\documentclass{mc}
%%%%% journal  info -- DO NOT CHANGE %%%%%%%%%
\setcounter{page}{197}
\renewcommand\thisnumber{x}
\renewcommand\thisyear {2018}
\renewcommand\thismonth{xxx}
\renewcommand\thisvolume{23}
\renewcommand\datereceived{May 11,2017}
\renewcommand\dateaccepted{October 24, 2017}
%%%%%%%%   end journal info   %%%%%%%%%%%%%
\usepackage[displaymath,mathlines,pagewise]{lineno}
% Rad zaprimljen dana:  11.05.2017.

% Rad prihvacen dana: 24.10.2017.

%%%%% author macros %%%%%%%%%
% place your own macros HERE
%%%%% end %%%%%%%%%

%%%%%%%%%    already loaded packages -- DO NOT RELOAD  %%%%%
% \usepackage{amssymb}
% \usepackage{amsmath,amsthm}
% \usepackage{latexsym}
% \usepackage{amsfonts}
% \usepackage{amsbsyt}
% \usepackage{graphicx}
%%%%%%    end   %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\begin{document}
%\begin{linenumbers}	
	%%%%% title and author(s):
	 \markboth{N.\,Irmak, K.\,Liptai and L.\,Szalay}{Factorial-like values in the balancing sequence}
	 \title{Factorial-like values in the balancing sequence}
	
	% multiple authors:
	% Please mark \corrauth after the name of the corresponding author.
	% different addresses:
	\author[N.\,Irmak, K.\,Liptai, L.\,Szalay]{Nurettin Irmak\affil{1}\comma\corrauth,  K\'alm\'an Liptai\affil{2} and L\'aszl\'o Szalay\affil{3,4}}
	\address{\affilnum{1}\ Art and Science Faculty,
		Mathematics Department,\"{O}mer Halisdemir University, Ni\u{g}de TR-51\,240, Turkey\\
		%Sakarya TR–54 187, Turkey
 \affilnum{2}\ Institute of Mathematics and Informatics, Facutly of Natural Sciences, University of Eger,
		 H-3\,300, Eger, Hungary\\
		 \affilnum{3}\ Department of Mathematics and Informatics, J.~Selye University, SK-501\,026 Komarno, Slovakia\\
		  \affilnum{4}\  Institute of Mathematics, University of West Hungary, H-9\,401 Sopron, Hungary}
	%
	%same address:
	%\author[AUTHOR1, AUTHOR2 and AUTHOR3]{AUTHOR1, AUTHOR2\corrauth and AUTHOR3}
	%\address{address of AUTHOR1, AUTHOR2 and AUTHOR3}
	%
	\emails{{\tt nirmak@ohu.edu.tr}\,\,(N.\,Irmak), {\tt liptaik@gemini.ektf.hu} (K.\,Liptai), {\tt szalay.laszlo@uni-sopron.hu} (L.\,Szalay)}
	%
	
	
	
	
	%%%%% Begin Abstract %%%%%%%%%%%
	\begin{abstract}
		In this paper, we solve a few Diophantine equations linked to balancing numbers and factorials. The basic problem consists of solving the equation $B_y=x!$ in positive integers $x$, $y$, which has only one nontrivial solution $B_2=6=3!$, as a direct consequence of the theorem of F.~Luca \cite{luca}. A more difficult problem is to solve $B_y=x_2!/x_1!$, but we were able to handle it under some conditions. Two related problems are also studied.
	\end{abstract}
	%%%%% end %%%%%%%%%%%
	
	%%%%% Keywords %%%%%%%%%%%
	\keywords{Factorials, balancing numbers, diophantine equation}
	
