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A307535
a(n) is the smallest k >= 0 such that 2^(2^n) + k*2^n + 1 is prime.
2
0, 0, 0, 0, 0, 12, 15, 3, 9, 202, 56, 304, 635, 11095, 8948, 6415, 14441, 877, 37436
OFFSET
0,6
COMMENTS
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.
FORMULA
a(n) == 1 (mod 2^n).
EXAMPLE
For n = 5, k = 12; 2^(2^5) + 12*2^5 + 1 = 4294967681 is prime, a(5) = 12.
MATHEMATICA
a[n_] := Module[{k = 0}, While[! PrimeQ[2^(2^n) + k*2^n + 1], k++];
k]; Array[a, 10, 0]
PROG
(PARI) 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
(Python)
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
CROSSREFS
KEYWORD
nonn,more,hard
AUTHOR
Amiram Eldar and Thomas Ordowski, Apr 13 2019
EXTENSIONS
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
STATUS
approved