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A122607
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Expansion of x*(8*x^5 + 5*x^4 - x^3 - 5*x^2 - 1)/(x^6 + 3*x^5 + 6*x^4 + 4*x^3 - 5*x^2 + x - 1).
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0
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1, 1, 1, 1, 1, 1, 10, 19, -17, -62, 163, 550, -548, -3050, 2665, 19450, -7550, -113534, 8308, 667423, 187462, -3800825, -2366747, 21303154, 21068938, -116488961, -162036530, 621601885, 1153785034, -3216794309, -7799929064, 16026195376, 50784142789, -75764359214, -320876463932
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OFFSET
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1,7
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COMMENTS
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Obtained as the top element of the vector resulting from multiplying the n-th power of the 6 X 6 matrix [[0, 1, 0, 0, 0, 0], [0, 0, 1, 0, 0, 0], [0, 0, 0, 1, 0, 0], [0, 0, 0, 0, 1, 0], [0, 0, 0, 0, 0, 1], [1, 3, 6, 4, -5, 1]] with the column vector which contains only 1's.
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LINKS
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FORMULA
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G.f.: x*(8*x^5+5*x^4-x^3-5*x^2-1)/(x^6+3*x^5+6*x^4+4*x^3-5*x^2+x-1). - Colin Barker, Nov 08 2012
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MATHEMATICA
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M = {{0, 1, 0, 0, 0, 0}, {0, 0, 1, 0, 0, 0}, {0, 0, 0, 1, 0, 0}, {0, 0, 0, 0, 1, 0}, {0, 0, 0, 0, 0, 1}, {1, 3, 6, 4, -5, 1}}; v[1] = {1, 1, 1, 1, 1, 1}; v[n_] := v[n] = M.v[n - 1] a = Table[Floor[v[n][[1]]], {n, 1, 50}]
LinearRecurrence[{1, -5, 4, 6, 3, 1}, {1, 1, 1, 1, 1, 1}, 40] (* Harvey P. Dale, Feb 17 2024 *)
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CROSSREFS
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KEYWORD
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sign,easy
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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