as before because when applying Lemma \ref{lem1} to the expression \eqref{eq42}, the corresponding quantity $\Vert\mu q\Vert$ appearing in Lemma \ref{lem1} is zero.
In these cases, the parameters $\gamma$ and $\mu$ are linearly dependent, which yields that the corresponding value of $\xi$ from Lemma \ref{lem1} is always negative and therefore the reduction method is not useful for reducing the bound on $k$ in these instances. However, one can see that if $(u,v) = (2,4)$, then the resulting inequality from \eqref{eq42} has the shape
$$0 &lt; \vert x \gamma - y \vert &lt; \frac{7}{\alpha^{k}},$$ 
with $\gamma$ 
being an irrational number and $x,y \in \mathbb{Z}$. So, one can use to the known properties of the convergents of the continued fractions to obtain a nontrivial lower bound for

$$\vert x \gamma - y \vert.$$
This clearly gives us an upper bound for $k$. For example, when $(u,v)= (2,4)$,
$$- \frac{\log(1 + \alpha^{-u} + \alpha^{-v})}{\log\alpha} = 2- 2\frac{\log2}{\log\alpha}$$
 and from \eqref{eq42}, we get that

\begin{equation}
\label{eq43}
0  &lt;  (d-2)\gamma - (k-2)  &lt; \frac{7}{\alpha^{k}}.
\end{equation}
Let $[a_0 ,a_1 ,a_2 ,a_3 ,a_4,a_5,a_6,a_7,\ldots] = [1,2,3,1,2,3,2,4\ldots]$ be the continued fraction expansion of $\gamma$, and let
denote $p_n /q_n$ its $n$th convergent. Recall also that $d &lt; 12\cdot10^{34}$ by Lemma \ref{lem2}.

Furthermore, $a_N := \max\lbrace{a_i; i = 0,1,\ldots,44}\rbrace = a_{17} = 134$. So, from the known properties
of continued fractions, we obtain that
\begin{equation}
\label{eq44}
\vert (d-2)\gamma - (k-2)\vert  &gt;  \frac{1}{(a_N + 2)d}.
\end{equation}
Comparing estimates \eqref{eq43} and \eqref{eq44}, we get  that
\begin{equation}
\label{eq45}
\alpha^{k}  &lt;  7\cdot 136\cdot d &lt; 12\cdot 10^{37},
\end{equation}
leading to $k &lt; 184$.
Using the above argument,  we obtain
$k&lt;184$ in the case when $(u,v) \in \varpi_1$ except $(2,4)$. We omit the details in order to avoid unnecessary repetitions. This completes the analysis of the cases when $(u,v) \in \varpi_1$. Consequently, $k&lt; 196$ always holds.

