A301738 a(n) is the least A for which there exists B with 0 < B < A so that (A^(2^n) + B^(2^n))/2 is prime.

3, 3, 3, 5, 3, 3, 3, 49, 7, 35, 67, 75, 157, 107, 71, 137, 275, 531

Offset: 0

Author

Jeppe Stig Nielsen, Mar 26 2018

Keywords

• nonn
• hard
• more

Comments

Both A and B will be odd, with gcd(A, B) = 1. B values can be seen in link section.

If we require B=1, we get A275530. Therefore a(n) <= A275530(n).

Examples

			a(10)=67 corresponds to the prime number (67^1024 + 57^1024)/2, the smallest prime number of the form (A^1024 + B^1024)/2 (or more precisely, it minimizes A).
		

Links

• H. Lifchitz and R. Lifchitz, PRP Top (275^65536+53^65536)/2.
• H. Lifchitz and R. Lifchitz, PRP Top (531^131072+25^131072)/2.
• Jeppe Stig Nielsen, First primes of the form (A^(2^n) + B^(2^n))/2 (gives B values).

Crossrefs

Cf. A291944, A275530.

Programs

Mathematica

Table[a=1; While[Or@@PrimeQ[Table[(a^(2^n)+b^(2^n))/2,{b,a++}]]==False];a,{n,0,9}] (* Giorgos Kalogeropoulos, Mar 31 2021 *)

PARI

for(n=0, 30, forstep(a=3, +oo, 2, forstep(b=1, a-2, 2, if(ispseudoprime((a^(2^n)+b^(2^n))/2), print1(a, ", "); next(3)))))

Python

from sympy import isprime
def a(n):
  A, p, Ap = 3, 2**n, 3**(2**n)
  while True:
    if any(isprime((Ap + B**p)//2) for B in range(1, A, 2)): return A
    A += 2; Ap = A**p
print([a(n) for n in range(10)]) # Michael S. Branicky, Mar 03 2021

Extensions

a(16)=275 with B=53, calculated by Kellen Shenton, added by Jeppe Stig Nielsen, Nov 10 2020

a(17)=531 with B=25, by Kellen Shenton, added by Jeppe Stig Nielsen, Mar 30 2021