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A131641
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Triangular sequence obtained from polynomials derived from: x^2-x/(2*b(n)-1=0 where b(n)->{n=3, theta0=1.32472},{n=4, theta1=1.38028} as a polynomial recursion: y(n) = 16 - 8 x - 48 x^2 + 18 x^3 + 48 x^4 - 8 x^5 - 16 x^6 - x* y(n - 1] + 2 x^2 *y(n - 1) + x^2 *y(n - 2).
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0
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-8, -3, 8, 4, -4, -10, 4, 4, -8, 4, 24, -9, -24, 4, 8, 16, 0, -64, -2, 95, 2, -64, 0, 16, 16, -24, -24, 86, -54, -116, 148, 72, -120, -16, 32, 16, -24, 24, -6, -150, 216, 87, -378, 160, 264, -208, -64, 64, 16, -24, 24, -78, 78, 216, -586, 229, 700, -844, -64, 720, -320, -192, 128, 16, -24, 24, -78, 198, -248, -226, 1234
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OFFSET
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1,1
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COMMENTS
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The name contains an unmatched parenthesis. - Editors, Mar 13 2024
FullSimplify[(1 + Sqrt[1 + 16b^2]/(4 b) /. b -> theta]gives the polynomial: -8 + 4 x + 24 x^2 - 9 x^3 - 24 x^4 + 4 x^5 + 8 x^6 FullSimplify[(1 + Sqrt[1 + 16b^2]/(4 b) /. b -> theta1]gives the polynomial: 16 - 64 x^2 - 2 x^3 + 95 x^4 + 2 x^5 - 64 x^6 + 16 x^8 The polynomial recursion back solved from these two.
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LINKS
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FORMULA
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Polynomial recursion in x: y(n) = 16 - 8 x - 48 x^2 + 18 x^3 + 48 x^4 - 8 x^5 - 16 x^6 - x* y(n - 1] + 2 x^2 *y(n - 1) + x^2 *y(n - 2); y(1) = -8 - 3 x + 8 x^2; y(2) = 4 - 4 x - 10 x^2 + 4 x^3 + 4 x^4;
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EXAMPLE
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{-8, -3, 8},
{4, -4, -10, 4, 4},
{-8, 4,24, -9, -24, 4, 8},
{16, 0, -64, -2, 95, 2, -64, 0, 16},
{16, -24, -24, 86, -54, -116, 148, 72, -120, -16, 32},
{16, -24, 24, -6, -150, 216, 87, -378, 160, 264, -208, -64, 64}
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MATHEMATICA
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y[1] = -8 - 3 x + 8 x^2; y[2] = 4 - 4 x - 10 x^2 + 4 x^3 + 4 x^4; y[3] = -8 + 4 x + 24 x^2 - 9 x^3 - 24 x^4 + 4 x^5 + 8 x^6 y[4] = 16 - 64 x^2 - 2 x^3 + 95 x^4 + 2 x^5 - 64 x^6 + 16 x^8 y[n_] := y[n] = 16 - 8 x - 48 x^2 + 18 x^3 + 48 x^4 - 8 x^5 - 16 x^6 - x* y[n - 1] + 2 x^2 *y[n - 1] + x^2 *y[n - 2] a0 = Table[CoefficientList[y[n], x], {n, 1, 10}]; Flatten[a0]
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CROSSREFS
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KEYWORD
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tabf,uned,sign
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AUTHOR
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STATUS
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approved
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