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\begin{center}
\vskip 1cm{\LARGE\bf  
On Tribonacci Numbers that are \\
\vskip .1in
Products of  Factorials}
\vskip 1cm
\large
Adel Alahmadi \\ 
Research Group in Algebraic Structures and Applications\\
King Abdulaziz University\\
Jeddah \\
Saudi Arabia \\
\href{mailto:analahmadi@kau.edu.sa }{\tt analahmadi@kau.edu.sa}\\
\ \\
Florian~Luca\\
School of Mathematics\\
University of the Witwatersrand \\
1 Jan Smuts Avenue \\
Braamfontein 2050\\
Johannesburg \\
South Africa\\
and\\
Research Group in Algebraic Structures and Applications\\
King Abdulaziz University\\
Jeddah \\
Saudi Arabia \\
and\\
Centro de Ciencias Matem\'aticas UNAM\\
Morelia \\
M\'exico\\
\href{mailto:florian.luca@wits.ac.za}{\tt florian.luca@wits.ac.za}\\

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\vskip .2 in

\begin{abstract}
We determine all Tribonacci numbers of positive or negative indices that
are products of factorials. 
\end{abstract}

\section{Introduction}

The Tribonacci numbers $\{T_n\}_{n\ge 0}$ are given by $T_0=0,~T_1=T_2=1$ and $T_{n+3}=T_{n+2}+T_{n+1}+T_n$ for $n\ge 0$. 
In \cite{ML}, Marques and Lengyel determined all Tribonacci numbers $T_n$ that are factorials. They showed that if $T_n=m!$ then $n\in \{1,2,3,7\}$. Here we take this a couple of steps further.

\begin{theorem}
\label{thm:1}
If $n\ge 1$ and 
$$
T_n=m_1!\, m_2! \, \cdots\, m_k!
$$
for some positive integers $m_1\le m_2\le \cdots \le m_k$, then $n\in \{1,2,3,4,7\}$. 
\end{theorem}  

Next, one can extend the sequence of Tribonacci numbers in the negative direction using the recurrence relation. Namely, since $T_0=0,~T_1=T_2=1$, one computes that $T_{-1}=0,~T_{-2}=1$ and for $n\ge 3$, we have 
$$
T_{-n}=-T_{-(n-1)}-T_{-(n-2)}+T_{-(n-3)}.
$$
The Tribonacci numbers with negative indices change signs infinitely often. Nevertheless, we can still ask 
what about the equation $|T_n|=m_1!\cdots m_k!$. Here is our result.

\begin{theorem}
\label{thm:2}
If $n\in {\mathbb Z}$ and $|T_n|=m_1!\, m_2! \, \cdots \, m_k!$ then 
$$
n\in \{-9,-8,-7,-5,-3,-2,1,2,3,4,7\}.
$$ 
\end{theorem}

Note the near miss $T_{-32}=-2^4\cdot 3\cdot 5^2\cdot 7=7!\cdot (5/3)$.  

\section{Preliminaries}

The main ingredient is an exact formula for the exponent of $2$ in $T_n$. Let $\nu_p(m)$ be the exponent of the prime $p$ in the factorization of the integer $m$ with convention that 
$\nu_p(0)=\infty$.  Here is Theorem 1 in \cite{ML}.

\begin{theorem}
\label{thm:ML}
For $n\ge 1$, we have
$$
\nu_2(T_n)=
\begin{cases} 0, &amp; \text{if }  n\equiv 1,2\pmod 4;\\
1, &amp; \text{if }  n\equiv 3,11\pmod {16};\\
2, &amp; \text{if }  n\equiv 4,8\pmod {16};\\
\nu_2(n)-1, &amp; \text{if }  n\equiv 0\pmod {16};\\
\nu_2(n+4)-1 &amp; \text{if }  n\equiv 12\pmod {16};\\
\nu_2(n+17)+1 &amp; \text{if }  n\equiv 15\pmod {32};\\
\nu_2(n+1)+1 &amp; \text{if }  n\equiv 31\pmod {32}.
\end{cases}
$$
\end{theorem}

