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%I #2 Mar 30 2012 17:34:28
%S 1,0,1,3,5,8,5,5,2,1,0,2,0,2,2,7,9,2,8,7,5,6,7,6,3,6,8,2,2,7,9,2,4,5,
%T 5,9,2,3,7,5,4,2,5,0,2,9,5,0,0,7,6,8,7,8,2,5,8,5,0,3,5,2,7,4,3,2,6,9,
%U 7,8,3,1,9,9,3,1
%N Marsaglia-Zaman type binet solution rationalized: f(n)=(11/20 - Sqrt[512]/20)^n/3 + (2/3)*(11/20 + Sqrt[512]/20)^n; a(n)=Mod[Floor[f(n)],10].
%C The binet solution came from Mathematica:
%C f[n_Integer] = Module[{a}, a[n] /. RSolve[{a[n] == a[n - 1] + a[n - 2] + a[n - 1]/10,a[0] == 1, a[1] == 1}, a[n], n][[1]] // FullSimplify].
%C I used coefficients{1/2,2/3} on the golden ration like roots to get my function.
%D Ivars Peterson, The Jungles of Randomness, 1998, John Wiley and Sons, Inc., page 207
%F f(n)=(11/20 - Sqrt[512]/20)^n/3 + (2/3)*(11/20 + Sqrt[512]/20)^n;
%F a(n)=Mod[Floor[f(n)],10].
%t g[n_] := (11/20 - Sqrt[512]/20)^n/3 + (2/3)*(11/20 + Sqrt[512]/20)^n;
%t Table[Mod[Floor[FullSimplify[Expand[g[n]]]], 10], {n, 0, 76}]
%K nonn
%O 0,4
%A _Roger L. Bagula_, Dec 02 2008
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