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\begin{center}
\vskip 1cm{\LARGE\bf 
Product of Consecutive Tribonacci Numbers \\
\vskip .13in
With Only One Distinct Digit
}
\vskip 1cm
Eric F. Bravo and Carlos A. G\'omez\\ 
Department of Mathematics\\
Universidad del Valle\\
Calle 13 No 100 -- 00 \\
Cali \\
Colombia \\
\href{mailto: eric.bravo@correounivalle.edu.co}{\tt eric.bravo@correounivalle.edu.co} \\
\href{mailto: carlos.a.gomez@correounivalle.edu.co}{\tt carlos.a.gomez@correounivalle.edu.co} \\
\ \\
Florian Luca\\ 
School of Mathematics\\
University of the Witwatersrand \\ 
Johannesburg \\
South Africa \\
and \\
Research Group in Algebraic Structures and Applications\\
King Abdulaziz University\\
Jeddah \\
Saudi Arabia\\
and\\ 
Department of Mathematics \\
University of Ostrava \\
30 Dubna 22, 701 03 \\
Ostrava 1 \\
Czech Republic \\
\href{mailto: florian.luca@wits.ac.za}{\tt florian.luca@wits.ac.za} \\
\end{center}

\vskip .2 in

\begin{abstract}
Let $\left(F_n\right)_{n\ge0}$ be the sequence of Fibonacci
numbers. Marques and Togb\'e proved that if the product $F_{n}\cdots
F_{n+\ell-1}$ is a repdigit (i.e., a number with only
distinct digit in its decimal expansion), with at least two digits,
then $\left(\ell,n \right) = \left(1,10\right)$. In this paper, we solve
the same problem with Tribonacci numbers instead of Fibonacci numbers.
\end{abstract}

\section{Introduction}

A positive integer is called a \textit{repdigit} if it has only one
distinct digit in its decimal expansion. The sequence of numbers with
repeated digits is included in Sloane's {\it On-Line Encyclopedia of
Integer Sequences} (OEIS) \cite{slone} as sequence 
\seqnum{A010785}. In 2000, Luca
\cite{FL} showed that the largest repdigit Fibonacci number is $F_{10}
= 55$ and the largest repdigit Lucas number is $L_{5} = 11$.

Motivated by the results of Luca \cite{FL}, several authors have explored repdigits in generalizations of Fibonacci numbers and Lucas numbers. For instance, Bravo and Luca \cite{BL} showed that the only repdigit in the $k$-{\it generalized Fibonacci sequence}\footnote{The $k$-{\it generalized Fibonacci sequence} $F^{(k)}$, for an integer $k\ge 2$, satisfies that its first $k$ terms are $0, \ldots, 0, 1$ and each term afterwards is the sum of the preceding $k$ terms. For $k = 2$, this reduces to the familiar {\it Fibonacci numbers}., while for $k = 3$ these are the {\it Tribonacci numbers}.}, is $F^{(3)}_{8} = 44$. On the other hand, Bravo and Luca \cite{BL1} showed that only repdigit in the $k$-{\it generalized Lucas sequence}\footnote{The $k$-{\it generalized Lucas sequence} $L^{(k)}$ satisfies that its first $k$ terms are $0, \ldots, 0, 2, 1$ and each term afterwards is the sum of the preceding $k$ terms. For $k = 2$, this reduces to the familiar {\it Lucas numbers}.}, is $L^{(4)}_{5} = 22$.

Recently, this problem has been extended to study the product of consecutive Fibonacci or Lucas numbers which are repdigits. Marques and Togb\'e \cite{MT} proved that no product of more of two consecutive Fibonacci numbers can be a repdigit with at least two digits, while Irmak and Togb\'e \cite{IT} proved that 77 is the only repdigit with at least two digits appearing as the product of two or more consecutive Lucas numbers.

