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%I #5 Jul 30 2015 23:18:46
%S 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,
%T 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,
%U 12,12,13,13,12,12,12,12,12,13,13,13,13,13,13,14,14,14,14,14,14,15,15,15
%N Number of cubes m = k^3 such that n <= m <= (n+1)^2.
%C a(n)>=1 because between n and (n+1)^2 there is always at least one cubic number.
%C a(6)=2 because between 5 and 6^2 there are two cubic numbers: 8 and 27
%t 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 *)
%K easy,nonn
%O 0,1
%A _Giovanni Teofilatto_, Apr 23 2005
%E More terms from _Robert G. Wilson v_, Apr 24 2005
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