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A270641 The sequence a of 1's and 2's starting with (1,1,1,1) such that a(n) is the length of the (n+1)st run of a.

%I A270641 #10 Nov 22 2017 01:15:06
%S A270641 1,1,1,1,2,1,2,1,2,2,1,2,2,1,2,2,1,1,2,1,1,2,2,1,2,2,1,1,2,1,2,2,1,2,
%T A270641 1,1,2,2,1,2,2,1,1,2,1,2,2,1,2,2,1,1,2,1,1,2,1,2,2,1,1,2,1,1,2,2,1,2,
%U A270641 1,1,2,1,1,2,2,1,2,2,1,1,2,1,2,2,1,2
%N A270641 The sequence a of 1's and 2's starting with (1,1,1,1) such that a(n) is the length of the (n+1)st run of a.
%C A270641 Guide to related sequences (with adjustments for initial terms):
%C A270641 1, 1, 1, 1;    a(n) = length of (n + 1)st run of a; A270641
%C A270641 1, 1, 1, 2;    a(n) = length of (n + 2)nd run of a; A270641
%C A270641 1, 1, 2, 1;    a(n) = length of (n + 3)rd run of a; A270641
%C A270641 1, 1, 2, 2;    a(n) = length of (n + 2)nd run of a; A270642
%C A270641 1, 2, 1, 1;    a(n) = length of (n + 3)rd run of a; A022300
%C A270641 1, 2, 1, 2;    a(n) = length of (n + 4)th run of a; A270641
%C A270641 1, 2, 2, 1;    a(n) = length of (n + 3)rd run of a; A270643
%C A270641 1, 2, 2, 2;    a(n) = length of (n + 2)nd run of a; A270644
%C A270641 2, 1, 1, 1;    a(n) = length of (n + 2)nd run of a; A270645
%C A270641 2, 1, 1, 2;    a(n) = length of (n + 3)rd run of a; A222300
%C A270641 2, 1, 2, 1;    a(n) = length of (n + 4)th run of a; A270641
%C A270641 2, 1, 2, 2;    a(n) = length of (n + 3)rd run of a; A000002 (Kolakoski)
%C A270641 2, 2, 1, 1;    a(n) = length of (n + 2)nd run of a; A270646
%C A270641 2, 2, 1, 2;    a(n) = length of (n + 3)rd run of a; A270647
%C A270641 2, 2, 2, 1;    a(n) = length of (n + 2)nd run of a; A270644
%C A270641 2, 2, 2, 2;    a(n) = length of (n + 1)st run of a; A270648
%H A270641 Clark Kimberling, <a href="/A270641/b270641.txt">Table of n, a(n) for n = 1..10000</a>
%e A270641 a(1) = 1, so the 2nd run has length 1, so a(5) must be 2 and a(6) = 1.
%e A270641 a(2) = 1, so the 3rd run has length 1, so a(7) = 2.
%e A270641 a(3) = 1, so the 4th run has length 1, so a(8) = 1.
%e A270641 a(4) = 1, so the 5th run has length 1, so a(9) = 2.
%e A270641 a(5) = 2, so the 6th run has length 2, so a(10) = 2 and a(11) = 1.
%e A270641 Globally, the runlength sequence of a is 4,1,1,1,1,2,1,2,1,2,2,1,...., and deleting the first term leaves a = A270641.
%t A270641 a = {1, 1, 1, 1};
%t A270641 Do[a = Join[a, ConstantArray[If[Last[a] == 1, 2, 1], {a[[n]]}]], {n,   200}]; a
%t A270641 (* _Peter J. C. Moses_, Apr 01 2016 *)
%Y A270641 Cf. A000002, A006928, A022300,
%K A270641 nonn,easy
%O A270641 1,5
%A A270641 _Clark Kimberling_, Apr 05 2016