A113061 Sum of cube divisors of n.

1, 1, 1, 1, 1, 1, 1, 9, 1, 1, 1, 1, 1, 1, 1, 9, 1, 1, 1, 1, 1, 1, 1, 9, 1, 1, 28, 1, 1, 1, 1, 9, 1, 1, 1, 1, 1, 1, 1, 9, 1, 1, 1, 1, 1, 1, 1, 9, 1, 1, 1, 1, 1, 28, 1, 9, 1, 1, 1, 1, 1, 1, 1, 73, 1, 1, 1, 1, 1, 1, 1, 9, 1, 1, 1, 1, 1, 1, 1, 9, 28, 1, 1, 1, 1, 1, 1, 9, 1, 1, 1, 1, 1, 1, 1, 9, 1, 1, 1, 1, 1, 1, 1, 9

Offset: 1

Author

Paul Barry, Oct 13 2005

Keywords

• nonn
• mult

Comments

Multiplicative with a(p^e) = (p^(3*(1+floor(e/3)))-1)/(p^3-1). The Dirichlet generating function is zeta(s)*zeta(3s-3). The sequence is the inverse Mobius transform of n*A010057(n). - R. J. Mathar, Feb 18 2011

Links

• Antti Karttunen, Table of n, a(n) for n = 1..19683
• R. J. Mathar, Survey of Dirichlet series of multiplicative arithmetic functions, arXiv:1106.4038 [math.NT] (2011), Remark 15.
• Index entries for sequences related to sums of divisors

Crossrefs

Cf. A004709, A010057, A035316.

Programs

Maple

A113061 := proc(n)
    local a,pe,p,e;
    a := 1;
    for pe in ifactors(n)[2] do
        p := pe[1] ;
        e := pe[2] ;
        e := 3*(1+floor(e/3)) ;
        a := a*(p^e-1)/(p^3-1) ;
    end do:
    a ;
end proc:
seq(A113061(n),n=1..100) ; # R. J. Mathar, Oct 08 2017

Mathematica

a[n_] := Sum[If[IntegerQ[d^(1/3)], d, 0], {d, Divisors[n]}];
Array[a, 100] (* Jean-François Alcover, Nov 25 2017 *)
f[p_, e_] := (p^(3*(1 + Floor[e/3])) - 1)/(p^3 - 1); a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100] (* Amiram Eldar, May 01 2020 *)

PARI

A113061(n) = sumdiv(n,d,ispower(d,3)*d); \\ Antti Karttunen, Oct 08 2017

Scheme

;; With memoization-macro definec, after the multiplicative formula of R. J. Mathar:
(definec (A113061 n) (if (= 1 n) n (let ((p (A020639 n)) (e (A067029 n))) (* (/ (+ -1 (expt p (* 3 (+ 1 (A002264 e))))) (+ -1 (expt p 3))) (A113061 (A028234 n)))))) ;; Antti Karttunen, Oct 08 2017

Formula

G.f.: Sum_{k>=1} k^3*x^(k^3)/(1 - x^(k^3)). - Ilya Gutkovskiy, Dec 24 2016

a(n) = Sum_{d|n} A010057(d)*d. - Antti Karttunen, Oct 08 2017

Sum_{k=1..n} a(k) ~ zeta(4/3)*n^(4/3)/4 - n/2. - Vaclav Kotesovec, Dec 01 2020