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A120749 Numbers k such that {k* sqrt(2)} > 1/2, where { } = fractional part.

%I A120749 #16 Mar 24 2025 12:44:59
%S A120749 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 A120749 52,53,55,57,60,62,64,65,67,69,70,72,74,77,79,81,82,84,86,89,91,93,94,
%U A120749 96,98,101,103,106,108,110,111,113,115,118,120,122,123,125,127,130,132,134
%N A120749 Numbers k such that {k* sqrt(2)} > 1/2, where { } = fractional part.
%C A120749 The complement of A120749 is A120243.
%H A120749 Clark Kimberling, <a href="/A120749/b120749.txt">Table of n, a(n) for n = 1..1000</a>
%H A120749 Henk Bruin and Robbert Fokkink, <a href="https://arxiv.org/abs/2503.11734">On the records and zeros of a deterministic random walk</a>, arXiv:2503.11734 [math.DS], 2025. See pp. 1, 3-4.
%e A120749 Call the present sequence b and its complement a. Then
%e A120749 {r} = {1.4142...} = 0.4142... < 1/2, so a(1) = 1;
%e A120749 {2r} = 0.828... > 1/2, so b(1) = 2;
%e A120749 {3r} = 0.242... < 1/2, so a(2) = 3.
%t A120749 z = 150; r = Sqrt[2]; f[n_] := If[FractionalPart[n*r] < 1/2, 0, 1]
%t A120749 Flatten[Position[Table[f[n], {n, 1, z}], 0]] (* A120243 *)
%t A120749 Flatten[Position[Table[f[n], {n, 1, z}], 1]] (* A120749 *)
%t A120749 Select[Range[200],FractionalPart[# Sqrt[2]]>1/2&] (* _Harvey P. Dale_, Aug 20 2024 *)
%Y A120749 Cf. A120243, A120750, A120751.
%K A120749 nonn
%O A120749 1,1
%A A120749 _Clark Kimberling_, Jul 01 2006
%E A120749 Updated by _Clark Kimberling_, Sep 16 2014