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The third largest prime divisor of an odd perfect number exceeds one hundred

Author: Douglas E. Iannucci
Journal: Math. Comp. 69 (2000), 867-879
MSC (1991): Primary 11A25, 11Y70
Published electronically: May 17, 1999
MathSciNet review: 1651762
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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.

It is unknown whether or not odd perfect numbers exist, although many conditions necessary for their existence have been found. For example, Cohen and Hagis have shown that the largest prime divisor of an odd perfect number must exceed $10^6$, and Iannucci showed that the second largest must exceed $10^4$. In this paper, we prove that the third largest prime divisor of an odd perfect number must exceed 100.

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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): December 12, 1997
Received by editor(s) in revised form: January 26, 1998, and June 2, 1998
Published electronically: May 17, 1999
Additional Notes: This paper presents the main result of the author’s doctoral dissertation completed at Temple University in 1995 under the direction of Peter Hagis, Jr.
Article copyright: © Copyright 2000 American Mathematical Society