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\begin{document}

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\theoremstyle{plain}
\newtheorem{theorem}{Theorem}
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\begin{center}
\vskip 1cm{\LARGE\bf 
Tribonacci Numbers and the \\
\vskip .11in
Brocard-Ramanujan Equation
}
\vskip 1cm
\large
Vin\' icius Fac\' o and Diego Marques \\
Departamento de Matem\' atica\\
Universidade  de Bras\' ilia\\
Bras\' ilia 70910-900 \\
Brazil\\
\href{mailto:vinicius@mat.unb.br}{\tt vinicius@mat.unb.br}\\
\href{mailto:diego@mat.unb.br}{\tt diego@mat.unb.br}\\
\end{center}

\vskip .2 in

\begin{abstract}
Let $(T_n)_{n \geq 0}$ be the Tribonacci sequence defined by the
recurrence $T_{n+2}=T_{n+1}+T_n+T_{n-1}$, with $T_0=0$ and $T_1=T_2=1$.
In this short note, we prove that there are no integer solutions $(u,m)$
to the Brocard-Ramanujan equation $m!+1=u^2$ where $u$ is a Tribonacci
number.
\end{abstract}



\section{Introduction}

In the past few years, several authors have considered Diophantine equations involving factorial numbers. For instance, 
Erd\H{o}s and Selfridge \cite{erdos} proved that $n!$ is a perfect power only when $n=1$. However, the most famous among such equations was posed by Brocard \cite{broc} in 1876 and 
independently by Ramanujan (\cite{ram}, \cite[p.\ 327]{ram2}) in 1913.
The Diophantine equation
\begin{equation}\label{1}
m!+1=u^2
\end{equation}
is now known as {\it Brocard-Ramanujan Diophantine equation}.

It is a simple matter to find the three known solutions, namely 
$m=4,\ 5$ and $7$. Recently, Berndt and Galway \cite{berndt} showed
that there are no further solutions with $m \leq 10^9$. An interesting
contribution to the problem is due to Overholt \cite{over}, who showed
that the equation \eqref{1} has only finitely many solutions if we
assume a weak version of the abc conjecture. However, the
Brocard-Ramanujan equation is still an open problem.

Let $(F_n)_{n\geq 0}$ be the \textit{Fibonacci sequence} (sequence 
\seqnum{A000045} in the OEIS \cite{oeis}) given by $F_0=0,\ F_1=1$ and $F_{n+2}=F_{n+1}+F_n$, for $n\geq 0$.

A number of mathematicians have been interested in Diophantine equations that involve both factorial and Fibonacci numbers. For example, Grossman and Luca \cite{gross} showed that if $k$ is fixed, then there are only finitely many positive integers $n$ such that
\[
F_n=m_1!+m_2!+\cdots +m_k!
\]
holds for some positive integers $m_1,\ldots,m_k$.
Also, all the solutions for the case $k\leq 2$ were determined. Later,
Bollman, Hern\' andez and Luca \cite{luca3} solved the case $k=3$. In a
recent paper, Luca and Siksek \cite{siksek} found all factorials
expressible as the sum of at least three Fibonacci numbers.

In 1999, Luca \cite{luca} proved that $F_n$ is a product of factorials
only when $n=1,2,3,6$ and $12$. Also, Luca and St\u anic\u a
\cite{stanica} showed that the largest product of distinct Fibonacci
numbers which is a product of factorials is
$F_1F_2F_3F_4F_5F_6F_8F_{10}F_{12}=11!$.

In 2012, Marques \cite{notes} proved that $(m,u)=(4,5)$ is the only
solution of Eq.~(\ref{1}) where $u$ is a Fibonacci number. His proof
depends on the primitive divisor theorem together with factorizations
formulas for $F_n\pm 1$.

Among the several generalizations of Fibonacci numbers, one of the best
known is the {\it Tribonacci} sequence $(T_n)_{n \geq 0}$ (sequence
\seqnum{A000073} in the OEIS). This is defined by the recurrence
$T_{n+1}=T_{n}+T_{n-1}+T_{n-2}$, with initial values $T_0=0$ and
$T_1=T_2=1$. The first few terms of this sequence are 
\[ 0, 1, 1, 2, 4,
7, 13, 24, 44, 81, 149, 274, 504, 927, 1705.  \]

Tribonacci numbers have a long history. They were first studied in 1914
by Agronomof \cite{rf1} and subsequently by many others. The name
Tribonacci was coined in 1963 by Feinberg \cite{rf2}.

