A247543 Numbers k such that d(r,k) = 0 and d(s,k) = 1, where d(x,k) = k-th binary digit of x, r = {e}, s = {1/e}, and { } = fractional part.

2, 5, 13, 14, 21, 25, 28, 29, 35, 36, 40, 44, 49, 50, 52, 54, 56, 60, 73, 77, 78, 79, 81, 86, 87, 88, 95, 117, 125, 129, 132, 133, 140, 141, 152, 153, 155, 161, 166, 168, 170, 173, 179, 182, 184, 187, 192, 196, 203, 211, 217, 218, 220, 225, 227, 230, 238

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) = 2 and a(2) = 5.

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, A247544, A247545.

Sequence in context: A111296 A089728 A127987 * A220268 A049476 A216889

Adjacent sequences: A247540 A247541 A247542 * A247544 A247545 A247546

Keyword

nonn,easy,base

Author

Clark Kimberling, Sep 21 2014

Status

approved