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%I #5 Nov 01 2012 07:52:56
%S 2,6,10,23,42,98,178,415,754,1758,3194,7447,13530,31546,57314,133631,
%T 242786,566070,1028458,2397911,4356618,10157714,18454930,43028767,
%U 78176338,182272782,331160282,772119895,1402817466,3270752362
%N Inverse score permutation of an Fibonacci -anti-Fibonacci zero sum game of 2 X 2 matrices.
%C 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}}
%F 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}]
%F 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]
%t 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}]
%K nonn,uned
%O 1,1
%A _Roger L. Bagula_, Jan 10 2008