A similar theorem is proved for the prime $p=3$ in \cite{Biluetal}. 

\begin{theorem}
\label{thm:Biluetal}
For $n\ge 1$, we have 
$$
\nu_3(T_n)= \begin{cases} 
0, &amp; \text{if }  n\equiv 1,2,3,4,5,6,8,10,11\pmod {13};\\
1, &amp; \text{if }  n\equiv 7\pmod {13};\\
\nu_3(n)+2,  &amp; \text{if }  n\equiv 0\pmod {13};\\
\nu_3(n+1)+2, &amp; \text{if }   n\equiv 12\pmod {13};\\
4, &amp; \text{if }  n\equiv 9\pmod {39};\\
\nu_3(n+17)+4, &amp; \text{if }  n\equiv 22\pmod {39};\\
\nu_3(n+4)+4, &amp; \text{if }  n\equiv 35\pmod {39}.
\end{cases}
$$
\end{theorem}

\subsection{The proof of Theorem \ref{thm:1}}

 Assume $k\ge 1$, $n\ge 3$ and 
 $$
 T_n=m_1!\cdots m_k!.
 $$
 We may assume that $2\le m_1\le \cdots\le m_k$. It is well-known that letting $\alpha$ be real root of $x^3-x^2-x-1=0$, then $\alpha&gt;1.83$ and $T_n\ge \alpha^{n-2}$ holds for all $n\ge 1$. Thus,
 $$
 \alpha^{n-2}\le T_n=m_1!\cdots m_k!&lt;m_1^{m_1}\cdots m_k^{m_k},
 $$
 so 
 \begin{equation}
 \label{eq:0}
 n&lt;2+\frac{m_1\log m_1+\cdots +m_k \log m_k}{\log(1.83)}.
 \end{equation}
Note that 
$$
\nu_2(m!)=\lfloor m/2\rfloor+\lfloor m/4\rfloor+\cdots\ge m/2
$$
holds for all $m\ge 2$, except when $m=3$, when $\nu_2(3!)=(3-1)/2\ge 3/3$. Hence, the inequality $\nu_2(m!)\ge m/3$ holds for all $m\ge 2$. 
Next, by Theorem \ref{thm:ML}, we see that
\begin{align*}
\nu_2(T_n) &amp; \le  \max\{2,\nu_2(n)-1,\nu_2(n+4)-1,\nu_2(n+1)+1,\nu_2(n+17)+1\}\\
&amp; \le  \frac{\log(n+17)}{\log 2}+1.
\end{align*}
This gives 
$$
\frac{\log(n+17)}{\log 2}+1\ge \nu_2(T_n)=\nu_2(m_1!)+\cdots+\nu_2(m_k!).
$$
Thus, putting 
$$
x:=\nu_2(m_1!)+\cdots+\nu_2(m_k!),
$$
we have 
\begin{equation}
\label{eq:1}
n&gt;2^{x-1}-17,
\end{equation}
and since $m_i\le 3\nu_2(m_i!)$ for $i=1,\ldots,k$, we have 
\begin{align*}
m_1\log m_1+\cdots +m_k\log m_k &amp; &lt;  (m_1+\cdots +m_k)\log (m_1+\cdots+m_k)\\
&amp; \le  3x\log(3x),
\end{align*}
so \eqref{eq:0} yields
\begin{equation}
\label{eq:2}
n&lt;2+\frac{3x\log(3x)}{\log(1.83)}.
\end{equation}
Combining \eqref{eq:1} and \eqref{eq:2}, we get
$$
2^{x-1}-17&lt;2+\frac{3x\log(3x)}{\log(1.83)},
$$
giving $x\le 8$. Since $\nu_2(11!)=8$, we conclude that either $T_n=11!$ and $k=1$, or $m_k\le 10$, in which case $P(T_n)\le 7$, where $P(m)$ is the largest prime factor of $m$. 
Since $T_n=11!$ is not possible, we conclude that $P(T_n)\le 7$. All Tribonacci numbers whose largest prime factor is at most $7$ have been determined in \cite{BrLu} and they are   
$T_1=T_2=1,~T_2=2,~T_3=4,~T_5=7,~T_7=2^3\cdot 3,~T_9=3^4,~T_{12}=2^3\cdot 3^2\cdot 7,~T_{15}=2^6\cdot 7^2$, and the only numbers from the previous list that are products of 
factorials correspond to $n\in \{1,2,3,4,7\}$. 

