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Search: id:A105256
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| A105256 |
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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). |
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+0
2
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| 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, 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, 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
(list; graph; listen)
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OFFSET
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0,2
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COMMENT
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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.
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REFERENCES
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"The Construction of Self-Similar Tilings", Richard Kenyon, Section 6
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FORMULA
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1->{4, 2} 2->{1, 3} 3->{1} 4->{1, 5} 5->{4, 6} 6->{4}
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MATHEMATICA
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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]
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CROSSREFS
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Cf. A073058, A105111.
Sequence in context: A156199 A135513 A016514 this_sequence A064127 A105397 A147973
Adjacent sequences: A105253 A105254 A105255 this_sequence A105257 A105258 A105259
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KEYWORD
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nonn,uned
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AUTHOR
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Roger Bagula (rlbagulatftn(AT)yahoo.com), Apr 14 2005
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