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The sequence a of 1's and 2's starting with (2,2,1,2) such that a(n) is the length of the (n+3)rd run of a.
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%I #8 Jun 27 2023 07:08:03

%S 2,2,1,2,1,1,2,2,1,2,2,1,2,1,1,2,2,1,2,2,1,1,2,1,1,2,1,2,2,1,1,2,1,1,

%T 2,2,1,2,1,1,2,1,2,2,1,2,2,1,1,2,1,2,2,1,2,1,1,2,2,1,2,2,1,2,1,1,2,1,

%U 1,2,2,1,2,2,1,1,2,1,2,2,1,2,2,1,1,2

%N The sequence a of 1's and 2's starting with (2,2,1,2) such that a(n) is the length of the (n+3)rd run of a.

%C See A270641 for a guide to related sequences.

%H Clark Kimberling, <a href="/A270647/b270647.txt">Table of n, a(n) for n = 1..10000</a>

%e a(1) = 2, so the 4th run has length 2, so a(5) must be 1 and a(6) = 1.

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

%e a(3) = 1, so the 6th run has length 1, so a(9) = 1 and a(10) = 2.

%e Globally, the runlength sequence of a is 2,1,1,2,2,1,2,1,1,2,2,1,2,2,1,2,1,1,2,..., and deleting the first 3 terms leaves a = A270647.

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

%Y Cf. A000002, A006928, A022300, A270641.

%K nonn,easy

%O 1,1

%A _Clark Kimberling_, Apr 07 2016