A105111 -id:A105111 - cp's OEIS frontend

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

Roger L. Bagula, Apr 14 2005

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.

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

Cf. A073058, A105111.

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]

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

A105113 Triangle read by rows, based on the morphism f: 1->2, 2->3, 3->{3,5,5,5,4}, 4->5, 5->6, 6->{6,2,2,2,1}. First row is 1. If current row is a,b,c,..., then the next row is a,b,c,...,f(a),f(b),f(c),...

Original entry on oeis.org

1, 1, 2, 1, 2, 2, 3, 1, 2, 2, 3, 2, 3, 3, 3, 5, 5, 5, 4, 1, 2, 2, 3, 2, 3, 3, 3, 5, 5, 5, 4, 2, 3, 3, 3, 5, 5, 5, 4, 3, 3, 5, 5, 5, 4, 3, 5, 5, 5, 4, 3, 5, 5, 5, 4, 6, 6, 6, 5, 1, 2, 2, 3, 2, 3, 3, 3, 5, 5, 5, 4, 2, 3, 3, 3, 5, 5, 5, 4, 3, 3, 5, 5, 5, 4, 3, 5, 5, 5, 4, 3, 5, 5, 5, 4, 6, 6, 6, 5, 2, 3, 3, 3, 5, 5

Offset: 0

Roger L. Bagula, Apr 07 2005

This substitution with the polynomial that goes with it gives a new tile, not predicted in the Kenyon paper.

q=3 version of bi-Kenyon 6-symbol substitution.

Richard Kenyon, The Construction of Self-Similar Tilings

Cf. A103684, A105112, A105111.

s[n_] := n /. {1 -> 2, 2 -> 3, 3 -> {3, 5, 5, 5, 4}, 4 -> 5, 5 -> 6, 6 -> {6, 2, 2, 2, 1}}; t[a_] := Join[a, Flatten[s /@ a]] p[0] = {1}; p[1] = t[{1}]; Flatten[ NestList[t, {1}, 5]]

A105258 Triangle of trajectory of 1 under the morphism 1->2, 2->13, 3->1.

Original entry on oeis.org

1, 1, 2, 1, 2, 2, 1, 3, 1, 2, 2, 1, 3, 2, 1, 3, 1, 3, 2, 1, 1, 2, 2, 1, 3, 2, 1, 3, 1, 3, 2, 1, 2, 1, 3, 1, 3, 2, 1, 1, 3, 2, 1, 2, 1, 1, 3, 2, 1, 2, 2, 1, 3, 2, 1, 3, 1, 3, 2, 1, 2, 1, 3, 1, 3, 2, 1, 1, 3, 2, 1, 2, 1, 1, 3, 2, 2, 1, 3, 1, 3, 2, 1, 1, 3, 2, 1, 2, 1, 1, 3, 2, 1, 3, 2, 1, 2, 1, 1, 3, 2, 2, 1, 1, 3

Offset: 0

Roger L. Bagula, Apr 14 2005

			The first steps are:
{1}
{1, 2}
{1, 2, 2, 1, 3}
{1, 2, 2, 1, 3, 2, 1, 3, 1, 3, 2, 1}

Index entries for sequences that are fixed points of mappings

Cf. A073058, A105111.

s[1] = {2}; s[2] = {1, 3}; s[3] = {1}; t[a_] := Join[a, Flatten[s /@ a]]; p[0] = {1}; p[1] = t[{1}]; p[n_] := t[p[n - 1]] aa = Flatten[Table[p[n], {n, 0, 6}]]

{a(n)=local(m, v, w); v=w=[1]; while(length(w)

Edited by the Associate Editors of the OEIS, Apr 07 2009

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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).

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A105113 Triangle read by rows, based on the morphism f: 1->2, 2->3, 3->{3,5,5,5,4}, 4->5, 5->6, 6->{6,2,2,2,1}. First row is 1. If current row is a,b,c,..., then the next row is a,b,c,...,f(a),f(b),f(c),...

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A105258 Triangle of trajectory of 1 under the morphism 1->2, 2->13, 3->1.

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