The third largest prime divisor of an odd perfect number exceeds one hundred
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- by Douglas E. Iannucci PDF
- Math. Comp. 69 (2000), 867-879 Request permission
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.
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.
References
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Additional Information
- Douglas E. Iannucci
- Affiliation: University of the Virgin Islands, 2 John Brewers Bay, St. Thomas, VI 00802
- Email: diannuc@uvi.edu
- 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.
- © Copyright 2000 American Mathematical Society
- Journal: Math. Comp. 69 (2000), 867-879
- MSC (1991): Primary 11A25, 11Y70
- DOI: https://doi.org/10.1090/S0025-5718-99-01127-8
- MathSciNet review: 1651762