Here, we are interested in solutions $(m,u)$ of the Brocard-Ramanujan equation where $u$ is a Tribonacci number. We point out that in this we have neither a primitive divisor theorem for $T_n$ nor a factorization formula for $T_n\pm 1$. 

More precisely, we shall prove the following theorem.
\begin{theorem}\label{thm1}
There is no solution $(m,u)$ for the Brocard-Ramanujan equation \eqref{1},
where $u$ is a Tribonacci number.
\end{theorem}

The idea behind the proof is as follows. The equation we are interested
in solving is $m! = (T_n-1)(T_n+1)$. The $2$-adic valuation of $m!$ is
$m + O(\log m)$. We show that the $2$-adic valuation of
$(T_n-1)(T_n+1)$ is $\lessapprox 8 \log n/\log 2$. Thus $m \lessapprox
8 \log n/\log 2$. This forces $m!$ to be smaller than $(T_n-1)(T_n+1)$,
for $m$ and $n$ sufficiently large, which allows us to complete the
proof.




\section{The proof of Theorem \ref{thm1}}\label{auxiliary}

\subsection{A key lemma}

The $p$-adic order, $\nu_p(r)$, of $r$ is the exponent of the highest
power of a prime $p$ which divides $r$. The $p$-adic order of a
Fibonacci number was completely characterized by Lengyel \cite{order}.
Also, very recently the $2$-adic order of Tribonacci numbers was made
explicit by Lengyel and Marques \cite{tamas}. Here, we shall prove the
following key result which will play an important role in the proof of
Theorem \ref{thm1}.

\begin{lemma}\label{key}
We have
\begin{displaymath}
\nu_2(T_n+1) = \begin{cases}
15, &amp; \text{if $n=61$;}\\
0, &amp; \text{if $n\equiv \modd{0,3} {4}$;}\\
1, &amp; \text{if $n\equiv \modd{1, 2, 6} {8}$;}\\
3, &amp; \text{if $n\equiv \modd{5} {16}$;}\\
\nu_2((n+3)^2)-3, &amp; \text{if $n\equiv \modd{13,29,45} {64}$;}\\
\nu_2((n-61)(n+3))-3, &amp; \text{if $n &gt; 61$ and $n\equiv \modd{61} {64}$}.
\end{cases}
\end{displaymath}


and, for $n\geq 5$,
\begin{displaymath}
\nu_2(T_n-1) = \begin{cases}
0, &amp; \text{if $n\equiv \modd{0,3} {4}$;}\\
1, &amp; \mbox{if $n\equiv \modd{5} {8}$;}\\
\nu_2(n+2)-1, &amp; \text{if $n\equiv \modd{6} {8}$;}\\
\nu_2(n-2)-1, &amp; \text{if $n\equiv \modd{2} {8}$;}\\
\nu_2((n-1)(n+7))-3, &amp; \text{if $n\equiv \modd{1} {8}$.}
\end{cases}
\end{displaymath}

\end{lemma}


\bigskip
\noindent
{\bf The case $T_n-1$:}

First, note that Lengyel and Marques \cite{tamas} proved that $T_n - 1$ is odd for every $n \equiv 0,3$ (mod $4$), which proves the first case.
Now, note that, in order to prove the second case, it suffices to prove that $T_n \equiv 3$ (mod $4$). In this case, we have $n= 8k + 5$, with $k \geq 0$. Then we proceed on induction on $k$.
For $k=0$, it follows directly, since $T_5 - 1 = 7 - 1 = 6 = 2 \cdot 3$. So, we suppose that $T_{8k + 5} \equiv 3$ (mod $4$). Using the sum formula for $T_n$ (proved by Feng \cite{feng}), we have that 

\begin{eqnarray*}
 T_{8(k+1) + 5} &amp; = &amp;  T_{(8k+5) + 8} \\
 &amp; = &amp; T_6 T_{8k+5} + (T_6 + T_5) T_{8k+ 6} + T_7 T_{8k+7} \\
 &amp; = &amp; 13 T_{8k+5} + 20 T_{8k+6} + 24 T_{8k+7} \\ 
 &amp; \equiv &amp;  3 \pmod{4}.
\end{eqnarray*}

In the third case, for $t \geq 6$ and $ s \geq 1$ odd, we write $n = 2^{t-3}s + 2$. Now, by a Lengyel and Marques result \cite[Lemma 3.1]{tamas}, we have that 
\begin{eqnarray*}
T_{2^{t-3}s + 2} &amp;  = &amp; T_{2^{t-3}s + 1} + T_{2^{t-3}s } + T_{2^{t-3}s -1} \\ 
&amp; \equiv &amp; 1 + 2^{t-4} + 0 \pmod{2^{t-3}} \\ 
&amp; \equiv &amp; 1 + 2^{t-4} \pmod{2^{t-3}}. 
\end{eqnarray*}
This yields  $\nu_2(T_n - 1) = t-4 = \nu_2(2^{t-3}s) - 1 = \nu_2(n-2) - 1$.

