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A194762 Number of k such that {k*2^(1/3)} < {n*2^(1/3)}, where { } = fractional part.

%I A194762 #6 Mar 30 2012 18:57:44
%S A194762 1,2,3,1,3,5,7,2,5,8,11,3,7,11,15,4,9,14,19,5,11,17,23,6,13,20,1,9,17,
%T A194762 25,3,12,21,30,5,15,25,35,7,18,29,40,9,21,33,45,11,24,37,50,13,27,41,
%U A194762 2,17,32,47,5,21,37,53,8,25,42,59,11,29,47,65,14,33,52,71,17,37
%N A194762 Number of k such that {k*2^(1/3)} < {n*2^(1/3)}, where { } = fractional part.
%t A194762 r = 2^(1/3); p[x_] := FractionalPart[x];
%t A194762 u[n_, k_] := If[p[k*r] <= p[n*r], 1, 0]
%t A194762 v[n_, k_] := If[p[k*r] > p[n*r], 1, 0]
%t A194762 s[n_] := Sum[u[n, k], {k, 1, n}]
%t A194762 t[n_] := Sum[v[n, k], {k, 1, n}]
%t A194762 Table[s[n], {n, 1, 100}]  (* A194762 *)
%t A194762 Table[t[n], {n, 1, 100}]  (* A194763 *)
%Y A194762 Cf. A194763, A194738.
%K A194762 nonn
%O A194762 1,2
%A A194762 _Clark Kimberling_, Sep 02 2011