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Number of numbers k <= n such that tau(k) == 2 (mod 3) where tau(k) = A000005(k) is the number of divisors of k.
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%I #22 Aug 29 2020 06:56:36

%S 0,1,2,2,3,3,4,4,4,4,5,5,6,6,6,7,8,8,9,9,9,9,10,11,11,11,11,11,12,13,

%T 14,14,14,14,14,14,15,15,15,16,17,18,19,19,19,19,20,20,20,20,20,20,21,

%U 22,22,23,23,23,24,24,25,25,25,25,25,26,27,27,27,28,29,29,30,30,30,30

%N Number of numbers k <= n such that tau(k) == 2 (mod 3) where tau(k) = A000005(k) is the number of divisors of k.

%H Amiram Eldar, <a href="/A074796/b074796.txt">Table of n, a(n) for n = 1..10000</a>

%F a(n) is asymptotic to c*n with c = 0.38....

%t Accumulate[Table[If[Mod[DivisorSigma[0,n],3]==2,1,0],{n,80}]] (* _Harvey P. Dale_, Apr 22 2018 *)

%o (PARI) a(n)=sum(k=1,n,if(numdiv(k)%3-2,0,1))

%Y Cf. A000005, A074794, A074795, A211338.

%K nonn

%O 1,3

%A _Benoit Cloitre_, Sep 07 2002