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
Examples
a(12) = 18 because 12*18 = 6^3 (and 12*13, 12*14, 12*15, 12*16, 12*17 are not perfect cubes).
Links
- Peter Kagey, Table of n, a(n) for n = 1..5000
Crossrefs
Programs
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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 *)
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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
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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
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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
Comments