A105256 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).

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

Offset: 0

Author

Roger L. Bagula, Apr 14 2005

Keywords

• nonn
• uned

Comments

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.

References

• "The Construction of Self-Similar Tilings", Richard Kenyon, Section 6

Crossrefs

Cf. A073058, A105111.

Programs

Mathematica

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]

Formula

1->{4, 2} 2->{1, 3} 3->{1} 4->{1, 5} 5->{4, 6} 6->{4}