A247631 Numbers k such that d(r,k) = 0 and d(s,k) = 0, where d(x,k) = k-th binary digit of x, r = {sqrt(2)}, s = {sqrt(8)}, and { } = fractional part.

8, 9, 10, 11, 14, 20, 24, 28, 37, 47, 51, 54, 57, 58, 59, 62, 63, 69, 81, 82, 85, 92, 106, 121, 128, 129, 147, 148, 149, 150, 161, 162, 165, 168, 181, 182, 183, 186, 190, 200, 201, 214, 217, 218, 219, 225, 226, 227, 228, 232, 236, 241, 245, 248, 249, 258

Offset: 1

Author

Clark Kimberling, Sep 23 2014

Keywords

• nonn
• easy
• base

Comments

Every positive integer lies in exactly one of these: A247631, A247632, A247633, A247634. Deleting the initial 1 from the representation of r gives the representation of s.

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) = 8 and a(2) = 9.
		

Links

• Clark Kimberling, Table of n, a(n) for n = 1..1181

Crossrefs

Cf. A247632, A247633, A247634, A247519.

Programs

Mathematica

z = 400; 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]]
t1 = Table[If[u[[n]] == 0 && v[[n]] == 0, 1, 0], {n, 1, z}];
t2 = Table[If[u[[n]] == 0 && v[[n]] == 1, 1, 0], {n, 1, z}];
t3 = Table[If[u[[n]] == 1 && v[[n]] == 0, 1, 0], {n, 1, z}];
t4 = Table[If[u[[n]] == 1 && v[[n]] == 1, 1, 0], {n, 1, z}];
Flatten[Position[t1, 1]]  (* A247631 *)
Flatten[Position[t2, 1]]  (* A247632 *)
Flatten[Position[t3, 1]]  (* A247633 *)
Flatten[Position[t4, 1]]  (* A247634 *)