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\begin{center}
\vskip 1cm{\LARGE\bf 
The $2$-adic Order of the Tribonacci Numbers \\
\vskip .1in
and the Equation $T_n=m!$
}
\vskip 1cm
\large
Diego Marques\\
Departamento de Matem\' atica\\
Universidade de Bras\' ilia\\
Bras\' ilia 70910-900 \\
Brazil\\
\href{mailto:diego@mat.unb.br}{\tt diego@mat.unb.br}\\
\ \\
Tam\'as Lengyel\\
Occidental College\\
1600 Campus Road \\
Los Angeles, CA 90041 \\
USA\\
\href{mailto:lengyel@oxy.edu}{\tt lengyel@oxy.edu}\\
\end{center}

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\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 paper, we characterize the $2$-adic valuation of $T_n$ and, as
an application, we completely solve the Diophantine equation $T_n=m!$.
\end{abstract}


\section{Introduction}\label{sec:1}

Let $(F_n)_{n\geq 0}$ be the Fibonacci sequence given by $F_{n+2}=F_{n+1}+F_n$, for $n\geq 0$, where $F_0=0$ and $F_1=1$.  The first few terms of this sequence are
\[
0,1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987, \ldots
\]

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; see
\cite{21,order,order2,22,23}. For instance, from the main theorem of Lengyel \cite{order}, we extract the following facts:

\begin{center}
$\nu_2(F_n) = %\left\{
\begin{cases}
0,  &amp;\mbox{if}\  n\equiv 1,2\pmod 3;\\
1,  &amp;\mbox{if}\ n\equiv 3 \pmod 6;\\
3,  &amp;\mbox{if}\ n\equiv 6 \pmod{12};\\
\nu_2(n)+2, &amp;\mbox{if}\ n\equiv 0\pmod{12}.
\end{cases}
%\right.
$
\end{center}

To prove this, many congruences involving Fibonacci numbers were
used (see for instance \cite{jacob}).

Among the several generalizations of Fibonacci numbers, one of the most known is the {\it Tribonacci} sequence $(T_n)_{n\geq 0}$, which 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
\[
0, 1, 1, 2, 4, 7, 13, 24, 44, 81, 149, 274, 504, 927, 1705, \ldots
\]

Recall that the 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}.
The basic properties of Tribonacci numbers can be found in
\cite{rf3,rf7,spick,rf5}. For recent papers, we refer the reader to
{\cite{rf11,rf8,rf9,bohe,rf10}} and to a collection
\cite{rf12,rf13,rf14}.

In this paper, we provide a complete description of the $2$-adic order of Tribonacci numbers. More precisely, we prove the following result:
\begin{theorem}\label{m1}
For $n\geq 1$, we have
\begin{center}
$\nu_2(T_n) = %\left\{
\begin{cases}
0, &amp;\mbox{if}\  n\equiv 1,2\pmod 4;\\
1, &amp;\mbox{if}\ n\equiv 3,11 \pmod {16};\\
2, &amp;\mbox{if}\ n\equiv 4,8 \pmod{16};\\
3, &amp;\mbox{if}\ n\equiv 7 \pmod{16};\\
\nu_2(n)-1, &amp;\mbox{if}\ n\equiv 0\pmod{16};\\
\nu_2(n+4)-1, &amp;\mbox{if}\ n\equiv 12\pmod{16};\\
\nu_2((n+1)(n+17))-3, &amp;\mbox{if}\ n\equiv 15\pmod{16}.
\end{cases}
%\right.
$
\end{center}
\end{theorem}

A number of mathematicians have been interested in Diophantine
equations involving factorial and Fibonacci numbers. For example, in
1999, Luca \cite{luca} proved that $F_n$ is a product of factorials
only when $n=1,2,3,6,12$. Also, the largest product of distinct
Fibonacci numbers which is a product of factorials is
$F_1F_2F_3F_4F_5F_6F_8F_{10}F_{12}=11!$; see \cite{stanica} (a proof of
this fact can be achieved by applying the primitive divisor theorem
\cite{char}).


Also, in \cite{gross} it is shown 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 $m_1,\ldots,m_k$; furthermore,
all the solutions for the case $k\leq 2$ have been determined. {Later} the case $k=3$ was also solved; see \cite{luca3}. In a recent paper, Luca and Siksek \cite{siksek} found all factorials expressible as the sum of {at most} three Fibonacci numbers.

