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\begin{center}
{\bf AN UPPER BOUND FOR ODD PERFECT NUMBERS} \vskip 20pt
{\bf Pace P. Nielsen}\\
{\smallit Department of Mathematics, University of California, Berkeley, CA 94720, U.S.A.}\\
{\tt pace@math.berkeley.edu}\\ \vskip 10pt
\end{center}
\vskip 30pt \centerline{\smallit Received:1/31/03, 
Revised: 9/29/03, Accepted: 10/21/03,
Published:10/22/03} \vskip 30pt

\centerline{\bf Abstract}

\noindent If $N$ is an odd perfect number with $k$ distinct prime
factors then we show that $N< 2^{4^{k}}$.  If some of the small
prime factors of $N$ are known then this bound can be further
improved.

\pagestyle{myheadings} \markright{\smalltt INTEGERS: \smallrm
ELECTRONIC JOURNAL OF COMBINATORIAL NUMBER THEORY \smalltt 3
(2003), \#A14\hfill}

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

\section*{\normalsize 1. Introduction}  Let $N$ be a positive
integer.  For $n, d\in \mathbb{Z}_{+}$, we say that $N$ is an
$n/d$-perfect number if $\sigma(N)/N=n/d$, where $\sigma$ is the
sum the divisors function.  In the special case when $n/d=2$, $N$
is called a perfect number. Currently there are 39 known perfect
numbers. They are all of the form $2^{p-1}(2^{p}-1)$ with
$2^{p}-1$ prime ($p$ must also be prime).  It is conjectured that
all perfect numbers are of this form.  Euler proved that this is
indeed true if we restrict ourselves to the case when $N$ is
even. The problem is still open when $N$ is odd, and this is
probably the oldest unsolved problem in mathematics.

There are a number of results that restrict the size and
arithmetic properties of possible odd perfect numbers. To give
some examples, suppose that $N$ is an odd perfect number with $k$
distinct prime factors. Euler proved that one can write
$N=a^{2}(4b+1)^{4c+1}$ with $a, b, c\in \mathbb{Z}_{+}$. E.Z.
Chein in his doctoral thesis \cite{Chein} and Hagis \cite{PHagis}
independently proved that $k\geqslant 8$. Jenkins \cite{Jenkins}
showed that the largest prime factor of $N$ must exceed $10^{7}$;
Iannuci \cite{Ian}, \cite{Ian2} proved that the second largest
exceeds $10^{4}$ and the third largest exceeds $10^{2}$. An
extensive computer search by Brent, Cohen, and te Riele
\cite{Brent} demonstrated that $N>10^{300}$.

In 1913, Dickson \cite{Di} proved that there are only finitely
many odd perfect numbers with $k$ distinct prime factors.  In
1977, Pomerance \cite{Pom} gave an explicit upper bound in terms
of $k$.  Heath-Brown \cite{HB} later improved the bound to
$N<4^{4^{k}}$, and Cook \cite{Cook} reduced this further to
$N<D^{4^{k}}$ with $D=(195)^{1/7}$.

The methods of this paper slightly improve and generalize Cook's
bound.  Throughout this paper we assume that $n,d\in
\mathbb{Z}_{+}$.  Further, if we say that $N$ is an $n/d$-perfect
number we do not assume that $n/d$ is written in lowest terms,
unless explicitly stated.

\vskip 30pt

\section*{\normalsize 2. A Lemma and Some Notation}

\noindent{\bf Lemma 1.}  Let $r, a, b\in \mathbb{Z}_{+}$ and let
$x_{1},\ldots,x_{r}$ be integers such that $1 < x_{1} < \ldots <
x_{r}$.  If
\begin{align*}
\prod_{i=1}^{r}\left(1-\frac{1}{x_{i}}\right) \leqslant
\frac{a}{b} < \prod_{i=1}^{r-1}\left(1-\frac{1}{x_{i}}\right) &
\qquad\qquad(\ast)
\end{align*}
then
\[
a\prod_{i=1}^{r}x_{i} < (a+1)^{2^{r}}.
\]

\noindent{\it Proof.}  This is just a summary of some results from
\cite{Cook}.  For completeness we include the proof, which is by
induction on $r$. The case $r=1$ is easy, so assume that
$r\geqslant 2$ and that the result holds (for any $a$ and $b$)
for each integer less than $r$.

