For those of you who enjoy math that is logically consistent and useful, using the fact that the number of primes less than x, Pi(x), is asymptotic to x/[log(x) - 1], we can calculate: Pi[M39] ~= (2^13466917 - 1)/[log(2^13466917-1) - 1] = (2^13466917-1)/(9334554.5493..)
So, in fact, M39 is nowhere near the 9 millionth prime number. If we say M39 is the N'th prime, then N has roughly 4,053,940 digits. And no, this won't make it easier to find primes. Even the best approximations are never close enough to make it useful for that. Also, better approximations require more CPU power anyway. Regards, PdoX danny fleming wrote: > To Whom It May Concern: > > I have devised a method of easily figuring out > > approximately how many prime numbers are before > > a given prime. Here it is: since the natural > > logarithm of a number increases +2.3 for every > > power of 10, the 39th Mersenne Prime, since it > > contains 4,053,946 digits, is the 4,053,945*2.3= > > 9,324,074th prime number. Even more sophisticated > > methods can come even closer, it might make it easier > > to find unknown primes. > > Sincerely yours, > Danny Karl Fleming > > > _______________________________________________ > Like this email address? Choose from 100's of > the BEST at http://www.wildemail.com > _________________________________________________________________________ > Unsubscribe & list info -- http://www.ndatech.com/mersenne/signup.htm > Mersenne Prime FAQ -- http://www.tasam.com/~lrwiman/FAQ-mers > > _________________________________________________________________________ Unsubscribe & list info -- http://www.ndatech.com/mersenne/signup.htm Mersenne Prime FAQ -- http://www.tasam.com/~lrwiman/FAQ-mers
