New paper on odd perfect numbers
Dec. 4th, 2021 02:18 pm![[personal profile]](https://www.dreamwidth.org/img/silk/identity/user.png)
Recall that a number is perfect if the sum of its divisors less than the number is equal to the number. For example, 6 is perfect since the relevant divisors are 1, 2 and 3, and 1+2+3=6. One of the oldest unsolved problems in all of math is whether there are any odd numbers which are perfect.
In the last few years, there's been a bit of work on bounds about the largest prime factors of an odd perfect number N. In particular, assume that the the distinct prime factors of N are p1, p2, p3 ... pk with p1 < p2 < p3 ... < pk .
Acquaah and Konyagin proved that one must have pk < (3N)(1/3) . Subsequently, using essentially the same techniques as them proved that p(k-1) < (2N)(1/5), and that p(k-1)pk < 3N(1/2). In a similar vein Luca and Pomerance using a similar technique showed that p1p2p3... pk < 2N(17/26). In this paper , we prove that p(k-2) < (2N)(1/6), and that p(k-2)p(k-1)pk < (2N)(3/5). Note that 17/26 > 3/5, so this is an actual improvement over Luca and Pomerance's bound, although at the cost of only applying to the product three largest distinct prime factors rather than the product of all of them.
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Date: 2021-12-06 11:59 am (UTC)