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Number of numbers k <= n such that tau(k) == 1 (mod 3) where tau(k) = A000005(k) is the number of divisors of k.
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%I #21 Aug 29 2020 06:56:53

%S 1,1,1,1,1,2,2,3,3,4,4,4,4,5,6,6,6,6,6,6,7,8,8,8,8,9,10,10,10,10,10,

%T 10,11,12,13,13,13,14,15,15,15,15,15,15,15,16,16,17,17,17,18,18,18,18,

%U 19,19,20,21,21,21,21,22,22,23,24,24,24,24,25,25,25,25,25,26,26,26,27,27

%N Number of numbers k <= n such that tau(k) == 1 (mod 3) where tau(k) = A000005(k) is the number of divisors of k.

%H Amiram Eldar, <a href="/A074794/b074794.txt">Table of n, a(n) for n = 1..10000</a>

%F a(n) is asymptotic to c*n with c = 0.36....

%t Accumulate[Table[Boole[Mod[DivisorSigma[0, n], 3] == 1], {n, 1, 100}]] (* _Amiram Eldar_, Aug 29 2020 *)

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

%Y Cf. A000005, A074795, A074796, A211337.

%K nonn

%O 1,6

%A _Benoit Cloitre_, Sep 07 2002