A104758 Number of cubes m = k^3 such that n <= m <= (n+1)^2.

2, 1, 1, 1, 1, 2, 2, 3, 3, 2, 2, 3, 3, 3, 4, 4, 4, 4, 5, 5, 5, 5, 6, 6, 6, 6, 7, 7, 6, 6, 6, 7, 7, 7, 7, 7, 8, 8, 8, 8, 8, 9, 9, 9, 9, 9, 10, 10, 10, 10, 10, 10, 11, 11, 11, 11, 11, 11, 12, 12, 12, 12, 12, 13, 13, 12, 12, 12, 12, 12, 13, 13, 13, 13, 13, 13, 14, 14, 14, 14, 14, 14, 15, 15, 15

Offset: 0

Author

Giovanni Teofilatto, Apr 23 2005

Keywords

• easy
• nonn

Comments

a(n)>=1 because between n and (n+1)^2 there is always at least one cubic number.

a(6)=2 because between 5 and 6^2 there are two cubic numbers: 8 and 27

Programs

Mathematica

f[n_] := Floor[(n + 1)^(2/3)] - Floor[n^(1/3)] + If[ IntegerQ[n^(1/3)], 1, 0]; Table[ f[n], {n, 0, 84}] (* Robert G. Wilson v, Apr 24 2005 *)

Extensions

More terms from Robert G. Wilson v, Apr 24 2005