A022300 - OEIS

%I #8 Nov 22 2017 01:15:01

%S 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,

%T 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,

%U 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

%N 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.

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

%H Clark Kimberling, <a href="/A022300/b022300.txt">Table of n, a(n) for n = 1..20000</a>

%e a(1) =1, so the 3rd run has length 1, so a(5) must be 2.

%e a(2) = 1, so the 4th run has length 1, so a(6) = 1.

%e a(3) = 2, so the 5th run has length 2, so a(7) = 1 and a(8) = 2.

%e a(4) = 1, so the 6th run has length 1, so a(9) = 1.

%e 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.

%t a = {1, 1, 2}; Do[a = Join[a, ConstantArray[If[Last[a] == 1, 2, 1], {a[[n]]}]], {n, 200}]; a

%t (* _Peter J. C. Moses_, Apr 01 2016 *)

%Y Cf. A022303, A006928, A000002.

%K nonn

%O 1,3

%A _Clark Kimberling_

%E Clarified and augmented by _Clark Kimberling_, Apr 02 2016