Factoring from the factor1
in 2^P2 you allways have P as a factor
Let F the number to test
first test if prime
if so then N=F1
Find all factor S of N
test if N  2^S1
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.
