A022300 The sequence a of 1's and 2's starting with (1,1,2,1) such that a(n) is the length of the (n+2)nd run of a.

1, 1, 2, 1, 2, 1, 1, 2, 1, 1, 2, 1, 2, 2, 1, 2, 1, 1, 2, 1, 1, 2, 2, 1, 2, 2, 1, 2, 1, 1, 2, 1, 2, 2, 1, 1, 2, 1, 1, 2, 2, 1, 2, 2, 1, 2, 1, 1, 2, 1, 1, 2, 2, 1, 2, 1, 1, 2, 1, 2, 2, 1, 1, 2, 1, 1, 2, 2, 1, 2, 2, 1, 2, 1, 1, 2, 1, 2, 2, 1, 1, 2, 1, 1, 2, 1, 2, 2, 1, 2, 2, 1, 1, 2, 1, 2, 2, 1, 2, 1, 1, 2, 2, 1, 2, 2, 1, 1

Offset

1,3

Comments

It appears that various properties and unsolved problems associated with the Kolakoski sequence, A000002, apply also to A022300.

Links

Clark Kimberling, Table of n, a(n) for n = 1..20000

Example

a(1) =1, so the 3rd run has length 1, so a(5) must be 2.
a(2) = 1, so the 4th run has length 1, so a(6) = 1.
a(3) = 2, so the 5th run has length 2, so a(7) = 1 and a(8) = 2.
a(4) = 1, so the 6th run has length 1, so a(9) = 1.
Globally, the runlength sequence of a is 2,1,1,1,2,1,2,1,1,2,1,1,2,...., and deleting the first two terms leaves a = A022300.

Mathematica

a = {1, 1, 2}; Do[a = Join[a, ConstantArray[If[Last[a] == 1, 2, 1], {a[[n]]}]], {n, 200}]; a
(* Peter J. C. Moses, Apr 01 2016 *)

Crossrefs

Cf. A022303, A006928, A000002.

Sequence in context: A175244 A206722 A245222 * A347552 A300983 A279205

Adjacent sequences: A022297 A022298 A022299 * A022301 A022302 A022303

Keyword

nonn

Author

Clark Kimberling

Extensions

Clarified and augmented by Clark Kimberling, Apr 02 2016

Status

approved