A247636 Numbers k such that d(r,k) != d(s,k), where d(x,k) = k-th binary digit of x, r = {sqrt(2)}, s = {sqrt(8)}, and { } = fractional part.

1, 3, 4, 5, 6, 7, 12, 13, 15, 19, 21, 23, 25, 27, 29, 36, 38, 41, 42, 46, 48, 50, 52, 53, 55, 56, 60, 61, 64, 65, 66, 68, 70, 71, 72, 77, 78, 80, 83, 84, 86, 88, 89, 91, 93, 95, 96, 99, 100, 101, 102, 103, 104, 105, 107, 108, 109, 110, 111, 112, 113, 118

Offset: 1

Author

Clark Kimberling, Sep 23 2014

Keywords

• nonn
• easy
• base

Comments

Every positive integer lies in exactly one of the sequences A247635 and A247636.

Examples

			r has binary digits 0, 1, 1, 0, 1, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 1, 1, 1, ...
s has binary digits 1, 1, 0, 1, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 1, 1, 1, 1, ...
so that a(1) = 1 and a(2) = 3.
		

Links

• Clark Kimberling, Table of n, a(n) for n = 1..1000

Crossrefs

Cf. A247635, A247632, A247524.

Programs

Mathematica

z = 200; r = FractionalPart[Sqrt[2]]; s = FractionalPart[Sqrt[8]];
u = Flatten[{ConstantArray[0, -#[[2]]], #[[1]]}] &[RealDigits[r, 2, z]];
v = Flatten[{ConstantArray[0, -#[[2]]], #[[1]]}] &[RealDigits[s, 2, z]];
t = Table[If[u[[n]] == v[[n]], 1, 0], {n, 1, z}];
Flatten[Position[t, 1]]  (* A247635 *)
Flatten[Position[t, 0]]  (* A247636 *)