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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),...
0
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
0,3
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.
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]]
CROSSREFS
KEYWORD
nonn,tabf
AUTHOR
Roger L. Bagula, Apr 07 2005
STATUS
approved