Suppose now that $\omega_2 &lt; 0$. First, note that $\frac{7}{\alpha^k} &lt; \frac{1}{2}$ since $k &gt; 250$. Then, from \eqref{eq24},
we have that 
$$\vert1 - e^{\omega_2}\vert &lt; \frac{1}{2},$$
thus
 $$\frac{1}{2} &lt; e^{\omega_2} &lt; \frac{3}{2}.$$
  	Therefore 
$$e^{\vert\omega_2\vert} &lt; 2.$$
 Since $\omega_2 &lt; 0$, we have
$$0 &lt; \vert\omega_2\vert \leq e^{\vert\omega_2\vert} -1 = e^{\vert\omega_2\vert}\vert e^{-\vert\omega_2\vert} -1\vert = e^{\vert\omega_2\vert}\vert e^{\omega_2} -1\vert &lt; \frac{6}{\alpha^{k}}.$$
Then we obtain
$$0 &lt; -d\log2 + k\log\alpha + \log(1 + \alpha^{-u}+ \alpha^{-v}) &lt; \frac{6}{\alpha^{k}}.$$
 By the same arguments used for proving \eqref{eq24}, we obtain
 \begin{equation}
\label{eq46}
 0  &lt;  k\left(\frac{\log\alpha}{\log2}\right) - d + \frac{\log(1 + \alpha^{-u}+ \alpha^{-v})}{\log2} &lt; \frac{9}{\alpha^k}.
 \end{equation}
 We now put
$$\gamma := \frac{\log\alpha}{\log2}, \quad \mu :=  \frac{\log(1 + \alpha^{-u} + \alpha^{-v})}{\log2}, \quad A := 9 \quad\hbox{and}\quad B := \alpha.$$
Clearly $\gamma$ is an irrational number. We also put $N := 12\cdot 10^{34}$ , which is an upper bound on $d$ by
Lemma \ref{lem2}. We therefore apply Lemma \ref{lem2} to inequality \eqref{eq46} for all choices of $u \in \lbrace1,\ldots,184\rbrace, v \in \lbrace1,\ldots ,184\rbrace$ except when 
\begin{align*}
(u,v) \in \varpi_2 &amp;= \lbrace (2, 1),(3, 3),(4, 2),(5, 2),(7, 7),
 (8, 2),(9, 5),(9, 6),(10, 2),(11, 4),(12, 1),(13, 1), \\
&amp; (14, 2),(17, 4),(19, 5),(19, 6),(20, 9),(22, 1),(23, 1),(24, 1),(24, 5),(25, 1), (25, 2), \\
&amp; (28, 2),(30, 1),(31, 4),(32, 7),(33, 1),(36, 3),(37, 2), (38, 1),(40, 2),(44, 1),(47, 2), \\
&amp; (48, 1),(48, 6),(50, 1),(50, 2), (51, 9),(53, 1),(54, 2),(56, 2),(57, 2),(58, 5),(58, 6), \\
&amp; (58, 8), (62, 1),(62, 4),(64, 1),(66, 1),(69, 2),(70, 6),(75, 1),(75, 2),
 (76, 1), \\
&amp; (76, 4),(77, 2),(77, 3),(78, 2),(79, 2),(83, 6),(87, 1), (87, 7),(89, 1),(89, 2),(91, 1), \\
&amp; (91, 2),(92, 1),(94, 1),(95, 1), (95, 7),(96, 2),(97, 1),(98, 1),(99, 5),(99, 6),(102, 2), \\
&amp; (102, 3), (112, 2),(113, 1),(113, 5),(113, 6),(120, 1),(122, 1),(122, 5),
 (122, 6), \\
&amp; (123, 2),(129, 2),(129, 3),(130, 4),(130, 5),(134, 7), (135, 2),(138, 1),(139, 1), \\
&amp; (141, 1),(141, 2),(142, 4),(143, 1), (147, 1),(148, 1),(149, 2) \rbrace,
\end{align*}
and get that
$$k &lt; \frac{\log(Aq/\xi)}{\log B},$$
where $q &gt; 6N$ is a denominator of a convergent of the continued fraction of $\gamma$ such that
$\xi = \Vert \mu q\Vert - N\Vert \gamma q\Vert &gt; 0$. Indeed, using  
{\tt Sagemath}, we have hat
$$q = q_{82} = 1234165504911193651820557190855668171489.$$
 We find that if $(k,l,t,d)$ is
a possible solution of the equation \eqref{eq2} with $\omega_2 &lt; 0$ and $ (u,v) \notin \varpi_2$, then $k &lt; 192$. This is false because our assumption is that $k &gt; 250$.
With the same arguments as in the case $\omega_2&gt;0$, when $(u,v) \in \varpi_2$, we obtain that $k&lt;192$.
Thus, Theorem \ref{thm3} is proven.

\section{Acknowledgments}
This work was partially supported by a grant from the Simons
Foundation.  Part of
this work was done during a very enjoyable visit of B. Edjeou 
at AIMS S\'en\'egal.
This author thanks AIMS S\'en\'egal for the hospitality and support. We also thank Bernadette Faye and Eva Goedhart for useful discussions.

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+ F_m + F_l = 2^d$, \textit{Lith. Math. J.} \textbf{55}
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\bibitem{2} J. J. Bravo and F. Luca, On the Diophantine equation $L_n
+ L_m = 2^a$, \textit{J. Integer Sequences} \textbf{17} (2014),
Article $14.8.3$.

\bibitem{3}A. Dujella and A. Peth\H{o}, A generalization of a theorem
of Baker and Davenport, \textit{Quart. J. Math. Oxford Ser. (2)} \textbf{49}
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\bibitem{5}A. C. G. Lomel\'i and S. H. Hern\'andez, Powers of two as
sums of two  Padovan numbers, \textit{INTEGERS} \textbf{18} (2018), A84.

\bibitem{6}  M. Laurent, M. Mignotte, and Y. Nesterenko, Formes lin\'eaires en deux logarithmes et d\'eterminants d'interpolation, \textit{J. Number Theory} \textbf{55} (1995), 285--321.

\bibitem{7}  F. Luca and S. Siksek, Factorials expressible as sums of two and three Fibonacci numbers, \textit{Proc. Edinb. Math. Soc.} \textbf{53} (2010), 747--763.

\bibitem{8} E. M. Matveev, An explicit lower bound for a homogeneous
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\end{thebibliography}


\bigskip
\hrule
\bigskip

\noindent 2010 {\it Mathematics Subject Classification}:
Primary 11B39, Secondary 11J86.

\noindent \emph{Keywords: }
Lucas number,  Diophantine equation, linear form in logarithms.

\bigskip
\hrule
\bigskip

\noindent
(Concerned with sequence
\seqnum{A000032}.)

\bigskip
\hrule
\bigskip

\vspace*{+.1in}
\noindent
Received August 21 2019; 
revised versions received August 9 2020; August 26 2020.
Published in {\it Journal of Integer Sequences}, October 13 2020.

\bigskip
\hrule
\bigskip

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


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