In this paper, we investigate the presence of repdigits in the product of consecutive Tribonacci numbers. More precisely, we prove the following result.
\begin{theorem}\label{main:thm}
	The only solution of the Diophantine equation
	\begin{equation}\label{main:eqn}
	T_{n}\cdots T_{n+\left(\ell-1\right)}=d\left(\frac{10^{m}-1}{9}\right), \quad {\it in ~positive ~integers} ~~ n, ~\ell,~ m,~ d,
	\end{equation}
	with $1\leq d\leq 9$ and $m\geq 2$ is $\left(n,\ell,m,d\right)=\left(8,1,2,4\right)$; i.e., $T_8 = 44$.
\end{theorem}
To solve equation \eqref{main:eqn}, we use the $2$-adic order of the Tribonacci numbers in order to bound the number of factors $\ell$, then linear forms in logarithms to bound $\max\{m, n\}$ and finally a version of the Baker-Davenport Lemma to reduce such bounds to manageable values and finish off with a few calculations.

\section{Preliminary results}

\subsection{The Tribonacci sequence} We consider ${T}:=\left(T_n\right)_{n \ge -1}$ given by $T_{-1} = T_0 = 0, ~T_1 = 1$ and $T_{n+3} = T_{n+2}+T_{n+1}+T_{n}$, for all $n\ge 0$. Its characteristic equation, $z^{3}-z^{2}-z-1=0$, has one real root $\alpha$ and two complex roots $\beta$ and $\gamma=\bar{\beta}$. In 1982, Spickerman \cite{WS} found the Binet formula for the Tribonacci numbers
\begin{equation}\label{BT}
T_{s}=a\alpha^{s}+b\beta^{s}+c\gamma^{s}, \quad \text{for all} \quad s\geq 0,
\end{equation}
where
\[
a=\frac{1}{\left(\alpha-\beta\right)\left(\alpha-\gamma\right)},\quad b=\frac{1}{\left(\beta-\alpha\right)\left(\beta-\gamma\right)}, \quad c=\frac{1}{\left(\gamma-\alpha\right)\left(\gamma-\beta\right)}=\bar{b}.
\]
It is easy to see that $\alpha\in \left(1.83, 1.84\right), |\beta|=|\gamma| \in \left(0.73, 0.74\right), a \in \left(0.18, 0.19\right)$ and $|b|=|c| \in \left(0.35, 0.36\right)$.
Since $|\beta|=|\gamma|<1$, setting $e_s:=T_{s}-a\alpha^{s}$, we have
\begin{equation}\label{error}
T_{s} = a\alpha^{s}+e_s, \qquad {\rm with} \quad |e_s|<\frac{1}{\alpha^{s/2}} \quad \text{for all} ~~ s\geq 1.
\end{equation}
Further,
\begin{equation}\label{E3}
\alpha^{s-2}\leq T_{s}\leq \alpha^{s-1} \quad \text{for all} \quad s \geq 1 \quad (\text{see \cite{BL}}).
\end{equation}
We recall a result of Marques and Lengyel \cite{ML} on the 2-adic order of a Tribonacci number. For a prime number $p$ and a nonzero integer $r$, the $p$-adic order $\upsilon_{p}(r)$ is the exponent of the highest power of a prime $p$ which divides $r$.
\begin{lemma}\label{orderp}
For $n\geq 1$, we have
$$
\upsilon_{2}\left(T_{n}\right) = \left \{
	\begin{array}{ll}
	0, & \mbox{if }n\equiv 1,2\pmod{4};\\
	1, & \mbox{if }n\equiv 3,11\pmod{16};\\
	2, & \mbox{if }n\equiv 4,8\pmod{16};\\
	3, & \mbox{if }n\equiv 7\pmod{16};\\
	\upsilon_{2}(n)-1, & \mbox{if }n\equiv 0\pmod{16};\\
	\upsilon_{2}(n+4)-1, & \mbox{if }n\equiv 12\pmod{16};\\
	\upsilon_{2}((n+1)(n+17))-3, & \mbox{if }n\equiv 15\pmod{16}.
	\end{array}\right.
$$
\end{lemma}
The Tribonacci sequence is included in the OEIS \cite{slone} as sequence 
\seqnum{A101292}.

