cp's OEIS Frontend

Also degree of minimal polynomial of function F(n)=(gamma(1/n)*gamma((n-1)/n))/Pi over the rationals. Roots of minimal polynomials of F(n) belonging to algebraic extension of sin(n/Pi) and vice versa (e.g. gamma(1/11)*gamma(10/11)/Pi = 20*sin(Pi/11) - 112*sin(Pi/11)^3 + 256*sin(Pi/11)^5 - 256*sin(Pi/11)^7 + (1024*sin(Pi/11)^9)/11). - Artur Jasinski, Oct 17 2011

The algebraic numbers sin(Pi/(2*l)) are given in A228783 in the power basis of the number field Q(2*cos(Pi/(2*l))) if n is even and of Q(2*cos(Pi/l)) if l is odd. In A228785, sin(Pi/(2*l+1)) is given in the power basis of Q(2*cos(Pi/(2*(2*l+1)))) (only odd powers appear). The minimal polynomials for 2*sin(Pi/n), n>=1, are given in A228786. - Wolfdieter Lang, Oct 10 2013

In the regular (2*l+1)-gon, l >= 1, inscribed in a circle of radius R the length ratio side/R is s(2*l+1) = 2*sin(Pi/(2*l+1)). This can be written as a polynomial in the length ratio (smallest diagonal)/side in the (2*(2*l+1))-gon given by rho(2*(2*l+1)) = 2*cos(Pi/(2*(2*l+1))). This leads, in a first step, to the signed triangle A111125. Because of the minimal polynomial of the algebraic number rho(2*(2*l+1)) of degree delta(2*(2*l+1)) = A055034(2*(2*l+1)), called C(2*(2*l+1),x) (with coefficients given in A187360), one can eliminate all powers rho(2*(2*l+1))^k with k >= delta(2*(2*l+1)) by using C(2*(2*l+1),rho(2*(2*l+1))) = 0. This leads to the present table expressing s(2*(l+1)) in terms of odd powers of rho(2*(2*l+1)) with maximal exponent delta(2*(2*l+1))-1.

This table gives the coefficients of s(2*l+1), related to the (2*l+1)-gon, in the power basis of the algebraic number field Q(rho(2*(2*l+1))) of degree delta(2*(2*l+1)), related to rho from the (2*(2*l+1))-gon, provided one inserts zeros for the even powers, starting each row with a zero and filling zeros at the end in order to obtain the row length delta(2*(2*l+1)). Note that some trailing zeros in the present table (e.g., row l = 10) have been given such that the row length for the s(2*l+1) coefficients in the power basis Q(rho(2*(2*l+1))) becomes just twice the one of this table.

Thanks go to Seppo Mustonen for telling me about his findings regarding the square of the sum of all length in the regular n-gon, which led me to consider this entry (even though for odd n this is not needed because only s(2*l+1)^2 = 4 - rho(2*l+1)^2 enters).

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.

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.

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).

Thanks go to Seppo Mustonen, who asked a question about the square of the sum of all lengths in the regular n-gon, which led to this computation of s(n) and its minimal polynomial.

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.

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

The row length of this table is delta(2*l) = A055034(2*l), l >= 1.

sqLhat(2*l) is the square of the sum of the lengths ratios of all lines/R (also called chords/R) divided by (2*l)^2 in a regular (2*l)-gon, l >= 1, inscribed in a circle of radius R. One may put R = 1 length unit.

sqLhat(2*l) is an algebraic number of degree delta(2*l) = A055034(2*l) and lives in the algebraic number field Q(rho(2*l)), with rho(n):= 2*cos(Pi/n). The power basis of Q(rho(n)) is <1, rho(n), rho(n)^2, ..., rho(n)^(delta(n)-1)>. This table gives the coefficients of sqLhat(2*l) in that basis: sqLhat(2*l) := Sum_{m=0..delta(2*l)-1} a(l,m)*rho(2*l)^m, l >= 1. See also A187360, and the W. Lang link below.

The formula to begin with is sqLhat(2*l) = (s(n)*Sum_{k=0..l-1} S(k,rho(n)))^2 with n=2*l, s(n) = 2*sin(Pi/n) (length ratio of the side to the radius R) and the Chebyshev S-polynomials (for the coefficients see A049310). sqLhat(2*l) = S2(2*l) + 1 - 2*s(2*l)*Sum_{k=0..l-1} S(k,rho(2*l)), with the square of the sums of the distinct length ratios S2(2*l) with power basis coefficients given in A228780(2*l). The power basis coefficients of s(2*l) are for even l given in A228783. For odd l = 2*L+1 one has to replace rho(l) by rho(2*l)^2 - 2 in the result for s(4*L+2) from A228783, in order to work in Q(rho(2*l)). One always computes modulo C(2*l,rho(2*l)) (which is zero) with the minimal polynomial C(n,x) of degree delta(n) for rho(n) known from A187360.

Thanks go to Seppo Mustonen, who asked me to look into this matter. The author thanks him for giving the below given link to his work about the square of the sum of all lengths in an n-gon, called there L(n)^2. Here n is even (n=2*l) and sqLhat(2*l) = (L(n)^2)/n^2. The odd n case is obtained from A228780 as L(2*l+1)^2 = n^2*S2(2*l+1) (observing that all distinct line lengths come precisely n times in the regular n-gon if n is odd).