	%%%% AMS subject classifications %%%%
	\ams{11B39, 11D72}
	
	%%%% maketitle %%%%%
	\maketitle

\section{Introduction}

The balancing sequence $\left\{ B_{n}\right\}$ is given by $B_{0}=0$ and $B_{1}=1$, and by the
recursive rule $B_{n}=6B_{n-1}-B_{n-2}$ for $n\geq2$. The $n^{th}$ element of the associate sequence of $\left\{ B_{n}\right\}$ is denoted by $C_{n}$, which satisfies the recurrence relation  $C_{n}=6C_{n-1}-C_{n-2}$\ ($n\ge2$), where the initial values are $C_{0}=2$ and $C_{1}=6$.
The elements of the sequence $\left\lbrace C_n\right\rbrace $ are often called Lucas-balancing numbers.
Note that
\begin{equation}\label{agnes}
B_{2n}=B_n C_n,
\end{equation}
and
\begin{equation}\label{by}
C_n^2-32B_n^2=4.
\end{equation}
These two identities can be obtained similarly to those for Fibonacci and Lucas numbers. Observe that
\begin{equation}\label{marton}
B_1=1=1!,\quad C_0=2=2!,\quad B_2=C_1=6=3!,
\end{equation}
so the question arises naturally whether there are other factorial values in $\left\{B_{n}\right\}$ or in $\left\{C_{n}\right\}$. More generally, one may claim the solutions to the Diophantine equations
\begin{equation}\label{kezdo}
B_y=x_1!\cdot x_2!\cdots x_r!, \qquad C_y=x_1!\cdot x_2!\cdots x_r!,
\end{equation}
in the positive integers $x_1,\dots,x_r$, and $y$.
Today it is an easy question since Luca (\cite{luca}, Theorem 4) proved the following theorem. A Lucas sequence $\{u_n\}$ is a non-degenerate binary recurrence with the initial values $u_0=0$, $u_1=1$. Let $\cal PF$ be the set of all positive integers which can be written as a product of factorials.
\begin{theorem}[see \cite{luca}, Theorem 4] Let $\left( u_n\right)_{n\ge0} $ be a Lucas sequence. Let $\alpha$ and $\beta$ denote the two roots of the characteristic equations. Suppose that $\left| \alpha\right| \ge \left| \beta\right|$. If $\left|u_n \right| \in {\cal PF} $, then

\begin{equation}\label{boundd}
y\le\max\left\lbrace 12, 2e|\alpha|+1\right\rbrace.
\end{equation}

\end{theorem}
The same upper bound is true for the associate sequence of $\{u_n\}$. Luca used deep algebraic number theoretical considerations and the Baker method.

In the case of balancing a sequence and its associate sequence, the zeros of the characteristic polynomial $x^2-6x+1$ are $\alpha=3+2\sqrt{2}$ and $\beta=3-2\sqrt{2}$. Thus, by (\ref{boundd}) we obtain $y\le32$. Using a brute force algorithm,   computer search provides only (\ref{marton}) as all the solutions to (\ref{kezdo}). (All of them are single factorial terms.)

For the Fibonacci sequence given by $F_0=0$, $F_1=1$, and $F_n=F_{n-1}+F_{n-2}$, Luca's bound
is only 53, and $u_y=x_1!\cdot x_2!\cdots x_r!$ is fulfilled in the cases
\[
F_1=F_2=1!,\quad F_3=2!,\quad F_6=(2!)^3,\quad F_{12}=(2!)^2(3!)^2=3!\cdot4!.
\]

Consider now the Tribonacci sequence defined by $T_0=0$, $T_1=T_2=1$ and by $T_n=T_{n-1}+T_{n-2}+T_{n-3}$. The equation
\[
T_y=x!
\]
was solved by Marques and Lengyel \cite{marques}, and it showed that the only solutions are
$(y,x)=(1,1)$, $(2,1)$, $(3,2)$, $(7,4)$.
Their proof is based on the determination of the $2$-adic order of Tribonacci numbers. The $p$-adic order of the non-zero integer $n$  denoted by $\nu _{p}\left( n\right)$  is defined
by the exponent of the highest power of prime $p$ dividing $n$.

