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

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, 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, 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

Offset: 1

Author

Antti Karttunen, Jan 24 2024

Keywords

• nonn

Examples

			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.
		

Links

• Antti Karttunen, Table of n, a(n) for n = 1..65537

Crossrefs

Inverse Möbius transform of A353557.

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

Programs

PARI

A353557(n) = ((n%2)&&(!(bigomega(n)%2)));
A369257(n) = sumdiv(n,d,A353557(d));

Formula

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

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

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

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

From Antti Karttunen, Jan 27 2024: (Start)

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

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

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

(End)