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The second largest prime divisor
of an odd perfect number
exceeds ten thousand

Author: Douglas E. Iannucci
Journal: Math. Comp. 68 (1999), 1749-1760
MSC (1991): Primary 11A25, 11Y70
Published electronically: May 17, 1999
MathSciNet review: 1651761
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Abstract | References | Similar Articles | Additional Information

Abstract: Let $\sigma(n)$ denote the sum of positive divisors of the natural number $n$. Such a number is said to be perfect if $\sigma(n)=2n$. It is well known that a number is even and perfect if and only if it has the form $2^{p-1} (2^p-1)$ where $2^p-1$ is prime.

No odd perfect numbers are known, nor has any proof of their nonexistence ever been given. In the meantime, much work has been done in establishing conditions necessary for their existence. One class of necessary conditions would be lower bounds for the distinct prime divisors of an odd perfect number.

For example, Cohen and Hagis have shown that the largest prime divisor of an odd perfect number must exceed $10^6$, and Hagis showed that the second largest must exceed $10^3$. In this paper, we improve the latter bound. In particular, we prove the statement in the title of this paper.

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Additional Information

Douglas E. Iannucci
Affiliation: University of the Virgin Islands, 2 John Brewers Bay, St. Thomas, VI 00802

Keywords: Perfect numbers, cyclotomic polynomials
Received by editor(s): June 16, 1997
Received by editor(s) in revised form: August 25, 1997
Published electronically: May 17, 1999
Article copyright: © Copyright 1999 American Mathematical Society

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