Are Mersenne primes randomly distributed? If so, then Prob(Mn is a Mersenne prime) = Prob(n is prime) * Prob(Mn is prime|n is prime). The first factor is 1/log(n), and the second factor is 2/log(2^n-1) = (2/log 2)*(1/n), where the extra factor of 2 comes from knowing that Mn is odd. Thus Prob(Mn is prime) = (2/log 2)*(1/(n log(n)), and the expected number of Mersenne primes Mp for n1 < p < n2 is (2/log 2)*int(1/(x log(x)), x=n1..n2) = (2/log 2)*(loglog(n2)-loglog(n1)). Since each Mp has a small probability of being prime, the variance equals the mean. So the expected standard deviation of the number of Mersenne primes equals the square root of the expected number of Mersenne primes. The known Mersenne primes are listed at http://www.utm.edu/research/primes/mersenne/index.html . Take n1 = 100 to avoid small number glitches, and n2 = 10^6 since 10^7 would take us into an incompletely explored gap region. Then the expected number of Mersenne primes is 3.17, while the actual number is 23. On the other hand, if we ignore the factor log(x) in the denominator of the integrand, then loglog is replaced by log, and then the expected number of Mersenne primes is 26.6 +/- 5.2. So it appears that the Mersenne numbers Mp = 2^p - 1 possess an enhanced likelihood of being prime by a factor of log(p) over randomly chosen odd numbers of similar size. Gene __________________________________ Do you Yahoo!? New and Improved Yahoo! Mail - Send 10MB messages! http://promotions.yahoo.com/new_mail