A120243 Numbers k such that {k*sqrt(2)} < 1/2, where { } = fractional part.

1, 3, 5, 6, 8, 10, 13, 15, 17, 18, 20, 22, 25, 27, 29, 30, 32, 34, 35, 37, 39, 42, 44, 46, 47, 49, 51, 54, 56, 58, 59, 61, 63, 66, 68, 71, 73, 75, 76, 78, 80, 83, 85, 87, 88, 90, 92, 95, 97, 99, 100, 102, 104, 105, 107, 109, 112, 114, 116, 117, 119, 121, 124, 126, 128, 129

Offset: 1

Author

Clark Kimberling, Jul 01 2006

Keywords

• nonn

Comments

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?

Examples

			{r} = {1.4142...} = 0.4142... < 1/2, so a(1)=1.
{2r} = 0.828... > 1/2, so b(1) = 2, where b = complement of a.
{3r} = 0.242... < 1/2, so a(2) = 3.
		

Links

• Clark Kimberling, Table of n, a(n) for n = 1..1000
• Henk Bruin and Robbert Fokkink, On the records and zeros of a deterministic random walk, arXiv:2503.11734 [math.DS], 2025. See pp. 1, 4.

Crossrefs

Cf. A120749, A120750, A120751.

Programs

Mathematica

z = 150; r = Sqrt[2]; f[n_] := If[FractionalPart[n*r] < 1/2, 0, 1]
Flatten[Position[Table[f[n], {n, 1, z}], 0]] (* A120243 *)
Flatten[Position[Table[f[n], {n, 1, z}], 1]] (* A120749 *)

Extensions

Updated by Clark Kimberling, Sep 16 2014