%I #44 Jan 06 2024 09:21:24
%S 1,6,8,10,14,15,21,22,26,27,33,34,35,38,39,46,48,51,55,57,58,62,64,65,
%T 69,74,77,80,82,85,86,87,91,93,94,95,106,111,112,115,118,119,120,122,
%U 123,125,129,133,134,141,142,143,145,146,155,158,159,161,162
%N Numbers k for which the number of divisors, tau(k), is congruent to 1 modulo 3.
%C Any term a(n) can be expressed as 1 term from A211484 times 1 nonzero term from A000578. - _Douglas Latimer_, Apr 20 2012
%C 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
%H Amiram Eldar, <a href="/A211337/b211337.txt">Table of n, a(n) for n = 1..10000</a> (terms 1..1000 from Douglas Latimer)
%F 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
%e 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.
%t Select[Range[162], Mod[DivisorSigma[0, #], 3] == 1 &] (* _T. D. Noe_, Apr 21 2012 *)
%o (PARI) {plnt=1 ; mxind=100 ;for(k=1, 10^6,
%o if(numdiv(k) % 3 == 1, print(k); plnt++; if(mxind+1 == plnt, break() )))}
%Y This is an extension of A030513 (numbers with 4 divisors).
%Y The union of A059269 and A211338 is the complementary sequence to this one.
%Y The definition of this sequence uses A000005 (the number of divisors of n).
%Y Cf. A000578, A074794, A211484, A346602.
%K nonn
%O 1,2
%A _Douglas Latimer_, Apr 07 2012