The main purpose of this paper is to investigate the solvability of three Diophantine equations linked to factorials and balancing numbers:
\[
B_y=\frac{x_2!}{x_1!},\qquad B_y=\frac{x_2!}{x_1},\qquad B_y=x_1x_2!,
\]
in positive integers $x_1$, $x_2$ and $y$, under some conditions on $x_1$ and $x_2$.

One important argument, which will be used later, is the characterization of the $2$-adic order of balancing numbers as it has been already obtained for Fibonacci numbers by Lengyel \cite{lengyel}.
\begin{theorem}\label{thm1}
For $n\geq 1,$ we have
\begin{equation*}\label{2adic.balancing}
\nu _{2}\left( B_{n}\right) =\left\{
\begin{tabular}{ll}
$0,$ & if $n\equiv 1\;\left(\bmod\,2\right)  $ \\[0.5ex]
$1,$ & if $n\equiv 2\;\left(\bmod\,4\right)  $ \\[0.5ex]
$\nu _{2}\left( n\right),$ & if $n\equiv 0\;\left(\bmod\,4\right)  $
\end{tabular}
\right.  .
\end{equation*}
\end{theorem}


Precise results are formulated in the next three theorems. We note that in Theorem \ref{th01} the value $\delta=0.98$  and in Theorem \ref{th03} the value $K=10^6$ were choosen to carry out   precise calculations. Our method, at least in theory, works for arbitrary $0<\delta<1$ and $K\ge1$.

\begin{theorem}\label{th01}
Unless $(x_1,x_2,y)=(1,3,2)$, the Diophantine equation
\[
B_y=\frac{x_2!}{x_1!},
\]
in positive integers $y$, $x_1$, $x_2$ with $x_1+2\le x_2$ has no solution in the folowing cases:
\begin{itemize}
	\item $x_1\le 0.98 x_2$, or
	\item $x_1=x_2-\eta$, where $\eta\in\left\lbrace 2,3,4\right\rbrace $.
\end{itemize}

\end{theorem}

\begin{theorem}\label{th02}
The only solutions of the Diophantine equation
$$
B_y=\frac{x_2!}{x_1}
$$
in positive integers $x_1$, $x_2$ and $y$ with $x_1\le x_2$ are
\begin{equation*}
(x_1,x_2,y)=(1,1,1),\;(2,2,1),\;(1,3,2),\;(4,4,2).
\end{equation*}
\end{theorem}

\begin{theorem}\label{th03}
The only solutions of the Diophantine equation
\[
B_y=x_1 x_2!
\]
in positive integers $x_1$, $x_2$ and $y$ with the condition $x_1\le10^6 x_2$ are
\begin{eqnarray*}
%(x_1,x_2,y)&=&(1,1,1),\;(3,2,2),\;(102,2,4),\;(3465,2,6),\\
%&&(117708,2,8),\;(1,3,2),\;(34,3,4),\;(1155,3,8),\\
%&&(39236,3,8),\;(1332869,3,10),\;(9809,4,8).
(x_1,x_2,y) =&&(1,1,1),\;(6,1,2),\;(3,2,2),\;(1,3,2),\;(35,1,3),\\
&&(204,1,4),\;(102,2,4),\;(34,3,4),\;(1189,1,5),\;(6930,1,6),\\
&& (3465,2,6),\;(1155,3,6),\;(40391,1,7),\;(235416,1,8),\\
&& (117708,2,8),\;(39236,3,8),\;(9809,4,8),\;(1332869,3,10).
\end{eqnarray*}
\end{theorem}



\section{Preliminaries}

In this section, we present several lemmas which help us to prove the theorems.

\begin{lemma}\label{lmm1}
If $n\geq 2$, then
\begin{equation*}
\alpha ^{n-1}<B_{n},
\end{equation*}
where $\alpha$ is the larger zero in absolute value
of the characteristic polynomial of the sequence $\left\lbrace B_n\right\rbrace $.
\end{lemma}

\begin{proof}
See Lemma 4 in \cite{irmak}.
\end{proof}

Theorem \ref{thm1} is an immediate consequence of the following lemma; it states a bit more when $4$ does not divide $n$.

\begin{lemma} \label{lmm2}
Let $n$ be a positive integer.
\begin{enumerate}
\item $B_{n}\equiv n\left(\bmod\,4\right)$.
\item Let $n=2^sr$ for some integers $s\geq 2$ and odd $r$. Then $B_{2^{s}r}\equiv 2^{s}r\;\left(\bmod\,2^{s+1}\right)$.
\end{enumerate}
\end{lemma}

\begin{proof}

(1) It is an easy consequence of considering the sequence of balancing numbers $B_n$ modulo 4.

