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A247455
Numbers k such that d(r,k) = 0 and d(s,k) = 0, where d(x,k) = k-th binary digit of x, r = {sqrt(2)}, s = {3*sqrt(2)}, and { } = fractional part.
6
1, 8, 9, 10, 11, 15, 21, 25, 29, 38, 42, 48, 51, 54, 57, 58, 59, 62, 64, 66, 70, 72, 78, 81, 82, 86, 89, 93, 96, 107, 109, 111, 113, 122, 128, 130, 134, 136, 139, 144, 147, 148, 149, 151, 153, 161, 162, 165, 169, 173, 181, 182, 183, 187, 191, 195, 200, 202
OFFSET
1,2
COMMENTS
Every positive integer lies in exactly one of these: A247455, A247456, A247457, A247758. Let s denote any of these; what can be said about lim(#s < n)/n, where (#s < n) represents the number of numbers in s that are < n?
LINKS
EXAMPLE
{1*sqrt(2)} has binary digits 0, 1, 1, 0, 1, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 1,...
{3*sqrt(2)} has binary digits 0, 0, 1, 1, 1, 1, 1, 0, 0, 0, 0, 1, 1, 1, 0, 1,...
so that a(1) = 2 and a(2) = 8.
MATHEMATICA
z = 400; r = FractionalPart[Sqrt[2]]; s = FractionalPart[3*Sqrt[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]] (* A247455 *)
Flatten[Position[t2, 1]] (* A247456 *)
Flatten[Position[t3, 1]] (* A247457 *)
Flatten[Position[t4, 1]] (* A247458 *)
CROSSREFS
KEYWORD
nonn,easy,base
AUTHOR
Clark Kimberling, Sep 18 2014
STATUS
approved