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\begin{document}
\vspace*{-60pt} 
\centerline{\smalltt INTEGERS: 
 \smallrm ELECTRONIC JOURNAL OF COMBINATORIAL NUMBER THEORY \smalltt 5 
(2005), \#A02} 
\vskip 50pt 
\begin{center}
 \uppercase{\bf On the number of ways of writing $t$ as a product of
factorials}
 \vskip 20pt
 {\bf Daniel M. Kane}\\
{\tt dankane@mit.edu}
 \end{center}
 \vskip 30pt
 \centerline{\smallit Received:10/6/04, Revised: 12/20/04, Accepted:
2/5/05, Published: 2/8/05}
 \vskip 30pt


\centerline{\bf Abstract}
\noindent
Let $\No$ denote the set of non-negative integers.  In this paper
we prove that
$$
\limsup_{t\rightarrow \infty} \left| \left\{ (n,m)\in \No^2 :
n!m!=t \right\} \right| = 6.
$$


\pagestyle{myheadings}
 \markright{\smalltt INTEGERS: \smallrm ELECTRONIC JOURNAL OF COMBINATORIAL NUMBER THEORY \smalltt 5 (2005),
\#A02\hfill}

 \thispagestyle{empty} 
 \baselineskip=15pt 
 \vskip 30pt


\section*{\normalsize 1. Introduction}
\addtocounter{section}{+1}
Let $\No$ denote the set of non-negative integers.  In this paper
we will prove that
$$
\limsup_{t\rightarrow \infty} \left| \left\{ (n,m)\in \No^2 :
n!m!=t \right\} \right| = 6.
$$
We use three techniques to prove this result.  First, it is not
difficult to generate an infinite set of $t$ each of which has at
least 6 representations as a product of factorials thus establishing
the lower bound. We then use considerations of the number of times
two divides $t$ in order to show that all of the solutions must be
near each other. Lastly we use some analytic techniques analogous to
those in \cite{kn:kane}.

The following three conjectures also seem likely.

\begin{conj}
$$
\max \left| \left\{ (n,m)\in \No^2 : n!m!=t \right\} \right| = 6.
$$
\end{conj}

\begin{conj}
$$
\limsup_{t\rightarrow \infty} \left| \left\{ (n,m)\in \N^2 : n!m!=t
\right\} \right| = 4.
$$
\end{conj}

\begin{conj}
$$
\max \left| \left\{ (n,m)\in \N^2 : n!m!=t \right\} \right| = 4.
$$
\end{conj}

It is true that conjecture 3 would imply both other conjectures, and
that any of these conjectures is stronger than our main theorem.

\section*{\normalsize 2. The lower bound}
\addtocounter{section}{+1}
Notice that for any integer $n>2$, we have that
$$
(n!)! = 0!\cdot (n!)! = 1!\cdot (n!)! = n!\cdot (n!-1)! =
(n!-1)!\cdot n! = (n!)!\cdot 1! = (n!)! \cdot 0!.
$$
Therefore, we have that
$$
\limsup_{t\rightarrow \infty} \left| \left\{ (n,m)\in \No^2 :
n!m!=t \right\} \right| \geq 6.
$$

\section*{\normalsize 3. The first technique}
\addtocounter{section}{+1}
For a positive integer $n$, let $e(n)$ denote the largest $k$ so
that $2^k$ divides $n$.  Notice that
\begin{eqnarray*}
e(n!) &=& \sum_{i=1}^{\left\lfloor \log_2 (n) \right\rfloor}
\left\lfloor
\frac{n}{2^i} \right\rfloor \\
& = & \sum_{i=1}^{\lfloor \log_2 (n) \rfloor} \frac{n}{2^i} -
\sum_{i=1}^{\lfloor \log_2 (n) \rfloor} O(1)\\
& = & n + O(\log n).
\end{eqnarray*}

Therefore we have that if $n!m! = t$, then $e(n!) + e(m!) =e(t)$,
and therefore, $n+m+O(\log n + \log m) = e(t)$.  Since
$(n/2)^{n/2}<n!<t$, we have that $n<\log t$ for sufficiently large
$t$.  Therefore, for sufficiently large $t$, $n+m+O(\log\log t) =
e(t)$.  Hence if $n_1!m_1! = n_2!m_2! = t$, then $n_1+m_1 = n_2+m_2
+ O(\log \log t)$.  This fact provides an elementary proof that for
fixed $t$ the number of solutions to $n!m!=t$ is $O(\log \log t)$
because by convexity of the log of the factorial function, at most
two solutions to $n!m!=t$ have a given sum of $n+m$, and this sum
cannot vary by more than $O(\log \log t)$.

