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

0,4

Comments

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];

Links

Table of n, a(n) for n=0..104.

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

Formula

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}

Mathematica

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}]

Crossrefs

Cf. A105056.

Sequence in context: A204420 A331570 A102410 * A058291 A132615 A238260

Adjacent sequences: A105120 A105121 A105122 * A105124 A105125 A105126

Keyword

nonn,uned

Author

Roger L. Bagula, Apr 08 2005

Status

approved