A247522 Numbers k such that d(r,k) = 1 and d(s,k) = 1, where d(x,k) = k-th binary digit of x, r = {golden ratio}, s = {(golden ratio)/2}, and { } = fractional part.

1, 5, 6, 7, 12, 15, 16, 19, 20, 21, 25, 28, 29, 35, 36, 37, 38, 39, 40, 51, 52, 53, 54, 65, 66, 67, 68, 72, 73, 77, 78, 82, 91, 101, 102, 106, 107, 110, 113, 114, 124, 151, 152, 155, 160, 161, 162, 163, 164, 168, 169, 179, 180, 193, 194, 195, 196, 197, 203

Offset: 1

Author

Clark Kimberling, Sep 19 2014

Keywords

• nonn
• easy
• base

Comments

Every positive integer lies in exactly one of these: A247519, A247520, A247521.

Examples

			r has binary digits 1, 0, 0, 1, 1, 1, 1, 0, 0, 0, 1, 1, 0, 1, 1, 1, ...
s has binary digits 1, 1, 0, 0, 1, 1, 1, 1, 0, 0, 0, 1, 1, 0, 1, 1, ...
so that a(1) = 1 and a(2) = 5.
		

Links

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

Crossrefs

Cf. A247519, A247520, A247521.

Programs

Mathematica

z = 400; r1 = GoldenRatio; r = FractionalPart[r1]; s = FractionalPart[r1/2];
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]] (* A247519 *)
Flatten[Position[t2, 1]] (* A247520 *)
Flatten[Position[t3, 1]] (* A247521 *)
Flatten[Position[t4, 1]] (* A247522 *)