|
|
A059269
|
|
Numbers m for which the number of divisors, tau(m), is divisible by 3.
|
|
18
|
|
|
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
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
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 - (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 *)
|
|
PROG
|
|
|
CROSSREFS
|
|
|
KEYWORD
|
nonn,easy
|
|
AUTHOR
|
Avi Peretz (njk(AT)netvision.net.il), Jan 24 2001
|
|
EXTENSIONS
|
|
|
STATUS
|
approved
|
|
|
|