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%I #6 Mar 12 2014 16:36:47
%S 1,4,2,4,2,1,5,1,3,4,2,1,5,1,3,1,5,1,3,4,2,4,6,4,2,1,4,2,1,5,1,3,1,5,
%T 1,3,4,2,4,6,4,2,1,1,5,1,3,4,2,4,6,4,2,1,4,2,4,6,4,2,1,1,5,1,3,1,5,4,
%U 1,5,1,3,4,2,4,2,1,5,1,3,1,5,1,3,4,2,4,6,4,2,1,1,5,1,3,4,2,4,6,4,2,1,4,2,4
%N Sign doubling substitution of the Rauzy: 1->{1,2},2->{1,3},3->1 using a digraphy symmetry to the bi-Kenyon version (not a triangular nest of nests, but a straight level 5).
%C The French/ Siegel Rauzy substitution is: 1->{1,2} 2->{1,3} 3->{1} This is digraph symmetrical to the Kenyon type substitution : 1->{2} 2->{3} 3->{3,2,1} Looking at the digraph of: 1->{2} 2->{3} 3->{6,2,1} 4->{5} 5->{6} 6->{3,5,4} I get the same linked two triangle structure for this six-symbol substitution. The problem with the digraph approach is that the order is not specific as it is in actual substitutions.
%D "The Construction of Self-Similar Tilings", Richard Kenyon, Section 6
%F 1->{4, 2} 2->{1, 3} 3->{1} 4->{1, 5} 5->{4, 6} 6->{4}
%t s[1] = {4, 2}; s[2] = {1, 3}; s[3] = {1}; s[4] = {1, 5}; s[5] = {4, 6}; s[6] = {4}; t[a_] := Join[a, Flatten[s /@ a]]; p[0] = {1}; p[1] = t[{1}]; p[n_] := t[p[n - 1]] aa = p[5]
%Y Cf. A073058, A105111.
%K nonn,uned
%O 0,2
%A _Roger L. Bagula_, Apr 14 2005
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