login
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
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
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
KEYWORD
nonn,easy,base
AUTHOR
Clark Kimberling, Sep 19 2014
STATUS
approved