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A134016
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Inverse score permutation of an Fibonacci -anti-Fibonacci zero sum game of 2 X 2 matrices.
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0
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2, 6, 10, 23, 42, 98, 178, 415, 754, 1758, 3194, 7447, 13530, 31546, 57314, 133631, 242786, 566070, 1028458, 2397911, 4356618, 10157714, 18454930, 43028767, 78176338, 182272782, 331160282, 772119895, 1402817466, 3270752362
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OFFSET
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1,1
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COMMENTS
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If the starting vector {0,1,1,0} means the matrix 2 X 2 MA gives A zero points and B one point, them the permutation: p = {{0, 0, 0, 1}, {0, 0, 1, 0}, {0, 1, 0, 0}, {1, 0, 0, 0}}; reverses that to give {1,0,0,1}. This method makes the output of matrix MA the feed for MB and the output of MB the feed of MA as: M={{MA,0},{0,MB}}
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LINKS
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FORMULA
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M = {{0, 1, 0, 0}, {1, 1, 0, 0}, {0, 0, 1, 0}, {0, 0, 3, 1}}; v[1] = {0, 1, 1, 0}; p = {{0, 0, 0, 1}, {0, 0, 1, 0}, {0, 1, 0, 0}, {1, 0, 0, 0}} v[n_] := v[n] = p.M.v[n - 1] a(n) = Sum[v[n][[i]],{i,1,4}]
Conjecture: a(n) = 4*a(n-2)+a(n-4). G.f.: x*(x^3-2*x^2-6*x-2)/(x^4+4*x^2-1). [Colin Barker, Nov 01 2012]
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MATHEMATICA
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M = {{0, 1, 0, 0}, {1, 1, 0, 0}, {0, 0, 1, 0}, {0, 0, 3, 1}}; v[1] = {0, 1, 1, 0}; p = {{0, 0, 0, 1}, {0, 0, 1, 0}, {0, 1, 0, 0}, {1, 0, 0, 0}}; v[n_] := v[n] = p.M.v[n - 1]; a = Table[Apply[Plus, v[n]], {n, 1, 50}]
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CROSSREFS
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KEYWORD
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nonn,uned
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AUTHOR
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STATUS
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approved
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