(2) We use induction on $s$. Assume $s=2$. The balancing sequence modulo 8 begins with
\[0,1,6,3,4,5,2,7,0,1,\dots.\]
Clearly, the length of the period is 8 and $B_{4r}\equiv4\equiv 4r\;(\bmod\,8)$ (note that here $r$ is odd).

Suppose that the statement is true for some $s\ge2$ and $r$ odd, i.e. $B_{2^{s}r}=2^{s}r+2^{s+1}k$ holds for some positive integer $k$. It is easy to see that $C_n\equiv2\;(\bmod\,4)$. Thus $C_n=4u_n+2$ for some positive integer sequence $\{u_n\}$. Applying   identity (\ref{agnes}), we have
\begin{eqnarray*}
B_{2^{s+1}r}&=&C_{2^{s}r}B_{2^{s}r}=(4u_{2^{s}r}+2)\left(2^{s}r+2^{s+1}k\right)\\
&=&2^{s+1}r+2^{s+2}\left(k+u_{2^{s}r}r+2u_{2^{s}r}k\right) \\
&\equiv& 2^{s+1}r\; (\bmod\,2^{s+2}),
\end{eqnarray*}
and the proof of the lemma is complete (and Theorem \ref{thm1} follows).
\end{proof}

Let $s_p(k)$ denote the sum of the base-$p$ digits of the positive integer $k$.
%As usual, $\lfloor z\rfloor$ will denote the floor function of the real number $z$.

\begin{lemma}[Legendre]
For any integer $k\ge1$ and $p$ prime, we have
\[
\nu_p(k!)=\frac{k-s_p(k)}{p-1}.
\]
\end{lemma}

\begin{proof}
See \cite{legendre}.
\end{proof}

The result of Legendre has the following consequence.	
																														
\begin{corollary}\label{cor4}
For any integer $k\geq 2$ and prime $p$ the inequalities
\begin{equation*}
\frac{k}{p-1}-\frac{\log k}{\log p}-1\;\leq\; \nu_{p}\left( k!\right) \;\leq \;\frac{k-1}{p-1}
\end{equation*}
hold.
\end{corollary}

\begin{proof}
Consider the maximal and minimal values of $s_p(k)$, respectively.
\end{proof}

\section{General approach to the proofs}

This approach does not affect the problem $B_y=x_2!/ x_1!$ with $x_2-x_1=c$, $c\in\{1,2,3\}$.

For a given  positive integer $r$ and the integer valued function
$f(x_1,x_2,\dots,x_r)$
we would like to solve the Diophantine equation
\begin{equation}\label{main}
B_y=f(x_1,x_2,\dots,x_r)
\end{equation}
in the positive integers $y,x_1,\dots,x_r$.
Recall Theorem \ref{thm1} to remind us that the value $\nu_2(B_n)$ is rather small. If we are able to give a ``good'' lower bound for the ``sufficiently large'' $\nu_2(f(x_1,x_2,\dots,x_r))$, meanwhile we can provide a ``good'' upper bound for $f(x_1,x_2,\dots,x_r)$, then there is a chance to bound the variables.
More precisely, Lemma \ref{lmm1} leads to
\[
y<1+\frac{\log f(x_1,x_2,\dots,x_r)}{\log\alpha}
\]
starting from (\ref{main}). Theorem \ref{thm1} implies
\[
\nu_2(f(x_1,x_2,\dots,x_r))=\nu_2(B_y)\le\nu_2(y)\le\frac{\log y}{\log 2}.
\]

