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%I #23 Aug 05 2021 21:45:12
%S 0,1,-2,1,-3,0,1,-2,0,1,5,0,-5,0,1,-1,1,-7,0,14,0,-7,0,1,2,0,-4,0,1,
%T -3,0,9,0,-6,0,1,-1,1,1,-11,0,55,0,-77,0,44,0,-11,0,1,1,0,-4,0,1,13,0,
%U -91,0,182,0,-156,0,65,0,-13,0,1,1,-2,-1,1,1,0,-8,0,14,0,-7,0,1,2,0,-16,0,20,0,-8,0,1,17,0,-204,0,714,0,-1122,0,935,0,-442,0,119,0,-17,0,1,1,-3,0,1
%N Table of coefficients of the minimal polynomials of 2*sin(Pi/n), n >= 1.
%C s(n) := 2*sin(Pi/n) is, for n >= 2, the length ratio side/R of the regular n-gon inscribed in a circle of radius R. This algebraic number s(n), n>=1, has the degree gamma(n) := A055035(n), and the row length of this table is gamma(n) + 1.
%C s(n) has been given in the power basis of the relevant algebraic number field in A228783 for even n (bisected into n == 0 (mod 4) and n == 2 (mod 4)), and in A228785 for odd n.
%C For the computation of the minimal polynomials ps(n,x), using the coefficients of s(n) in the relevant number field, and the conjugates of the corresponding algebraic numbers rho (giving the length ratios (smallest diagonal)/side in the relevant regular polygons see a comment on A228781. Note that the product of the gamma(n) linear factors (x - conjugates) has to be computed modulo the minimal polynomial of the relevant rho(k) = 2*cos(Pi/k) (called C(k,x=rho(k)) in A187360).
%C Thanks go to Seppo Mustonen who asked a question about the square of the sum of all length in the regular n-gon which led to this computation of s(n) and its minimal polynomial.
%C It would be interesting to find out which length ratios in the regular n-gon give the other positive zeros of the minimal polynomial ps(n,x). See some examples below.
%C The zeros of the row polynomials ps(n,x) are 2*cos(2*Pi*k/c(2*n))) for gcd(k, c(2*n)) = 1, where c(n) = A178182(n), and k from {0, ..., floor(c(2*n)/2)}, for n >= 1. The number of these solutions is gamma(n) = A055035(n). See the formula section. This results from the zeros of the minimal polynomials of sin(2*Pi/n), with coefficients given in A181872/A181873. - _Wolfdieter Lang_, Oct 30 2019
%F a(n, m) = [x^m](minimal polynomial ps(n, x) of 2*sin(Pi/n) over the rationals), n >= 1, m = 0, ..., gamma(n), with gamma(n) = A055035(n).
%F ps(n,x) = Product{k=0.. floor(c(2*n)/n) and gcd(k, c(2*n)) = 1} (x - 2*cos(2*Pi*k/c(2*n)), with c(2*n) = A178182(2*n), for n >= 1. There are gamma(n) = A055035(n) zeros. - _Wolfdieter Lang_, Oct 30 2019
%e The table a(n, m) starts:
%e n\m 0 1 2 3 4 5 6 7 8 9 10 12 13 14 15 16 17
%e 1: 0 1
%e 2: -2 1
%e 3: -3 0 1
%e 4: -2 0 1
%e 5: 5 0 -5 0 1
%e 6: -1 1
%e 7: -7 0 14 0 -7 0 1
%e 8: 2 0 -4 0 1
%e 9: -3 0 9 0 -6 0 1
%e 10: -1 1 1
%e 11: -11 0 55 0 -77 0 44 0 -11 0 1
%e 12: 1 0 -4 0 1
%e 13: 13 0 -91 0 182 0 -156 0 65 0 -13 0 1
%e 14: 1 -2 -1 1
%e 15: 1 0 -8 0 14 0 -7 0 1
%e 16: 2 0 -16 0 20 0 -8 0 1
%e 17: 17 0 -204 0 714 0 -1122 0 935 0 -442 0 119 0 -17 0 1
%e 18: 1 -3 0 1
%e ...
%e n = 19: [-19, 0, 285, 0, -1254, 0, 2508, 0, -2717, 0, 1729, 0, -665, 0, 152, 0, -19, 0, 1],
%e n = 20: [1, 0, -12, 0, 19, 0, -8, 0, 1]
%e n = 5: ps(5,x) = 5 -5*x^2 +1*x^4, with the zeros s(5) = sqrt(3 - tau), sqrt(2 + tau) = tau*s(5) and their negative values, where tau =rho(5) is the golden section. tau*s(5) is the length ratio diagonal/radius in the pentagon.
%e n = 7: ps(7,x) = -7 + 14*x^2 -7*x^4 + 1*x^6, with the positive zeros s(7) (side/R) about 0.868, s(7)*rho(7) (smallest diagonal/R) about 1.564, and s(7)*(rho(7)^2-1) (longer diagonal/R) about 1.950 in the heptagon inscribed in a circle with radius R.
%e n = 8: ps(8,x) = 2 -4*x^2 + x^4, with the positive zeros s(8) = sqrt(2-sqrt(2)) and rho(8) = sqrt(2+sqrt(2)) (smallest diagonal/side).
%e n = 10: ps(10,x) = -1 + x + x^2 with the positive zero s(10) = tau - 1 (the negative solution is -tau).
%Y Cf. A187360, A055035, A228781, A228783, A228785.
%K sign,tabf
%O 1,3
%A _Wolfdieter Lang_, Oct 07 2013
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