\section{The proof of Theorem \ref{thm:2}}

This is trickier, since the lower bounds on $|T_{-n}|$ are quite weak. However, we can follow the arguments from \cite{BrLu1} and \cite{BrLu2}. Namely, we put $\Lambda:=\{\alpha,\beta,\gamma\}$ for the set of roots of $P(X)=X^3-X^2-X-1$. We assume $\beta=\alpha^{-1/2} e^{i\theta}$ where  $\theta\in (0,\pi)$ and $\gamma=\alpha^{-1/2}e^{-i\theta}$. So, $\beta$ is the complex nonreal root of $P(X)$ 
in the upper half-plane.  Then 
$$
T_n=c_{\alpha} \alpha^n+c_{\beta} \beta^n+c_{\gamma} \gamma^n\qquad {\text{\rm holds for all}}\quad n\in {\mathbb Z}.
$$
Here
$$
c_{\lambda}=\frac{\lambda}{P'(\lambda)}\quad {\text{\rm for}}\quad \lambda\in \Lambda.
$$
The Tribonacci sequence  $\{T_n\}_{n\in {\mathbb Z}}$ is periodic modulo $2^k$ with period $2^{k+2}$ for all $k\ge 1$. To see this, it is easier to work with
$$
22T_n=(22c_{\alpha})\alpha^n+(22c_{\beta})\beta^n+(22c_{\gamma})\gamma^n.
$$
The numbers $c_{\lambda}$ are not algebraic integers, but $22c_{\lambda}$ are algebraic integers of minimal polynomial $X^3-22X-242$ for $\lambda\in \Lambda$. Since $X^3-X^2-X-1$ divides $X^4-2X+1$, it follows that $\lambda^4\equiv 1\pmod {2}$ for $\lambda\in \Lambda$. Here, for algebraic integers $\gamma,~\delta$ and an integer $m\ge 1$, we write 
$\gamma\equiv \delta\pmod {m}$ if $(\gamma-\delta)/m$ is an algebraic integer. By induction we get $\lambda^{2^{k+1}}\equiv 1\pmod {2^k}$ for all $k\ge 1$. In particular, for all $n\in {\mathbb Z}$, we have 
$$
22T_{n+2^{k+1}}=\sum_{\lambda\in \Lambda} (22c_{\lambda})\lambda^{n+2^{k+1}}\equiv \sum_{\lambda\in \Lambda} (22c_{\lambda})\lambda^n\equiv 22T_n\pmod {2^{k}}.
$$
The above congruence implies that $T_{n+2^{k+1}}\equiv T_n\pmod {2^{k-1}}$ holds for all $k\ge 2$. 
Hence, $\{T_n\}_{n\in {\mathbb Z}}$ is periodic modulo $2^k$ with period $2^{k+2}$ for all $k\ge 1$. In particular, the Marques--Lengyel formulas from Theorem \ref{thm:ML} hold for all integers $n$, not only for the positive ones. Let us see why. 