The fourth case follows by proceeding in the same way as the third one. For $ t \geq 6$ and $ s \geq 1$ odd, we write $ n = 2^{t-3}s - 2$. Then, by the Lengyel and Marques result \cite[Lemma 3.1]{tamas}, we have that 
\begin{eqnarray*}
T_{2^{t-3}s - 2}  &amp; = &amp; T_{2^{t-3}s + 1} - T_{2^{t-3}s } - T_{2^{t-3}s -1} \\ 
&amp; \equiv &amp; 1 - 2^{t-4} - 0 \pmod{2^{t-3}} \\ 
&amp; \equiv &amp; 1 + 2^{t-4} \pmod{2^{t-3}}. 
\end{eqnarray*}
This yields $\nu_2(T_n - 1) = t-4 = \nu_2(2^{t-3}s) - 1 = \nu_2(n+2) - 1$.

Now, for the last case, we know that $16$ divides exactly one among $n-1$ and $n+7$. Suppose that $16 | (n + a)$, for some $a \in \left\{-1, 7\right\}$. Then $\nu_2(n+b) = 3$ for $b \in \left\{ -1 , 7 \right\} \backslash \left\{a\right\}$. So, we desire to prove that 
\[
 \nu_2(T_n-1) = \nu_2(n+a).
 \]
For that, we write $n = 2^{t-2}s - a$, for $t \geq 5$ and $s \geq 1$ odd, and proceed as in Lengyel and Marques \cite[Lemma 3.1]{tamas} to prove that 
\[
T_{2^{t-2}s - a}-1 \equiv 2^{t-2} \pmod{2^{t-1}}.
\]
Therefore 
\[
\nu_2(T_n - 1) = t-2 = \nu_2(n+a) + 1,
\]
and this completes the proof.  

\qed

\bigskip
\noindent
{\bf The case $T_n+1$:}

The first two cases are trivial. The third and the fourth cases follow in the same way. Note that, in order to prove them, it suffices to show that $T_n \equiv 1$ (mod $4$) when $ n \equiv 1, 2, 6$ (mod $8$) and to show that $T_n = 7 \pmod{16}$ when $ n  \equiv 5$ (mod $16$). In order to avoid unnecessary repetitions, we shall prove only one of these cases. So, let us write $n = 8k + 6$ and apply induction on $k \geq 0$.
For $k=0$, it follows directly, since $T_6 + 1 = 13 + 1 = 14 = 2 \cdot 7$. Now, suppose that $T_{8k+6}\equiv 1$ (mod $4$). Then, we have that 
\begin{eqnarray*}
T_{8(k+1) +6} &amp; = &amp;  T_{(8k+6) + 8} \\ 
&amp; = &amp; T_6 T_{8k+6} + (T_6 + T_5)T_{8k+7} + T_7 T_{8k+8} \\
&amp; = &amp; 13 T_{8k+6} + 20 T_{8k+7} + 24 T_{8k+8} \\
&amp; \equiv &amp; 1 \pmod 4.
\end{eqnarray*}

Now, for the the fifth case, note that, if $n=64k+13$,  
\begin{eqnarray*}
\nu_2((n+3)^2) - 3 &amp; = &amp; 2\nu_2(n+3) - 3 \\ 
&amp;= &amp; 2\nu_2(64k + 13 + 3) - 3 = 2\nu_2(16(4k+1)) - 3 = 2 \cdot 4 - 3 \\
&amp; = &amp; 5.
\end{eqnarray*}
So, it suffices to prove that $T_n \equiv 31$ (mod $64$). Again, we proceed on induction. First, observe that $T_{13} = 927 \equiv 31$ (mod $64$). Now, we have that 
\begin{eqnarray*}
T_{64(k+1) +13} &amp; = &amp; T_{(64k+13) + 64} \\ 
&amp; = &amp;  T_{62} T_{64k+13} + (T_{62} + T_{61})T_{64k+14} + T_{63} T_{64k+15} \\
&amp; \equiv &amp; -1 + 32T_{64k+14} \pmod{64}.
\end{eqnarray*}
But, from the previous case, we have that $T_{64k+14} \equiv 1$ (mod $4$). Then,  
\begin{eqnarray*}
T_{64(k+1) +13}  &amp; \equiv &amp;  -1 + 32T_{64k+14} \pmod{64} \\
&amp;  \equiv &amp; 32 - 1 \pmod{64} \\
&amp; \equiv &amp; 31 \pmod{64}.
\end{eqnarray*}
When $n \equiv 29, 45$ (mod $64$), we proceed in the same way.

