%I #22 Aug 07 2019 07:42:13
%S 1,1,1,2,2,3,3,3,5
%N Number of odd prime factors (with multiplicity) of generalized Fermat number 5^(2^n) + 1.
%H factordb.com, <a href="http://factordb.com/index.php?query=5%5E%282%5En%29%2B1">Status of 5^(2^n)+1</a>.
%F a(n) = A001222(A199591(n)) - 1. - _Felix Fröhlich_, Jul 25 2016
%e b(n) = (5^(2^n) + 1)/2.
%e Complete Factorizations
%e b(0) = 3
%e b(1) = 13
%e b(2) = 313
%e b(3) = 17*11489
%e b(4) = 2593*29423041
%e b(5) = 641*75068993*241931001601
%e b(6) = 769*3666499598977*96132956782643741951225664001
%e b(7) = 257*23653200983830003298459393*P62
%e b(8) = 1655809*101199664791578113*4563566430220614493697*
%e 12025702000065183805751513732616276516181800961*P88
%t Table[PrimeOmega[(5^(2^n) + 1)/2], {n, 0, 6}] (* _Michael De Vlieger_, Jul 26 2016 *)
%o (PARI) a(n) = bigomega(factor((5^(2^n)+1)/2))
%Y Cf. A199591, A273946.
%K nonn,hard,more
%O 0,4
%A _Arkadiusz Wesolowski_, Jul 25 2016
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