A247913 Numbers k such that A247911(k+1) = A247911(k) + 1.

1, 3, 5, 6, 8, 10, 11, 13, 14, 16, 18, 19, 21, 23, 24, 26, 28, 29, 31, 33, 34, 36, 38, 39, 41, 43, 44, 46, 48, 49, 51, 53, 54, 56, 58, 59, 61, 63, 65, 66, 68, 70, 71, 73, 75, 76, 78, 80, 82, 83, 85, 87, 88, 90, 92, 94, 95, 97, 99, 100, 102, 104, 106, 107

Offset

1,2

Comments

Complement of A247912.

Links

Clark Kimberling, Table of n, a(n) for n = 1..500

Example

A247911(n+1) - A247911(n) = (1,0,1,0,1,1,0,1,0,1,1,0,1,1,0,1,0,1,1,0,...), and a(n) is the position of the n-th 1.

Mathematica

$RecursionLimit = 1000; $MaxExtraPrecision = 1000;
z = 300; u[1] = 0; u[2] = 1; u[n_] := u[n] = u[n - 1] + u[n - 2]/(n - 2);
f[n_] := f[n] = Select[Range[z], (2 # + 1)/u[2 # + 1] - E < n^-n &, 1];
u = Flatten[Table[f[n], {n, 1, z}]]  (* A247911 *)
w = Differences[u]
Flatten[Position[w, 0]] (* A247912 *)
Flatten[Position[w, 1]] (* A247913 *)

Crossrefs

Cf. A247911, A247912, A247908, A247914.

Sequence in context: A184734 A189662 A186320 * A188046 A244644 A047220

Adjacent sequences: A247910 A247911 A247912 * A247914 A247915 A247916

Keyword

nonn,easy

Author

Clark Kimberling, Sep 27 2014

Status

approved