A247545 Numbers k such that d(r,k) = 1 and d(s,k) = 1, where d(x,k) = k-th binary digit of x, r = {e}, s = {1/e}, and { } = fractional part.

4, 6, 7, 11, 16, 18, 20, 26, 33, 39, 41, 43, 45, 46, 53, 55, 61, 63, 65, 67, 68, 69, 71, 74, 76, 82, 89, 97, 100, 106, 108, 110, 113, 114, 115, 116, 120, 126, 128, 130, 135, 137, 150, 157, 159, 163, 164, 171, 174, 178, 180, 183, 188, 191, 195, 206, 209, 212

Offset

1,1

Comments

Every positive integer lies in exactly one of these: A247542, A247543, A247544, A247545.

Links

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

Example

{e/1} has binary digits 1, 0, 1, 1, 0, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 1, 0, 1, ...
{1/e} has binary digits 0, 1, 0, 1, 1, 1, 1, 0, 0, 0, 1, 0, 1, 1, 0, 1, 0, 1, ...
so that a(1) = 4 and a(2) = 6.

Mathematica

z = 400; r = FractionalPart[E]; s = FractionalPart[1/E];
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]]  (* A247542 *)
Flatten[Position[t2, 1]]  (* A247543 *)
Flatten[Position[t3, 1]]  (* A247544 *)
Flatten[Position[t4, 1]]  (* A247545 *)

Crossrefs

Cf. A247542, A247543, A247544.

Sequence in context: A022436 A102139 A167228 * A310655 A102145 A102132

Adjacent sequences: A247542 A247543 A247544 * A247546 A247547 A247548

Keyword

nonn,easy,base

Author

Clark Kimberling, Sep 21 2014

Status

approved