A389832 Multiplicative sequence a(n) with a(p^e) = abs(2 - (e mod 4)) for prime p and e > 0.

1, 1, 1, 0, 1, 1, 1, 1, 0, 1, 1, 0, 1, 1, 1, 2, 1, 0, 1, 0, 1, 1, 1, 1, 0, 1, 1, 0, 1, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 1, 0, 0, 1, 1, 2, 0, 0, 1, 0, 1, 1, 1, 1, 1, 1, 1, 0, 1, 1, 0, 0, 1, 1, 1, 0, 1, 1, 1, 0, 1, 1, 0, 0, 1, 1, 1, 2, 2, 1, 1, 0, 1, 1, 1, 1, 1, 0, 1, 0, 1, 1, 1, 1, 1, 0, 0, 0

Offset

1,16

Comments

Dirichlet inverse b(n) is multiplicative with b(p^e) = (2 - e^2 mod 3) * (-1)^e for prime p and e > 0. - Werner Schulte, Oct 25 2025

Links

Andrew Howroyd, Table of n, a(n) for n = 1..10000

Formula

Dirichlet g.f.: zeta(s) * zeta(3*s) * zeta(4*s) / (zeta(2*s) * zeta(6*s)).

Sum_{k=1..n} a(k) ~ (63*zeta(3)/Pi^4) * n. - Amiram Eldar, Oct 16 2025

Maple

a:= n-> mul(abs(2-irem(i[2], 4)), i=ifactors(n)[2]):
seq(a(n), n=1..100);  # Alois P. Heinz, Oct 16 2025

Mathematica

f[p_, e_] := Abs[2 - Mod[e, 4]]; a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100] (* Amiram Eldar, Oct 16 2025 *)

Prog

(PARI) a(n) = factorback(apply(e -> abs(2 - (e%4)), factor(n)[, 2]))
(Python)
from sympy import factorint
def A389832(n): return 0 if 2 in (a:=[e&3 for e in factorint(n).values()]) else 1<<sum(1 for e in a if not e) # Chai Wah Wu, Oct 22 2025

Crossrefs

Sequence in context: A339872 A280269 A321894 * A355824 A355826 A355819

Adjacent sequences: A389829 A389830 A389831 * A389833 A389834 A389835

Keyword

nonn,easy,mult

Author

Werner Schulte, Oct 16 2025

Status

approved