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Old 2021-12-13, 01:11   #1
ThomRuley
 
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May 2003

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Default Question about OPN sieving

I thought of a random question today about the process of sieving of odd perfect numbers. As I have understood the process, we take prime numbers less than an arbitrary limit and generate sigma chains based on powers of those prime numbers. It occurred to me that we can't possibly check the sigma chains on ALL the primes less than the limit, so I now have two questions:

1. Since we can't check them all, how do we know that we've eliminated all possible OPNs below 10^2000 or whatever other limit we have chosen?
2. How do we pick the right prime numbers to start the sigma chains?
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Old 2021-12-13, 12:31   #2
henryzz
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Quote:
Originally Posted by ThomRuley View Post
I thought of a random question today about the process of sieving of odd perfect numbers. As I have understood the process, we take prime numbers less than an arbitrary limit and generate sigma chains based on powers of those prime numbers. It occurred to me that we can't possibly check the sigma chains on ALL the primes less than the limit, so I now have two questions:

1. Since we can't check them all, how do we know that we've eliminated all possible OPNs below 10^2000 or whatever other limit we have chosen?
2. How do we pick the right prime numbers to start the sigma chains?

This is explained in the end game section of http://www.lirmm.fr/~ochem/opn/ and also in https://maths-people.anu.edu.au/~brent/pd/rpb116.pdf


Some of the larger numbers are included because it shortens the proof. For example excluding 5 without first considering 3 effectively requires excluding 3 as 5^1+1 is 6=2*3 so it is better to do this first. There may be others worth adding for this purpose. I am currently experimenting to see if there are any small primes that can be excluded easily that would shorten future sections of the proof. Unfortunately this is a nontrivial problem as so much effort has been expanded on the standard set of primes.
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