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),...

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

Author

Roger L. Bagula, Apr 07 2005

Keywords

• nonn
• tabf

Comments

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.

Links

• Richard Kenyon, The Construction of Self-Similar Tilings

Crossrefs

Cf. A103684, A105112, A105111.

Programs

Mathematica

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]]