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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 fromula section. This results from the zeros of the minimal polynomials of sin(2*Pi/n), with coeffcients given in A181872/A181873. - Wolfdieter Lang, Oct 30 2019
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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 fromula section. This results from the zeros of the minimal polynomials of sin(2*Pi/n), with coeffcients given in A181872/A181873.- Wolfdieter Lang, Oct 30 2019
n = 20: [1, 0, -12, 0, 19, 0, -8, 0, 1]
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.
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).
n = 10: ps(10,x) = -1 + x + x^2 with the positive zero s(10) = tau - 1 (the negative solution is -tau).
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From _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 fromula section. This results from the zeros of the minimal polynomials of sin(2*Pi/n), with coeffcients given in A181872/A181873.- _Wolfdieter Lang_, Oct 30 2019: (Start)
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 fromula section. This results from the zeros of the minimal polynomials of sin(2*Pi/n), with coeffcients given in A181872/A181873.
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From Wolfdieter Lang, Oct 30 2019: (Start)
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 fromula section. This results from the zeros of the minimal polynomials of sin(2*Pi/n), with coeffcients given in A181872/A181873.
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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).
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
The table a(n, m) starts:
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