%I #33 Feb 02 2024 11:10:45

%S 1,1,2,1,2,1,3,1,2,1,3,2,2,1,3,1,3,1,4,1,2,2,3,1,2,2,5,1,2,1,3,2,3,1,

%T 4,1,4,1,3,2,2,2,4,1,2,1,6,2,2,1,4,2,2,2,3,1,4,1,5,1,4,2,3,1,2,1,6,2,

%U 2,2,3,2,2,2,6,1,4,1,3,1,3,3,4,1,2,1,6,1,4,1

%N a(n) = tau(3*n-1)/2, where tau(n) = number of divisors of n, cf. A000005.

%C Also number of divisors of 3*n-1 of form 3*k+1 (or 3*k+2).

%H Seiichi Manyama, <a href="/A359211/b359211.txt">Table of n, a(n) for n = 1..10000</a>

%F G.f.: Sum_{k>0} x^k/(1 - x^(3*k-1)).

%F G.f.: Sum_{k>0} x^(2*k-1)/(1 - x^(3*k-2)).

%F Sum_{k=1..n} a(k) = (log(n) + 2*gamma - 1 + 2*log(3))*n/3 + O(n^(1/3)*log(n)), where gamma is Euler's constant (A001620). - _Amiram Eldar_, Dec 26 2022

%t a[n_] := DivisorSigma[0, 3*n-1]/2; Array[a, 100] (* _Amiram Eldar_, Dec 21 2022 *)

%o (PARI) a(n) = numdiv(3*n-1)/2;

%o (PARI) a(n) = sumdiv(3*n-1, d, d%3==1);

%o (PARI) a(n) = sumdiv(3*n-1, d, d%3==2);

%o (PARI) my(N=100, x='x+O('x^N)); Vec(sum(k=1, N, x^k/(1-x^(3*k-1))))

%o (PARI) my(N=100, x='x+O('x^N)); Vec(sum(k=1, N, x^(2*k-1)/(1-x^(3*k-2))))

%Y Cf. A078703, A099774.

%Y Cf. A000005, A001620, A001817, A001822, A359212.

%K nonn,easy

%O 1,3

%A _Seiichi Manyama_, Dec 21 2022