

A332847


a(n) is the smallest k such that exactly one of k*2^(2^n)  2*k + 1 and k*2^(2^n) + 2*k  1 is a prime.


0



1, 4, 1, 1, 1, 3, 3, 10, 17, 8, 83, 92, 525, 1888, 20
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OFFSET

0,2


COMMENTS

Conjecture: a(n) > 1 for all n > 4.
a(n) = 1, n > 1, is equivalent to F(n) = 2^(2^n) + 1 being a Fermat prime, because the Mersenne number M(2^n) = 2^(2^n)  1 is prime only for n = 1 (since divisible by 3 for all n >= 1), where F(1) is also prime.  The two considered numbers can also be written 2*k*A077585(n) + 1 resp. 2*k*A000051(A000225(n))  1.  M. F. Hasler, Mar 05 2020


LINKS

Table of n, a(n) for n=0..14.


EXAMPLE

a(0) = 1 because 1*2^(2^0)  2*1 + 1 = 1 is a nonprime and 1*2^(2^0) + 2*1  1 = 3 is a prime.
a(1) = 4 because 4*2^(2^1)  2*4 + 1 = 9 is a composite and 4*2^(2^1) + 2*4  1 = 23 is a prime.
a(2) = 1 because 1*2^(2^2)  2*1 + 1 = 15 is a composite and 1*2^(2^2) + 2*1  1 = 17 is a prime.
a(3) = 1 because 1*2^(2^3)  2*1 + 1 = 255 is a composite and 1*2^(2^3) + 2*1  1 = 257 is a prime.
a(4) = 1 because 1*2^(2^4)  2*1 + 1 = 65535 is a composite and 1*2^(2^4) + 2*1  1 = 65537 is a prime.


PROG

(PARI) a(n) = {my(k=1, m=2^2^n); while(ispseudoprime(k*m2*k+1)ispseudoprime(k*m+2*k1)==0, k++); k; } \\ Jinyuan Wang, Feb 26 2020


CROSSREFS

Cf. A019434, A331487.
Cf. A000215 (Fermat numbers 2^2^n + 1), A000225 (Mersenne numbers 2^n  1).
Sequence in context: A321592 A031278 A010328 * A245501 A247026 A193512
Adjacent sequences: A332844 A332845 A332846 * A332848 A332849 A332850


KEYWORD

nonn,more


AUTHOR

JuriStepan Gerasimov, Feb 26 2020


EXTENSIONS

Offset changed to 0 and a(11)a(14) from Jinyuan Wang, Feb 26 2020


STATUS

approved