\section*{\normalsize 4. The second technique}
\addtocounter{section}{+1}
Our second technique is similar to that used in \cite{kn:kane}. We
begin with the following lemma:

\begin{lem}
\label{rolleslemma} If $F(x):\R \rightarrow \R$ is an infinitely
differentiable function and if $F(x)=0$ for
$x=x_1,x_2,...,x_{n+1}$ (where $x_1<x_2<...<x_{n+1}$), then
$F^{(n)}(y)=0$ for some $y\in(x_1,x_{n+1}).$
\end{lem}
\begin{proof} We proceed by induction on $n$.  The
case of $n=1$ is Rolle's Theorem.  Given the statement of Lemma
2.1 for $n-1$, if there exists such an $F$ with $n+1$ zeroes,
$x_1<x_2<...<x_{n+1}$, then by Rolle's theorem, there exist points
$y_i\in(x_i,x_{i+1})$ ($1\leq i \leq n$) so that $F'(y_i)=0$. Then
since $F'$ has at least $n$ roots, by the induction hypothesis
there exists a $y$ with $x_1< y_1 < y < y_n < x_{n+1}$, and
$F^{(n)}(y)=(F')^{(n-1)}(y)=0$. \end{proof}

We now state a lemma that helps us to  count the number of integer
points on smooth curves.

\begin{lem}
\label{continuoslemma} Let $f:\R\rightarrow \R$ be a $C^k$
function. Suppose that for $x\in(a,b)$, that
$$
0<\left|\frac{1}{k!}\frac{\partial^k}{\partial
x^k}f(x)\right|<\alpha.
$$
Then if we have $a<x_0<x_1<...<x_k<b$ where $x_i\in\Z$ and
$f(x_i)\in\Z$ for all $0\leq i \leq k$, then $x_k-x_0\geq
\alpha^{\frac{-2}{k(k+1)}}.$
\end{lem}
\begin{proof}
Let
$$
g(x) = \sum_{i=0}^k f(x_i)\prod_{\substack{0\leq j\leq k \\ i \neq
j}} \frac{x-x_j}{x_i-x_j}
$$
be the polynomial of degree $k$ that interpolates $f$ at the
$x_i$.  Let $h(x)=f(x)-g(x)$.  Then $h(x_i)=0$.  Hence by Lemma
\ref{rolleslemma} we have that for some $a<x_0<y<x_k<b$ that
$\frac{\partial^k}{\partial x^k}h(y)=0.$ Or that
$$
\left(\frac{1}{k!}\frac{\partial^k}{\partial x^k}f(x)\right)_{x=y}
=\left(\frac{1}{k!}\frac{\partial^k}{\partial x^k} g(x)\right)_{x=y}
= \sum_{i=0}^k f(x_i)\prod_{\substack{0\leq j\leq k \\ j \neq i}}
\frac{1}{x_i-x_j}.
$$
Therefore,
$$
s=\left(\frac{1}{k!}\frac{\partial^k}{\partial
x^k}f(x)\right)_{x=y}
$$
is an integer multiple of $M = \prod_{0\leq i < j \leq k}
\frac{1}{x_j-x_i}$.  Therefore, either $s=0$ or else $|s| \geq M$.
But by assumption, $0<|s|<\alpha$.  Therefore, $\alpha \geq |s|
\geq M$.  Hence $\alpha \geq (x_k-x_0)^{\frac{-k(k+1)}{2}}$, and
hence we have that $\alpha^{\frac{-2}{k(k+1)}}\leq x_k -x_0$ as
desired.
\end{proof}

We will also make use of a generalization of Stirling's formula
which states that:
$$
\log(\Gamma(z+1)) = (z+\half)\log(z) - z + \half \log(2\pi) +
O(z^{-1})
$$
uniformly for $\Re(z)>0$. This follows readily from the $m=2$ case
of
\begin{eqnarray*}
\log\Gamma(z+1) & = &  \frac{1}{2}\log(2\pi) +
\left(z+\frac{1}{2}\right)\log(z)-z \\
&&+\sum_{j=1}^{m}\frac{B_{2j}}{(2j-1)(2j)z^{2j-1}}-\frac{1}{2m}\int_0^\infty
\frac{B_{2m}(x-[x])}{(x+z)^{2m}}dx.
\end{eqnarray*}
where $B_{2j}$ and $B_{2m}$ are the Bernoulli numbers and Bernoulli
polynomials (see \cite{kn:Rademacher}).

