>(1) if you consider the prime number theorem to approximate the probability
>that 2^p-1 is prime and sum that over all primes p, you get an infinite number
>which means that you expect an infinite number of Mersenne primes.
True, the probability of a given n being prime is ~1/log(n), and
the sum from 1 to infinity of 1/log(2^p-1)~=log(2)*sum from 1 to infinity 1/p
which euler proved is infinite. I think that the probability of 2^p-1 being
prime is considerably higher than most numbers that big, because they can
only be divisable by numbers of the form 2*k*p+1, and they must be ==+/-1
mod 8. Thus
1/4*(1/(2*p))*sqrt(2^p-1)~=2^(p/2-3)/p possible divisors (less when you look
at the probability that 2*k*p+1 is prime) as opposed to
~sqrt(2^p)/log(sqrt(2^p))=2^(p/2)/(log(2)*p/2)
which is 2^3*2/log(2)~=23 times fewer possible factors than a normal number
that size (it's not much, but hey, it's something).
>(2) there are several conjectures concerning the growth rate of successive
>Mersenne primes. They all suggest that on average, one exponent resulting in
>a Mersenne prime is no more than twice the previous one. This implies an
>infinite number of Mersenne primes. The known Mersenne primes are in very good
>agreement with the conjecture that, on the average, an exponent resulting in a
>Mersenne prime is about 3/2 as large as the previous one. That, of course,
>would imply an infinite number of Mersenne primes.
Well, I don't know about that. Using conjectured behavior of mersenne primes
to argue other conjectures... We must tread carefully...
-Lucas Wiman
________________________________________________________________
Unsubscribe & list info -- http://www.scruz.net/~luke/signup.htm
