mersenneforum.org - View Single Post - 5dpf, 5 digits from the end partial factors

Maybeso · 2004-04-26, 03:05

Using brute force on the end 5dpf's:
~ 50000 odds less than 100000
~ 40000 with 5's removed, there should be
~ 20000 pairs, but we lose those matched with the missing 5's, so actual pairs
= 11145.

So the plan could be to sieve a block of k*10^5 + 45903 and k*10^5 + 46303. Removing a number from the low end of one group should also remove one from the high end of the other.

If handled properly, that will reduce the work by a factor of 7.

2004-04-26, 03:05	#6
Maybeso Aug 2002 Portland, OR USA 2×137 Posts	Using brute force on the end 5dpf's: ~ 50000 odds less than 100000 ~ 40000 with 5's removed, there should be ~ 20000 pairs, but we lose those matched with the missing 5's, so actual pairs = 11145. So the plan could be to sieve a block of k10^5 + 45903 and k10^5 + 46303. Removing a number from the low end of one group should also remove one from the high end of the other. If handled properly, that will reduce the work by a factor of 7.