%I #5 Mar 12 2014 16:36:48
%S 1,2,3,1,2,1,1,2,3,1,1,3,1,2,3,1,2,1,1,2,1,2,3,1,1,2,3,1,2,1,1,2,3,1,
%T 1,3,1,2,3,1,1,2,3,1,3,1,1,3,1,2,3,1,2,1,1,2,3,1,1,3,1,2,3,1,2,1,1,2,
%U 1,2,3,1,1,2,3,1,2,1,1,2,1,2,3,1,2,1,1,2,3,1,1,3,1,2,3,1,1,2,3,1,2,1,1,2,3
%N Triangle Loop von Koch substitution: characteristic polynomial:x^3-6x^2+8*x.
%C To get the fractal: bb = aa /. 1 -> {1, 0} /. 2 -> {-1, N[Sqrt[3]]}/2 /. 3 -> {-1, -N[Sqrt[3]]}/2; ListPlot[FoldList[Plus, {0, 0}, bb], PlotJoined -> True, PlotRange -> All, Axes -> False];
%F 1->{1, 2, 3, 1}, 2->{2, 1, 1, 2}, 3->{3, 1, 1, 3}
%t s[1] = {1, 2, 3, 1}; s[2] = {2, 1, 1, 2}; s[3] = {3, 1, 1, 3};; t[a_] := Flatten[s /@ a]; p[0] = {1}; p[1] = t[p[0]]; p[n_] := t[p[n - 1]] aa = p[4]
%K nonn,uned
%O 0,2
%A _Roger L. Bagula_, May 05 2005