Quote:
Originally Posted by Citrix
1) Why do we have to use all the primes in B1. A smaller subset set of primes in B1 might have an lower hammer weight N. We then take this smaller subset and the remaining B1 primes (or move them to the B2 stage). Have you tried this?

Thinking about this further... all primes in B1 are not necessary and it does not need to be an absolute multiple of powersmooth(1000000). Some primes can be twice, some three times etc.
Turning the problem around  instead of fixing a number powersmooth (x) and then searching we leave both input and output variable.
If we are flexible with above  then we could generate a 2^n+c that has a lot of factors (smooth) and c is relatively small and we can control number of bits in n. This should have a low hammer weight.
OR generate a number with low hammer weight that has a lot of factors.