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%I #10 Apr 21 2021 03:47:22
%S 1,3,5,6,8,10,13,15,17,18,20,22,25,27,29,30,32,34,35,37,39,42,44,46,
%T 47,49,51,54,56,58,59,61,63,66,68,71,73,75,76,78,80,83,85,87,88,90,92,
%U 95,97,99,100,102,104,105,107,109,112,114,116,117,119,121,124,126,128,129
%N Numbers k such that {k*sqrt(2)} < 1/2, where { } = fractional part.
%C The complement of a is b=A120749. Is a(n) < b(n) for all n? If k is a positive integer, then is b(n) - a(n) = k for infinitely many n?
%H Clark Kimberling, <a href="/A120243/b120243.txt">Table of n, a(n) for n = 1..1000</a>
%e {r} = {1.4142...} = 0.4142... < 1/2, so a(1)=1.
%e {2r} = 0.828... > 1/2, so b(1) = 2, where b = complement of a.
%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. A120749, A120750, A120751.
%K nonn
%O 1,2
%A _Clark Kimberling_, Jul 01 2006
%E Updated by _Clark Kimberling_, Sep 16 2014