\section*{\normalsize 5. The Strategy}
\addtocounter{section}{+1}
We have yet to prove that for sufficiently large $t$
$$
\left| \left\{ (n,m)\in \No^2 : n!m!=t \right\} \right| \leq 6.
$$
It is
sufficient to show that $\left| \left\{ (n,m)\in \No^2 : n\geq m,
n!m!=t \right\} \right| \leq 3$ for all sufficiently large $t$. We
will split solutions of this form into three overlapping cases:

\noindent 1. $m<\exp(2\sqrt{\log \log t})$

\noindent 2. $m>\exp(\sqrt{\log \log t}), n-m > (\log t) ^
{25/36}$

\noindent 3. $n-m < (\log t)^{26/36}$

Furthermore, we will show by our results from sections 3 and 4,
that for all sufficiently large $t$, that all integer solutions to
$n!m! =t $ lie in one of these regions.

Define the function $f:\R \rightarrow \R$ implicitly by
$\Gamma(f(x)+1)\Gamma(x+1) = t$.  It is clear that
$$
f'(x) = -\frac{g(x)}{g(f(x))}
$$
where
$$
g(x)=\frac{\partial}{\partial x}\log \Gamma(x+1) = \log (x) +
O(x^{-1})
$$
(by our strong form of Stirling's formula).  So
$$
f'(x) = -\frac{\log x+O(1)}{\log (f(x))}.
$$
So if we have two pairs of solutions $(n_1,m_1),(n_2,m_2)$ to
$n\geq m, n!m!=t$, where $m_2>m_1$, then
$$
O(\log \log t) > n_2+m_2 - (n_1 + m_1) = \int_{m_1}^{m_2}
1+f'(x)dx.
$$

We need to show that if there are solutions with $m$ too big for
region 1, there are none with $m$ too small for region 2, and that
if there are solutions with $m$ too small for region 3, there are
none with $m$ too big for region 2.

We can show the first of these by verifying that for sufficiently
large $t$
\begin{eqnarray*}
\int_{\exp(\sqrt{\log \log t})}^{\exp(2\sqrt{\log \log t})} 1+f'(x)dx & > &
\int_{\exp(\sqrt{\log \log t})}^{\exp(2\sqrt{\log \log t})}
\frac{\log t - 4\sqrt{\log \log t}+O(1)}{\log t} dx\\ & = &
\exp(2\sqrt{\log \log t})+ O(1) \\& \gg & \log \log t.
\end{eqnarray*}
This shows that for sufficiently large $t$, it is impossible to have
two solutions, one of which has $m$ too small to be in region 2, and
the other of which has $m$ too large to be in region 1 since this
would imply that $n_1+m_1-(n_2+m_2) \gg \log \log t$.

Now if $x_1$ and $x_2$ are the numbers so that $f(x_2)-x_2 = (\log
t)^{25/36}$ and $f(x_1)-x_1 = (\log t)^{26/36}$, we notice that
since the log of the gamma-function is convex that $x_2-x_1 >
\frac{1}{3} (\log t)^{26/36}$. Thus we verify the second of these
statements by noticing that
\begin{eqnarray*}
\int_{x_1}^{x_2} 1+f'(x)dx & > & 
\frac{1}{3} (\log t)^{26/36} (1 + f'(x_2))\\ & > & 
\frac{1}{3} (\log t)^{26/36} \frac{\log
(f(x_2)/x_2)+O(x_2^{-1})}{\log(f(x_2))}\\ & > &
\Omega\left((\log t)^{26/36} \frac{(f(x_2) -x_2)/x_2}{\log (f(x_2))}
\right)\\ & = & 
\Omega\left( \frac{(\log t)^{15/36}}{\log \log t}  \right)\\ &
\gg & 
 \log \log t.
\end{eqnarray*}
  Recall that $a(t) = \Omega(b(t))$ means that there exists a
constant $c>0$ so that for all sufficiently large $t$, $a(t)>cb(t)$,
and that $a(t) = \Theta(b(t))$ means that there exist $c_1>0$ and
$c_2>0$ so that for all sufficiently large $t$, $c_1 a(t) > b(t) >
c_2 a(t)$.

In section 6, we will cover the case where there are solutions in
the first region.  In section 7, we will cover the case where
there are solutions in the second region.  In section 8, we will
cover the case where there are solutions in the third region.