\subsection{Linear forms in logarithms} Let $\eta$ be an algebraic number of degree $d$ over $\mathbb{Q}$
with minimal primitive polynomial $f(X) := a_0 \prod_{i=1}^{d}(X-\eta^{(i)}) \in \mathbb{Z}[X],$
where the leading coefficient $a_0$ is positive. The {\it logarithmic height} of $\eta$ is given by
$$
h(\eta) := \dfrac{1}{d}\left(\log a_0 +
\sum_{i=1}^{d}\log\max\{|\eta^{(i)}|,1\}\right).
$$
In particular, if $\eta = p/q \in \mathbb{Q}$ with $ \gcd (p,q) = 1 $ and $ q>0 $, then $ h(\eta) = \log\max\{|p|, q\} $. The following are some of the properties of the logarithmic height function $ h(\cdot) $, which will be used in  this paper: $h(\eta\pm \gamma) \leq h(\eta) +h(\gamma) +\log 2$,
$h(\eta\gamma^{\pm 1}) \leq h(\eta) + h(\gamma)~$ and $~h(\eta^{s}) = |s|h(\eta) ~~ (s\in\mathbb{Z})$.

Many Diophantine problems can be solved by reducing them to an instance in which one can apply lower bounds for linear forms in logarithms of algebraic numbers. We will use the following theorem, which is a variation of a result of Matveev \cite{M}, proved by Bugeaud, Mignotte and Siksek \cite[Theorem 9.1]{BMS}.
\begin{theorem}
Let $\mathbb{L}$ be a number field of degree $d_{\mathbb{L}}$ over $\mathbb{Q},\,\,$  $\eta_1, \ldots, \eta_t$ nonzero elements of $\mathbb{L}$, and $b_1, \ldots,  b_t$ rational integers. Put
$\Lambda := \eta_1^{b_1} \cdots \eta_t^{b_t} - 1$ and $B \geq \max\{|b_1|, \ldots ,|b_t|\}$. Let $A_i \geq \max\{d_{\mathbb{L}}h(\eta_i), |\log \eta_i|, 0.16\}$ be real numbers, for
$i = 1, \ldots, t$. Then, assuming that $\Lambda \not = 0$, we have
%\small
$$
|\Lambda| > \exp(-3 \cdot 30^{t+4} \cdot (t+1)^{5.5} \cdot d_{\mathbb{L}}^2(1 + \log d_{\mathbb{L}})(1 + \log tB)A_1 \cdots A_t).
$$
\end{theorem}

\section{Absolute bounds on the variables}

\subsection{On the number \texorpdfstring{$\ell$}{Lg} of factors} We claim that $\ell\leq 6$. Indeed, from Lemma~\ref{orderp} we have the following table for all possible values of $n\equiv x\pmod{16}$, $x\in \left\{0,1,2,\ldots,15\right\}$.
\begin{table}[H]
\centering
	\begin{tabular}[t]{|l|l|l|}
		\hline
		$\ell$ & $x$ & $\upsilon_{2}(T_{n}T_{n+1}\cdots T_{n+(\ell-1)})$ \\
		\hline
		1 & 15 & $\geq 5$\\
		\hline
		2 & 7, 11, 14 & $\geq 5$, $\geq 4$, $\geq 5$\\
		\hline
		3 & 6, 10, 13 & $\geq 5$, $\geq 4$, $\geq 5$\\
		\hline
		4 & 0, 4, 5, 9, 12 & $\geq 4$, $\geq 5$, $\geq 5$, $\geq 4$, $\geq 8$\\
		\hline
		5 & 3, 8 & $\geq 6$, $\geq 4$\\
		\hline
		6 & 2 & $\geq 6$\\
		\hline
		7 & 1 & $\geq 6$\\
		\hline
	\end{tabular}
	\caption{2-adic order of product of consecutive Tribonacci numbers}
\end{table}
We analyze a pair of cases to illustrate the above table. The remaining cases are similar.
\begin{itemize}
\item $n\equiv 1\pmod{16}$. Here, $n+2\equiv 3\pmod{16}$, $n+3\equiv 4\pmod{16}$ and $n+6\equiv 7\pmod{16}$. So, by Lemma~\ref{orderp} we have $\upsilon_{2}(T_{n+2})=1$, $\upsilon_{2}(T_{n+3})=2$ and $\upsilon_{2}(T_{n+6})=3$. Hence, with $\ell=7$, $\upsilon_{2}(T_{n}T_{n+1}\cdots T_{n+6})\geq 6$.
\item $n\equiv 12\pmod{16}$. Then, from Lemma~\ref{orderp} we have $\upsilon_{2}(T_{n})=\upsilon_{2}(n+4)-1\geq 3$, since $n+4\equiv 0 \pmod{16}$. On other hand, $n+3\equiv 15 \pmod{16}$ and $\upsilon_{2}(T_{n+3})=\upsilon_{2}((n+4)(n+20))-3\geq 5$, where we used Lemma~\ref{orderp} and the fact that $n+20\equiv 0 \pmod{16}$. Therefore, with $\ell=4$, $\upsilon_{2}(T_{n}T_{n+1}T_{n+2}T_{n+3})\geq 8$.
\end{itemize}
Since $\upsilon_{2}\left(d\left(\frac{10^{m}-1}{9}\right)\right) = \upsilon_{2}(d) \leq 3$ for all $1\leq d\leq 9$, it then follows that $\ell\leq 6$.

