A019555 Smallest number whose cube is divisible by n.

1, 2, 3, 2, 5, 6, 7, 2, 3, 10, 11, 6, 13, 14, 15, 4, 17, 6, 19, 10, 21, 22, 23, 6, 5, 26, 3, 14, 29, 30, 31, 4, 33, 34, 35, 6, 37, 38, 39, 10, 41, 42, 43, 22, 15, 46, 47, 12, 7, 10, 51, 26, 53, 6, 55, 14, 57, 58, 59, 30, 61, 62, 21, 4, 65, 66, 67, 34, 69, 70, 71, 6, 73, 74, 15, 38, 77, 78

Offset: 1

Author

R. Muller

Keywords

• nonn
• easy
• mult

Comments

This can be thought as an "upper 3rd root" of a positive integer. Upper k-th roots were studied by Broughan (2002, 2003, 2006). The sequence of "lower 3rd root" of positive integers is given by A053150. - Petros Hadjicostas, Sep 15 2019

Links

• Peter Kagey, Table of n, a(n) for n = 1..10000
• Henry Bottomley, Some Smarandache-type multiplicative sequences.
• Kevin A. Broughan, Restricted divisor sums, Acta Arithmetica, 101(2) (2002), 105-114.
• Kevin A. Broughan, Relationship between the integer conductor and k-th root functions, Int. J. Pure Appl. Math. 5(3) (2003), 253-275.
• Kevin A. Broughan, Relaxations of the ABC Conjecture using integer k'th roots, New Zealand J. Math. 35(2) (2006), 121-136.
• Vaclav Kotesovec, Graph - the asymptotic ratio (1000000 terms).
• Ana Rechtman, Mai 2021, 4e défi, Images des Mathématiques, CNRS, 2021 (in French).
• Florentin Smarandache, Collected Papers, Vol. II, Tempus Publ. Hse, Bucharest, 1996.
• Eric Weisstein's World of Mathematics, Smarandache Ceil Function.

Crossrefs

Cf. A000188 (inner square root), A019554 (outer square root), A053150 (inner 3rd root), A053164 (inner 4th root), A053166 (outer 4th root), A015052 (outer 5th root), A015053 (outer 6th root).

Cf. A000189, A015050.

Programs

Maple

f:= n -> mul(t[1]^ceil(t[2]/3), t = ifactors(n)[2]):
map(f, [$1..100]); # Robert Israel, Sep 22 2015

Mathematica

cubes=Range[85]^3; Table[Position[Divisible[cubes,i],True,1,1][[1,1]],{i,85}] (* Harvey P. Dale, Jan 12 2011 *)
f[p_, e_] := p^Ceiling[e/3]; a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100]  (* Amiram Eldar, Jan 06 2024 *)

PARI

a(n)=my(r=1);while(r^3%n!=0,r++);r \\ Anders Hellström, Sep 22 2015

PARI

for(n=1, 100, print1(direuler(p=2, n, (1 + p*X + p*X^2)/(1 - p*X^3))[n], ", ")) \\ Vaclav Kotesovec, Aug 30 2021

PARI

a(n) = {my(f = factor(n)); prod(i = 1, #f~, f[i,1]^ceil(f[i,2]/3));} \\ Amiram Eldar, Jan 06 2024

Python

from math import prod
from sympy import factorint
def A019555(n): return prod(p**((q%3 != 0)+(q//3)) for p, q in factorint(n).items()) # Chai Wah Wu, Aug 18 2021

Sage

[prod([t[0]^(ceil(t[1]/3)) for t in factor(n)]) for n in range(1,79)] # Danny Rorabaugh, Sep 22 2015

Formula

Replace any cubic factors in n by their cube roots.

a(n) = n/A000189(n).

Multiplicative with a(p^e) = p^ceiling(e/3). - R. J. Mathar, May 29 2011

From Vaclav Kotesovec, Aug 30 2021: (Start)

Dirichlet g.f.: zeta(3*s-1) * Product_{p prime} (1 + p^(1 - s) + p^(1 - 2*s)).

Dirichlet g.f.: zeta(3*s-1) * zeta(s-1) * Product_{p prime} (1 - p^(2 - 3*s) + p^(1 - 2*s) - p^(2 - 2*s)).

Sum_{k=1..n} a(k) ~ c * zeta(5) * n^2 / 2, where c = Product_{p prime} (1 - 1/p^2 + 1/p^3 - 1/p^4) = 0.684286924186862318141968725791218083472312736723163777284618226290055... (End)

Extensions

Corrected and extended by David W. Wilson