Assume $n\ne 0,1,4,17$, since  for these values of $b$ we have
$T_{-n}=0$, so the formula from Theorem \ref{thm:ML} holds with both sides equal $\infty$. Let $k$ be large ($k\ge n+10$). Then $T_{-n}\equiv T_{2^{k}-n}\pmod {2^{k-2}}$ and $2^k-n$ is positive. Unless $-n\equiv 0,1,4,17\pmod {16}$, the formulas from Theorem \ref{thm:ML} show that $\nu_2(2^k-n)\in \{0,1,2\}$ and $16\mid 2^k$. It then follows that $\nu_2(T_{2^k-n})=0,1,2$ and since $T_{2^k-n}\equiv T_{-n}\pmod {2^{k-2}}$, it follows that $\nu_2(T_{-n})=\nu_2(T_{2^k-n})$. Hence, the formula from Theorem \ref{thm:ML} holds for the residue classes for $-n$ modulo $16$ that are not one of $0,1,4,17$. For the rest of the residue classes, $\nu_2(T_{2^k-n})=\nu_2(2^k-n+a)+b$ for some $a\in \{0,1,4,17\}$ and $b\in \{-1,1\}$. But 
$$
\nu_2(2^k-n+a))=\nu_2(2^k-(n-a))=\nu_2(n-a)\quad {\text{\rm  for large}}\quad k\quad {\text{\rm since}}\quad n-a\ne 0.
$$ 
Since $T_{-n}\equiv T_{2^k-n}\pmod {2^{k-2}}$ and
$\nu_2(T_{-n})=\nu_2(T_{2^k-n})=\nu_2(-n+a)+b$, the formula from Theorem \ref{thm:ML} holds for $-n$ that are congruent to one of $\{0,1,4,17\}$ modulo $16$. Similar (and in fact easier) considerations modulo $13\cdot 3^k$ show that the sequence $\{T_n\}_{n\in {\mathbb Z}}$ is periodic modulo 
$3^{k+1}$ with period $13\cdot 3^k$ and that the formula from Theorem \ref{thm:Biluetal} holds for all integers $n$ not only for the positive ones.  

