Any other primes in this sequence?

[Citrix]

Cant you use PFGW and find all the primes of this form?

Citrix

[R.D. Silverman]

Originally Posted by akruppa

This is, in fact, a good question and well posed. Promoted to "Math". Unfortunately, I don't know the answer.

Alex

Almost certainly there are finitely many.

We want N= 2^p-1 and N1 = 2^(p-1)(2^p - 1) + 1 to be simultaneously
prime. The probability of this is O(1/ ((log N)(log N1))) and the sum over
all p converges. It converges fairly rapidly.

[akruppa]

Agreed. For all known Mersenne primes we now know that there aren't any more primes of this form and larger such primes are extremely unlikely. So the answer to the original poster's question is: almost certainly not.

As for the question for an efficient primality proof: since we have the complete factorisation of P(n) by definition of P(n) (for n a Mersenne prime exponent), a simple P-1 primality test will work. Find some integer a so that a^{[I]P[/I]([I]n[/I])}≡1 (mod P(n)+1) and for each p|P(n), a^{[I]P[/I]([I]n[/I])/[I]p[/I]}!≡1 (mod P(n)+1). The test can be further simpified but this is the basic idea.

Alex

[flava]

I looked for the primes satisfying your equation and for the following primes p (2^p-1)*(2^(p-1))+1 is prime or probable prime:

p in {2,3,13,19,271,601,4729,34039}

Tested up to p=67261

As you can see they are quite scarce ...

[R.D. Silverman]

Originally Posted by flava

I looked for the primes satisfying your equation and for the following primes p (2^p-1)*(2^(p-1))+1 is prime or probable prime:

p in {2,3,13,19,271,601,4729,34039}

Tested up to p=67261

As you can see they are quite scarce ...

This is not the original sequence. For the sequence you propose, we
expect infinitely many. The original sequence ALSO requires that 2^p-1
is prime.

[flava]

Originally Posted by R.D. Silverman

This is not the original sequence. For the sequence you propose, we expect infinitely many. The original sequence ALSO requires that 2^p-1 is prime.

Yes I know. I was just satisfying my own curiosity and made this quick check to see if the distribution of (2^p-1)*(2^(p-1))+1 primes is not higher than expected. What makes you expect infinitely many primes of this form? Is it the fact that sum(1/p) diverges over primes sufficient?