A030239 -id:A030239 - cp's OEIS frontend

A307535 a(n) is the smallest k >= 0 such that 2^(2^n) + k*2^n + 1 is prime.

0, 0, 0, 0, 0, 12, 15, 3, 9, 202, 56, 304, 635, 11095, 8948, 6415, 14441, 877, 37436

Offset: 0

Amiram Eldar and Thomas Ordowski, Apr 13 2019

2^(2^n) + a(n)*2^n + 1 = A019434(n) for n <= 4, the known Fermat primes.
Conjecture: 2^(2^n) + a(n)*2^n + 1 = A307532(n) for all n > 4.
Note that a(9) = A030239(9) = 202.

			For n = 5, k = 12; 2^(2^5) + 12*2^5 + 1 = 4294967681 is prime, a(5) = 12.

Cf. A019434, A030239, A307532.

a[n_] := Module[{k = 0}, While[! PrimeQ[2^(2^n) + k*2^n + 1], k++];
  k]; Array[a, 10, 0]

isok(k, n) = isprime(2^(2^n) + k*2^n + 1);
a(n) = my(k=0); while (!isok(k, n), k++); k; \\ Michel Marcus, Apr 15 2019

from sympy import isprime
def A307535(n):
    r = 2**n
    m, k = 2**r+1, 0
    w = m
    while not isprime(w):
        k += 1
        w += r
    return k # Chai Wah Wu, Apr 29 2019

a(n) == 1 (mod 2^n).

a(15) from Daniel Suteu, Apr 14 2019
a(16)-a(17) from Chai Wah Wu, Apr 30 2019
a(18) from Michael S. Branicky, Jun 05 2024

cp's OEIS Frontend

A307535 a(n) is the smallest k >= 0 such that 2^(2^n) + k*2^n + 1 is prime.

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