OFFSET
0,5
COMMENTS
A generalized Fermat prime b^(2^n)+1 can be thought of as belonging to the "family" n. Then a(n) counts how many generalized Fermat primes in family n precede the first generalized Fermat prime in family n+1.
Each family as defined here is a subset of its preceding family, in the sense that b^(2^n) + 1 = (b^2)^(2^(n-1)) + 1.
a(12) is expected to be near 97000.
LINKS
EXAMPLE
To find a(5), find all primes b^32 + 1 until you reach a base b that is a perfect square. In this case you find 152 nonsquare b values { 30, 54, 96, 112, ..., 10396 }, but the 153rd b is 10404, a perfect square. So 10404^32 + 1 = 102^64 + 1 belongs to the next family. Therefore a(5)=152.
PROG
(PARI) b=2; for(n=0, 100, x=0; until(, if(ispseudoprime(b^(2^n)+1), if(issquare(b, &b), break, x++)); b+=2); print("a(", n, ")=", x, ", next b is ", b))
CROSSREFS
KEYWORD
nonn,more,hard
AUTHOR
Jeppe Stig Nielsen, Jul 06 2015
EXTENSIONS
a(12) via b-file of A088362 from Jeppe Stig Nielsen, Feb 16 2022
STATUS
approved