A078708 - OEIS

A078708

Sum of divisors d of n such that n/d is not congruent to 0 mod 3.

1, 3, 3, 7, 6, 9, 8, 15, 9, 18, 12, 21, 14, 24, 18, 31, 18, 27, 20, 42, 24, 36, 24, 45, 31, 42, 27, 56, 30, 54, 32, 63, 36, 54, 48, 63, 38, 60, 42, 90, 42, 72, 44, 84, 54, 72, 48, 93, 57, 93, 54, 98, 54, 81, 72, 120, 60, 90, 60, 126, 62, 96, 72, 127, 84, 108, 68, 126, 72, 144

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OFFSET

1,2

LINKS

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

FORMULA

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

a(n) = (A000203(3*n)-A000203(n))/3. - Vladeta Jovovic, Dec 22 2003

G.f.: Sum_{k>=1} k*x^k*(1 + x^k)/(1 - x^(3*k)). - Ilya Gutkovskiy, Sep 13 2019

From R. J. Mathar, May 25 2020: (Start)

a(n) = A326399(n) + A326400(n).

a(n) = A000203(n) - A000203(n/3), where A000203(.) = 0 for non-integer arguments. (End)

From Amiram Eldar, Oct 30 2022: (Start)

Multiplicative with a(3^e) = 3^e and a(p^e) = (p^(e+1)-1)/(p-1) if p != 3.

Sum_{k=1..n} a(k) ~ c * n^2, where c = 2*Pi^2/27 = 0.731081... (A346933). (End)

Dirichlet g.f.: zeta(s)*zeta(s-1)*(1-1/3^s). - Amiram Eldar, Dec 30 2022

For k >= 0, a(n) = (1/3^k) * ( sigma((3^k)*n) - sigma(3^(k-1)*n) ) = 3^(v_3(n)) * sigma(n/ 3^(v_3(n))) is independent of the value of k, where v_3(n) is the exponent of 3 in the prime factorization of n. The cases k = 0 and 1 are given above. - Peter Bala, May 19 2026

MAPLE

with(numtheory):

seq((1/3)*(sigma[1](3*n) - sigma[1](n)), n = 1..100); # Peter Bala, May 19 2026

MATHEMATICA

f[p_, e_] := If[p == 3, 3^e, (p^(e+1)-1)/(p-1)]; a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100] (* Amiram Eldar, Oct 30 2022 *)

PROG

(PARI) for(n=1, 70, d=divisors(n); s=0; for(j=1, matsize(d)[2], if((n/d[j])%3>0, s=s+d[j])); print1(s, ", "))

(PARI) a(n)=sumdiv(n, d, if((n/d)%3, 1, 0)*d)

CROSSREFS

Cf. A000203, A035191, A046913, A346933.

Cf. A002131 (k=2), this sequence (k=3), A285895 (k=4), A285896 (k=5).

Cf. A326399, A326400.