Fix $a\geqslant 1$.  Suppose that $1<x_{1}<\ldots <x_{r}$ are
integers satisfying $(\ast)$ (for some $b$) with $x_{1}\cdots
x_{r}$ maximal. Set $n_{i}=(a+1)^{2^{i-1}}+1$ for $i<r$ and put
$n_{r}=(a+1)^{2^{r-1}}$.  It is easy to see that
$1<n_{1}<\ldots<n_{r}$ and
\[
\prod_{i=1}^{r}\left(1-\frac{1}{n_{i}}\right) = \frac{a}{a+1} <
\prod_{i=1}^{r-1}\left(1-\frac{1}{n_{i}}\right).
\]
Therefore $\prod_{i=1}^{r}x_{i}\geqslant \prod_{i=1}^{r}n_{i}$.

If $ax_{1}<an_{1}=(a+1)^{2}-1$ then multiplying $(\ast)$ through
by $\frac{x_{1}}{x_{1}-1}$ yields
\[
\prod_{i=2}^{r}\left(1-\frac{1}{x_{i}}\right) \leqslant
\frac{ax_{1}}{b(x_{1}-1)} <
\prod_{i=2}^{r-1}\left(1-\frac{1}{x_{i}}\right).
\]
Our induction hypothesis then implies $(ax_{1})x_{2}\cdots x_{r}<
(ax_{1}+1)^{2^{r-1}}<(a+1)^{2^{r}}$.  So we may as well assume
that $x_{1}\geqslant n_{1}$.

If $ax_{1}x_{2}<an_{1}n_{2}=(a+1)^{4}-1$ then multiplying $(\ast)$
through by $\frac{x_{1}x_{1}}{(x_{1}-1)(x_{2}-1)}$ yields
\[
\prod_{i=3}^{r}\left(1-\frac{1}{x_{i}}\right) \leqslant
\frac{ax_{1}x_{2}}{b(x_{1}-1)(x_{2}-1)} <
\prod_{i=3}^{r-1}\left(1-\frac{1}{x_{i}}\right).
\]
Our induction hypothesis implies $(ax_{1}x_{2})x_{3}\cdots
x_{r}<(ax_{1}x_{2}+1)^{2^{r-2}}<(a+1)^{2^{r}}$.  So we may as
well assume that $x_{1}x_{2}\geqslant n_{1}n_{2}$.

Continuing in this manner, we may as well assume
$\prod_{i=1}^{u}x_{i}\geqslant \prod_{i=1}^{u}n_{i}$ for each
$u<r$.  We already saw that this inequality also holds for $u=r$.
But then Lemma 2 of \cite{Cook} contradicts $(\ast)$ (for any $b$
we might choose).
Hence, in all cases, $a\prod_{i=1}^{r}x_{i}<(a+1)^{2^{r}}$. \hfill $\Box$ \\

\noindent{\bf Notation.}  Let $S=\{x_{1},\ldots,x_{r}\}$ be a set
of $r$ distinct integers, each greater than $1$, ordered under
the usual ordering.  We use the notation $\Pi(S)=\prod_{i=1}^{r}
x_{i}$, with the convention that $\Pi(\emptyset)=1$.  If $a$ is a
positive integer and $r>0$ then we put
\[
B(r,a)= ((a+1)/a)^{2^{r}/(2^{r}-1)}.
\]
By convention, we set $B(0,a)=1$.

Let $r,a,b >0$ and suppose we have the following inequalities:
\[
\prod_{i=1}^{r}\left(1-\frac{1}{x_{i}}\right) \leqslant
\frac{a}{b} < \prod_{i=1}^{r-1}\left(1-\frac{1}{x_{i}}\right).
\]
Lemma 1 implies $a\prod_{i=1}^{r}x_{i}<(a+1)^{2^{r}}$.  A quick
calculation shows that this is equivalent to $\Pi(S) <
(B(r,a)a)^{2^{r}-1}$.


