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%I #11 Jun 16 2015 19:13:43
%S 1,1,1,6,3,1,1,1,6,3,1,1,6,3,1,1,6,3,1,6,8,1,6,3,1,2,4,1,6,3,1,1,1,6,
%T 3,1,1,6,3,1,1,6,3,1,6,8,1,6,3,1,2,4,1,6,3,1,1,6,3,1,1,6,3,1,6,8,1,6,
%U 3,1,2,4,1,6,3,1,1,6,3,1,6,8,1,6,3,1,2,4,1,6,3,1,1,6,3,1,6,8,1,6,3,1,2,4,1
%N Substitution Group for a von Koch curve.
%C To get a von Koch set to show up you need level 6 and: bb = aa /. 1 -> {-0.5,0.8660254037844386} /. 2 -> {-0.5, 0.8660254037844386} /. 3 -> {1, 0} /. 4 -> {-1, 0} /. 5 -> {0.5, -0.8660254037844386} /. 6 -> {0.5, -0.8660254037844386} /. 7 -> {-1, 0 } /. 8 -> {1, 0}; ListPlot[FoldList[Plus, {0, 0}, bb], PlotJoined -> False, PlotRange -> All];
%H F. M. Dekking, <a href="http://dx.doi.org/10.1016/0001-8708(82)90066-4">Recurrent sets</a>, Advances in Mathematics, vol. 44, no. 1 (1982), 78-104; page 96, section 4.11.
%F 1->{1, 6, 3, 1} 2->{2, 4, 5, 2} 3->{3, 1, 2, 4} 4->{3, 1, 2, 4} 5->{5, 2, 7, 5} 6->{6, 8, 1, 6} 7->{7, 5, 6, 8} 8->{7, 5, 6, 8}
%t s[1] = {1, 6, 3, 1}; s[2] = {2, 4, 5, 2}; s[3] = {3, 1, 2, 4}; s[4] = {3, 1, 2, 4}; s[5] = {5, 2, 7, 5}; s[6] = {6, 8, 1, 6}; s[7] = {7, 5, 6, 8}; s[8] = {7, 5, 6, 8}; t[a_] := Join[a, Flatten[s /@ a]]; p[0] = {1}; p[1] = t[{1}]; p[n_] := t[p[n - 1]] aa=Table[p[n], {n, 0, 3}]
%Y Cf. A105056.
%K nonn,uned
%O 0,4
%A _Roger L. Bagula_, Apr 08 2005
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