We next assume that $n\ge 18$.
Then \cite[Inequality (2.3)]{BrLu1} 
(also see \cite[Inequality (4.2)]{BrLu2}) shows that
$$
T_{-n}&gt;\alpha^{n/2-1.2\times 10^{16}\log n}.
$$
The condition $n\ge 18$ is needed since $T_{-17}=0$. 
Thus, the analogue of inequality \eqref{eq:0} when $|T_{-n}|=m_1! \cdots m_k!$ for $n\ge 18$ is 
\begin{equation}
\label{eq:4}
n/2-1.2\times 10^{16}\log n&lt;\frac{m_1\log m_1+\cdots+m_k\log m_k}{\log(1.83)}.
\end{equation}
Further, using that Theorem \ref{thm:ML}  holds for $n\in {\mathbb Z}$, we get that 
\begin{align*}
\nu_2(T_{-n}) &amp; \le  \max\{2, \nu_2(n)-1,~\nu_2(n-4)-1,\nu_2(n-1)+3,\nu_2(n-17)+3\}\\
&amp; \le  \frac{\log (n-1)}{\log 2}+3,
\end{align*}
so the analogue of inequality \eqref{eq:1} is 
\begin{equation}
\label{eq:13}
n\ge 2^{x-3}+1.
\end{equation}
The function $y\mapsto y/2-1.2\times 10^{16} \log y$ is increasing for $y&gt;2.4\times 10^{16}$, so if $x\ge 58$, then $2^{x-3}+1\ge 2^{55}+1&gt;2.4\times 10^{16}$, so 
\begin{equation}
\label{eq:6}
n/2-1.2\times 10^{16}\log n\ge 2^{x-4}+0.5-1.2\times 10^{16} (x-3)\log 2.
\end{equation}
We thus get
$$
2^{x-4}+0.5-1.2\times 10^{16} (x-3)\log 2&lt;\frac{3x\log(3x)}{\log(1.83)},
$$
which gives $x\le 62$. Hence, $x\le 62$, so $m_1+m_2+\cdots+m_k\le 3x\le 186$. Thus, 
$$
m_1!\, m_2! \, \cdots \, m_k!\mid (m_1+\cdots+m_k)! \mid 186!.
$$
Further, from \eqref{eq:4} we get
$$
n/2-1.2\times 10^{16}\log n&lt;\frac{3x\log (3x)}{\log(1.83)}&lt;\frac{186\log 186}{\log(1.83)},
$$
so $n&lt;10^{18}$. We are ready to do some Baker-Davenport reduction. This  has been explained and used in many places. For our application, we refer the reader to \cite[Lemma 5.1]{BrLu2} and \cite[Eqs.~(3.4), (3.5)]{BrLu1}.
We write
$$
c_{\beta}\beta^{-n}+c_{\gamma}\gamma^{-n}=\pm m_1!\cdots m_k!-c_{\alpha}\alpha^{-n}.
$$
Thus, assuming $n&gt;6000$, we get
$$
\left|\left(-\frac{c_{\gamma}}{c_{\beta}}\right)\left(\frac{\beta}{\gamma}\right)^{n}-1\right|=\frac{|\pm m_1!\cdots m_k!-c_{\alpha}\alpha^{-n}|}{\alpha^{n/2}}&lt;\frac{186!+1}{\alpha^{n/2}}&lt;\frac{1}{\alpha^{n/4}}.
$$
The last inequality holds for $n&gt;6000$. Writing $\beta/\gamma=e^{2i\theta}$ and also $-c_{\gamma}/c_{\beta}=-e^{2i \omega}$ for some $\omega\in (0,2\pi)$, we get
$$
|e^{i (2n\theta+\pi-2\omega)}-1|&lt;\frac{1}{\alpha^{n/4}}.
$$
The argument from \cite{BrLu1}  ((3.2)--(3.4)) shows that if $\l:=\lfloor (2n\theta+\pi-2\omega)/\pi\rceil$, then the left-hand side above is at least 
$$
2\left| n\left(\frac{2\theta}{\pi}\right)-(l-1)-\frac{2\omega}{\pi}\right|.
$$
Thus, we get
$$
\left|n\left(\frac{2\theta}{\pi}\right)-(l-1)-\frac{2\omega}{\pi}\right|&lt;\frac{2}{\alpha^{n/4}}.
$$
The left-hand side above is $n\tau-m+\mu$, where $\tau=2\theta/\pi$, $\mu=-2\omega/\pi$, $m=l-1$ and the right-hand side above is of the form $A/B^k$, with the parameters $A=2,~B=\alpha$, $k=n/4$. The continued fraction of $\tau$ starts as follows:
\begin{align*}
&amp; [1, 2, 1, 1, 2, 6, 1, 5, 1, 1, 1, 11, 25, 2, 21, 1, 2, 1, 5, 4, 60, 8, 2, 1, 2, 8, 2, 1, 1, 60,\\
&amp; 1, 5, 3, 1, 4, 29, 2, 24, 19, 1,\ldots]
\end{align*}
We take $M:=10^{18}$, which is an upper bound on $n$. We take the $39$th convergent $p_{39}/q_{39}$ and with $q:=q_{39}$, calculate $\varepsilon:=\| q \mu\|-M\| q\tau\|&gt;0.4$, and we get 
by the Baker--Davenport reduction method that 
$$
n/4=k&lt;\frac{\log( Aq/\varepsilon)}{\log B}&lt;\frac{\log(2 q_{39}/0.4)}{\log(1.83)},
$$
which gives $n\le 361$, contradicting our assumption that $n&gt;6000$. So, $n\le 6000$. 

Now inequality \eqref{eq:13} gives $x\le 16$. Since $\nu_2(20!)=17&gt;x$, it follows that $m_k\le 19$. But we can do better. Assume that $m_k\ge 12$.  Then $\nu_2(T_{-n})\ge 
\nu_2(12!)=11$ and $\nu_3(T_{-n})\ge \nu_3(12!)\ge 5$. Theorems \ref{thm:ML} and \ref{thm:Biluetal} show that $n$ is congruent to one of $0,1,4,17$ modulo $2^{10}$ and also modulo $13\cdot 3$. 
Solving the above $4^2$ possibilities with the Chinese remainder lemma we get that the only possibility for which $n&lt;6000$ is $n\equiv 4096\pmod {2^{10}\cdot 3\cdot 13}$, so $n=4096$, but $\nu_3(T_{-4096})=4$ (not $5$). This shows that $m_k\le 11$. 

