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Old 2003-06-21, 00:40   #1
Sep 2002

1000001102 Posts
Default Factoring from the factor-1

in 2^P-2 you allways have P as a factor

Let F the number to test
first test if prime
if so then N=F-1
Find all factor S of N
test if N | 2^S-1

wouldn`t that be faster than to go about than try to divide one mersenne by all primes.

You only have a few to test for each factor.
So when you get to 2^64 you have found all factor less than 2^64 for all Mersenne.
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