A254767 a(n) is the least k > n such that k*n is a cube.

8, 4, 9, 16, 25, 36, 49, 27, 24, 100, 121, 18, 169, 196, 225, 32, 289, 96, 361, 50, 441, 484, 529, 72, 40, 676, 64, 98, 841, 900, 961, 54, 1089, 1156, 1225, 48, 1369, 1444, 1521, 200, 1681, 1764, 1849, 242, 75, 2116, 2209, 288, 56, 160, 2601, 338, 2809, 108

Offset

1,1

Comments

a(n) <= n^2 for all n > 1 because n * n^2 = n^3.

Links

Peter Kagey, Table of n, a(n) for n = 1..5000

Example

a(12) = 18 because 12*18 = 6^3 (and 12*13, 12*14, 12*15, 12*16, 12*17 are not perfect cubes).

Mathematica

f[n_] := Block[{k = n + 1}, While[! IntegerQ@ Power[k n, 1/3], k++]; k]; Array[f, 54] (* Michael De Vlieger, Mar 17 2015 *)

Prog

(Ruby)
def a(n)
  min = (n**(2/3.0)).ceil
  (min..n+1).each { |i| return i**3/n if i**3 % n == 0 && i**3 > n**2 }
end
(PARI) a(n)=if (n==1, 8, for(k=n+1, n^2, if(ispower(k*n, 3), return(k))))
vector(100, n, a(n)) \\ Derek Orr, Feb 07 2015
(PARI) a(n) = {f = factor(n); for (i=1, #f~, if (f[i, 2] % 3, f[i, 2] = 3 - f[i, 2]); ); cb = factorback(f); cbr = sqrtnint(cb*n, 3); cb = cbr^3; k = cb/n; while((type(k=cb/n) != "t_INT") || (k<=n), cbr++; cb = cbr^3; ); k; } \\ Michel Marcus, Mar 14 2015

Crossrefs

Cf. A072905 (an analogous sequence for squares).

Cf. A048798 (similar sequence, no restriction that a(n) > n).

Sequence in context: A205184 A105144 A277781 * A194184 A194217 A239971

Adjacent sequences: A254764 A254765 A254766 * A254768 A254769 A254770

Keyword

nonn,easy

Author

Peter Kagey, Feb 07 2015

Status

approved