However, we point out that, to the best of the authors' knowledge, the
problem of finding all factorials among Tribonacci numbers has not yet
been solved. For this problem, the approach of using the primitive
divisor theorem does not work, simply because such a theorem for higher
order recurrence sequences seems to be out of reach.

Here we use Theorem \ref{m1} to solve completely the equation $T_n=m!$. We have
\begin{theorem}\label{m2}
The only solutions of the Diophantine equation
\begin{equation}\label{m}
T_n=m!
\end{equation}
in positive integers $n,m$ are
\[
(n,m)\in \{(1,1),(2,1),(3,2),(7,4)\}.
\]
\end{theorem}


\section{Auxiliary results}

Before proceeding further, some considerations will be needed for the convenience of the reader.

From \cite[Lemma 1]{BL}, we can extract the following result.

\begin{lemma}\label{l1}
For all $n\geq 1$, we have that
\begin{equation}\label{Lu}
\phi^{n-2}\leq T_n\leq \phi^{n-1},
\end{equation}
where $\phi=1.839286\cdots$.
\end{lemma}

The next result provides an addition formula for Tribonacci numbers
(see \cite[Section 3]{feng});  it plays a crucial role in proving
Theorem~\ref{m1}.

\begin{lemma}\label{l2}
For all integers $n,m$, with $n\geq 0$ and $m\geq 2$, we have that
\[
T_{n+m}=T_{m-2}T_n+(T_{m-3}+T_{m-2})T_{n+1}+T_{m-1}T_{n+2}.
\]
\end{lemma}

We require one last fact about $\nu_p$ in order to complete our proof of Theorem \ref{m2}.

\begin{lemma}\label{l3}
For any integer $k\geq 1$ and $p$ prime, we have
\begin{equation}\label{est}
\frac{k}{p-1}-\left\lfloor \frac{\log k}{\log p}\right\rfloor -1\leq \nu_p(k!)\leq \frac{k-1}{p-1},
\end{equation}
where $\lfloor x\rfloor$ denotes the largest integer less than or equal to $x$.
\end{lemma}

We refer the reader to \cite[Lemma 2.4]{d21} for a proof of this result.

Now we are ready to deal with the proof of the theorems.

\section{Proof of Theorem \ref{m1}}

In order to prove the first 4 cases of the $2$-adic valuation, it suffices to prove the following congruences
\begin{itemize}
\item[(i)] $T_n\equiv 1\pmod 2$, if $n\equiv 1,2\pmod 4$.
\item[(ii)] $T_n\equiv 2\pmod 4$, if $n\equiv 3,11\pmod{16}$.
\item[(iii)] $T_n\equiv 4\pmod 8$, if $n\equiv 4,8\pmod{16}$.
\item[(iv)] $T_n\equiv 8\pmod{16}$, if $n\equiv 7\pmod{16}$.
\end{itemize}

To avoid unnecessary repetitions we shall prove only that $T_n\equiv
4\pmod 8$ if $n\equiv 4\pmod{16}$ (the other cases can be handled in
much the same way and we {leave} them as an exercise to the reader).

Thus, we want to prove that $T_{16t+4}\equiv 4\pmod {8}$, for all $t\geq 0$. For that, we shall proceed by induction on $t$. The basis case, $t=0$, follows because $T_4=4$. Thus, we may suppose that $T_{16t+4}\equiv 4\pmod {8}$. Therefore, we use Lemma \ref{l2} to $T_{16(t+1)+4}=T_{(16t+4)+16}$ to obtain
\[
T_{16(t+1)+4}=1705T_{16t+4}+2632T_{16t+5}+3136T_{16t+6}.
\]
Then $T_{16(t+1)+4}\equiv 1705\cdot 4\equiv 4\pmod 8$, where we used that $8$ divides both $2632$ and $3136$.

Now, we shall split our proof into three cases:

\bigskip

\noindent
\textbf{Case 1.} $n\equiv 0\pmod{16}$. To deal with this case, we shall prove the following result:

\begin{lemma}\label{t1}
For all $s\geq 1$ and $t\geq 6$, we have that $T_{2^{t-3}s}\equiv s2^{t-4}\pmod{2^{t-3}}$.
\end{lemma}

\begin{proof}
We shall use induction on $s$ to prove simultaneously the following congruences
\begin{equation}\label{3}
T_{2^{t-3}s}\equiv s2^{t-4}\pmod{2^{t-3}}, T_{2^{t-3}s-1}\equiv 0 \pmod{2^{t-3}}, T_{2^{t-3}s+1}\equiv 1\pmod{2^{t-3}}.
\end{equation}