\subsection{An absolute bound for \texorpdfstring{$m$}{Lg} and \texorpdfstring{$n$}{Lg}} First, assume that $n\geq 20$. Combining \eqref{main:eqn} and \eqref{E3}, we get
\[
10^{m-1}<\alpha^{\ell(n-1) +\frac{\ell(\ell-1)}{2}}.
\]
Thus,
\begin{equation}
\label{cotam}
m<\ell n +\ell(\ell-1)/2.
\end{equation}
Now, by \eqref{error}, we get that
\begin{align*}
T_{n} \cdots T_{n+(\ell-1)}&=(a\alpha^{n}+e_n)\cdots(a\alpha^{n+(\ell-1)}+e_{n+\ell-1})\\
                            &=a^{\ell}\alpha^{\ell n+\ell(\ell-1)/2} + r(a,\alpha,n,\ell)	
\end{align*}
where $r(a,\alpha,n,\ell)$ involves the part of the expansion of the previous line that contains the product of powers of $a,~\alpha$ and the errors $e_i$, for $i=n,\ldots n+(\ell-1)$.
Moreover, $r(a,\alpha,n,\ell)$ is the sum of $63$ terms with maximum absolute value $a^6\alpha^{(\ell-1)n +\ell(\ell-1)/2}\alpha^{-n/2}$.

