A074943 a(n) = tau(n) mod 3.

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

Offset

1,2

Links

Antti Karttunen, Table of n, a(n) for n = 1..10000

Index entries for sequences computed from exponents in factorization of n.

Formula

From Amiram Eldar, Apr 16 2024: (Start)

a(n) = A010872(A000005(n)).

a(A059269(n)) = 0; a(A211337(n)) = 1; a(A211338(n)) = 2.

Conjecture: Asymptotic mean: Limit_{m->oo} (1/m) * Sum_{k=1..m} a(k) = 9*zeta(3)/Pi^2 = 1.0961... . The conjecture is true if A211337 and A211338 have the same asymptotic density (see also A059269). (End)

Mathematica

Mod[DivisorSigma[0, Range[110]], 3] (* Harvey P. Dale, Dec 22 2013 *)

Prog

(PARI) a(n)=numdiv(n)%3
(Scheme) (define (A074943 n) (modulo (A000005 n) 3)) ;; Antti Karttunen, Jul 26 2017

Crossrefs

Cf. A000005 (tau), A002117, A010872, A059269, A211337, A211338.

Sequence in context: A049850 A319954 A050949 * A272011 A352362 A236374

Adjacent sequences: A074940 A074941 A074942 * A074944 A074945 A074946

Keyword

easy,nonn,changed

Author

Benoit Cloitre, Oct 04 2002

Status

approved