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Smallest number whose cube is divisible by n.
16

%I #64 Oct 19 2024 15:57:32

%S 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,

%T 29,30,31,4,33,34,35,6,37,38,39,10,41,42,43,22,15,46,47,12,7,10,51,26,

%U 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

%N Smallest number whose cube is divisible by n.

%C 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

%H Peter Kagey, <a href="/A019555/b019555.txt">Table of n, a(n) for n = 1..10000</a>

%H Henry Bottomley, <a href="http://fs.gallup.unm.edu/Bottomley-Sm-Mult-Functions.htm">Some Smarandache-type multiplicative sequences</a>.

%H Kevin A. Broughan, <a href="https://doi.org/10.4064/aa101-2-2">Restricted divisor sums</a>, Acta Arithmetica, 101(2) (2002), 105-114.

%H Kevin A. Broughan, <a href="http://ijpam.eu/contents/2003-5-3/2/2.pdf">Relationship between the integer conductor and k-th root functions</a>, Int. J. Pure Appl. Math. 5(3) (2003), 253-275.

%H Kevin A. Broughan, <a href="http://nzjm.math.auckland.ac.nz/images/d/d6/Relaxations_of_the_ABC_Conjecture_using_integer_k%27th_roots.pdf">Relaxations of the ABC Conjecture using integer k'th roots</a>, New Zealand J. Math. 35(2) (2006), 121-136.

%H Vaclav Kotesovec, <a href="/A019555/a019555.jpg">Graph - the asymptotic ratio (1000000 terms)</a>.

%H Ana Rechtman, <a href="http://images-archive.math.cnrs.fr/Mai-2021-4e-defi.html">Mai 2021, 4e défi</a>, Images des Mathématiques, CNRS, 2021 (in French).

%H Florentin Smarandache, <a href="http://www.gallup.unm.edu/~smarandache/CP2.pdf">Collected Papers, Vol. II</a>, Tempus Publ. Hse, Bucharest, 1996.

%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/SmarandacheCeilFunction.html">Smarandache Ceil Function</a>.

%F Replace any cubic factors in n by their cube roots.

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

%F Multiplicative with a(p^e) = p^ceiling(e/3). - _R. J. Mathar_, May 29 2011

%F From _Vaclav Kotesovec_, Aug 30 2021: (Start)

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

%F 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)).

%F 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)

%p f:= n -> mul(t[1]^ceil(t[2]/3), t = ifactors(n)[2]):

%p map(f, [$1..100]); # _Robert Israel_, Sep 22 2015

%t cubes=Range[85]^3; Table[Position[Divisible[cubes,i],True,1,1][[1,1]],{i,85}] (* _Harvey P. Dale_, Jan 12 2011 *)

%t f[p_, e_] := p^Ceiling[e/3]; a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100] (* _Amiram Eldar_, Jan 06 2024 *)

%o (PARI) a(n)=my(r=1);while(r^3%n!=0,r++);r \\ _Anders Hellström_, Sep 22 2015

%o (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

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

%o (Sage) [prod([t[0]^(ceil(t[1]/3)) for t in factor(n)]) for n in range(1,79)] # _Danny Rorabaugh_, Sep 22 2015

%o (Python 3.8+)

%o from math import prod

%o from sympy import factorint

%o def A019555(n): return prod(p**((q%3 != 0)+(q//3)) for p, q in factorint(n).items()) # _Chai Wah Wu_, Aug 18 2021

%Y 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).

%Y Cf. A000189, A015050.

%K nonn,easy,mult

%O 1,2

%A R. Muller

%E Corrected and extended by _David W. Wilson_