Combining the last two formulas, we obtain
\begin{equation}\label{fo}
\nu_2(f(x_1,x_2,\dots,x_r))<\frac{1}{\log 2}\log\left(1+\frac{\log f(x_1,x_2,\dots,x_r)}{\log\alpha}\right).
\end{equation}
We succeed if the comparison of the two sides bounds the variables. This will happen in the following cases:

\begin{enumerate}
	\item $f(x_1,x_2)=x_2!/ x_1!$, with the condition $x_1\le \delta x_2$ for some $0<\delta<1$,
	\item $f(x_1,x_2)=x_2!/x_1$ with $x_1\le x_2$,
	\item $f(x_1,x_2)=x_1x_2!$ with the restriction $x_1\le Kx_2$ for some positive integer $1\le K$.
	
\end{enumerate}



\section{Proof of the theorems}

\subsection{Proof of Theorem \ref{th01}}
\medskip

{\bf \noindent Case 1. $x_1+2<x_2$ and $x_1\le \delta x_2$ with a fixed $0<\delta<1.$}

Assume that the positive integer solutions $x_1$, $x_2$ and $y$ satisfy $x_1+2<x_2$ and $x_1\le\delta x_2$ with a fixed $0< \delta<1$.

Corollary \ref{cor4} provides
\begin{eqnarray*}
\nu_2\left(\frac{x_2!}{x_1!}\right)&=&\nu_2(x_2!)-\nu_2(x_1!)\ge x_2-\frac{\log x_2}{\log 2}-1-(x_1-1) \\
&\ge& (1-\delta)x_2-\frac{\log x_2}{\log 2}.
\end{eqnarray*}
On the other hand,
\[
\frac{x_2!}{x_1!}\le x_2!\le \left(\frac{x_2}{2}\right)^{x_2}
\]
follows  where we applied the well-known identity $k!\le(k/2)^k$.
The preparation till now enables us to apply (\ref{fo}). It leads to
\begin{equation}\label{eren}
(1-\delta)x_2-\frac{\log x_2}{\log 2}<\frac{1}{\log 2}\log\left(1+\frac{x_2\log(x_2/2)}{\log \alpha}\right).
\end{equation}
For fixed $\delta$, it provides an upper bound for $x_2$. Indeed, if $x_2$ is large enough, the left-hand side of (\ref{eren}) is positive, further the leading term is linear, while the right-hand side is approximately logarithmic in $x_2$. For instance, if $\delta=49/50$, then $x_2\le1102$.
Making a simple computer verification in the range $3< x_2\le 1102$, $1\le x_1\le x_2-2$, $x_1\le49/50x_2$, according to (\ref{by}), we find a balancing number if
\[
\sqrt{8\left(\frac{x_2!}{x_1!}\right)^2+1}
\]
is an integer. It occurs only in the case $(x_1,x_2)=(1,3)$, which gives $B_y=6$, and then $y=2$. Taking another example, say $\delta=1-10^{-6}$, we obtain $x_2<5.5\cdot 10^7$. This bound is too large, even to check possible cases by a computer!
\medskip

{\bf \noindent Case 2. $x_1=x_2-2$.}

We have to solve $B_y=x_2\left(x_2-1 \right)$. Put $z=C_n$. Then $z^2=32x_2^2(x_2-1)^2+4$ via $z_1=z/2$ leads to the equation
\[
z_1^2=8x_2^4-16x_2^3+8x_2^2+1.
\]
To this equation, the Magma procedure
\begin{center}
{\tt IntegralQuarticPoints([8,-16,8,0,1]);}
\end{center}
determines the solutions
\[
(x_2,z_1)=(-2,\pm17),\,(0,\pm1),\,(1,\pm1),\,(3,\pm17).
\]
Only the last one provides solution to $B_y=x_2(x_2-1)$, namely $B_2=6=3\cdot2$, i.e.~$(x_2,y)=(3,2)$.
\medskip

