A048798 Smallest k > 0 such that n*k is a perfect cube.

1, 4, 9, 2, 25, 36, 49, 1, 3, 100, 121, 18, 169, 196, 225, 4, 289, 12, 361, 50, 441, 484, 529, 9, 5, 676, 1, 98, 841, 900, 961, 2, 1089, 1156, 1225, 6, 1369, 1444, 1521, 25, 1681, 1764, 1849, 242, 75, 2116, 2209, 36, 7, 20, 2601, 338, 2809, 4, 3025, 49, 3249

Offset: 1

Author

Charles T. Le (charlestle(AT)yahoo.com)

Keywords

• nonn
• easy
• mult

Comments

Note that in general the smallest number k(>0) such that n*k is a perfect m-th power (rather obviously) = (the smallest m-th power divisible by n)/n and also (slightly less obviously) =n^(m-1)/(the number of solutions of x^m==0 mod n)^m. - Henry Bottomley, Mar 03 2000

Examples

			a(12) = a(2*2*3) = 2*3*3 = 18 since 12*18 = 6^3.
a(28) = a(2*2*7) = 2*7*7 = 98 since 28*98 = 14^3.
		

Links

• Amiram Eldar, Table of n, a(n) for n = 1..10000 (terms 1..5000 from Peter Kagey)
• Krassimir T. Atanassov, On the 22nd, the 23rd and 24th Smarandache Problems, Notes on Number Theory and Discrete Mathematics, Sophia, Bulgaria, Vol. 5, No. 2 (1998), pp. 80-82.
• Krassimir T. Atanassov, On Some of Smarandache's Problems, American Research Press, 1999, 16-21.
• Henry Bottomley, Some Smarandache-type multiplicative sequences.
• Marcela Popescu and Mariana Nicolescu, About the Smarandache Complementary Cubic Function, Smarandache Notions Journal, Vol. 7, No. 1-2-3, 1996, pp. 54-62.
• Florentin Smarandache, Only Problems, Not Solutions! (see Unsolved Problem: 28, p. 26).

Crossrefs

Cf. A000189, A007913, A053149.

Cf. A254767 (analogous sequence with the restriction that k > n).

Cf. A002117, A013667.

Programs

Mathematica

a[n_] := For[k = 1, True, k++, If[ Divisible[c = k^3, n], Return[c/n]]]; Table[a[n], {n, 1, 60}] (* Jean-François Alcover, Sep 03 2012 *)
f[p_, e_] := p^(Mod[-e, 3]); a[n_] := Times @@ (f @@@ FactorInteger[n]); Array[a, 100] (* Amiram Eldar, Sep 10 2020 *)
With[{cbs=Range[3300]^3},Table[SelectFirst[cbs,Mod[#,n]==0&]/n,{n,60}]] (* Harvey P. Dale, May 10 2024 *)

PARI

a(n)=my(f=factor(n));prod(i=1,#f[,1],f[i,1]^(-f[i,2]%3)) \\ Charles R Greathouse IV, Feb 27 2013

PARI

a(n)=for(k=1,n^2,if(ispower(k*n,3),return(k)))
vector(100,n,a(n)) \\ Derek Orr, Feb 07 2015

Python

from math import prod
from sympy import factorint
def A048798(n): return prod(p**(-e%3) for p, e in factorint(n).items()) # Chai Wah Wu, Aug 05 2024

Formula

a(n) = A053149(n)/n = n^2/A000189(n)^3.

Multiplicative with a(p^e) = p^((-e) mod 3). - Mitch Harris, May 17 2005

Sum_{k=1..n} a(k) ~ c * n^3, where c = (zeta(9)/(3*zeta(3))) * Product_{p prime} (1 - 1/p^2 + 1/p^3) = 0.2079875504... . - Amiram Eldar, Oct 28 2022

Extensions

More terms from Patrick De Geest, Feb 15 2000