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A369257 a(n) = number of odd divisors of n that have an even number of prime factors with multiplicity.

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%I A369257 #23 Jan 28 2024 09:20:19
%S A369257 1,1,1,1,1,1,1,1,2,1,1,1,1,1,2,1,1,2,1,1,2,1,1,1,2,1,2,1,1,2,1,1,2,1,
%T A369257 2,2,1,1,2,1,1,2,1,1,3,1,1,1,2,2,2,1,1,2,2,1,2,1,1,2,1,1,3,1,2,2,1,1,
%U A369257 2,2,1,2,1,1,3,1,2,2,1,1,3,1,1,2,2,1,2,1,1,3,2,1,2,1,2,1,1,2,3,2,1,2,1,1,4
%N A369257 a(n) = number of odd divisors of n that have an even number of prime factors with multiplicity.
%H A369257 Antti Karttunen, <a href="/A369257/b369257.txt">Table of n, a(n) for n = 1..65537</a>
%F A369257 a(n) = Sum_{d|n} A353557(d).
%F A369257 a(n) = A001227(n) - A369258(n).
%F A369257 a(n) = a(2*n) = a(A000265(n)).
%F A369257 For n >= 1, a(2n-1) = A038548(2n-1); for n > 1, a(2n) < A038548(2n).
%F A369257 From _Antti Karttunen_, Jan 27 2024: (Start)
%F A369257 a(n) = A038548(A000265(n)).
%F A369257 a(n) = (A001227(n)+A053866(n))/2.
%F A369257 Dirichlet g.f.: (zeta(s)^2*(1-2^-s) + zeta(2s)*(1+2^-s)) / 2.
%F A369257 (End)
%e A369257 Of the eight odd divisors of 105, the four divisors 1, 15, 21, 35 all have an even number of prime factors (A001222(d) is even), therefore a(105) = 4.
%o A369257 (PARI)
%o A369257 A353557(n) = ((n%2)&&(!(bigomega(n)%2)));
%o A369257 A369257(n) = sumdiv(n,d,A353557(d));
%Y A369257 Inverse Möbius transform of A353557.
%Y A369257 Cf. A000265, A001227, A038548, A046337, A053866, A353557, A369258, A369454 (Dirichlet inverse).
%K A369257 nonn
%O A369257 1,9
%A A369257 _Antti Karttunen_, Jan 24 2024