Combining the above equality with \eqref{main:eqn}, we obtain
\[
a^{\ell}\alpha^{\ell n+\ell(\ell-1)/2}-\frac{d}{9}10^{m}=-\frac{d}{9}-r(a,\alpha,n,\ell).
\]
Dividing both sides of the above equality by $a^{\ell}\alpha^{\ell n+\ell(\ell-1)/2}$ and taking the absolute value, we conclude that
\begin{align}
\label{cota1}
\left|\frac{d}{9a^{\ell}}\alpha^{-(\ell n+\ell(\ell-1)/2)}10^{m}-1\right|&\leq \left(\frac{d}{9}+|r(a,\alpha,n,\ell)|\right)\cdot a^{-\ell}\alpha^{-(\ell n+\ell(\ell-1)/2)}\\
&< (1+63a^{\ell-1}\alpha^{(\ell-1)n+\ell(\ell-1)/2}\alpha^{-n/2})\cdot a^{-\ell}\alpha^{-(\ell n+\ell(\ell-1)/2)}\nonumber\\
&\leq 64a^{-1}\alpha^{-3n/2},\nonumber
\end{align}
Below, we use Matveev's theorem to find a lower bound for the left-hand side of \eqref{cota1}, with the parameters
\[
t:=3, ~~ (\eta_{1}, b_1):=((d/9)a^{-\ell},1), ~~ (\eta_{2},b_2):=(\alpha,-(\ell n+\ell(\ell-1)/2))\quad {\rm and} \quad(\eta_{3},b_3):=(10,m).
\]
The number field containing $\eta_{1},\eta_{2},\eta_{3}$ is $\mathbb{L}:=\mathbb{Q}(\alpha,\beta)$, which has $d_{\mathbb{L}}:=6$. We claim that $\Lambda:=\eta_{1}^{b_{1}}\eta_{2}^{b_{2}}\eta_{3}^{b_{3}}-1 \neq 0$. Otherwise, we get
\[
a^{\ell}\alpha^{\ell n+\ell(\ell-1)/2}=d\cdot10^{m}/9.
\]
Conjugating the above relation by the automorphism $\sigma : \alpha \rightarrow \beta,\beta \rightarrow \alpha, \gamma \rightarrow \gamma$ (here, we use the fact that the Galois group of $\mathbb{L}$ over $\mathbb{Q}$ is isomorphic to $S_{3}$), and then taking absolute values on both sides of the resulting equality, we obtain
\[
|b|^{\ell}|\beta|^{\ell n+\ell(\ell-1)/2}=d\cdot10^{m}/9,
\]
which is not possible because $|b|^{\ell}|\beta|^{\ell n+\ell(\ell-1)/2}<1$ and $d\cdot10^{m}/9>10$. Thus, $\Lambda\neq 0$. Next, $h(\eta_{1})\leq h(d)+h(9a^{\ell})\leq \log 9+h(9)+\ell h(a)$, $h(\eta_{2})=\frac{1}{3}\log \alpha$ and $h(\eta_{3})=\log 10$. Now we need to estimate $h(a)$. For it, the minimal polynomial of $a$ is $44X^{3}+4X-1$. So, $h(a)=\frac{1}{3}\log 44$ and $h(\eta_{1})\leq 2\log 9+2\log 44$. Thus, we can take $A_{1}:=72$, $A_{2}:=2$ and $A_{3}:=14$. According to \eqref{cotam}, we take $B:=\ell n +\ell(\ell-1)/2$. Applying Matveev's theorem we get a lower bound for $|\Lambda|$, which by comparing it to \eqref{cota1}, leads to
\[
\exp\left(-2.72251\times 10^{19}(1+\log 3(\ell n+\ell(\ell-1)/2))\right)<\frac{356}{\alpha^{3n/2}}.
\]
Taking logarithms in the above inequality, we get
\begin{align*}
\frac{3n}{2}\log \alpha -\log 356&<2.72251\times 10^{19}(1+\log3(6n+15))\\
&<5.44503\times 10^{19}\log (6n+15),
\end{align*}
where we used the fact that $1+\log 3h<2\log h$, for all $h \geq 9$. Hence,
\[
n<6.1\times 10^{19}\log (6n+15).
\]
Therefore, we obtain $n<3.1\times 10^{21}$ and record what we have proved so far.
\begin{lemma}\label{redn}
If $(n,\ell,m,d)$ is a positive integer solution of \eqref{main:eqn} with $n\geq 20$, $m\geq 2$ and $1\leq d\leq 9$, then $1\leq \ell \leq 6$ and
	\[
	\max\{m, n\} < 3.1\times 10^{21}.
	\]
\end{lemma}

\section{Reducing \texorpdfstring{$\max\{m, n\}$}{Lg}}

To lower the bound of $n$, we will use the following result of diophantine approximation, see \cite{BD}.
\begin{lemma}\label{bd}
	Let $\kappa$ be an irrational number, $M$ be a positive integer, and $p/q$ be a convergent of the continued fraction of $\kappa$ such that $q>6M$. Let $A,B,\mu$ be some real numbers with $A>0$ and $B>1$. Let $\epsilon:=\left\|\mu q\right\|-M\left\|\kappa q\right\|$, where $\left\|\cdot \right\|$ denotes the distance from the nearest integer. If $\epsilon>0$, then there is no solution of the inequality
	\[
	0<|r\kappa-s+\mu|<AB^{-w}
	\]
	in positive integers $r,s$ and $w$ with $r\leq M$ and $w\geq \log (Aq/\epsilon)/\log B$.
\end{lemma}
Let $\Gamma :=m\log 10-(\ell n+\ell(\ell-1)/2)\log \alpha + \log (d/9a^{\ell})$. Therefore, \eqref{cota1} can be rewritten as $|e^{\Gamma}-1| < 356/\alpha^{3n/2}$. Since $|e^{\Gamma}-1|<1/2$ for all $n\geq 20$ (because $365/\alpha^{3n/2}<1/2$), it follows that $e^{|\Gamma|}<2$ and so $0<|\Gamma|\leq e^{|\Gamma|}-1=e^{|\Gamma|}|e^{\Gamma}-1| < 712/\alpha^{3n/2}$. Hence, $0 < |\Gamma| < 712/\alpha^{3n/2}$ holds for $n \geq 20$.

