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

1,1

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.

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) = 3.

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. A247520, A247521, A247522.

Sequence in context: A085459 A092169 A093108 * A344295 A077560 A326003

Adjacent sequences: A247516 A247517 A247518 * A247520 A247521 A247522

Keyword

nonn,easy,base

Author

Clark Kimberling, Sep 19 2014

Status

approved