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%I #10 Apr 21 2021 03:50:38
%S 2,4,7,9,11,12,14,16,19,21,23,24,26,28,31,33,36,38,40,41,43,45,48,50,
%T 52,53,55,57,60,62,64,65,67,69,70,72,74,77,79,81,82,84,86,89,91,93,94,
%U 96,98,101,103,106,108,110,111,113,115,118,120,122,123,125,127,130,132,134
%N Numbers k such that {k* sqrt(2)} > 1/2, where { } = fractional part.
%C The complement of A120749 is A120243.
%H Clark Kimberling, <a href="/A120749/b120749.txt">Table of n, a(n) for n = 1..1000</a>
%e Call the present sequence b and its complement a. Then
%e {r} = {1.4142...} = 0.4142... < 1/2, so a(1) = 1;
%e {2r} = 0.828... > 1/2, so b(1) = 2;
%e {3r} = 0.242... < 1/2, so a(2) = 3.
%t z = 150; r = Sqrt[2]; f[n_] := If[FractionalPart[n*r] < 1/2, 0, 1]
%t Flatten[Position[Table[f[n], {n, 1, z}], 0]] (* A120243 *)
%t Flatten[Position[Table[f[n], {n, 1, z}], 1]] (* A120749 *)
%Y Cf. A120243, A120750, A120751.
%K nonn
%O 1,1
%A _Clark Kimberling_, Jul 01 2006
%E Updated by _Clark Kimberling_, Sep 16 2014
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