Finally, let us note that $\nu_3(T_{-n})\le 9$. Indeed, if $\nu_3(T_{-n})\ge 10$, then Theorem \ref{thm:Biluetal} shows that $n\equiv 0,1,4,17\pmod {13\cdot 3^{6}}$ and $13\cdot 3^6&gt;6000$
(certainly $n$ cannot be one of $0,1,4,17$ since then $T_{-n}=0$). So,
$$
m_1!\cdots m_k!=2^a\cdot 3^b\cdot 5^c\cdot  7^d\cdot 11^e,
$$
where $a\le 16,~b\le 9$, $c\le 5,~d\le 4,~e\le 2$. To see the upper bounds on the exponents of the  primes larger  than $3$ above, note that if $c\ge 6$, then $5!^6\mid m_1!\cdots m_k!$, which makes $\nu_2(T_{-n})\ge 18$, a contradiction. The rest are proved in the same way. We created the list of all the numbers of the above form and intersected it with the list of absolute values $\{|T_{-n}|\}_{1\le n\le 6000}$, obtaining 
some values in the intersection with the largest index $n=33$ for which $|T_{-33}|=2^6\cdot 7^2$. From here, we recovered the solutions $n$ for which $|T_{-n}|$ is a product of factorials namely $T_{-2}=1,~T_{-3}=-1,~T_{-5}=2,~T_{-7}=1,~T_{-8}=2^2,~T_{-9}=-2^3$.  This finishes the proof. 

\section{Acknowledgments} 

We thank the anonymous referee for a careful reading of the paper and for comments that improved the quality of our manuscript.  F. L. worked on this paper during a visit to the Max Planck Institute for Software Systems in Saarbr\"ucken, Germany in 2022. This author thanks Professor J. Ouaknine for the invitation and the MPI-SWS for hospitality and support. 


\begin{thebibliography}{9}

\bibitem{Biluetal} Y. Bilu, F. Luca, J. Nieuwveld, J. Ouaknine, and J. B. Worrell, 
On the $p$-adic zeros of the Tribonacci sequence,
arxiv preprint arXiv:2210.16959 [math.NT], 2022. Available
at \url{https://arxiv.org/abs/2210.16959}.

\bibitem{BrLu} J. J. Bravo and F. Luca, 
On the largest prime factor of the $k$-Fibonacci numbers, 
\textit{Int. J. Number Theory} \textbf{9} (2013), 1351--1366.

\bibitem{BrLu1} E. Bravo, C. A. G\'omez, and F. Luca, 
Total multiplicity of the tribonacci sequence, \textit{Colloq. Math.} \textbf{159} (2020), 71--76.

\bibitem{BrLu2} E. Bravo, C. A. G\'omez, B. Kafle, F. Luca, and A. Togb\'e, 
On a conjecture concerning the multiplicity of the tribonacci sequence, \textit{Colloq. Math.} \textbf{159} (2020), 61--69.

 \bibitem{ML} D. Marques and T. Lengyel, 
 On the $2$-adic order of Tribonacci numbers and the equation $T_n=m!$, 
 \textit{J. Integer Sequences} \textbf{17} (2014), 
 \href{https://cs.uwaterloo.ca/journals/JIS/VOL17/Lengyel/lengyel21.html}{Article 14.10.1}.

\end{thebibliography}

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\noindent 2020 {\it Mathematics Subject Classification}: Primary 11B39; Secondary 11A07, 11B50, 11D61. 

\noindent \emph{Keywords: } Tribonacci number, factorial,
Baker--Davenport reduction method.

\bigskip
\hrule
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\noindent (Concerned with sequence \seqnum{A000073}.)

\bigskip
\hrule
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\vspace*{+.1in}
\noindent
Received November 3 2022;
revised versions received  February 3 2023; February 18 2023.
Published in {\it Journal of Integer Sequences}, February 18 2023.

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\noindent
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\htmladdnormallink{Journal of Integer Sequences home page}{https://cs.uwaterloo.ca/journals/JIS/}.
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