\vskip 30pt

\section*{\normalsize 3. The Main Result}
Before we prove the theorem we give an independent, but
interesting, result:

\noindent{\bf Proposition 1.}  Let $N$ be an odd perfect number
with $k$ distinct prime factors, and let
\[
P=\prod_{p\,| N}p .
\]
Then
\[
N<P^{2^{k}-1}.
\]

\noindent{\it Proof.}  Let $N=\prod_{i=1}^{k}p_{i}^{e_{i}}$ be
the prime factorization of $N$.  Then $P=\prod_{i=1}^{k}p_{i}$,
and we set $P'=\prod_{i=1}^{k}(p_{i}-1)$.  Since $N$ is an odd
perfect number we find
$\prod_{i=1}^{k}(p_{i}^{e_{i}+1}-1)/(p_{i}-1) =
\sigma(N)=2N=2\prod_{i=1}^{k}p_{i}^{e_{i}}$.  Rewriting this, we
obtain
\[
\prod_{i=1}^{k}\left(1-\frac{1}{p_{i}^{e_{i}+1}}\right)=\frac{2
P'}{P}<1.
\]
Thus, the hypotheses of Lemma 1 hold with $a=2P'$, $b=P$, and
$x_{i}=p_{i}^{e_{i}+1}$ (reordering the $p_{i}$ if necessary). So
Lemma 1 gives the second inequality below:
\[
PN<2P'PN=2P'\prod_{i=1}^{k}p_{i}^{e_{i}+1} < (2P'+1)^{2^{k}}
\leqslant P^{2^{k}}.
\]
This implies $N < P^{2^{k}-1}$.\hfill $\Box$ \\


We now proceed with the proof of:


\noindent{\bf Theorem 1.}  If $N$ is an odd $n/d$-perfect number
with $k$ distinct prime factors then
\[
N<(d+1)^{4^{k}}.
\]

\noindent{\bf Proof.}  Much of our notation is taken from
Heath-Brown \cite{HB}.  Most of this proof is a direct
strengthening of the algorithm given there.

If $N=1$ then everything holds, so assume $N> 1$.  Let
$N=\prod_{p\,| N}p^{e(p)}$ be the prime factorization of $N$.  Let
$S$ be any set (possibly empty) of distinct primes dividing $N$.
Since $N$ is odd and $d/n=N/\sigma(N)\neq 1$, it follows that
$d/n$ has an odd numerator when written in lowest terms.  Hence
\[
\prod_{p \in S}\left(1-\frac{1}{p}\right) \neq \frac{d}{n},
\]
where, by convention, the product equals 1 if $S=\emptyset$.
\\
\\
{\bf Case 1:}\,\,\,\,\,$\displaystyle \prod_{p \in
S}\left(1-\frac{1}{p}\right) > \frac{d}{n}.$


Set $d' = d \prod_{p \in S}p = d \Pi(S)$ and $n' = n \prod_{p\in
S}(p-1)$.   Since $\prod_{p\in S}(1-(1/p))>d/n$ we have $d'/n'
<1$.  We also calculate
\[
\prod_{p\, |
N}\left(1-\frac{1}{p}\right)<\prod_{p\,|N}\frac{p-1}{p-\frac{1}{p^{e(p)}}}
= \prod_{p\,| N}\frac{p^{e(p)}}{\frac{p^{e(p)+1}-1}{p-1}} =
\frac{N}{\sigma(N)} = \frac{d}{n}.
\]
So we have
\[
\prod_{p \, | N,\,p \notin S}\left(1-\frac{1}{p}\right) <
\frac{d'}{n'} < 1.
\]
These last two inequalities imply that we can pick a non-empty,
ordered set of distinct prime divisors of $N$, say
$S'=\{p_{1},\ldots, p_{w}\}$, such that $S'\cap S=\emptyset$ and
such that the following inequalities hold:
\[
\prod_{i=1}^{w}\left(1-\frac{1}{p_{i}}\right) \leqslant
\frac{d'}{n'} < \prod_{i=1}^{w-1}\left(1-\frac{1}{p_{i}}\right).
\]
Fix such a set $S'$. Then, letting $B=B(w,d')$ where $w=|S'|$,
Lemma 1 implies
\[
\Pi(S') < (Bd')^{2^{w}-1}
\]
and hence
\begin{equation}
\label{eq:1} \Pi(S \cup S') \leqslant
(Bd)^{2^{w}-1}\left(\Pi(S)\right)^{2^{w}}.
\end{equation}
Also, from the definition of $S'$ we have
\begin{equation}
\label{eq:2}
\prod_{p \in S \cup S'}\left(1-\frac{1}{p}\right) < \frac{d}{n}.
\end{equation}
Notice, as remarked above, that equality can't hold in
(\ref{eq:2}).
\\
\\
{\bf Case 2:}\,\,\,\,\,$\displaystyle \prod_{p \in
S}\left(1-\frac{1}{p}\right) < \frac{d}{n}$.


