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%I #44 Jan 06 2024 09:21:29
%S 2,3,5,7,11,13,16,17,19,23,24,29,30,31,37,40,41,42,43,47,53,54,56,59,
%T 61,66,67,70,71,73,78,79,81,83,88,89,97,101,102,103,104,105,107,109,
%U 110,113,114,127,128,130,131,135,136,137,138,139,149,151,152,154
%N Numbers k for which the number of divisors, tau(k), is congruent to 2 modulo 3.
%C The product of any 2 terms a(i)*a(j) is not a member of the sequence.
%C 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
%C 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
%H Amiram Eldar, <a href="/A211338/b211338.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 16 are: 1, 2, 4, 8, 16 (5 divisors). 5 is congruent to 2 modulo 3. Thus 16 is a member of this sequence.
%t Select[Range[154], Mod[DivisorSigma[0, #], 3] == 2 &] (* _T. D. Noe_, Apr 21 2012 *)
%o (PARI) {plnt=1 ; mxind=100 ;for(k=1, 10^6,
%o if(numdiv(k) % 3 == 2, print(k); plnt++; if(mxind+1 == plnt, break() )))}
%Y This is an extension of A000040 (the prime numbers, which each have 2 divisors).
%Y The union of A059269 and A211337 is the complementary sequence to this one.
%Y The definition of this sequence uses A000005 (the number of divisors of n).
%Y Cf. A000578, A074796, A211485, A346602.
%K nonn
%O 1,1
%A _Douglas Latimer_, Apr 07 2012
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