First, let us deal with the basis case $s=1$. So, we desire to prove that, for all $t\geq 6$, we have that
\begin{equation}\label{3.1}
T_{2^{t-3}}\equiv 2^{t-4}\pmod{2^{t-3}}, T_{2^{t-3}-1}\equiv 0 \pmod{2^{t-3}}, T_{2^{t-3}+1}\equiv 1\pmod{2^{t-3}}.
\end{equation}
We will use again induction on $t$. Clearly the congruences hold for $t=6$. So, suppose that they are true for $t$. Then we use Lemma \ref{l2} for $T_{2^{t-2}}=T_{(2^{t-3}-1)+(2^{t-3}+1)}$ to obtain
\begin{equation}\label{T}
T_{2^{t-2}}=T_{2^{t-3}-1}^2+(T_{2^{t-3}-2}+T_{2^{t-3}-1})T_{2^{t-3}}+T_{2^{t-3}}T_{2^{t-3}+1}.
\end{equation}
By using the Tribonacci recurrence, we obtain $T_{2^{t-3}-2}\equiv 2^{t-4}+1\pmod{2^{t-3}}$. Now, we write $T_{2^{t-3}-2}=1+2^{t-4}+a2^{t-3}$, $T_{2^{t-3}-1}=b2^{t-3}$, $T_{2^{t-3}}=2^{t-4}+c2^{t-3}$ and $T_{2^{t-3}+1}=1+d2^{t-3}$. Then, we get
\begin{eqnarray*}
T_{2^{t-2}} &amp; = &amp; b^22^{2t-6}+(1+2^{t-4}+(a+b)2^{t-3})(2^{t-4}+c2^{t-3})+(2^{t-4}+c2^{t-3})(1+d2^{t-3})\\
&amp; \equiv &amp; 2^{t-4}+c2^{t-3}+2^{t-4}+c2^{t-3}\pmod{2^{t-2}}\\
&amp; \equiv &amp; 2^{t-3}\pmod{2^{t-2}},
\end{eqnarray*}
as desired. Here we used that $2t-8\geq t-2$, since $t\geq 6$. We proceed similarly to prove the other congruences in \eqref{3.1}. Now, by induction hypothesis, we suppose that the congruences in \eqref{3} hold for $s$. Then, we use exactly the same procedure as before (and Lemma \ref{l2}) for $T_{2^{t-3}(s+1)}=T_{(2^{t-3}s-1)+(2^{t-3}+1)}$. We omit the details.
\end{proof}

Since $16\mid n$, then $n=2^{t-3}s$, with $s$ odd and $t\geq 7$. By Lemma \ref{t1}, we obtain that $\nu_2(T_{2^{t-3}s})=t-4$ and then
\[
\nu_2(T_n)=\nu_2(T_{2^{t-3}s})=t-4=\nu_2(2^{t-3}s)-1=\nu_2(n)-1
\]
as desired.

\bigskip
\noindent
\textbf{Case 2.} $n\equiv 12\pmod{16}$. By \eqref{3} together with the Tribonacci recurrence we obtain that $T_{2^{t-3}s-4}\equiv s2^{t-4}\pmod{2^{t-3}}$ holds for all $s\geq 1$ and $t\geq 6$. Since $n\equiv -4\pmod{16}$, then we can write $n=2^{t-3}s-4$, for some $t\geq 7$ and $s\equiv 1\pmod 2$. Then, we apply the previous congruence to get
\[
\nu_2(T_{n})=t-4=\nu_2(2^{t-3}s-4+4)-1=\nu_2(n+4)-1.
\]

\bigskip
\noindent
\textbf{Case 3.} $n\equiv 15\pmod{16}$. In this case, we know that $32$ divides {exactly} one among $n+1$ and $n+17$. Suppose that $32\mid n+a$, for some $a\in \{1,17\}$. Then $\nu_2(n+b)=4$, for $b\in \{1,17\}\backslash \{a\}$, and so, we desire to prove that
\[
\nu_2(T_n)=\nu_2(n+a)+1.
\]
For that, we proceed as in the other cases to prove that the following congruence holds, for all $s\geq 1, t\geq 8$ and $a\in \{1,17\}$:
\[
T_{2^{t-3}s-a}\equiv s2^{t-2}\pmod{2^{t-1}}
\]
with $n$ being in the form $n=2^{t-3}s-a$.
Therefore,
\[
\nu_2(T_n)=\nu_2(T_{2^{t-3}s-a})=t-2=\nu_2(n+a)+1.
\]
This completes the proof.
\qed

\begin{remark}
We note that as a starting point one can use a multisection approach (cf. \cite{order2}) to discover the underlying structure of the above analysis. Here we applied the 4-section of the ordinary generating function of the sequence
$(T_n)_{n\geq 0}$.
\end{remark}


