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A105123
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Substitution Group for a von Koch curve.
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0
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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
(list; graph; refs; listen; history; internal format)
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OFFSET
| 0,4
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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];
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REFERENCES
| F. M. Dekking, "Recurrent Sets", Advances in Mathematics, vol. 44, no.1, April 1982, page 96, section 4.11
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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}
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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}]
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CROSSREFS
| Cf. A105056.
Sequence in context: A119923 A204420 A102410 * A058291 A132615 A021617
Adjacent sequences: A105120 A105121 A105122 * A105124 A105125 A105126
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KEYWORD
| nonn,uned
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AUTHOR
| Roger Bagula (rlbagulatftn(AT)yahoo.com), Apr 08 2005
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