A359211 a(n) = tau(3*n-1)/2, where tau(n) = number of divisors of n, cf. A000005.

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, 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, 2, 2, 3, 2, 2, 2, 6, 1, 4, 1, 3, 1, 3, 3, 4, 1, 2, 1, 6, 1, 4, 1

Offset

1,3

Comments

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

Links

Seiichi Manyama, Table of n, a(n) for n = 1..10000

Formula

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

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

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

Mathematica

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

Prog

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

Crossrefs

Cf. A078703, A099774.

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

Sequence in context: A336099 A290090 A066075 * A072347 A368684 A351034

Adjacent sequences: A359208 A359209 A359210 * A359212 A359213 A359214

Keyword

nonn,easy

Author

Seiichi Manyama, Dec 21 2022

Status

approved