A373591 Number of primes congruent to 1 modulo 3 dividing n (with multiplicity).

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

Offset: 1

Author

Antti Karttunen, Jun 13 2024

Keywords

• nonn

Links

• Antti Karttunen, Table of n, a(n) for n = 1..100000

Crossrefs

Cf. A001222, A007949, A248909, A373592.

Cf. also A065339, A083025.

Programs

Mathematica

f[p_, e_] := If[Mod[p, 3] == 1, e, 0]; f[3, e_] := 0; a[1] = 0; a[n_] := Plus @@ f @@@ FactorInteger[n]; Array[a, 100] (* Amiram Eldar, Jun 17 2024 *)

PARI

A373591(n) = sum(i=1, #n=factor(n)~, (1==n[1, i]%3)*n[2, i]); \\ After code in A083025

Formula

a(n) = A001222(A248909(n)).

a(n) = A001222(n) - (A007949(n)+A373592(n)).

Totally additive with a(3) = 0, a(p) = 1 if p == 1 (mod 3), and a(p) = 0 if p == 2 (mod 3). - Amiram Eldar, Jun 17 2024