A211338 - OEIS

A211338

Numbers k for which the number of divisors, tau(k), is congruent to 2 modulo 3.

2, 3, 5, 7, 11, 13, 16, 17, 19, 23, 24, 29, 30, 31, 37, 40, 41, 42, 43, 47, 53, 54, 56, 59, 61, 66, 67, 70, 71, 73, 78, 79, 81, 83, 88, 89, 97, 101, 102, 103, 104, 105, 107, 109, 110, 113, 114, 127, 128, 130, 131, 135, 136, 137, 138, 139, 149, 151, 152, 154

(list; graph; refs; listen; history; text; internal format)

OFFSET

1,1

COMMENTS

The product of any 2 terms a(i)*a(j) is not a member of the sequence.

Any term a(n) can be expressed as 1 term (required to be greater than 1) from A211485 times 1 nonzero term from A000578. - Douglas Latimer, Apr 20 2012

The numbers of terms not exceeding 10^k, for k = 1, 2, ..., are 4, 37, 368, 3681, 36596, 365336, 3653499, 36537962, 365381169, 3653826361, ... . Conjecture: the asymptotic density of this sequence exists and equals 3*zeta(3)/Pi^2 = 0.3653814847007... (A346602), so, a(n) ~ k*n with k = Pi^2/(3*zeta(3)) = 2.73686555524... . This conjecture is true if this sequence and A211337 have the same density (see A059269). - Amiram Eldar, Jan 06 2024

LINKS

Amiram Eldar, Table of n, a(n) for n = 1..10000 (terms 1..1000 from Douglas Latimer)

FORMULA

Conjecture: a(n) ~ k*n where k = 2/prod(1 - (p-1)/(p^(3*k))) = 2.7290077... where p ranges over the primes and k ranges over the positive integers. - Charles R Greathouse IV, Apr 13 2012

EXAMPLE

The divisors of 16 are: 1, 2, 4, 8, 16 (5 divisors). 5 is congruent to 2 modulo 3. Thus 16 is a member of this sequence.

MATHEMATICA

Select[Range[154], Mod[DivisorSigma[0, #], 3] == 2 &] (* T. D. Noe, Apr 21 2012 *)

PROG

(PARI) {plnt=1 ; mxind=100 ; for(k=1, 10^6,

if(numdiv(k) % 3 == 2, print(k); plnt++; if(mxind+1 == plnt, break() )))}

CROSSREFS

This is an extension of A000040 (the prime numbers, which each have 2 divisors).

The union of A059269 and A211337 is the complementary sequence to this one.

The definition of this sequence uses A000005 (the number of divisors of n).

Cf. A000578, A074796, A211485, A346602.