A247521 Numbers k such that d(r,k) = 1 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.

4, 11, 14, 18, 24, 27, 32, 34, 42, 45, 47, 50, 60, 62, 64, 71, 76, 81, 90, 98, 100, 105, 109, 112, 117, 123, 126, 132, 137, 143, 147, 150, 154, 157, 159, 167, 171, 175, 178, 183, 186, 188, 192, 202, 205, 210, 213, 215, 220, 224, 228, 233, 240, 245, 249, 252

Offset

1,1

Comments

Every positive integer lies in exactly one of these: A247519, A247520, A247522.

Links

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

Example

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) = 4.

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 *)

Crossrefs

Cf. A247519, A247520, A247522.

Sequence in context: A095797 A205846 A204542 * A285979 A299975 A271508

Adjacent sequences: A247518 A247519 A247520 * A247522 A247523 A247524

Keyword

nonn,easy,base

Author

Clark Kimberling, Sep 19 2014

Status

approved