A247519 Numbers k such that d(r,k) = 0 and d(s,k) = 0, where d(x,k) = k-th binary digit of x, r = {golden ratio}, s = {(golden ratio)/2}, and { } = fractional part.

3, 9, 10, 23, 31, 44, 49, 56, 57, 58, 59, 70, 75, 80, 84, 85, 86, 87, 88, 89, 93, 94, 95, 96, 97, 104, 116, 119, 120, 121, 122, 128, 129, 130, 131, 134, 135, 136, 139, 140, 141, 142, 145, 146, 149, 166, 173, 174, 177, 182, 185, 190, 191, 199, 200, 201, 208

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, A247522. Prefixing the binary digits for r by 1 gives the binary digits for s.

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) = 3.
		

Links

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

Crossrefs

Cf. A247520, A247521, A247522.

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 *)