20211213, 01:11  #1 
May 2003
3·97 Posts 
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? 
20211213, 12:31  #2  
Just call me Henry
"David"
Sep 2007
Liverpool (GMT/BST)
2^{3}×7×107 Posts 
Quote:
This is explained in the end game section of http://www.lirmm.fr/~ochem/opn/ and also in https://mathspeople.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. 

Thread Tools  
Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
Question about sieving  David703  Proth Prime Search  27  20201018 16:01 
Sieving Question  __HRB__  Math  1  20190428 05:47 
Dumb sieving question  fivemack  Software  7  20171127 22:48 
A question on lattice sieving  joral  Factoring  5  20080403 08:01 
Sieving question  jasong  Sierpinski/Riesel Base 5  9  20070723 00:03 