More on the total number of prime factors of an odd perfect number
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- by Kevin G. Hare PDF
- Math. Comp. 74 (2005), 1003-1008
Abstract:
Let $\sigma (n)$ denote the sum of the positive divisors of $n$. We say that $n$ is perfect if $\sigma (n) = 2 n$. Currently there are no known odd perfect numbers. It is known that if an odd perfect number exists, then it must be of the form $N = p^\alpha \prod _{j=1}^k q_j^{2 \beta _j}$, where $p, q_1, \ldots , q_k$ are distinct primes and $p \equiv \alpha \equiv 1 \pmod {4}$. Define the total number of prime factors of $N$ as $\Omega (N) := \alpha + 2 \sum _{j=1}^k \beta _j$. Sayers showed that $\Omega (N) \geq 29$. This was later extended by Iannucci and Sorli to show that $\Omega (N) \geq 37$. This paper extends these results to show that $\Omega (N) \geq 47$.References
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Additional Information
- Kevin G. Hare
- Affiliation: Department of Pure Mathematics, University of Waterloo, Waterloo, Ontario, Canada, N2L 3G1
- Email: kghare@math.uwaterloo.ca
- Received by editor(s): October 24, 2003
- Received by editor(s) in revised form: December 2, 2003
- Published electronically: June 29, 2004
- Additional Notes: This research was supported, in part, by NSERC of Canada.
- © Copyright 2004 by the author
- Journal: Math. Comp. 74 (2005), 1003-1008
- MSC (2000): Primary 11A25, 11Y70
- DOI: https://doi.org/10.1090/S0025-5718-04-01683-7
- MathSciNet review: 2114661