For the last case, we proceed as for the last case of the previous theorem. Note that $128$ divides exactly one among $n-61$ and $n+3$. Suppose that $128 | (n + a)$, for some $a \in \left\{-61, 3\right\}$. Then $\nu_2(n+b) = 6$ for $b \in \left\{ -61 , 3 \right\} \backslash \left\{a\right\}$. So, we desire to prove that 
\[
\nu_2(T_n+1) = \nu_2(n+a)+3.
\]
For that, we write $ n = 2^{t-2}s - a$, for $t \geq 8$ and $s \geq 1$ odd, and proceed as in Lengyel and Marques \cite[Lemma 3.1]{tamas} to prove that 
\[
T_{2^{t-2}s - a}+1 \equiv 2^{t+1} \pmod{2^{t+2}}.
\]
Therefore 
\[
\nu_2(T_n + 1) = t+1 = \nu_2(n+a) + 3.
\]
This completes the proof.  
\qed


Now, we are ready to deal with the proof of the main theorem.

\subsection{The proof}\label{proof}

If $n\leq 61$, a straightforward search shows that there are no solutions. So  we shall suppose that $n&gt;61$. Then $m\geq 30$.  Next we use the fact
that $\nu_2(m!)\geq m-\left\lfloor \log m/\log 2\right\rfloor-1$ (which is a consequence of the De Polignac's formula) together with Lemma \ref{key}. Then
\begin{eqnarray*}
m-\left\lfloor \frac{\log m}{\log 2}\right\rfloor-1 &amp;\leq &amp; \nu_2(m!)=\nu_2(T_n-1)+\nu_2(T_n+1)\\
 &amp; &lt; &amp; \nu_2((n+2)(n-2)(n-1)(n+7)(n+3)^3(n-61))+5\\
 &amp; \leq &amp; 8\nu_2(n+\omega)+5, 
\end{eqnarray*}
for some $\omega\in \{-61,-2,-1,2,3,7\}$. Thus $\nu_2(n+\omega)\geq (m-\left\lfloor \log m/\log 2\right\rfloor-6)/8$. Therefore, $2^{\lfloor (m-\left\lfloor \log m/\log 2\right\rfloor-6)/8\rfloor}\mid (n+\omega)$. In particular, $2^{\lfloor (m-\left\lfloor \log m/\log 2\right\rfloor-6)/8\rfloor}\leq |n+\omega|\leq n+61$ (here we used that $n+\omega\neq 0$). By applying the $\log$ function, we obtain
\begin{equation}\label{I}
\left\lfloor  \frac{1}{8}\left(m-\left\lfloor \frac{\log m}{\log 2}\right\rfloor-6\right) \right\rfloor\leq \dfrac{\log (n+61)}{\log 2}.
\end{equation}

On the other hand, $(1.83)^{2n-4}&lt;T_n^2=m!+1&lt; 2(m/2)^m$ (the first inequality was proved by Bravo and Luca \cite{BL}). So $n&lt;0.9m\log (m/2)+2.6$. Substituting this in equation (\ref{I}), we obtain
\[
\left\lfloor  \frac{1}{8}\left(m-\left\lfloor \frac{\log m}{\log 2}\right\rfloor-6\right) \right\rfloor\leq \dfrac{\log (0.9m\log (m/2)+63.6)}{\log 2}.
\]
This inequality yields $m\leq 78$. Then $n&lt;0.9\cdot 78\log (78/2)+2.6=259.782\ldots$. Now, we use a simple routine written in \textit{Mathematica} which does not return any solution in the range $30\leq m\leq 78$ and $62\leq n\leq 259$. The proof is complete.\qed

\section{Acknowledgment}
The authors would like to thank CNPq for the financial support and to
the editor and the referee for helpful comments.


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\end{thebibliography}

\bigskip
\hrule
\bigskip

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

\noindent \emph{Keywords: }
Fibonacci number, $p$-adic order, Tribonacci number, Brocard-Ramanujan
equation.


\bigskip
\hrule
\bigskip

\noindent (Concerned with sequences
\seqnum{A000045} and
\seqnum{A000073}.)

\bigskip
\hrule
\bigskip

\vspace*{+.1in}
\noindent
Received December 22 2015;
revised versions received March 15 2016; April 26 2016.
Published in {\it Journal of Integer Sequences}, April 26 2016.

\bigskip
\hrule
\bigskip

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