|
|
A059269
|
|
Numbers n for which the number of divisors, tau(n), is divisible by 3.
|
|
7
|
|
|
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.
The union of A211337 and A211338 is the complementary sequence to this one. [Douglas Latimer, Apr 12 2012]
|
|
LINKS
|
Charles R Greathouse IV, Table of n, a(n) for n = 1..10000
|
|
FORMULA
|
Conjecture: a(n) ~ k*n where k = 1/(1 - prod(1 - (p-1)/(p^(3*k)))) = 3.743455... where p ranges over the primes and k ranges over the positive integers. - Charles R Greathouse IV, Apr 13 2012
|
|
EXAMPLE
|
a(7) = 28 because the number of divisors of 28 d(28) = 6
|
|
MAPLE
|
with(numtheory): for n from 1 to 1000 do if tau(n) mod 3 = 0 then printf(`%d, `, n) fi: od:
|
|
PROG
|
(PARI) is(n)=vecmax(factor(n)[, 2]%3)==2 \\ Charles R Greathouse IV, Apr 10 2012
(PARI) is(n)=numdiv(n)%3==0 \\ Charles R Greathouse IV, Sep 18 2015
|
|
CROSSREFS
|
Cf. A038109, A000005, A211337, A211338.
Sequence in context: A276885 A089910 A177880 * A081619 A038109 A067259
Adjacent sequences: A059266 A059267 A059268 * A059270 A059271 A059272
|
|
KEYWORD
|
nonn,easy
|
|
AUTHOR
|
Avi Peretz (njk(AT)netvision.net.il), Jan 24 2001
|
|
EXTENSIONS
|
More terms from James A. Sellers, Jan 24 2001
|
|
STATUS
|
approved
|
|
|
|