%I
%S 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,
%T 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,
%U 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
%N 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),...
%C This substitution with the polynomial that goes with it gives a new tile, not predicted in the Kenyon paper.
%C q=3 version of biKenyon 6symbol substitution.
%H Richard Kenyon, <a href="http://arXiv.org/abs/math.MG/9505210">The Construction of SelfSimilar Tilings</a>
%t 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]]
%Y Cf. A103684, A105112, A105111.
%K nonn,tabf
%O 0,3
%A _Roger L. Bagula_, Apr 07 2005