{\bf \noindent Case 3. $x_1=x_2-3$.}

Now, our task is to solve $B_y=x_2(x_2-1)(x_2-2)$. Let $z=C_n$ and $t=x_2-1$. Then we have
\[z^2=32(t-1)^2t^2(t+1)^2+4=32(t^2-1)^2t^2+4.\]
Applying $z=2z_1$ and $t_1=t^2$, and multiplying the equation by $3^6$, together with $t_1=(T-4)/6$, we arrive at
the elliptic equation
\begin{equation}\label{szombat}
(27z_1)^2=T^3-108T+1161.
\end{equation}

We used Magma ({\tt E:=EllipticCurve([-108,1161]);IntegralPoints(E);}) to solve (\ref{szombat}), and we got
\[
(T,27z_1)=(-12,\pm27),\,(-2,\pm37),\,(6,\pm27),\,(15,\pm54),\,(60,\pm459).
\]
None of them gives a solution to $B_y=x_2(x_2-1)(x_2-2)$ with the given conditions.

\medskip

{\bf \noindent Case 4. $x_1=x_2-4$.}

The corresponding equation is $B_y=x_2(x_2-1)(x_2-2)(x_2-3)$. Put $z=C_n$. Then $z^2=32x_2^2(x_2-1)^2(x_2-2)^2(x_2-3)^2+4$ via $z_1=z/2$ and $t=x_2^2-3x_2+1$ leads  to
\[
z_1^2=8t^4-16t^2+9.
\]
{\tt IntegralQuarticPoints([8,0,-16,0,9]);}
returns with
\[
(t,z_1)=(\pm6,\pm99),\,(\pm1,\pm1),\,(0,\pm3).
\]
Clearly, none of them leads to a solution of $B_y=x_2(x_2-1)(x_2-2)(x_2-3)$.


\subsection{Proof of Theorem \ref{th02}}

Here $f(x_1,x_2)=x_2!/x_1$ assuming $x_1\le x_2$. Thus
\begin{equation*}
\nu_2(B_y)=\nu_2(x_2!/x_1)=\nu_2(x_2!)-\nu_2(x_1)\ge x_2-1-2\frac{\log x_2}{\log 2}.
\end{equation*}
Further
\[
\frac{x_2!}{x_1}\le x_2!\le \left(\frac{x_2}{2}\right)^{x_2}
\]
follows. Putting them together to apply (\ref{fo}),  we obtain
\begin{equation*}
x_2-1-2\frac{\log x_2}{\log 2}<\frac{1}{\log 2}\log\left(1+\frac{x_2\log(x_2/2)}{\log \alpha}\right).
\end{equation*}
It provides $2 \le x_2\le11$. Lastly, we checked the possible values of $x_1$ and $x_2$, and found three solutions.

\subsection{Proof of Theorem \ref{th03}}

Now we study the function $f(x_1,x_2)=x_1x_2!$ with the restriction $x_1\le Kx_2$, where $K=10^6$.
\begin{equation*}
\nu_2(B_y)=\nu_2(x_1x_2!)=\nu_2(x_1)+\nu_2(x_2!)\ge x_2-1-\frac{\log x_2}{\log 2}
\end{equation*}
follows by Corollary \ref{cor4}. Also,
\[
x_1x_2!\le x_1\left(\frac{x_2}{2}\right)^{x_2}
\]
holds, so together with (\ref{fo}) we have
\begin{eqnarray*}
x_2-1-\frac{\log x_2}{\log 2}&<&\frac{1}{\log 2}\log\left(1+\frac{\log x_1+x_2\log(x_2/2)}{\log \alpha}\right)\\
&\le&\frac{1}{\log 2}\log\left(1+\frac{\log K+\log x_2+x_2\log(x_2/2)}{\log \alpha}\right).
\end{eqnarray*}
The solution of the inequality above for $K=10^6$ is $x_2\le8$. A computer verification for $B_y=x_1x_2!$ returns   18 solutions described in the theorem.


\begin{thebibliography}{10}

%\bibitem{bollman} M. Bollman, H. S. Hern´andez, and F. Luca, Fibonacci numbers which are sums of three factorials, Publ. Math. Debrecen 77 (2010), 211–224.

