%I
%S 0,1,1,0,1,0,1,0,1,1,0,1,0,1,1,0,1,0,1,1,0,1,0,1,0,1,1,0,1,0,1,1,0,1,
%T 0,1,0,1,1,0,1,0,1,1,0,1,0,1,0,1,1,0,1,0,1,1,0,1,0,1,1,0,1,0,1,0,1,1,
%U 0,1,0,1,1,0,1,0,1,0,1,1,0,1,0,1,1,0,1,0,1,1,0,1,0,1,0,1,1,0,1,0,1,1,0,1,0,1,0,1,1,0,1,0,1,1,0,1,0,1,1,0
%N Fixed point of the morphism 0>011, 1>01.
%C From _Danny Rorabaugh_, Mar 14 2015: (Start)
%C Let x(i) and y(i) be the number of 0s and 1s, respectively, after the ith stage of generating this word, so x(0) = 1, y(0) = 0, x(i+1) = x(i) + y(i), and y(i+1) = 2x(i) + y(i). Equivalently: x(0) = 1, x(1) = 1, x(i+2) = 2x(i+1) + x(i), y(0) = 0, y(1) = 2, and y(i+2) = 2y(i+1) + y(i).
%C The number of 0s after the ith stage is x(i) = A001333(i).
%C The number of 1s after the ith stage is y(i) = 2*A000129(i) = A163271(i+1) = A001333(i+1)  A001333(i).
%C Let S(n) = Sum_{j<=n} a(j) be the partial sums of this sequence, so S(x(i)+y(i)) = y(i). Consequently, if the Cesàro sum of a(n) exists, then it is lim_{n>infinity} S(n)/n = lim_{i>infinity} A163271(i+1)/A001333(i+1) = 2  sqrt(2).
%C (End)
%e 0>011>0110101>01101010110101101>
%t t = Nest[Flatten[# /. {0>{0,1,1}, 1>{0,1}}] &, {0}, 5] (*A189687*)
%t f[n_] := t[[n]]
%t Flatten[Position[t, 0]] (* A086377 conjectured *)
%t Flatten[Position[t, 1]] (* A081477 conjectured *)
%t s[n_] := Sum[f[i], {i, 1, n}]; s[0] = 0;
%t Table[s[n], {n, 1, 120}] (*A189688*)
%Y Cf. A189688, A086377, A189688.
%Y Fixed points of similar morphisms: A004641, A005614, A080764, A159684, A171588, A189572.
%K nonn
%O 1
%A _Clark Kimberling_, Apr 25 2011
