A247542 Numbers k such that d(r,k) = 0 and d(s,k) = 0, where d(x,k) = k-th binary digit of x, r = {e}, s = {1/e}, and { } = fractional part.

12, 15, 17, 19, 22, 23, 30, 32, 34, 38, 47, 57, 62, 64, 66, 83, 90, 91, 92, 93, 94, 96, 98, 99, 103, 104, 109, 111, 112, 118, 123, 124, 134, 136, 145, 146, 147, 149, 154, 156, 162, 167, 175, 176, 185, 189, 194, 197, 202, 204, 205, 207, 208, 214, 215, 219

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) = 12 and a(2) = 15.

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. A247543, A247544, A247545.

Sequence in context: A368996 A368995 A188766 * A281880 A153047 A265128

Adjacent sequences: A247539 A247540 A247541 * A247543 A247544 A247545

Keyword

nonn,easy,base

Author

Clark Kimberling, Sep 21 2014

Status

approved