A341211 Smallest prime p such that (p^(2^n) + 1)/2 is prime.

3, 3, 3, 13, 3, 3, 3, 113, 331, 3631, 827, 3109, 4253, 7487, 71

Offset: 0

Author

Jon E. Schoenfield, Feb 06 2021

Keywords

• nonn
• hard
• more

Comments

Expressions of the form m^j + 1 can be factored (e.g., m^3 + 1 = (m + 1)*(m^2 - m + 1)) for any positive integer j except when j is a power of 2, so (p^j + 1)/2 for prime p cannot be prime unless j is a power of 2.

a(12) <= 4253, a(13) <= 7487, a(14) <= 71. - Daniel Suteu, Feb 07 2021

a(13) > 2500 and a(14) = 71. - Jinyuan Wang, Feb 07 2021

Examples

			No term is smaller than 3 (since 2 is the only smaller prime, and (2^(2^n) + 1)/2 is not an integer).
(3^(2^0) + 1)/2 = (3^1 + 1)/2 = (3 + 1)/2 = 4/2 = 2 is prime, so a(0)=3.
(3^(2^1) + 1)/2 = (3^2 + 1)/2 = 5 is prime, so a(1)=3.
(3^(2^2) + 1)/2 = (3^4 + 1)/2 = 41 is prime, so a(2)=3.
(3^(2^3) + 1)/2 = (3^8 + 1)/2 = 3281 = 17*193 is not prime, nor is (p^8 + 1)/2 for any other prime < 13, but (13^8 + 1)/2 = 407865361 is prime, so a(3)=13.
		

Links

• Dario Alpern, Integer factorization calculator

Crossrefs

Cf. A005383, A048161, A176116, A340480.

Cf. A093625 and A171381 (both for when p=3).

Programs

Alpertron

x=3;x=N(x);NOT IsPrime((x^8192+1)/2);N(x)
# Martin Ehrenstein, Feb 08 2021

PARI

a(n) = my(p=3); while (!isprime((p^(2^n) + 1)/2), p=nextprime(p+1)); p; \\ Michel Marcus, Feb 07 2021

Python

from sympy import isprime, nextprime
def a(n):
  p, pow2 = 3, 2**n
  while True:
    if isprime((p**pow2 + 1)//2): return p
    p = nextprime(p)
print([a(n) for n in range(9)]) # Michael S. Branicky, Mar 03 2021

Extensions

a(11) from Daniel Suteu, Feb 07 2021

a(12) from Jinyuan Wang, Feb 07 2021

a(13)-a(14), using Dario Alpern's integer factorization calculator and prior bounds, from Martin Ehrenstein, Feb 08 2021