\section*{\normalsize 6. The First Region}
\addtocounter{section}{+1}
In this section, we will prove that for sufficiently large $t$,
that there are at most 2 solutions to $n!m!=t$ with $0<m\leq
\exp(\sqrt{\log \log t}).$

Notice that
\begin{eqnarray*}
e\left( \frac{(n+x)!}{n!}\right) & = & e((n+x)!) - e(n!) \\
& = & \sum_{i=1}^\infty \left\lfloor \frac{n+x}{2^i} \right\rfloor -
\left\lfloor
\frac{n}{2^i} \right\rfloor\\
& = & \sum_{i=1}^{\lfloor \log x \rfloor} \frac{x}{2^i} + O(1) +
\max_{n<c\leq n+x} e(c) - \log x\\
& = & x + \max_{n<c\leq n+x} e(c) + O (\log x).
\end{eqnarray*}
Therefore, if we have any two such solutions, $n_1!m_1! = n_2!m_2!
= t$, with $n_1>n_2$ then $e\left(\frac{n_1!}{n_2!}\right) =
e\left(\frac{m_2!}{m_1!}\right)$.  Therefore,
$$
n_1-n_2 + O(\log (n_1 -n_2)) + \max_{n_1<c\leq n_2} e(c) \geq
m_2-m_1 + O(\log(m_2-m_1)).
$$
Which implies that
$$
\max_{n_1<c\leq n_2} e(c) > (m_2-m_1) - (n_1 - n_2)  + O(\log
(m_2-m_1)).
$$

Notice that if $n_1!m_1! = n_2!m_2!$, then $\frac{n_1!}{n_2!} =
\frac{m_2!}{m_1!}$, and therefore, $m_2-m_1 > (n_1-n_2) \cdot
\frac{\log n_2}{\log m_2}$.  Now, $n_2\log n_2 > \log t -
2\sqrt{\log \log t} \exp(2\sqrt{\log \log t})$.  Therefore, $n_2 =
\Omega(\frac{\log t}{\log \log t})$, so $\log n_2 = \Omega (\log
\log t)$.  Hence $m_2-m_1 = \Omega(\sqrt{\log \log t}) (n_1-n_2)$.
Therefore, we have that
\begin{eqnarray*}
\max_{n_1<c\leq n_2} e(c) & > &  (m_2-m_1)(1+O((\log \log
t)^{-1/2})) + O (\log (m_2 - m_1))\\  & = & \Omega (m_2 - m_1) \\
&=& \Omega (\sqrt{\log \log t}).
\end{eqnarray*}

Therefore, if we have three solutions in region 1,
$(n_1,m_1),(n_2,m_2),(n_3,m_3)$ with $0<m_1<m_2<m_3$, then we have
that there exist $n_3<c_1\leq n_2 < c_2 \leq n_1$ with $e(c_i) =
\Omega (\sqrt{\log \log t})$.  Therefore, since $\min(e(x),e(y))
\leq e(x-y)$, we have that $e(c_2-c_1) = \Omega(\sqrt{\log \log
t})$.  Therefore, $n_1-n_3 > c_2-c_1 > \exp(\Omega(\sqrt{\log \log
t}))$.  But we notice that this and previous inequalities imply
that
$$
m_3 + n_3 - (m_1 + n_1) = \Omega(\sqrt{\log \log
t})\exp(\Omega(\sqrt{\log \log t})).
$$
Since this cannot be $O(\log \log t)$, we have that for
sufficiently large $t$, there are at most 2 solutions with $m\neq
0$ in region 1.  Hence there are at most 3 solutions in region 1.

