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%I #8 Jun 24 2014 01:08:36
%S 0,1,2,0,1,0,2,0,1,2,1,0,2,1,2,0,2,1,0,1,2,0,1,0,2,0,1,2,1,0,1,2,0,1,
%T 0,2,1,2,0,2,1,0,1,2,0,1,0,2,0,1,2,1,0,2,1,2,0,2,1,0,2,0,1,2,1,0,1,2,
%U 0,1,0,2,1,2,0,2,1,0,2,0,1,2,1,0,2,1,2,0,2,1,0,1,2,0,1,0,2,1,2,0,2,1,0,2,0
%N A squarefree sequence: define a mapping from the set of strings over the alphabet {0,1,2} by f(0)=01201, f(1)=020121, f(2)=0212021 and f of the concatenation of s and t is the concatenation of f(s) and f(t). Then each of 0, f(0), f(f(0)), ... is an initial substring of the next; their limit is the infinite sequence given above.
%C f is a "squarefree morphism"; i.e. f(s) is squarefree iff s is squarefree.
%C For any i>0, f^i(0) has the same number of 0's and 1's and one less 2. The length of f^i(0) is A083066(i) = (4*6^i + 1)/5.
%D Jean Berstel and Christophe Reutenauer, Squarefree words, p. 31.
%D M. Lothaire, Combinatorics on Words, Cambridge University Press, 1997.
%e f(f(0))=01201020121021202101201020121
%t f[s_] := Flatten[{{0, 1, 2, 0, 1}, {0, 2, 0, 1, 2, 1}, {0, 2, 1, 2, 0, 2, 1}}[[ #+1]]&/@s]; f[f[f[{0}]]]
%K easy,nonn
%O 0,3
%A Claude Lenormand (claude.lenormand(AT)free.fr), Jul 29 2003
%E Edited by _Dean Hickerson_, Oct 19 2003
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