

A059269


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


17



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
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OFFSET

1,1


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.


LINKS



FORMULA

Conjecture: a(n) ~ k*n where k = 1/(1  Product(1  (p1)/(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/(16*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 *)


PROG



CROSSREFS



KEYWORD

nonn,easy


AUTHOR

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


EXTENSIONS



STATUS

approved



