OFFSET
1,2
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.
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)
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 *)
PROG
(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
(Sage) [prod([t[0]^(ceil(t[1]/3)) for t in factor(n)]) for n in range(1, 79)] # Danny Rorabaugh, Sep 22 2015
(Python 3.8+)
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
CROSSREFS
KEYWORD
nonn,easy,mult
AUTHOR
R. Muller
EXTENSIONS
Corrected and extended by David W. Wilson
STATUS
approved