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A114723
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G.f.: x*(1 - 2*x^2)/(1 - x - 3*x^2 - 3*x^3 - x^4).
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0
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0, 1, 1, 2, 8, 18, 49, 129, 338, 890, 2340, 6153, 16181, 42550, 111892, 294238, 773745, 2034685, 5350526, 14070054, 36999432, 97295857, 255854841, 672810762, 1769262288, 4652554954
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OFFSET
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0,4
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COMMENTS
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The first three of the sequence of polynomials: x^n-(x+1)^n are Pisots, suggesting the name Pascal-Pisots as the (x+1)^n polynomials give the Pascal triangular sequence. r = Abs[Table[x /. NSolve[Det[M - IdentityMatrix[4]*x] == 0, x][[n]], {n, 1, 4}]] gives:{0.5497, 0.831739, 0.831739, 2.62966}
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LINKS
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FORMULA
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M = {{0, 1, 0, 0}, {0, 0, 1, 0}, {0, 0, 0, 1}, {1, 3, 3, 1}}; w[0] = {0, 1, 1, 2}; w[n_] := w[n] = M.w[n - 1] a(n) = w[n][[1]].
a(n) = a(n-1)+3*a(n-2)+3*a(n-3)+a(n-4). G.f.: x*(2*x^2 -1)/(x^4 +3*x^3 +3*x^2 +x -1). [Colin Barker, Oct 11 2012]
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MATHEMATICA
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M = {{0, 1, 0, 0}, {0, 0, 1, 0}, {0, 0, 0, 1}, {1, 3, 3, 1}}; w[0] = {0, 1, 1, 2}; w[n_] := w[n] = M.w[n - 1] a = Flatten[Table[w[n][[1]], {n, 0, 25}]]
LinearRecurrence[{1, 3, 3, 1}, {0, 1, 1, 2}, 30] (* Harvey P. Dale, Jan 13 2015 *)
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PROG
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CROSSREFS
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KEYWORD
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nonn,easy
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AUTHOR
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STATUS
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approved
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