odds of a Mers. prime being found in a range of n.

[jasong]

Is the following true? Or, at least, assumed to be true?

odds of a random n yielding a Mersenne prime:

ln(n)*ln(2^n-1)

I had a hell of a time coming up with the qualifier(the second line). Corrections are appreciated and encouraged.

Btw, I put this in Information and Answers because it's fairly simple math, and I feel that a lot of people that might not normally visit the Math forum might be interested in this, since it has to do with Mersenne numbers.

Edit: Another question: Assuming GIMPS throughput increases at a sustained rate, basically a parabolical curve, will our speed at finding primes be likely to increase, decrease, or stay exactly the same? (Maybe this one should go in Math, and the answer should come back here when it's decided upon)

[TimSorbet]

Originally Posted by jasong

Is the following true? Or, at least, assumed to be true?

odds of a random n yielding a Mersenne prime:

ln(n)*ln(2^n-1)

I had a hell of a time coming up with the qualifier(the second line). Corrections are appreciated and encouraged.

Btw, I put this in Information and Answers because it's fairly simple math, and I feel that a lot of people that might not normally visit the Math forum might be interested in this, since it has to do with Mersenne numbers.

Edit: Another question: Assuming GIMPS throughput increases at a sustained rate, basically a parabolical curve, will our speed at finding primes be likely to increase, decrease, or stay exactly the same? (Maybe this one should go in Math, and the answer should come back here when it's decided upon)

Originally Posted by http://www.mersenne.org/math.htm

What are the chances that the Lucas-Lehmer test will find a new Mersenne prime number? A simple approach is to repeatedly apply the observation that the chance of finding a factor between 2^X and 2^X+1 is about 1/x. For example, you are testing 2^10000139-1 for which trial factoring has proved there are no factors less than 2⁶⁴. The chance that it is prime is the chance of no 65-bit factor * chance of no 66 bit factor * ... * chance of no 5000070 bit factor. That is:
64 65 5000069
-- * -- * ... * -------
65 66 5000070
This simplifies to 64 / 5000070 or 1 in 78126. This simple approach isn't quite right. It would give a formula of how_far_factored divided by (exponent divided by 2). However, more rigorous work has shown the formula to be (how_far_factored-1) / (exponent times Euler's constant (0.577...)). In this case, 1 in 91623. Even these more rigourous formulas are unproven.

.

[R.D. Silverman]

Originally Posted by Mini-Geek

.

This question is anything BUT simple math. The question involves
VERY deep questions in analytic number theory and the true answer
isn't even known, let alone proven. It is *conjectured* on the basis
of some heuristics that Mersenne primes are Poisson distributed and that
the expected number of Mersenne primes between 2^n and 2^(n+1) is
exp(gamma) as n --> oo. However this conjecture is not very "firm".
It is more of an open question than a true conjecture (like (say) the
twin prime conjecture).

[jasong]

I apologize,

odds of a random n yielding a Mersenne prime

should've been

odds of a random n being a Mersenne prime