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\begin{center}
\vskip 1cm{\LARGE\bf 
Repdigits as Sums of Two Padovan Numbers
}
\vskip 1cm
\large
Ana Cecilia Garc\'ia Lomel\'i and
Santos Hern\'andez Hern\'andez \\
Unidad Acad\'emica de Matem\'aticas\\
Universidad Aut\'onoma de Zacatecas, Campus II\\
Calzada Solidaridad entronque Paseo a la Bufa\\
C.P. 98000\\
Zacatecas, Zac.\\
Mexico\\
\href{mailto:aceciliagarcia.lomeli@gmail.com}{\tt aceciliagarcia.lomeli@gmail.com} \\
\href{mailto:shh@uaz.edu.mx}{\tt shh@uaz.edu.mx}
\end{center}

\vskip .2 in


\begin{abstract}
In this paper, we determine all repdigits in base 10 that can be written as a sum of two 
Padovan numbers.
\end{abstract}




\section{Introduction}
We recall that a \textit{repdigit} is a positive integer $r$ that
has only one distinct 
digit when written in base 10. That is, $r$ is of the form
$
d\left(\frac{10^\ell-1}{9}\right)
$
for some positive integers $d$, $\ell$ with $1\leq d\leq 9$. 

Recently, the problem of looking for repdigits in a linear recurrence sequence has been studied. 
For example, Luca \cite{luca} and Marques~\cite{marques} prove that $55$ and $44$ 
are the largest repdigits in the Fibonacci and Tribonacci sequences, respectively. More generally, let $k\geq 2$ be an integer. 
Let $\mathcal{F}^{(k)}:=(F_n^{(k)})_{n\geq 0}$ be the $k$-\textit{Fibonacci sequence} given by
$$F_0^{(k)}=\cdots=F_{k-2}^{(k)}=0, \quad F_{k-1}^{(k)}=1$$ 
and the recurrence formula
$$
F_{n+k}^{(k)}=F_{n+k-1}^{(k)}+F_{n+k-2}^{(k)}+\cdots +F_n^{(k)}\quad\text{for all}\quad n\geq 0.
$$
Observe that the cases $k=2$, $3$ are the Fibonacci and Tribonacci sequences, respectively.
Bravo and Luca~\cite{bravoluca4} extended the results mentioned
above by showing that the only repdigits 
with at least two digits in a $k$-Fibonacci sequence are, for $k=2,3$ the aforementioned ones, 
while for $k>3$ there are no such repdigits belonging to $\mathcal{F}^{(k)}$. 
Actually, Bravo and Luca ~\cite{bravoluca} considered and solved the more general Diophantine equation
\begin{equation}\label{bl}
F_n^{(k)}+ F_{m}^{(k)} =d\left(\frac{10^\ell-1}{9}\right),
\end{equation}
in positive integers $k\geq 2$, $n\geq m$, $\ell\geq 2$, and  $1\leq d\leq 9$. From their result
we see that, for all $k$ there are at most 4 solutions and the largest one is $777$, which is found in $\mathcal{F}^{(4)}$.
Motivated by these results, in this note we study the similar problem with the Padovan sequence.