%\bibitem{grossman} G. Grossman and F. Luca, Sums of factorials in binary recurrence sequences, J. Number Theory 93 (2002), 87--107.

\bibitem{irmak} {\sc M.\,Alp, N.\,Irmak, L.\,Szalay,} {\em Balancing Diophantine
	Triples}, Acta Univ. Sapientiae {\bf 4}(2012), 11--19.

\bibitem{legendre} {\sc A.\,M.\,Legendre},  {\em Theorie des Nombres.} Firmin Didot Freres,
Paris, 1830.

\bibitem{lengyel} {\sc T.\,Lengyel,} {\em The order of the Fibonacci and Lucas numbers,}
Fibonacci Quart. {\bf 33}(1995), 234--239.

\bibitem{marques} {\sc T.\,Lengyel and D. Marques,} {\em The $2$-adic order of the
	Tribonacci numbers and the equation $T_{n}=m!,~$}  J. Integer Seq. {\bf 17}(2014), Article 14.10.1.

\bibitem{luca}  {\sc F.\,Luca,} {\em Products of factorials in binary recurrence
	sequences}, Rocky Mountain J. Math. {\bf 29}(1999), 1387--1411.

%\bibitem{siksek} F. Luca and S. Siksek, Factorials expressible as sums of at most three Fibonacci numbers, Proc. of the Edinburgh Math. Soc. 53 (2010), 679--729.

%\bibitem{luca2} F. Luca and P. Stanica, $ F_{1}F_{2}F_{3}F_{4}F_{5}F_{6}F_{8}F_{10}F_{12}=11!$, Port. Math. 63 (2006), 251--260.

%\bibitem{ray} P. K. Ray, Balancing and Cobalancing Numbers, PhD. thesis, Rourkela, India, 2009.

\end{thebibliography}
%\end{linenumbers}
\end{document}

-----------------------------

The Binet-type formulae of these sequences are
\begin{equation*}
B_{n}=\frac{\alpha ^{n}-\beta ^{n}}{\alpha -\beta }\qquad\text{ and }\qquad C_{n}=\alpha^{n}+\beta ^{n},
\end{equation*}%
where are the distinct roots of
the characteristic polynomial $x^2-6x+1$ of the balancing sequence. (For details, see
\cite{ray}.)

---------------------------

 It was
shown that there are only finitely many positive integers $n$%
\begin{equation}
F_{n}=m_{1}!+m_{2}!+\cdots +m_{k}!  \label{fact}
\end{equation}%
for fixed integer $k$ by Grossman and Luca \cite{grossman}. Moreover, all solutions were
given for the case $k\leq 2.$ Bollman, Hernandes and Luca \cite{bollman} solved the
equation (\ref{fact}) for the case $k=3.$ Luca and Siksek found all
factorials as the sum of at most three Fibonacci number (see \cite{siksek}).

---------------------------

The history of diophantine equation including factorials and the members of
linear recurrence is very rich. Luca \cite{luca} showed that $F_{n}$ is a
product of factorials for $n=1,2,3,6,12.$ Moreover, the largest product of
distinct Fibonacci numbers which is a product of factorials is $%
F_{1}F_{2}F_{3}F_{4}F_{5}F_{6}F_{8}F_{10}F_{12}=10!$ in \cite{luca2}.

