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

0,2

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

Links

Table of n, a(n) for n=0..104.

Formula

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

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]

Crossrefs

Cf. A073058, A105111.

Sequence in context: A016514 A224360 A269592 * A064127 A178253 A209272

Adjacent sequences: A105253 A105254 A105255 * A105257 A105258 A105259

Keyword

nonn,uned

Author

Roger L. Bagula, Apr 14 2005

Status

approved