Replacing $\Gamma$ in the above inequality and dividing both sides of the resulting inequality by $\log \alpha$, we obtain
\begin{equation}\label{dujella}
0<|m\kappa-(\ell n+\ell(\ell-1)/2)+\mu|<1180(\alpha^{3/2})^{-n},
\end{equation}
with $\kappa:=\log 10/\log \alpha$ and $\mu:=\log (d/9a^{\ell})/\log \alpha$. Here, we took $M:=1.9\times 10^{22}$, which is an upper bound on $m$ by Lemma~\ref{redn}, and we applied Lemma~\ref{bd} to inequality \eqref{dujella} for each $1\leq \ell \leq 6$ and $1\leq d\leq 9$. With the help of the
computer algebra system \texttt{Mathematica}, we found that
\[
q_{47}=938425170962281070635339>6M
\]
is a denominator of a convergent of the continued fraction of $\kappa$ such that the minimum value of $\epsilon:=\left\|\mu q_{47}\right\|-M\left\|\kappa q_{47}\right\|$ is greater than $0.00437116$. The conditions of Lemma \ref{bd} are fulfilled for $A:=1180$ and $B:=\alpha^{3/2}$. Then, there are no solutions of \eqref{main:eqn} on the interval
\[
\left[\left\lfloor \dfrac{\log(1180q_{47}/\epsilon)}{\log B}\right\rfloor+1,M\right)=[75,1.9\times 10^{22}).
\]
So, $m\leq 459$. Now, we start again the entire process using this much smaller bound of $m$. In this application of Lemma \ref{bd} we found that the assumption $n\geq 20$ implies $n\leq 26$. Thus, $n\leq 26$ holds. Hence, it remains to check equation \eqref{main:eqn} for $1\leq n \leq 26$, $1\leq \ell \leq 6$, $2\leq m \leq 171$ and $1\leq d \leq 9$. For this, we use \textit{Mathematica} and conclude that the quadruple $(n,\ell,m,d)=(8,1,2,4)$ is the only solution of the diophantine equation \eqref{main:eqn}. This completes the proof of Theorem \ref{main:thm}.

\section{Acknowledgments}

The authors are grateful to the referee for the useful comments to improve of this paper. E. F. B thanks Colciencias for support during his Ph.D. studies. F. L. was supported in part  by NRF (South Africa) Grant CPRR160325161141, RTNUM18 from CoEMaSS at Wits and by CGA (Czech Republic) Grant 17-02804S.

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\bibitem{IT} N. Irmak and A. Togb\'e, On repdigits as product of
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\bibitem{FL} F. Luca, Fibonacci and Lucas numbers with only one distinct
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one distinct digit, {\it Utilitas Math.} {\bf 98} (2015), 23--31.

\bibitem{ML} D. Marques and T. Lengyel, The 2-adic order of the Tribonacci
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\href{https://cs.uwaterloo.ca/journals/JIS/VOL17/Lengyel/lengyel21.html}{Article 14.10.1}.

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\bibitem{M} E. M. Matveev, An explicit lower bound for a homogeneous
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\bibitem{slone} N. J. A. Sloane et al., {\em The On-Line Encyclopedia of
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\end{thebibliography}

\bigskip
\hrule
\bigskip

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

\noindent \emph{Keywords: } 
Tribonacci number, lower bound, linear form in logarithms of algebraic numbers, repdigit.

\bigskip
\hrule
\bigskip

\vspace*{+.1in}
\noindent
Received January 10 2019;
revised versions received January 11 2019; August 14 2019.
Published in {\it Journal of Integer Sequences}, August 24 2019.

\bigskip
\hrule
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\noindent
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\htmladdnormallink{Journal of Integer Sequences home page}{http://www.cs.uwaterloo.ca/journals/JIS/}.
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