We conclude with a
\begin{conjecture}
For $n\ge n(p)$ with some multiple $\pi'(p)$ of the modulo $p$ period $\pi(p)$ of the Tribonacci sequence $T_n$ and some integer constants $n(p)&gt;0$, $0\leq r=r(p)\leq p$, $n_i=n_i(p)$, $c'_i=c'_i(p)$, $c_{i,j}=c_{i,j}(p), 1\leq i\leq \pi'(p), 1\leq j\leq  r$, we have that
\[\nu_p(T_n)=c'_i +\nu_p\left(\prod_{j=1}^{r} (n+c_{i,j})\right)   \ \mbox{ if }\ n\equiv n_i\pmod{\pi'(p)}.\]
(Note that it is easy to pick $c_{i,j}$s for $i$ so that $\nu_p(n+c_{i,j})=0$.) \end{conjecture}


\section{Proof of Theorem \ref{m2}}

If {$m\leq 4$}, one can see that the only solutions are the ones listed in the Theorem \ref{m2}. So we shall suppose that {$m\geq 5$}. By using Lemma {\ref{l3}} (for $p=2$) together with Theorem \ref{m1}, we deduce that
\begin{eqnarray*}
m-\left\lfloor \frac{\log m}{\log 2}\right\rfloor-1 &amp;\leq &amp; \nu_2(m!)=\nu_2(T_n)\\
 &amp; &lt; &amp; \nu_2(n(n+1)(n+4)(n+17))-5\leq 4\nu_2(n+\delta)-5,
\end{eqnarray*}
for some $\delta\in \{0,1,4,17\}$. Thus $\nu_2(n+\delta)\geq (m-\left\lfloor \log m/\log 2\right\rfloor+4)/4$ and therefore, $2^{\lfloor (m-\left\lfloor \log m/\log 2\right\rfloor+4)/4\rfloor}$ divides $n+\delta$. In particular, $2^{\lfloor (m-\left\lfloor \log m/\log 2\right\rfloor+4)/4\rfloor}\leq n+\delta\leq n+17$ and by applying the $\log$ function, we obtain
\begin{equation}\label{I}
\left\lfloor  \frac{1}{4}\left(m-\left\lfloor \frac{\log m}{\log 2}\right\rfloor+4\right) \right\rfloor\leq \dfrac{\log (n+17)}{\log 2}.
\end{equation}

On the other hand, by Lemma \ref{l1}, $(1.83)^{n-2}&lt;T_n=m!&lt; (m/2)^m$ and so $n&lt;1.66m\log (m/2)+2$. Substituting this in (\ref{I}), we arrive at
\[
\left\lfloor  \frac{1}{4}\left(m-\left\lfloor \frac{\log m}{\log 2}\right\rfloor+4\right) \right\rfloor\leq \dfrac{\log (1.66m\log (m/2)+19)}{\log 2}.
\]
This inequality yields {$m\le 32$} and then {$n&lt;1.66\cdot 32\log (32/2)+2=149.27\cdots$}. Now, we use a simple routine written in \textit{Mathematica} which (in a few seconds) does not return us any solution in the range {$5\leq m\leq 32$} and {$1\leq n\leq 149$}. The proof is complete.\qed

\section{Acknowledgment}
The {first} author is grateful to FAP-DF and CNPq for the financial support. He also thanks to Matheus Bernardini for nice discussions on the subject. The authors wish to thank the editor and the referee for their helpful comments.


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\end{thebibliography}


\bigskip
\hrule
\bigskip

\noindent 2010 {\it Mathematics Subject Classification}:
Primary 11B39; Secondary 11B50, 11A07.


\noindent \emph{Keywords: }
Tribonacci number, divisibility, 2-adic valuation.


\bigskip
\hrule
\bigskip

\noindent (Concerned with sequences
\seqnum{A000073} and
\seqnum{A248174}.)

\bigskip
\hrule
\bigskip

\vspace*{+.1in}
\noindent
Received June 30 2014;
revised version received October 1 2014.
Published in {\it Journal of Integer Sequences}, October 3 2014.

\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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