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%I #7 Mar 13 2024 19:25:31
%S -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,
%T -24,-24,86,-54,-116,148,72,-120,-16,32,16,-24,24,-6,-150,216,87,-378,
%U 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
%N 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).
%C The name contains an unmatched parenthesis. - Editors, Mar 13 2024
%C 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.
%F 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;
%e {-8, -3, 8},
%e {4, -4, -10, 4, 4},
%e {-8, 4,24, -9, -24, 4, 8},
%e {16, 0, -64, -2, 95, 2, -64, 0, 16},
%e {16, -24, -24, 86, -54, -116, 148, 72, -120, -16, 32},
%e {16, -24, 24, -6, -150, 216, 87, -378, 160, 264, -208, -64, 64}
%t 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]
%K tabf,uned,sign
%O 1,1
%A _Roger L. Bagula_, Sep 08 2007
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