We define $n'$ and $d'$ as in Case 1.  Let $S'=\emptyset$ and
$w=0$, and set $B=B(w,d')=1$.  Then both inequalities (\ref{eq:1})
and (\ref{eq:2}) hold in this case also.  (In fact, equality holds
in (\ref{eq:1}), but we don't need this fact.)


We have shown in the two cases above that for any choice of $S$
we can pick another set of prime divisors of $N$ disjoint from
$S$, namely $S'$, such that inequalities $(1)$ and $(2)$ hold. We
now continue with the proof in the general case.

Let $d'' = d \Pi(S \cup S')$ and $n'' = n \prod_{p \in S \cup
S'}(p-1)$. Inequality (\ref{eq:2}) says $n''/d''<1$. Now,
\[
\prod_{p \in S \cup S'}\frac{1-\frac{
1}{p^{e(p)+1}}}{1-\frac{1}{p}} = \prod_{p\in S\cup
S'}\frac{\frac{p^{e(p)+1}-1}{p-1}}{p^{e(p)}} \leqslant
\frac{\sigma(N)}{N} = \frac{n}{d}
\]
so that
\begin{equation}
\label{eq:3} \prod_{p \in S \cup S'}{1-\frac{1}{p^{e(p)+1}}}
\leqslant \frac{n''}{d''} < 1.
\end{equation}
Then, just as above, we can pick a non-empty subset $S'' \subseteq
S \cup S'$, for which Lemma 1 implies:
\begin{equation}
\label{eq:4}
\prod_{p\in S''}p^{e(p)+1} \leqslant
(B'n'')^{2^{v}-1}
\end{equation}
with $v=|S''|$, $B'=B(v,n'')$.

As in Heath Brown's paper we let $V=\prod_{p\in S''}p^{e(p)}$,
$U=N/V$, and $T=(S \cup S')-S''$. However we differ in our
definition of $\delta$ by setting $\delta=d\prod_{p \in
S''}(p^{e(p)+1}-1)/(p-1)$.  Finally, we let $\nu =
\delta\sigma(U)/U$.


