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%I #10 Feb 25 2024 02:17:02
%S 0,0,1,1,1,2,1,2,3,2,1,4,1,2,3,3,1,5,1,5,3,3,1,6,3,2,5,5,2,6,1,4,4,2,
%T 4,9,2,2,3,7,2,7,1,6,6,3,2,9,2,6,5,6,2,8,5,7,4,5,3,12,1,2,6,6,5,8,2,6,
%U 4,8,2,13,2,4,6,6,5,7,1,9,9,4,5,13,3,4
%N Number of prime factors of A001045(n) (Jacobsthal numbers) (counted with multiplicity).
%H Amiram Eldar, <a href="/A366770/b366770.txt">Table of n, a(n) for n = 1..1122</a>
%F a(n) = bigomega(Jacobsthal(n)) = A001222(A001045(n)).
%e a(9) = 3 because Jacobsthal(9) = 171 = 3^2 * 19.
%Y Cf. A001045, A001222, A366769, A366771, A038575.
%K nonn
%O 1,6
%A _Sean A. Irvine_, Oct 21 2023
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