----------------------------

\begin{lemma}\label{lmm1general}
Let $A$ and $B\ne0$, further $G_0$ and $G_1$ are arbitrary complex numbers. Define the binary recurrence sequence $\{G_n\}$ by $G_0$ and $G_1$, and by $G_n=AG_{n-1}+BG_{n-2}$ for $n\ge2$. Then
$$
G_n=(A^2+2B)G_{n-2}-B^2G_{n-4}
$$
holds for any $n\ge4$.
\end{lemma}
\begin{proof}
For $n\ge2$ we have $AG_{n-1}=G_n-BG_{n-2}$, therefore for $n\ge4$
\begin{eqnarray*}
G_n&=&AG_{n-1}+BG_{n-2}=A(AG_{n-2}+BG_{n-3})+BG_{n-2}\\
&=&(A^2+B)G_{n-2}+ABG_{n-3}\\
&=&(A^2+B)G_{n-2}+B(G_{n-2}-BG_{n-4})\\
&=&(A^2+2B)G_{n-2}-B^2G_{n-4}
\end{eqnarray*}
follows.
\end{proof}

\begin{remark}
If $\{H_n\}$ is the associate sequence of $\{G_n\}$, then $H_0=2G_1-AG_0$ and $H_1=AG_1+2BG_0$. Assuming $G_0=0$, $G_1=1$, we have $H_0=2$, $H_1=A$. In this specific case Lemma \ref{lmm1general} can be rewritten as
$$
G_n=H_2G_{n-2}-B^2G_{n-4}.
$$
Indeed, $H_2=AH_1+BH_0=A^2+2B$.
\end{remark}

\begin{corollary} (of Lemma \ref{lmm1general} and the previous remark)
$$G_n=H_4G_{n-4}-B^4G_{n-8}.$$
\end{corollary}
\begin{proof}
Obviously, $H_4=A^4+4A^2B+2B^2$.
Put $V_{n-j}=G_{n-2j}$ for non-negative integers $j$. Lemma \ref{lmm1general} says that $V_n=(A^2+2B)V_{n-1}-B^2V_{n-2}$. Apply now Lemma \ref{lmm1general} for the sequence $\{V_n\}$ to have
\begin{eqnarray*}
G_n&=&V_n=(((A^2+2B)^2+2(-B^2))V_{n-2}-(-B^2)^2V_{n-4}\\
&=&(A^4+4A^2B+2B^2)V_{n-2}-B^4V_{n-4}\\
&=&H_4G_{n-4}-B^4G_{n-8}.
\end{eqnarray*}
\end{proof}

If we assume that the sequence $\{G_n\}$ is the balancing sequence $\{B_n\}$, then $\{H_n\}$ is $\{C_n\}$. Thus
\begin{equation}\label{C2}
B_n=C_2B_{n-2}-B_{n-4},
\end{equation}
further
\begin{equation}\label{C4}
B_n=C_4B_{n-4}-B_{n-8}.
\end{equation}







\begin{equation*}
	\nu _{2}\left( T_{n}\right) =\left\{
	\begin{tabular}{ll}
		$0,$ & if $n\equiv 1,2\;\left({\bmod}\,4\right)$; \\
		$1,$ & if $n\equiv 3,11\;\left({\bmod}\,16\right)$; \\
		$2,$ & if $n\equiv 4,8\;\left({\bmod}\,16\right)$; \\
		$3$ & if $n\equiv 7\;\left({\bmod}\,16\right)$; \\
		$\nu _{2}\left( n\right) -1$ & if $n\equiv 0\;\left({\bmod}\,16\right)$; \\
		$\nu _{2}\left( n+4\right) -1$ & if $n\equiv 12\;\left({\bmod}\,16\right)$; \\
		$\nu _{2}\left( \left( n+1\right) \left( n+17\right) \right) -3$ & if $n\equiv15\;\left({\bmod}\,16\right)$.
	\end{tabular}
	\right.
\end{equation*}




each of them contains conditions on the unknown $x_1$.
The first question possesses the trivial solutions $x_1=x_2$, and $x_1=x_2-1$, therefore we will assume $x_1+2\le x_2$. Further restriction will also be supposed to handle different cases since the problem in its generality is rather difficult.
The second equation is considered under the assumption $x_1\le x_2$. Finally, in case of $B_y=x_1x_2!$ we take $x_1\le Kx_2$ for some large $K$.
