20060221, 20:12  #12 
Jun 2003
1576_{10} Posts 
Cant you use PFGW and find all the primes of this form?
Citrix 
20060221, 20:22  #13  
Nov 2003
2^{2}·5·373 Posts 
Quote:
We want N= 2^p1 and N1 = 2^(p1)(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. 

20060221, 21:47  #14 
"Nancy"
Aug 2002
Alexandria
2,467 Posts 
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 P1 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 pP(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 
20060225, 20:00  #15 
Feb 2003
2·59 Posts 
I looked for the primes satisfying your equation and for the following primes p (2^p1)*(2^(p1))+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 ... Last fiddled with by flava on 20060225 at 20:01 
20060227, 12:01  #16  
Nov 2003
2^{2}×5×373 Posts 
Quote:
expect infinitely many. The original sequence ALSO requires that 2^p1 is prime. 

20060227, 21:23  #17  
Feb 2003
2·59 Posts 
Quote:


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