A211337 - OEIS

A211337

Numbers k for which the number of divisors, tau(k), is congruent to 1 modulo 3.

1, 6, 8, 10, 14, 15, 21, 22, 26, 27, 33, 34, 35, 38, 39, 46, 48, 51, 55, 57, 58, 62, 64, 65, 69, 74, 77, 80, 82, 85, 86, 87, 91, 93, 94, 95, 106, 111, 112, 115, 118, 119, 120, 122, 123, 125, 129, 133, 134, 141, 142, 143, 145, 146, 155, 158, 159, 161, 162

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OFFSET

1,2

COMMENTS

Any term a(n) can be expressed as 1 term from A211484 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, 36, 366, 3635, 36499, 365456, 3654240, 36538501, 365382167, 3653804173, ... . 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 A211338 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 10 are: 1, 2, 5, 10 (4 divisors). 4 is congruent to 1 modulo 3. Thus 10 is a member of this sequence.

MATHEMATICA

Select[Range[162], Mod[DivisorSigma[0, #], 3] == 1 &] (* T. D. Noe, Apr 21 2012 *)

PROG

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

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

CROSSREFS

This is an extension of A030513 (numbers with 4 divisors).

The union of A059269 and A211338 is the complementary sequence to this one.

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

Cf. A000578, A074794, A211484, A346602.