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%I #44 Jun 05 2024 14:04:45
%S 0,0,0,0,0,12,15,3,9,202,56,304,635,11095,8948,6415,14441,877,37436
%N a(n) is the smallest k >= 0 such that 2^(2^n) + k*2^n + 1 is prime.
%C 2^(2^n) + a(n)*2^n + 1 = A019434(n) for n <= 4, the known Fermat primes.
%C Conjecture: 2^(2^n) + a(n)*2^n + 1 = A307532(n) for all n > 4.
%C Note that a(9) = A030239(9) = 202.
%F a(n) == 1 (mod 2^n).
%e For n = 5, k = 12; 2^(2^5) + 12*2^5 + 1 = 4294967681 is prime, a(5) = 12.
%t a[n_] := Module[{k = 0}, While[! PrimeQ[2^(2^n) + k*2^n + 1], k++];
%t k]; Array[a, 10, 0]
%o (PARI) isok(k, n) = isprime(2^(2^n) + k*2^n + 1);
%o a(n) = my(k=0); while (!isok(k, n), k++); k; \\ _Michel Marcus_, Apr 15 2019
%o (Python)
%o from sympy import isprime
%o def A307535(n):
%o r = 2**n
%o m, k = 2**r+1, 0
%o w = m
%o while not isprime(w):
%o k += 1
%o w += r
%o return k # _Chai Wah Wu_, Apr 29 2019
%Y Cf. A019434, A030239, A307532.
%K nonn,more,hard
%O 0,6
%A _Amiram Eldar_ and _Thomas Ordowski_, Apr 13 2019
%E a(15) from _Daniel Suteu_, Apr 14 2019
%E a(16)-a(17) from _Chai Wah Wu_, Apr 30 2019
%E a(18) from _Michael S. Branicky_, Jun 05 2024
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