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A134265
Coefficients of the polynomials of a three level Hadamard matrix substitution set based on the game matrix set: MA={{0,1},{1,1}};MB={{1,0},{3,1}} Substitution rule is for m[n]:If[m[n - 1][[i, j]] == 0, {{0, 0}, {0, 0}}, If[m[n - 1][[i, j]] == 1, MA, MB]] Based on the Previte idea of graph substitutions as applied to matrices of graphs in the Fibonacci/ anti-Fibonacci game.
0
1, 1, -1, 1, -2, 1, 1, 2, -1, -2, 1, 1, -2, -7, 6, 20, 6, -7, -2, 1, 1, 2, -25, -10, 225, -184, -498, 500, 610, -500, -498, 184, 225, 10, -25, -2, 1
OFFSET
1,5
COMMENTS
m[n_] := Table[Table[If[m[n - 1][[i, j]] == 0, {{0, 0}, {0, 0}}, If[m[n - 1][[i, j]] == 1, ma, {{1, 0}, {3, 1}}]], {j, 1, 2^(n - 1)}], {i, 1, 2^(n - 1)}]
Michelle Previte and Sean Yang say Have you ever wanted to build your own fractal? This article will describe a procedure called a vertex replacement rule that can be used to construct fractals. We also show how one can easily compute the topological and box dimensions of the fractals resulting from vertex replacements.
LINKS
FORMULA
m[n] = If[m[n - 1][[i, j]] == 0, {{0, 0}, {0, 0}}, If[m[n - 1][[i, j]] == 1, MA, MB]] m[0] = {{1}} m[1] = {{1, 0}, {3, 1}} m[2] = {{0, 1, 0, 0}, {1, 1, 0, 0}, {1, 0, 0, 1}, {3, 1, 1, 1}} m[3] = {{0, 0, 0, 1, 0, 0, 0, 0}, {0, 0, 1, 1, 0, 0, 0, 0}, {0, 1, 0, 1, 0, 0, 0, 0}, {1, 1, 1, 1, 0, 0, 0, 0}, {0, 1, 0, 0, 0, 0, 0, 1}, {1, 1, 0, 0, 0, 0, 1, 1}, {1, 0, 0, 1, 0, 1, 0, 1}, {3, 1, 1, 1, 1, 1, 1, 1}} m[4] = {{0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0}, {0, 0, 0, 0, 0, 0, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0}, {0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0}, {0, 0, 0, 0, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0}, {0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0}, {0, 0, 1, 1, 0, 0, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0}, {0, 1, 0, 1, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0}, {1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0}, {0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1}, {0, 0, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1}, {0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 1}, {1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1}, {0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1}, {1, 1, 0, 0, 0, 0, 1, 1, 0, 0, 1, 1, 0, 0, 1, 1}, {1, 0, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1}, {3, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1}}
EXAMPLE
{1},
{1, -1},
{1, -2, 1},
{1, 2, -1, -2, 1},
{1, -2, -7, 6, 20, 6, -7, -2,1},
{1, 2, -25, -10, 225, -184, -498, 500, 610, -500, -498,184, 225, 10, -25, -2, 1}
MATHEMATICA
m[0] = {{1}} m[1] = {{1, 0}, {3, 1}} m[2] = {{0, 1, 0, 0}, {1, 1, 0, 0}, {1, 0, 0, 1}, {3, 1, 1, 1}} m[3] = {{0, 0, 0, 1, 0, 0, 0, 0}, {0, 0, 1, 1, 0, 0, 0, 0}, {0, 1, 0, 1, 0, 0, 0, 0}, {1, 1, 1, 1, 0, 0, 0, 0}, {0, 1, 0, 0, 0, 0, 0, 1}, {1, 1, 0, 0, 0, 0, 1, 1}, {1, 0, 0, 1, 0, 1, 0, 1}, {3, 1, 1, 1, 1, 1, 1, 1}} m[4] = {{0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0}, {0, 0, 0, 0, 0, 0, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0}, {0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0}, {0, 0, 0, 0, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0}, {0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0}, {0, 0, 1, 1, 0, 0, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0}, {0, 1, 0, 1, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0}, {1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0}, {0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1}, {0, 0, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1}, {0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 1}, {1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1}, {0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1}, {1, 1, 0, 0, 0, 0, 1, 1, 0, 0, 1, 1, 0, 0, 1, 1}, {1, 0, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1}, {3, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1}}; Table[CharacteristicPolynomial[m[i], x], {i, 0, 4}]; a = Join[{{1}}, Table[CoefficientList[CharacteristicPolynomial[m[i], x], x], {i, 0, 4}]]; Flatten[a] (* visualization*) Table[ListDensityPlot[m[i]], {i, 0, 4}]
CROSSREFS
Sequence in context: A103921 A115623 A279044 * A182858 A175077 A001030
KEYWORD
tabf,uned,sign
AUTHOR
Roger L. Bagula, Jan 24 2008
STATUS
approved