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A306611
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The middle coefficient in the minimal polynomial for (2*cos(Pi/15))^n.
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2
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-4, 26, -49, 246, -619, 2621, -7774, 30126, -97879, 363131, -1237504, 4497801, -15702574, 56538746, -199764994, 716265246, -2545683874, 9110943101, -32474838004, 116135818131, -414537600379, 1481979727826, -5293483738474, 18921861083121, -67610126265619, 241664630238746
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OFFSET
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1,1
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COMMENTS
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rho(15) = 2*cos(Pi/15) = 2*A019887 gives the length ratio of the smallest diagonal and the side of a regular 15-gon. The minimal polynomial of rho(15) is C(n, x) = x^4 + x^3 - 4*x^2 - 4*x + 1, with zeros x_0 = rho(15), x_1 = 2*cos(7*Pi/15), x_2 = 2*cos(11*Pi/15) and x_3 = 2*cos(13*Pi/15). See A187360, also for a W. Lang link.
The minimal polynomial of rho(1)^n, for n >= 1, considered here, is C(15,n,x) = Product_{j=0..3} (x - x_j^n) = x^4 - A_1(n)x^3 + A_2(n)*x^2 - A_3(n)*x + A_4(n). The coefficients are the elementary symmetric functions A_j(n) = sigma_j((x_0)^n, (x_1)^n, (x_2)^n, (x_3)^n), for j = 1, 2, 3, and A_4(n) = (A_4(1))^n = 1. A_1(n) = A306603(n), A_2(n) = a(n), and A_3(n) = A306610(n), for n >= 1.
Thanks to Greg Dresden for sending me a proof that C(15,n,x) has integer coefficients and does not factor over the rationals for n >= 1. (End)
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LINKS
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FORMULA
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a(n) = -4*a(n-1) + 5*a(n-2) + 25*a(n-3) + 5*a(n-4) - 4*a(n-5) - a(n-6).
G.f.: -x*(4 - 10*x - 75*x^2 - 20*x^3 + 20*x^4 + 6*x^5) / ((1 + 3*x + x^2)*(1 + x - 9*x^2 + x^3 + x^4)). - Colin Barker, Feb 28 2019
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MATHEMATICA
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Table[Coefficient[MinimalPolynomial[(2Cos[Pi/15])^n, x], x, 2], {n, 1, 40}]
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PROG
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(PARI) Vec(-x*(4 - 10*x - 75*x^2 - 20*x^3 + 20*x^4 + 6*x^5) / ((1 + 3*x + x^2)*(1 + x - 9*x^2 + x^3 + x^4)) + O(x^30)) \\ Colin Barker, Feb 28 2019
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CROSSREFS
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Cf. A306603 (which gives the negative coefficient of x^3 in minimal polynomial for (2 cos(Pi/15))^n) and A306610 (likewise for the coefficient of x).
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KEYWORD
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sign,easy
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AUTHOR
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STATUS
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approved
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