We now deduce a long chain of inequalities.  First, we calculate
\[
\begin{array}{lll}
\displaystyle \frac{\delta+1}{\delta}\delta\Pi(T) & \displaystyle
= \frac{\delta+1}{\delta}d\prod_{p \in
S''}\frac{p^{e(p)+1}-1}{(p-1)p}\Pi(S \cup S') & \\[.25in]
& \displaystyle = \frac{\delta+1}{\delta}d\prod_{p \in S''}
\frac{p^{e(p)+1}
\left(1-\frac{1}{p^{e(p)+1}}\right)}{(p-1)p} \Pi(S \cup S') & \\[.25in]
& \displaystyle  \leqslant \frac{\delta+1}{\delta}
\frac{n''}{d''}d\prod_{p \in
S''} \frac{p^{e(p)+1}}{(p-1)p}\Pi(S \cup S') & \shoveright{\text{by inequality (3).}} \\[.25in]
\end{array}
\]
Since $S''\neq \emptyset$ and $\delta\in \mathbb{Z}_{+}$, it is
clear that $\frac{\delta +1}{\delta}\prod_{p\in
S''}\frac{1}{(p-1)p}<1$. Therefore, continuing the previous
inequality, we find
\[
\begin{array}{lll}
\displaystyle \frac{\delta+1}{\delta} \frac{n''}{d''}d\prod_{p \in
S''} \frac{p^{e(p)+1}}{(p-1)p}\Pi(S \cup S') & \displaystyle <
\frac{n''}{d''}d\left(\prod_{p \in S''}p^{e(p)+1}\right)\Pi(S
\cup S') & \\[.25in]
& \displaystyle < \frac{n''}{d''}d(B'n'')^{2^{v}-1}\Pi(S \cup S') & \text{by inequality (4)} \\[.25in]
& \displaystyle = \frac{1}{d''}d(n''+1)^{2^{v}}\Pi(S \cup S') &
\text{by definition of } B'.
\end{array}
\]
We again appeal to Lemma 1, in the form of inequality
(\ref{eq:1}), to obtain
\[
\begin{array}{lll}
\displaystyle \frac{1}{d''}d(n''+1)^{2^{v}}\Pi(S \cup S') & \displaystyle \leqslant \frac{d}{d''}(n''+1)^{2^{v}} (Bd)^{2^{w}-1}\Pi(S)^{2^{w}} & \text{by inequality (1)}\\[.25in]
&\displaystyle = \frac{1}{Bd''}(n''+1)^{2^{v}}(Bd\Pi(S))^{2^{w}} & \\[.25in]
&\displaystyle = \frac{1}{B} \left(\frac{n''+1}{d''}\right)^{2^{v}} (d'')^{2^{v}-1}(Bd\Pi(S))^{2^{w}}. & \\[.25in]
\end{array}
\]
As remarked above, inequality (\ref{eq:2}) is equivalent to
$n''/d'' < 1$.  Therefore, we know that $(n''+1)/d'' \leqslant 1$.
So our chain of inequalities continues as follows:
\begin{multline*}
\shoveright{\frac{1}{B} \left(\frac{n''+1}{d''}\right)^{2^{v}}
(d'')^{2^{v}-1}(Bd\Pi(S))^{2^{w}} \leqslant \frac{1}{B}
(d'')^{2^{v}-1}(Bd\Pi(S))^{2^{w}}}
\end{multline*}
\[
\begin{array}{lll}
&\displaystyle = \frac{1}{B} (d\Pi(S\cup S') )^{2^{v}-1}(Bd\Pi(S))^{2^{w}} & \qquad \text{from the definition of } d''\\[.25in]
&\displaystyle \leqslant \frac{1}{B} (d(Bd)^{2^{w}-1}\Pi(S)^{2^{w}} )^{2^{v}-1}(Bd\Pi(S))^{2^{w}} & \qquad \text{by inequality (1).}\\[.25in]
\end{array}
\]
This last quantity is just
$(B^{2^{w}-1})^{2^{v}}(d\Pi(S))^{2^{w+v}}$.  A short calculation,
using the definition of $B$, yields
\[
(B^{2^{w}-1})^{2^{v}}(d\Pi(S))^{2^{w+v}} \leqslant
\left(\left(\frac{d'+1}{d'}\right)d\Pi(S)\right)^{2^{w+v}}.
\]
Equality holds above, except in the case $w=0$. Now, since
$d'\geqslant d$, we also have
\[
\left(\left(\frac{d'+1}{d'}\right)d\Pi(S)\right)^{2^{w+v}}
\leqslant
\left(\left(\frac{d+1}{d}\right)d\Pi(S)\right)^{2^{w+v}}.
\]
Putting the last two pages of inequalities together, we arrive at
\begin{equation}
\label{eq:5} \frac{\delta+1}{\delta}\delta\Pi(T) < \left
(\left(\frac{d+1}{d}\right)d\Pi(S)\right)^{2^{w+v}}.
\end{equation}

In essence, we have re-proven Lemma 2 from Heath-Brown \cite{HB},
but with our inequality (\ref{eq:5}) replacing his inequality
(iv). One can easily check that all of the necessary conditions
of his lemma hold under our construction.  To finish the proof we
follow the inductive process described on page 196 of \cite{HB},
with the only change being that we put inequality (\ref{eq:5}) in
place of Heath-Brown's inequality (iv), everywhere it occurs.
This yields
\[
N<\left(\left(\frac{d+1}{d}\right)d\right)^{4^{k}}=(d+1)^{4^{k}}.
\]
\hfill $\Box$

