%I #8 Dec 09 2019 08:18:34
%S 2,1,2,3,2,4,2,3
%N Number of prime factors (with multiplicity) of generalized Fermat number 14^(2^n) + 1.
%C a(8) >= 5. - _Chai Wah Wu_, Dec 09 2019
%H <a href="http://factordb.com/index.php?query=14%5E256%2B1">factordb.com query on 14^256+1</a>.
%F a(n) = A001222(A152587(n)).
%e b(n) = 14^(2^n) + 1
%e Complete factorizations:
%e b(0) = 3*5
%e b(1) = 197
%e b(2) = 41*937
%e b(3) = 17*5393*16097
%e b(4) = 193*11284732320255809
%e b(5) = 7489*1204905857*1667461121*315256811699009
%e b(6) = 8633886977*P64
%e b(7) = 257*100497382788383295179961898289105815085380571534081*P95
%o (PARI) a(n) = bigomega(14^(2^n)+1)
%Y Cf. A275377, A275378, A275379, A275380, A275381, A275382, A275383, A302097, A152587.
%K nonn,hard,more
%O 0,1
%A _Jeppe Stig Nielsen_, Apr 01 2018