The \emph{Padovan sequence} $(P_n)_{n\geq 0}$, named  after the architect  R. Padovan, is the ternary recurrence sequence 
given by  $P_0=0$, $P_1=P_2=1$ and the recurrence  formula
\begin{equation}\label{padovanrecurrence}
P_{n+3}=P_{n+1} +P_n,\quad\text{for all}\quad n\geq 0.
\end{equation}
This is the sequence \seqnum{A000931} in~\textit{Sloane's Encyclopedia}~\cite{sloane} and it was introduced by Stewart~\cite{stewart}, 
where he relates it to the Fibonacci sequence. Its first few terms are
\[
0, 1, 1, 1, 2, 2, 3, 4, 5, 7, 9, 12, 16, 21, 28, 37, 49, 65, 86, 114,\ldots
\]
In this note we study the problem of writing repdigits as sums of two Padovan numbers. More precisely, we solve the Diophantine 
equation
\begin{equation}\label{princiequation}
P_n+P_{m}= d\left(\frac{10^\ell-1}{9}\right),
\end{equation}
in non-negative integers $n, m, \ell, d$ with $n\geq m$, $\ell\geq 2$ and $1\leq d\leq 9$. 
To establish our result we make the following conventions. Since $P_1=P_2=P_3=1$ and $P_4=P_5=2$ we will assume that $n, m\neq 1, 2,4$. 
That is, whenever we think of $1$ and $2$ as members of the Padovan sequence, we think of them as being $P_3$ and $P_5$, respectively. 
Thus, our result is
\begin{theorem}\label{princitma}
All non-negative integer solutions $(n,m,\ell,d)$ of equation~\eqref{princiequation} 
with $n\geq m$, $\ell\geq 2$ and $1\leq d \leq 9$ belong to the set
$$
\left\{
\begin{array}{cccc}
(10,5,2,1), &(9,7,2,1), &(13,3,2,2), &(14,8,2,3)\\
(13,11,2,3), &(15,9,2,4), &(14,12,2,4), &(17,3,2,6)\\
(17,11,2,7),&(16,14,2,7), &(18,5,2,8), &(31,7,4,3) 
\end{array}
\right\}.
$$
The set of repdigits $r$ that can be written as sums of two Padovan numbers is
$$
\{11, 22, 33, 44, 66, 77, 88, 3333\},
$$
and their respective representations are
\begin{align*}
11&= P_{10}+P_5 =P_9+P_7;\\
22&= P_{13}+P_3;\\
33&= P_{14}+P_8=P_{13}+P_{11};\\
44&= P_{15}+P_9=P_{14}+P_{12};\\
66&= P_{17}+P_3;\\
77&= P_{17}+P_{11}=P_{16}+P_{14};\\
88&= P_{18}+P_5;\\
3333&= P_{31}+P_7.
\end{align*}
\end{theorem}

In particular, we obtain that in the Padovan sequence there are no repdigits with more than one digit.

We also note that the problem of writing repdigits as sum of at most three Fibonacci numbers has been solved by Luca~\cite{luca11}. 
The problem of writing Fibonacci numbers as sum of two repdigits has been studied by Alvarado and Luca~\cite{diazluca}.

The method of proof of our result is the classical one with linear forms in logarithms and the reduction method of Baker-Davenport
as used in~\cite{bravogomezluca,bravoluca,luca11}, for example. 

\section{Tools} 
In this section, we gather the tools we need to prove Theorem~\ref{princitma}.
Let $\alpha$ be an algebraic number of degree $d$, let $a>0$ be the leading coefficient of its minimal polynomial over $\mathbb Z$ and 
let $\alpha=\alpha^{(1)} ,\ldots, \alpha^{(d)}$ denote its conjugates. The \textit{logarithmic height} of $\alpha$  is defined by
$$
h(\alpha)=\frac{1}{d} \left(\log a + \sum_{i=1}^{d}\log\max\left\{|\alpha^{(i)}|,1\right\}\right).
$$
This height has the following  basic properties. For $\alpha, \beta$ algebraic numbers
and $m\in\mathbb Z$ we have
\begin{enumerate}
\item[$\bullet$] $h(\alpha+\beta)\leq h(\alpha)+h(\beta)+\log 2$;
\item[$\bullet$] $h(\alpha \beta)\leq h(\alpha)+h(\beta)$;
\item[$\bullet$] $h(\alpha ^m)= |m|h(\alpha)$.
\end{enumerate}
Now let $\mathbb L$ be a real number field of degree $d_{\mathbb L}$, 
$\alpha_1 ,\ldots ,\alpha_s\in\mathbb L$ and $b_1,\ldots,
b_s\in\mathbb Z\setminus \{0\}$. Let  $B\geq \max \{|b_1|,\ldots,|b_s|\}$
and
$$
\Lambda=\alpha_1^{b_1} \cdots \alpha_s ^{b_s} -1.
$$
Let $A_1,\ldots, A_s$ be real numbers with 
$$
A_i \geq \max\{d_{\mathbb L}\,h(\alpha_i), |\log \alpha_i |,0.16\}, ~\qquad i=1,2,\ldots, s.
$$
The first tool we need is the following result due to
Matveev~\cite{matveev}; also see Bugeaud, Mignotte and
Siksek~\cite[Theorem 9.4]{bumisik}.

