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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.
6
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
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
KEYWORD
nonn,easy,base
AUTHOR
Clark Kimberling, Sep 19 2014
STATUS
approved