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A152304
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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].
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0
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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, 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, 7, 8, 3, 1, 9, 9, 3, 1
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OFFSET
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0,4
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COMMENTS
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The binet solution came from Mathematica:
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].
I used coefficients{1/2,2/3} on the golden ration like roots to get my function.
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REFERENCES
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Ivars Peterson, The Jungles of Randomness, 1998, John Wiley and Sons, Inc., page 207
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LINKS
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FORMULA
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f(n)=(11/20 - Sqrt[512]/20)^n/3 + (2/3)*(11/20 + Sqrt[512]/20)^n;
a(n)=Mod[Floor[f(n)],10].
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MATHEMATICA
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g[n_] := (11/20 - Sqrt[512]/20)^n/3 + (2/3)*(11/20 + Sqrt[512]/20)^n;
Table[Mod[Floor[FullSimplify[Expand[g[n]]]], 10], {n, 0, 76}]
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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