\begin{theorem} \label{Matveev}
Assume that $\Lambda\neq 0$. Then 
$$
\log |\Lambda|>-1.4\cdot 30^{s+3} \cdot s^{4.5} \cdot d_{\mathbb L}^2 \cdot(1+\log 
d_{\mathbb L})\cdot
(1+\log B) A_1 \cdots A_s.
$$
\end{theorem}

In this note, we always use $s=3$.
Further, $\mathbb L=\mathbb Q(\gamma)$,
where $\gamma$ is defined at the beginning of Section~\ref{sectionproof},
has degree $d_{\mathbb L}=3$. Thus, once and for all we fix the constant
$$
C:= 2.70444\times 10^{12}>1.4\cdot 30^{3+3}\cdot 3^{4.5}\cdot 3^2 \cdot(1+\log 3).
$$

Our second tool is a version of the reduction method of Baker and Davenport~\cite{bada}  based on 
their Lemma. We use the one given by Bravo, G\'omez and 
Luca ~\cite{bravogomezluca}. Also see Dujella and Peth\H o~\cite{dujellapetho}.
For a real number $x$, we write
$\|x\|$ for the distance from $x$ to the nearest integer.

\begin{lemma}\label{reductionlemma}
Let $M$ be a positive integer. Let $\tau, \mu, A>0, B>1 $ be given real numbers.
Assume that $p/q$ is a convergent of $\tau$ such that $q>6M$ and $\varepsilon :=\|\mu\, q\|-M\|\tau\, q\|>0$.
If $(n,m,w)$ is a positive solution to the inequality
$$
0<|n\,\tau-m+\mu|<\frac{A}{B^w}
$$
with $n\leq M$, then
$$
w< \frac{\log(Aq/\varepsilon)}{\log B}.
$$
\end{lemma}

Finally, the following result of Guzm\'an and Luca~\cite{guzmanluca} will be very useful. 
\begin{lemma}\label{toboundn}
If $m\geq 1$, $T>(4m^2)^m$ and $T>x/(\log x)^m$. Then 
$$
x<2^m T(\log T)^m.
$$
\end{lemma}  








\section{Proof of Theorem~\ref{princitma}}\label{sectionproof} 
To start with, we recall some properties of the Padovan sequence. For a complex number
$z$ we write $\overline{z}$ for its complex conjugate. Let $\omega\neq 1$ be a cubic root of 1. 
Let us consider the $\mathbb Q$-irreducible polynomial $X^3-X-1$. It is clear that its zeros are $\gamma, \delta, \overline{\delta}$ 
where
$$
\gamma:=\sqrt[3]{\frac{9+\sqrt{69}}{18}}+\sqrt[3]{\frac{9-\sqrt{69}}{18}},\quad
\delta:=\omega\sqrt[3]{\frac{9+\sqrt{69}}{18}}+\overline{\omega}\sqrt[3]{\frac{9-\sqrt{69}}{18}}.
$$
It can be proved, by induction for example, the Binet formula
\begin{equation}\label{Binetformula}
P_n=c_1 \gamma^n+c_2\delta^n+c_3\overline{\delta}^n \quad\text{for all}\quad n\geq 0,
\end{equation}
where
$$
c_1=\frac{\gamma(\gamma+1)}{2\gamma+3},\qquad c_2=\frac{\delta(\delta+1)}{2\delta+3},\qquad c_3=\overline{c_2}.
$$
Formula~\eqref{Binetformula} follows from the general theorem on linear recurrence sequences
since the above polynomial is the characteristic polynomial of the Padovan sequence. We note that
$$
\gamma=1.32471\ldots,\, |\delta|=0.86883\ldots,\, c_1=0.54511\ldots,\, |c_2|=0.28241\ldots
$$
Further, the inequality
\begin{equation}\label{cotpn}
\gamma^{n-3}\leq P_n\leq \gamma^{n-1}
\end{equation}
holds for all $n\geq 1$. This can also be proved by induction.