\section*{\normalsize 7. The Second Region}
\addtocounter{section}{+1}
In this section we will show that there are at most 2 solutions with
$m>\exp(\sqrt{\log \log t})$ and $n-m > (\log t)^{25/36}$. Recall
that $f:\R \rightarrow \R$ so that $\Gamma(f(x)+1)\Gamma(x+1)=t$.
This is defined in the range we are interested in, because the
gamma-function is increasing.  Let $g(x) = \log \Gamma(x+1)$.  So
$g(f(x))+g(x) =\log t$. Differentiating implicitly, we get that
$$
f'(x)=-\frac{g'(x)}{g'(f(x))}.
$$
Therefore,
\begin{eqnarray*}
 f''(x) & = & -\frac{g''(x)}{g'(f(x))} + f'(x)
\frac{g'(x)g''(f(x))}{(g'(f(x)))^2}\\
& = & -\frac{g''(x)(g'(f(x)))^2+(g'(x))^2g''(f(x))}{(g'(f(x)))^3}.
\end{eqnarray*}
By differentiating our strong form of Stirling's formula, we find
that for $f(x)>x$
$$
f''(x) = -\frac{(\log f(x))^2 x^{-1}+(\log x)^2
(f(x))^{-1}(1+O(x^{-1}))}{((\log f(x))+O(f(x)^{-1}))^3}.
$$
Therefore, for all sufficiently large $t$, for $x>\exp(\sqrt{\log
\log t})$ and $f(x)-x>(\log t)^{25/36}$ we have that
$$
0<\left| \frac{1}{2}f''(x) \right| < O\left(\frac{1}{x (\log
f(x))}\right).
$$
Assume for sake of contradiction that we have three solutions to
$n!m!=t$ in region 2 ($n_i!m_i! = t$ for $1\leq i\leq3$ where
$m_i<m_{i+1}$).  Then we have that $m_i$ is an integer and that
$f(m_i)$ is an integer.  Since between $m_1$ and $m_3$ we have
that
$$
0<\left| \frac{1}{2}f''(x) \right| < O\left(\frac{1}{m_1}\right).
$$
Therefore, by Lemma \ref{continuoslemma},
$m_3-m_1>\Omega(m_1^{1/3})$.  But we also have that
$(m_3+n_3)-(m_1+n_1)=O(\log \log t)$.  Therefore
$$
\int_{m_1}^{m_1+\Omega(m_1^{1/3})} \frac{(\log f(x))-\log x}{\log
f(x) }dx = O(\log \log t).
$$
Now we have that for $x$ in the range we are concerned with that
$$
\frac{(\log f(x))-\log x}{\log f(x) } =
\Omega\left(\frac{(f(x)-x)/x}{\log f(x)}\right) = \Omega((\log
t)^{-11/36}).
$$
Therefore, it must be that $m_1 = O((\log \log t)^3 (\log
t)^{11/12})$.  But in this range, the integrand we are concerned
with is at least $\frac{1}{12}+o(1)$.  Therefore, it must be that
$m_1 = O((\log \log t)^3)$ which does not hold.  Therefore, for
sufficiently large $t$, there are at most 2 solutions in region 2.

\section*{\normalsize 8. Region Three}
\addtocounter{section}{+1}
In this section we will show that there are at most 3 solutions in
region 3 for sufficiently large $t$.  This proof depends on the
fact that if $n$ and $m$ are integers, then so are $n+m$ and
$(n-m)^2$ and applications of Lemma \ref{continuoslemma} and
results from section 3.

