Quote:
Originally Posted by carpetpool
I want to bring up the of the Mersenne Prime 2^n1 being the second prime p+2 in a twin prime pair {p, p+2} are there finitely many Mersenne Primes which hold this condition (this is the same as primes p such that 2^p1 and 2^p3 are prime).

The question of 2^p  3 and 2^p  1 both being prime, doesn't seem very interesting. After all, 2^p  1 is very seldom prime.
The question of when 2^n  3 alone might be prime may be of some interest in its own right. It wouldn't surprise me at all if someone had compiled a factor table for n into the hundreds, and a list of pseudoprimes for larger n's.
By way of keeping in practice with this sort of thing, I note the following:
Values of n < 1000 for which 2^n  3 tests as a pseudoprime:
n = 3, 4, 5, 6, 9, 10, 12, 14, 20, 22, 24, 29, 94, 116, 122, 150, 174, 213, 221, 233, 266, 336, 452, 545, 689, 694, 850
Prime values of 2^n  3 seem to occur for composite n much more often than for prime n.
2^p  3 is (pseudo)prime, but 2^p  1 is not, for the primes p = 29 and 233.
Recurrent prime divisors p < 100 of 2^n  3. Primes followed by an asterisk only divide 2^n  3 for composite exponents n.
5 (n = 4*k + 3)
11* (n = 10*k + 8)
13* (n = 12*k + 4)
19 (n = 18*k + 13)
23 (n = 11*k + 8)
29 (n = 28*k + 5)
37* (n = 36*k + 26)
47 (n = 23*k + 19)
53 (n = 52*k + 17)
59* (n = 58*k + 50)
61* (n = 60*k + 6)
67* (n = 66*k + 39)
71 (n = 35*k + 16)
83* (n = 82*k + 72)
97 (n = 48*k + 19)
Nondivisors p of 2^n  3: There are the obvious cases p = 2 and 3. For p > 3, we have the following: If 2 is a qth power residue of the prime p but 3 is not, then no power of 2 can be congruent to 3 (mod p). In the case q = 2 we can say this is the case for p congruent to 7 or 17 (mod 24), e.g. p = 7, 17, 31, 41, 79, 89. An example with q = 3 is p = 43.