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Number of prime factors (with multiplicity) of generalized Fermat number 14^(2^n) + 1.
1

%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