A105056 -id:A105056 - cp's OEIS frontend

A105123 Substitution Group for a von Koch curve.

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, 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, 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

Offset: 0

Roger L. Bagula, Apr 08 2005

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];

F. M. Dekking, Recurrent sets, Advances in Mathematics, vol. 44, no. 1 (1982), 78-104; page 96, section 4.11.

Cf. A105056.

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}]

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}

cp's OEIS Frontend

A105123 Substitution Group for a von Koch curve.

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