%I #19 Jun 09 2023 10:57:38
%S 0,0,2,2,3,3,3,2,4,4,3,2,5,2,4,3,6,2,4,3,5,3,3,3,5,3,4,5,4,3,5,3,7,4,
%T 3,4,5,4,3,3,6,2,4,2,6,4,3,3,7,3,4,4,4,3,8,4,5,4,5,2,6,3,3,4,7,5,5,3,
%U 5,3,5,3,7,2,4,5,4,4,6,3,6,5,5,3,6,3,4,3,6,4,6,3,4,5,4,3,8,3,4,5,6
%N Number of prime factors (with multiplicity) of A007588(n).
%F a(n) = A001222(A007588(n)).
%e 73*(2*73^2 - 1) = 777961 = 73 * 10657, which has two prime factors, so a(73) = 2.
%e 100*(2*100^2 - 1) = 1999900 = 2^2 * 5^2 * 7 * 2857 has 6 prime factors.
%p a:= n-> numtheory[bigomega](n*(2*n^2-1)):
%p seq(a(n), n=0..100); # _Alois P. Heinz_, Mar 03 2023
%Y Cf. A001222, A007588, A106483, A106484.
%K easy,nonn
%O 0,3
%A _Jonathan Vos Post_, May 03 2005
%E a(78) corrected by _Sean A. Irvine_, Mar 03 2023