%I
%S 1,0,2,0,1,0,0,1,0,2,0,1,0,2,0,1,0,0,1,0,2,0,1,0,1,0,2,0,1,0,0,1,0,2,
%T 0,1,0,0,1,0,2,0,1,0,1,0,2,0,1,0,0,1,0,2,0,1,0,2,0,1,0,0,1,0,2,0,1,0,
%U 1,0,2,0,1,0,0,1,0,2,0,1,0,2,0,1,0,0
%N 1limiting word of the morphism 0>10, 1>20, 2>0.
%C Starting with 0, the first 4 iterations of the morphism yield words shown here:
%C 1st: 10
%C 2nd: 2010
%C 3rd: 0102010
%C 4th: 1020100102010
%C The 1limiting word is the limit of the words for which the number of iterations is congruent to 1 mod 3.
%C Let U, V, W be the limits of u(n)/n, v(n)/n, w(n)/n, respectively. Then 1/U + 1/V + 1/W = 1, where
%C U = 1.8392867552141611325518525646532866...,
%C V = U^2 = 3.3829757679062374941227085364...,
%C W = U^3 = 6.2222625231203986266745611011....
%C If n >=2, then u(n)  u(n1) is in {1,2}, v(n)  v(n1) is in {2,3,4}, and w(n)  w(n1) is in {4,6,7}.
%C From _Michel Dekking_, Mar 29 2019: (Start)
%C This sequence is one of the three fixed points of the morphism alpha^3, where alpha is the defining morphism
%C 0>10, 1>20, 2>0.
%C The other two fixed points are A286998 and A287174.
%C We have alpha = rho(tau), where tau is the Tribonacci morphism in A080843
%C 0>01, 1>02, 2>0,
%C and rho is the rotation operator.
%C An eigenvector computation of the incidence matrix of the morphism gives that 0,1, and 2 have frequencies 1/t, 1/t^2 and 1/t^3, where t is the tribonacci constant A058265.
%C Apparently (u(n)) := A287113. Thus U, the limit of u(n)/n, equals 1/(1/t), the tribonacci constant t. Also, V = A276800, and W = A276801.
%C (End)
%H Clark Kimberling, <a href="/A287112/b287112.txt">Table of n, a(n) for n = 1..10000</a>
%e 1st iterate: 10
%e 4th iterate: 1020100102010
%e 7th iterate: 102010010201020100102010102010010201001020101020100102010201001020101020100102010.
%t s = Nest[Flatten[# /. {0 > {1, 0}, 1 > {2, 0}, 2 > 0}] &, {0}, 10] (* A287112 *)
%t Flatten[Position[s, 0]] (* A287113 *)
%t Flatten[Position[s, 1]] (* A287114 *)
%t Flatten[Position[s, 2]] (* A287115 *)
%Y Cf. A287113, A287114, A287115, A286998, A287174.
%K nonn,easy
%O 1,3
%A _Clark Kimberling_, May 22 2017
