A059269 - OEIS (original) (raw)

A059269

Numbers m for which the number of divisors, tau(m), is divisible by 3.

4, 9, 12, 18, 20, 25, 28, 32, 36, 44, 45, 49, 50, 52, 60, 63, 68, 72, 75, 76, 84, 90, 92, 96, 98, 99, 100, 108, 116, 117, 121, 124, 126, 132, 140, 144, 147, 148, 150, 153, 156, 160, 164, 169, 171, 172, 175, 180, 188, 196, 198, 200, 204, 207, 212, 220, 224, 225, 228

COMMENTS

tau(n) is divisible by 3 iff at least one prime in the prime factorization of n has exponent of the form 3*m + 2. This sequence is an extension of the sequence A038109 in which the numbers has at least one prime with exponent 2 (the case of m = 0 here ) in their prime factorization.

FORMULA

Conjecture: a(n) ~ k*n where k = 1/(1 - Product(1 - (p-1)/(p^(3*i)))) = 3.743455... where p ranges over the primes and i ranges over the positive integers. - Charles R Greathouse IV, Apr 13 2012

The asymptotic density of this sequence is 1 - zeta(3)/zeta(2) = 1 - 6*zeta(3)/Pi^2 = 0.2692370305... (Sathe, 1945). Therefore, the above conjecture, a(n) ~ k*n, is true, but k = 1/(1-6*zeta(3)/Pi^2) = 3.7141993349... - Amiram Eldar, Jul 26 2020

EXAMPLE

a(7) = 28 is a term because the number of divisors of 28, d(28) = 6, is divisible by 3.

MAPLE

with(numtheory): for n from 1 to 1000 do if tau(n) mod 3 = 0 then printf(`%d, `, n) fi: od:

MATHEMATICA

Select[Range[230], Divisible[DivisorSigma[0, #], 3] &] (* Amiram Eldar, Jul 26 2020 *)

CROSSREFS

Characteristic function: A353470.

AUTHOR

Avi Peretz (njk(AT)netvision.net.il), Jan 24 2001