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A301738
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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.
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2
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3, 3, 3, 5, 3, 3, 3, 49, 7, 35, 67, 75, 157, 107, 71, 137, 275, 531
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internal format)
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OFFSET
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0,1
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COMMENTS
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Both A and B will be odd, with gcd(A, B) = 1. B values can be seen in link section.
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LINKS
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EXAMPLE
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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).
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MATHEMATICA
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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 *)
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PROG
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(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
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CROSSREFS
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KEYWORD
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nonn,hard,more
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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