How do I interpret Prime95 factoring benchmarks?

jasong · 2006-10-22, 21:04

Okay, I know that, for instance 2^x is half as big as 2^(x+1), so 2^60 is half as big as 2^61 and 1/4th as big as 2^62. Unfortunately, Prime95 gives the amount of time to test one factor, but I have no way of knowing how many factors are in a range.

Have I simply not managed to pick up on something that should be obvious?

dsouza123 · 2006-10-23, 12:08

For a 61 bit trial factor, it's value is between 2^60 and 2^61,
Given a Mersenne number m which is 2^p - 1 with p the prime exponent.

roughly 0.05*[int(2^61/(2*p)) - int(2^60/(2*p)]

because trial factors q are of the form 2*k*p + 1 with k = 1,2,3 etc
also they are 1 or 7 mod 8 and are also must be prime themselves.

The help file also mentions most get seived out leaving about 5 %
to test.

2006-10-22, 21:04	#1
jasong "Jason Goatcher" Mar 2005 3×7×167 Posts	How do I interpret Prime95 factoring benchmarks? Okay, I know that, for instance 2^x is half as big as 2^(x+1), so 2^60 is half as big as 2^61 and 1/4th as big as 2^62. Unfortunately, Prime95 gives the amount of time to test one factor, but I have no way of knowing how many factors are in a range. Have I simply not managed to pick up on something that should be obvious?

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2006-10-23, 12:08	#2
dsouza123 Sep 2002 2·331 Posts	For a 61 bit trial factor, it's value is between 2^60 and 2^61, Given a Mersenne number m which is 2^p - 1 with p the prime exponent. roughly 0.05[int(2^61/(2p)) - int(2^60/(2p)] because trial factors q are of the form 2k*p + 1 with k = 1,2,3 etc also they are 1 or 7 mod 8 and are also must be prime themselves. The help file also mentions most get seived out leaving about 5 % to test.