Thanks, now I understand. log(p)^2 is about 768 for p = 2^40, so q shouldn't be much bigger than about 2^50, which gives plenty of room.
On a 32 bit machine I guess we can happily get to 2^24 bits as log(p)^2 is then about 8 bits. Given the memory usage of the algorithm this is not much of a limit, I think. Bill. 2009/6/21 Robert Gerbicz <robert.gerb...@gmail.com>: > OK, I haven't known about that group, now I'm a member of that also. > > "I'm slightly confused. Isn't the first q of this form precisely q = 2p + 1? > If so, it cannot be as large as your limit." > > See, we are searching for primes of the form q=2kp+1, if p is prime then it > isn't sure that 2p+1 is also prime, and if it is composite, then prod=1 > remains, so we are still in the while loop, and we'll test q=4p+1 and so on. > If I find a prime then prod*=p will be true also. If prod>bound_for_p/p then > I'm exit from the while loop, I've found enough prime(s) for p (at least one > prime, however for small p more than one). > > The following pari-gp code gives the first such q number: > f(p)=q=2*p+1;while(isprime(q)==0,q+=2*p);return(q) > > for example f(97)=389, some "extreme" values: > f(5227)=397253, so p=5227, q=397253, for this: q=1.0368*p*log(p)^2 > f(170167)=24504049, so q=0.9926*p*log(p)^2 > > > > --~--~---------~--~----~------------~-------~--~----~ You received this message because you are subscribed to the Google Groups "mpir-devel" group. To post to this group, send email to mpir-devel@googlegroups.com To unsubscribe from this group, send email to mpir-devel+unsubscr...@googlegroups.com For more options, visit this group at http://groups.google.com/group/mpir-devel?hl=en -~----------~----~----~----~------~----~------~--~---
