A247544 Numbers k such that d(r,k) = 1 and d(s,k) = 0, where d(x,k) = k-th binary digit of x, r = {e}, s = {1/e}, and { } = fractional part.

1, 3, 8, 9, 10, 24, 27, 31, 37, 42, 48, 51, 58, 59, 70, 72, 75, 80, 84, 85, 101, 102, 105, 107, 119, 121, 122, 127, 131, 138, 139, 142, 143, 144, 148, 151, 158, 160, 165, 169, 172, 177, 181, 186, 190, 193, 198, 199, 200, 201, 210, 222, 226, 228, 233, 236

Offset

1,2

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, A247543, A247545.

Sequence in context: A051208 A211223 A350776 * A080524 A024550 A173179

Adjacent sequences: A247541 A247542 A247543 * A247545 A247546 A247547

Keyword

nonn,easy,base

Author

Clark Kimberling, Sep 21 2014

Status

approved