Now we start the study of equation~\eqref{princiequation} in non-negative 
integers $(n,m,\ell,d)$
where, as we have said, $n, m\neq 1,2,4$ and $\ell\geq 2$. Assume that $n\geq m$. Since 
$\ell\geq 2$ we assume that
$n\geq 1$. From~\eqref{cotpn} and~\eqref{princiequation}, we obtain 
$$
\gamma^{n-3}\leq P_n \leq P_n+P_m =d\left(\frac{10^\ell-1}{9}\right)\leq 10^\ell,
$$
and
$$
\gamma^{n+2}\geq 2P_n\geq P_n+P_m = d\left(\frac{10^\ell-1}{9}\right)\geq 10^{\ell-1},
$$
where we use $\gamma^3> 2$.
Thus,
\begin{equation}\label{disiprinci}
(n-3)\frac{\log \gamma}{\log 10}\leq \ell  \quad \text{and}\quad
(n+2)\frac{\log \gamma}{\log 10}\geq \ell-1.
\end{equation}
Since $\log \gamma/\log 10 = 0.122123\ldots$ we have, from~\eqref{disiprinci}, that if $n\leq
 360$ then $\ell\leq 45$. 
Running a \emph{Mathematica} program  in the range $0\leq m\leq n\leq 360$, 
$1\leq d\leq 9$ and $2\leq \ell\leq 45$ we obtain, with our conventions,  all solutions listed 
in Theorem~\ref{princitma}. We will prove that these are all of them.

From now on, we assume $n>360$. From~\eqref{disiprinci} we get that $\ell> 43$ and also 
that $n> \ell$.
From the Binet formula~\eqref{Binetformula}, we rewrite  our equation~\eqref{princiequation} and obtain 
$$
\left|c_1 \gamma^n-\frac{d 10^\ell}{9}\right|< 2|c_2|+\gamma^{m -1}+1<\gamma^2 +\gamma^{m -1}<\gamma^{m +5}.
$$    
Dividing through by $d10^\ell/9$ we get
\begin{equation}\label{l}
\left|\frac{9c_1}{d}\gamma^n 10^{-\ell}-1\right|< \frac{9\gamma^{m +5}}{d10^\ell}<\frac{1}{\gamma^{n- m -16}},
\end{equation}
where we use $\gamma^{n-3}\leq 10^\ell$ and  $9<\gamma^8$. Let $\Lambda$ be the expression inside the absolute value 
in the left-hand side of \eqref{l}.
Observe that $\Lambda\neq 0$. To see this, we consider the $\mathbb Q$-automorphism 
$\sigma$ of the Galois extension $\mathbb Q(\gamma, \delta)$ 
over $\mathbb Q$ given by $\sigma(\gamma):=\delta$ and $\sigma(\delta):=\gamma$. 
Now, if $\Lambda=0$,
then $\sigma(\Lambda)=0$. Thus,
$$
10^{\ell-1}\leq \frac{d10^\ell}{9}=\left|\sigma\left(c_1\gamma^n\right)\right|=\left|c_2\right|\left|\delta\right|^n< |c_2|
<\frac{1}{3},
$$ 
which is a contradiction since $\ell>43$. 
Hence, $\Lambda \neq 0$ and we apply Matveev's inequality to it. To do this we consider,
$$
\alpha_1=\frac{9 c_1}{d},\, \alpha_2=\gamma, \,\alpha_3=10, \quad b_1=1, \, b_2=n, \, b_3=-\ell.
$$
Thus, $B=n$. We have that $h(\alpha_2)=\log \gamma/3$ and $h(\alpha_3)=\log 10$. For $\alpha_1$ we use the height properties
to conclude that 
$$
h(\alpha_1)\leq \log \gamma+13\log 2.
$$
So, we choose $A_1=27.9$, $A_2 =0.3$ and $A_3=7$. From Matveev's inequality we obtain
$$
\log |\Lambda|>-C(1+\log n)\cdot 27.9\cdot 0.3\cdot 7, 
$$
which, combined  with~\eqref{l} gives us
\begin{equation}\label{cotann1}
(n-m)\log \gamma<1.58454\times 10^{14}(1+\log n).
\end{equation}