\vskip 20pt

\section*{\normalsize 4. Explicit Calculations and Examples}
\noindent{\bf Corollary 1.}  If $N$ is an odd perfect number (OPN)
then
\[
N < 2^{4^{k}}.
\]


\noindent{\bf Corollary 2.}  If $N$ is an OPN and $5 || N$ then
\[
N < 5\cdot 4^{4^{k-1}}.
\]

\noindent{\bf Proof.}  Let $N'=N/5$.  Then $N'$ is an odd
$5/3$-perfect number. Applying the theorem to $N'$ gives the
result.\hfill $\Box$ \\


\noindent{\bf Corollary 3.}  If $N$ is an OPN with $5 ||N$ and
$3^{2} || N$ then
\[
N < 45\cdot 14^{4^{k-2}}.
\]


Our first corollary is only a slight improvement on the bounds
given by Cook in \cite{Cook}.  However, our last two corollaries
demonstrate a method that further decreases the upper bound on
OPN's when one knows which small prime factors divide $N$.  These
methods, in conjunction with \cite{Brent}, \cite{Jenkins},
\cite{Ian}, and \cite{Ian2} (and some brute force), are
sufficient to produce a new proof that odd perfect numbers must
have at least seven factors. Unfortunately, these bounds are
still not strong enough to give better results.


One might be able to lower these bounds further if the inequality
given in Lemma 1 could be improved.  There are two clear (but
difficult) ways this might be done.  First, the proof of Lemma 1
doesn't take full advantage of the integer $b$.  Rather, it just
uses the fact that $b>a$, and reduces to the case $b=a+1$ to
prove the needed inequalities.  One might be able to generalize
the methods of \cite{Cook} to take advantage of this.

Second, if one looks carefully at the proof of the theorem given
above, we only invoke Lemma 1 when the integers in question are
relatively prime and are prime powers.  This assumes much more
stringent requirements on the $x_{i}$ than given in the statement
of Lemma 1.

To demonstrate both the strength and weakness of the bound above,
we have the following:

\noindent{\bf Example.}  Let $N=3^{2}7^{2}11^{2}13^{2}22021$.
Clearly $22021=19^{2}61$ is not prime.  We want to pretend that
$22021$ is prime by redefining $\sigma$.  Let
$\sigma'(x)=\sigma(x)$ for all $x$ such that $\gcd(x,22021)=1$,
and set $\sigma'(22021^{n})=(22021^{n+1}-1)/(22021-1)$ (parodying
the rule $\sigma(p^{n})=(p^{n+1}-1)/(p-1)$ for any prime $p$). As
pointed out by Descartes, $\sigma'(N)=2N$, so that $N$ is a spoof
perfect number!

If we restricts ourselves to numbers that can be expressed as a
power of 22021 and other factors relatively prime to 19 and 61,
and if we replace $\sigma$ by $\sigma'$, then the proof of the
above theorem still goes through.  So it should be true that for
$N=3^{2}7^{2}11^{2}13^{2}22021$ we have $N<2^{4^{5}}$.  In fact
$2^{4^{5}}/N \thickapprox 4.407 \times 10^{317}$.  This
demonstrates one of the difficulties in proving the non-existence
of OPN's.

\vskip 20pt

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\bibitem{Chein} Joseph E. Z. Chein, {\it An odd perfect number has
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University, 1979.

\bibitem{Cook} R. J. Cook, {\it Bounds for odd perfect numbers},
Number theory (Ottowa, ON, 1996), 67-71, CRM Proc. Lecture Notes
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\bibitem{Di} L. E. Dickson {\it Finiteness of odd perfect and primitive abundant
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\bibitem{PHagis} P. Hagis, Jr., {\it Outline of a proof that every
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\bibitem{HB} D. R. Heath-Brown, {\it Odd perfect numbers}, Math.
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\bibitem{Ian} D. E. Ianucci, {\it The second largest prime divisor
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\bibitem{Ian2} D. E. Ianucci, {\it The third largest prime divisor
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\bibitem{Jenkins} P. Jenkins, {\it Odd perfect numbers have a
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\bibitem{Pom} C. Pomerance, {\it Multiple perfect numbers, Mersenne primes and effective
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\end{thebibliography}

\end{document}