A247912 Numbers k such that A247911(k+1) = A247911(k).

2, 4, 7, 9, 12, 15, 17, 20, 22, 25, 27, 30, 32, 35, 37, 40, 42, 45, 47, 50, 52, 55, 57, 60, 62, 64, 67, 69, 72, 74, 77, 79, 81, 84, 86, 89, 91, 93, 96, 98, 101, 103, 105, 108, 110, 113, 115, 117, 120, 122, 125, 127, 129, 132, 134, 137, 139, 141, 144, 146

Offset

1,1

Comments

Complement of A247913.

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 0.

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, A247913, A247908, A247914.

Sequence in context: A165994 A163293 A188045 * A186321 A026356 A184735

Adjacent sequences: A247909 A247910 A247911 * A247913 A247914 A247915

Keyword

nonn,easy

Author

Clark Kimberling, Sep 27 2014

Status

approved