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%I #15 Apr 21 2018 21:01:03
%S 1,2,2,3,3,3,4,4,5,5,5,5,6,6,6,7,7,7,7,8,8,8,8,9,9,9,9,10,10,10,10,10,
%T 11,11,11,11,11,12,12,12,12,12,13,13,13,13,13,14,14,14,14,14,14,15,15,
%U 15,15,15,15,16,16,16,16,16,17,17,17,17,17,17,17,18,18,18,18,18,18,19
%N Number of integer cubes <= n^2.
%C a(n) is the least number m such that m^3 > n^2. - _Zak Seidov_, May 03 2005
%F a(n) = floor(n^(2/3))+1.
%F a(n) = [x^(n^2)] (1/(1 - x))*Sum_{k>=0} x^(k^3). - _Ilya Gutkovskiy_, Apr 20 2018
%e Cubes <= 10^2: 0, 1, 8, 27 and 64, hence a(10)=5;
%t Table[Floor[n^(2/3) + 1], {n, 0, 100}] (* _Zak Seidov_, May 03 2005 *)
%Y Cf. A077106, A026409, A026414, A054071, A077121.
%K nonn
%O 0,2
%A _Reinhard Zumkeller_, Oct 29 2002
%E Edited by _N. J. A. Sloane_, Aug 29 2008 at the suggestion of _R. J. Mathar_