A247634 Numbers k such that d(r,k) = 1 and d(s,k) = 1, where d(x,k) = k-th binary digit of x, r = {sqrt(2)}, s = {sqrt(8)}, and { } = fractional part.

2, 16, 17, 18, 22, 26, 30, 31, 32, 33, 34, 35, 39, 40, 43, 44, 45, 49, 67, 73, 74, 75, 76, 79, 87, 90, 94, 97, 98, 114, 115, 116, 117, 123, 124, 125, 126, 131, 132, 137, 140, 141, 142, 145, 154, 155, 170, 171, 174, 175, 188, 192, 193, 196, 205, 206, 207, 212

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.

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) = 4.
		

Links

• Clark Kimberling, Table of n, a(n) for n = 1..1122

Crossrefs

Cf. A247631, A247632, A247633.

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 *)