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A369257
a(n) = number of odd divisors of n that have an even number of prime factors with multiplicity.
4
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
1,9
LINKS
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)
EXAMPLE
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.
PROG
(PARI)
A353557(n) = ((n%2)&&(!(bigomega(n)%2)));
A369257(n) = sumdiv(n, d, A353557(d));
CROSSREFS
Inverse Möbius transform of A353557.
Cf. A000265, A001227, A038548, A046337, A053866, A353557, A369258, A369454 (Dirichlet inverse).
Sequence in context: A300820 A356936 A257990 * A257743 A033272 A324907
KEYWORD
nonn
AUTHOR
Antti Karttunen, Jan 24 2024
STATUS
approved