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Number of prime factors (with multiplicity) of A007588(n).
2

%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