Now we find an upper bound on $n$. To do this, from the Binet formula ~\eqref{Binetformula} 
we rewrite our equation~\eqref{princiequation} as 
$$
\left|c_1\left(\gamma^{n-m}+1\right)\gamma^{m}-\frac{d 10^\ell}{9}\right|\leq 4|c_2|+1<\gamma^3.
$$      
Dividing through by $d10^\ell/9$ we obtain
\begin{equation}\label{1}
\left|\frac{9c_1(\gamma^{n-m}+1)}{d}\gamma^{m} 10^{-\ell}-1\right|< \frac{9\gamma^{3}}{d 10^\ell}< \frac{1}{\gamma^{n-14}}.
\end{equation}
Let $\Lambda_1$ be the expression inside
the absolute value in the left-hand side of~\eqref{1}. 
Note that with a similar argument as the above one, it can be proved that $\Lambda_1\neq 0$. So, we apply Matveev's inequality to it.  
We consider 
$$
\alpha_1=\frac{9c_1(\gamma^{n-m}+1)}{d}, \, \alpha_2=\gamma,\, \alpha_3=10, \quad b_1=1, \,
b_2=m, \, b_3=-\ell.
$$
Thus, $B=n$. The heights of $\alpha_2$ and $\alpha_3$ have already been 
calculated. From the properties of the heights
we get that
$$
h(\alpha_1)\leq \log\gamma+14\log 2+ (n-m)\frac{\log\gamma}{3}
<\frac{1.58455\times 10^{14}(1+\log n)}{3},
$$
where we have used \eqref{cotann1}.
We choose $A_1=1.58455\times 10^{14}(1+\log n)$, $A_2$ and $A_3$ as above. Therefore, from Theorem~\ref{Matveev} we obtain
$$
\log |\Lambda|>-C(1+\log n)\cdot\left( 1.58455\times 10^{14}(1+\log n)\right)\cdot 0.3\cdot 7,
$$
which,
combined with~\eqref{1} gives us $n<1.28011\times 10^{28}(\log n)^2$.
Now, from Lemma~\ref{toboundn} we get
\begin{equation}\label{theboundonn}
n<2.14474\times 10^{32}.
\end{equation}
\

Now we will reduce the above bound on $n$. To do this, we first consider 
$$
\Gamma=\ell\log 10-n\log \gamma+\log\left(\frac{d}{9c_1}\right),\quad 1\leq d\leq 9.
$$
Assume $n-m\geq 20$ and go to ~\eqref{l}. Note that $e^{-\Gamma}-1=\Lambda\neq 0$.
Thus, $\Gamma\neq 0$. If $\Gamma<0$ then
$$
0<|\Gamma|<e^{|\Gamma|}-1=|\Lambda|<\frac{1}{\gamma^{n-m -16}}.
$$
If $\Gamma >0$ we have that $1-e^{-\Gamma}=|e^{-\Gamma}-1|<1/2$. Hence, 
$e^{\Gamma}<2$. Thus, we get 
$$
0<\Gamma<e^{\Gamma}-1=e^{\Gamma}|\Lambda|<\frac{2}{\gamma^{n-m -16}}.
$$
So, in both cases we have 
$$
0<|\Gamma|<\frac{2}{\gamma^{n-m -16}}.
$$
Dividing through by $\log\gamma$ we get
\begin{equation}\label{2}
|\ell\,\tau-n+\mu_d|<\frac{640}{\gamma^{n-m}},
\end{equation}
where 
$$
\tau:=\frac{\log 10}{\log \gamma}\quad \text{and}\quad \mu_d:=\frac{\log(d/9c_1)}
{\log\gamma}, \quad 1\leq d\leq 9.
$$
Now we apply Lemma~\ref{reductionlemma}. Put $M:=2.14474\times 10^{32}$. Since $\ell<n$, from \eqref{theboundonn} we have that 
$M$ is the upper bound on $\ell$. A quick computation with \emph{Mathematica} reveals that the
convergent 
\[
\frac{p_{81}}{q_{81}}=\frac{7680712497306530625416042950269134710}{
937994904644870281019845134758679913}
\]
of $\tau$ is such that  $q_{81}>6M$ and $\varepsilon_d\geq 0.0440913>0$. Thus, with $A:=640$, $B:=\gamma$ we calculated
each value $\log(q_{81} 640/\varepsilon_d)/\log \gamma$ and found that all of them are at most $328$. Thus we have that
$$
n-m\leq 328.
$$
Next, we consider 
$$
\Gamma_1=\ell\log 10-m \log \gamma+\log\left(\frac{d}{9c_1(\gamma^{n-m} + 1)}\right),
\quad 1\leq d\leq 9,
$$
and go to inequality~\eqref{1}.  
Note that $e^{-\Gamma_1}-1=\Lambda_1\neq 0$. Thus, $\Gamma_1\neq 0$.
Since $n>360$, with an argument similar to the above one 
we get 
$$
0<|\Gamma_1|<\frac{2}{\gamma^{n-14}}.
$$
Dividing through by $\log \gamma$ the we obtain
\begin{equation}\label{3}
\left|\ell\,\tau-m+\mu_d\right|<\frac{365}{\gamma^n},
\end{equation}
where $\tau$ is as above, and
$$ 
\mu_d:=\frac{\log(d/9c_1(\gamma^{n-m} + 1))}
{\log \gamma} \quad \text{for}\quad 1\leq d\leq 9.
$$
Consider 
$$ 
\mu_{d,k}=\frac{\log(d/9c_1(\gamma^k+1))}{\log\gamma}, \qquad d=1,\ldots, 9, \qquad
k=1,\ldots, 328.
$$
Note that $m\neq 0$, since if it is not the case we would have $n\leq 328$, which contradicts $n>360$.
Thus, we apply Lemma~\ref{reductionlemma} again. A quick computation with \emph{Mathematica} gives that the $81$-th convergent
of $\tau$ again well works, that is, it is such that $q_{81}>6M$ and $\varepsilon_{d,k}\geq 0.0000152967>0$ 
for all $d=1,\ldots,9$, $k=1,\ldots,328$
except for the case $\varepsilon_{9,11}$, which is always negative. Thus, for all of these values we calculated 
$\log(q_{81}\,365/\varepsilon_{d,k})$ and we find that all of them are at most $354$. Thus,
$n\leq 354$ for all of these values.

