%I #6 Apr 07 2016 02:13:02
%S 1,1,2,2,1,2,1,1,2,2,1,2,2,1,2,1,1,2,2,1,2,2,1,1,2,1,1,2,1,2,2,1,1,2,
%T 1,1,2,2,1,2,1,1,2,1,2,2,1,2,2,1,1,2,1,2,2,1,2,1,1,2,2,1,2,2,1,2,1,1,
%U 2,1,1,2,2,1,2,2,1,1,2,1,2,2,1,2,2,1
%N The sequence a of 1's and 2's starting with (1,1,2,2) such that a(n) is the length of the (n+2)nd run of a.
%C See A270641 for a guide to related sequences.
%H Clark Kimberling, <a href="/A270642/b270642.txt">Table of n, a(n) for n = 1..10000</a>
%e a(1) = 1, so the 3rd run has length 1, so a(5) must be 1.
%e a(2) = 1, so the 4th run has length 1, so a(6) = 2 and a(7) = 1.
%e a(3) = 2, so the 5th run has length 2, so a(8) = 1 and a(9) = 2.
%e a(4) = 2, so the 6th run has length 2, so a(10) = 2 and a(11) = 1.
%e Globally, the runlength sequence is 2,2,1,1,2,2,1,2,1,1,2,2,1,2,..., and deleting the first 2 terms leaves the same sequence.
%t a = {1, 1, 2, 2}; Do[a = Join[a, ConstantArray[If[Last[a] == 1, 2, 1], {a[[n]]}]], {n, 200}]; a (* _Peter J. C. Moses_, Apr 01 2016 *)
%Y Cf. A000002, A006928, A022300, A270641.
%K nonn,easy
%O 1,3
%A _Clark Kimberling_, Apr 06 2016