Suppose that $\Gamma((a+\sqrt{x}+2)/2)\Gamma((a-\sqrt{x}+2)/2)=t$,
where $x = O(a^{3/2})$.  Then we have by our strong form of
Stirling's formula that
\begin{eqnarray*}
\log t & = & \frac{a+\sqrt{x}+1}{2}\log\left(\frac{a+\sqrt{x}}{2}
\right) + \frac{a-\sqrt{x}+1}{2}\log\left(\frac{a-\sqrt{x}}{2}
\right)\\ && - a + \log{2\pi} + O(a^{-1})\\
& = & (a+1)\log(a/2) -a + \log(2\pi)\\  &&+
\frac{a+\sqrt{x}+1}{2}\log\left(1+\frac{\sqrt{x}}{a} \right)
+\frac{a-\sqrt{x}+1}{2}\log\left(1-\frac{\sqrt{x}}{a} \right) +
O(a^{-1})\\
& = & (a+1)\log(a/2) -a + \log(2\pi) + \frac{x}{a}+O(a^{-1}).
\end{eqnarray*}
Therefore, we have that
$$
x = a\log\left(\frac{t}{2\pi} \right) - a(a+1)\log
\left(\frac{a}{2} \right) +a^2 + O(1).
$$
Let $a_0$ be the positive real value so that
$(\Gamma((a_0/2)+1))^2 = t$.  So then we have that
$$
a_0\log\left(\frac{t}{2\pi} \right) - a_0(a_0+1)\log
\left(\frac{a_0}{2} \right) +a_0^2 = O(1).
$$
It is also true that $a_0 = O(\log t)$. If we pick an $a$ so that
$$
a\log\left(\frac{t}{2\pi} \right) - a(a+1)\log \left(\frac{a}{2}
\right) +a^2 + O(1) < (\log t)^{16/11}.
$$
Then there must be a unique complex $x$ with $|x|\leq (\log
t)^{16/11}$ so that
$\Gamma((a+\sqrt{x}+2)/2)\Gamma((a-\sqrt{x}+2)/2)=t$.  Since the
derivative of
$$
a\log\left(\frac{t}{2\pi} \right) - a(a+1)\log \left(\frac{a}{2}
\right) +a^2
$$
is
$$
\log\left(\frac{t}{2\pi} \right) -(2a+1)\log \left(\frac{a}{2}
\right) + a -1
$$
and since its second derivative is $O(\log a)$, this should hold as
long as $|a-a_0| < O((\log t)^{9/20})$.  Furthermore, for $a$ in
this range, $x$ attains all values with $|x|\leq (\log t)^{13/9}$.
This allows use to define an analytic function $h(a)$ defined on
$|a-a_0|<O((\log t)^{29/20})$ so that
$$
\Gamma\left(\frac{a+\sqrt{h(a)}}{2} +1\right)
\Gamma\left(\frac{a-\sqrt{h(a)}}{2} +1\right)=t.
$$
Furthermore, $h(a)$ attains all values of absolute value at most
$(\log t)^{13/9}$ when $|a-a_0| = O(\log t)^{5/9}$.  Additionally,
we have that
$$
h(a)=a\log\left(\frac{t}{2\pi} \right) - a(a+1)\log
\left(\frac{a}{2} \right) +a^2 + O(1).
$$
Since the $O(1)$ is uniform in the region stated, its third
derivative when $|a-a_0|=O(\log t)^{13/29}$ (notice that
$4/9<13/29<9/20$) can be written using Cauchy's Integral formula as
$$
\int_C O((z-a)^{-4})dz,
$$
where $C$ is a contour that traverses a circle centered at $a$ of
radius $\Omega((\log t)^{9/20})$ once in the counter-clockwise
direction.  This is $O((\log t)^{-27/20})$.  Therefore, when
$|h(a)|<(\log t)^{13/9}$ we have that
\begin{eqnarray*}
h'''(a) & = & \frac{\partial^3}{\partial a^3}\left(
a\log\left(\frac{t}{2\pi} \right) - a(a+1)\log \left(\frac{a}{2}
\right) +a^2 \right) + O((\log t)^{-27/20})\\
& = & -\frac{1}{a} + O((\log t)^{-27/20}).
\end{eqnarray*}
Since it is clear by Stirling's formula that $a_0 = \Theta\left(
\frac{\log t}{\log \log t} \right)$ we have that for sufficiently
large $t$, when $|h(a)| \leq (\log t)^{13/9}$ then
$0<|h'''(a)|<O\left(\frac{\log \log t}{\log t} \right)$.

Now suppose for sake of contradiction that $(n_i,m_i)$ are distinct
region 3 solutions for $1\leq i \leq 4$.  Then $n_i+m_i \in \Z$ and
$h(n_i+m_i) = (n_i-m_i)^2 \in \Z$.  Furthermore, $|h(n_i+m_i)| \leq
(\log t)^{13/9}$.  Then since in the range between the $n_i+m_i$ we
have that $0<|h'''(a)|<O\left(\frac{\log \log t}{\log t} \right)$,
Lemma \ref{continuoslemma} implies that the difference between the
largest and smallest of the $n_i+m_i$ is at least
$$
\Omega\left(\left(\frac{\log t}{\log \log t} \right)^{1/6}
\right).
$$
Since this is larger than $O(\log \log t)$, we have from results
in section 3, that for sufficiently large $t$, this is impossible.

Hence there are at most three solutions in region 3 for
sufficiently large $t$.

\section*{\normalsize 9. Conclusions}
\addtocounter{section}{+1}
Hence we have proved our result that
$$
\limsup_{t\rightarrow \infty} \left| \left\{ (n,m)\in \No^2 :
n!m!=t \right\} \right| = 6.
$$
Notice that all of our statements about there being at most 6
solutions for ``sufficiently large $t$" can be made effective,
although this was not done in this paper.  I do not believe that the
effective bound that is achieved would be small enough to allow for
a reasonable proof that there are at most 6 solutions for any $t$,
at least without further insight.

\begin{thebibliography}{[9]} \footnotesize


\bibitem{kn:kane} D. Kane,
\emph{New bounds on the number of representations of $t$ as a
binomial coefficient},
Integers 4 (2004).

\bibitem{kn:Rademacher} Hans Rademacher,
\emph{Topics in Analytic Number
Theory} Springer-Verlag, Berlin-Heidelberg-New York 1970.

\end{thebibliography}

\end{document}