The problem in case of the pair $(9,11)$ is that we have the identity
\[
\frac{2\gamma+3}{\gamma(\gamma+1)(\gamma^{11}+1)}=\frac{1}{\gamma^9}.
\]
Thus, inequality~\eqref{3} is
\[
\left|\ell\,\tau-(m+9)\right|<\frac{365}{\gamma^n}
\] 
and we use the theory of continued fractions to study it. Recall, for example, that if
\[
\left|\alpha-\frac{p}{q}\right|<\frac{1}{2q^2}
\]
then $p/q$ is a convergent of $\alpha$. See~\cite[Thm.\ 184, p.\ 153]{hardywright}. This result is known as Legendre's theorem. 
Since $n>360$ and $M>\ell$, we have $730 \ell<\gamma^n$.
From Legendre's theorem, this implies that $(m+9)/\ell$ is a $i$-th convergent of $\tau$. Let $[a_0,a_1,a_2,\ldots]=[8,5,3,\ldots]$ be 
the continued fraction expression of $\tau$. A quick computation with \emph{Mathematica} gives that
$$
q_{71}\leq M < q_{72} \quad \text{and}\quad
b:=\max\{a_j : j=1,2,\ldots, 72\}=49.
$$
In particular, we have that $b\geq a_{i+1}$. Then, from the properties of the convergents of a continued fraction we obtain
$$
\frac{1}{(b+2)\ell}<\left|\ell\,\tau-(m+9)\right|<\frac{365}{\gamma^n},
$$
which yields 
$$
\gamma^n<365\cdot 51\cdot M.
$$
Thus $n\leq 299$. So, together with the results above,
this shows that in 
all cases we have $n\leq 354$,
which contradicts what we assumed about $n$. 
This finishes the
proof of Theorem~\ref{princitma}.

\section{Acknowledgments} 
We wish to express our gratitude to the anonymous referee for careful reading and for the numerous suggestions and
remarks that improved the presentation of this work.
The first author was supported by a CONACyT Doctoral Fellowship and partly supported by
Fundaci\'on Kovalevskaia de la Sociedad Matem\'atica Mexicana.
 

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\end{thebibliography}


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\noindent 2010 {\it Mathematics Subject Classification}:
Primary 11J86; Secondary 11D61, 11D45. 

\noindent \emph{Keywords: } repdigit, Padovan sequence, linear form
in logarithms, reduction method.


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\noindent (Concerned with sequence
\seqnum{A000931}.)

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\vspace*{+.1in}
\noindent
Received September 28 2018;
revised version received  October 8 2018; January 11 2019.
Published in {\it Journal of Integer Sequences}, February 23 2019.

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