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a(n) = number of odd divisors of n that have an even number of prime factors with multiplicity.
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%I #23 Jan 28 2024 09:20:19

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

%H Antti Karttunen, <a href="/A369257/b369257.txt">Table of n, a(n) for n = 1..65537</a>

%F a(n) = Sum_{d|n} A353557(d).

%F a(n) = A001227(n) - A369258(n).

%F a(n) = a(2*n) = a(A000265(n)).

%F For n >= 1, a(2n-1) = A038548(2n-1); for n > 1, a(2n) < A038548(2n).

%F From _Antti Karttunen_, Jan 27 2024: (Start)

%F a(n) = A038548(A000265(n)).

%F a(n) = (A001227(n)+A053866(n))/2.

%F Dirichlet g.f.: (zeta(s)^2*(1-2^-s) + zeta(2s)*(1+2^-s)) / 2.

%F (End)

%e 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 (PARI)

%o A353557(n) = ((n%2)&&(!(bigomega(n)%2)));

%o A369257(n) = sumdiv(n,d,A353557(d));

%Y Inverse Möbius transform of A353557.

%Y Cf. A000265, A001227, A038548, A046337, A053866, A353557, A369258, A369454 (Dirichlet inverse).

%K nonn

%O 1,9

%A _Antti Karttunen_, Jan 24 2024