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The row length sequence of this table is delta(2*k+1), with the degree delta(n) = A055034(n) of the algebraic number rho(n):= 2*cos(Pi/n), k >= 1.

The numbers S2(n) have been given in A228780 in the power basis of the degree delta(n) number field Q(rho(n)), with rho(n):= 2*cos(Pi/n), n >= 2. Here the odd n case, n = 2*k + 1 is considered. S2(n) is the square of the sum of the distinct length ratios side/radius or diagonal/radius with the radius of the circle in which a regular n-gon is inscribed. For two formulas for S2(n) in terms of powers of rho(n) see the comment section of A228780.

The minimal (monic) polynomial of S2(2*k+1) has degree delta(2*k+1) and is given by

p(2*k+1,x) = Product_{j=1..delta(2*k+1)} (x - S2(2*k+1)^{(j-1)} (mod C(2*k+1,delta(n))) = sum(a(k, m)*x^m, m = 0..delta(2*k+1)), where S2(2*k+1)^{(0)} = S2(2*k+1) and S2(2*k+1)^{(j-1)} is the (j-1)-th conjugate of S2(2*k+1). The conjugate of a number alpha(n) = Sum_{j=0..(delta(n)-1)} b(n, j)*rho(n)^j in Q(rho(n)) is obtained from the conjugates of rho(n), given in turn by the zeros x(n, j) of the minimal polynomial C(n, x) (see A187360 and the link to the W. Lang Galois paper, tables 2 and 3) as rho(n)^{(j-1)} = x(n, j), j = 1..delta(n), with rho(n)^{(0)} = rho(n).

The motivation to look into this problem originated from emails by Seppo Mustonen, who found experimentally polynomials which had as one zero the square of the total length/radius of all chords (sides and diagonals) in the regular n-gon. See his paper given as a link below. The author thanks Seppo Mustonen for sending his paper.

If the minimal polynomial of the algebraic number S2(n) in the n-gon with n = 2*k+1 is p(n, x) then the minimal polynomial of the square of the sum of the length of all n sides and n*(n-3)/2 diagonals is P(n, x) = n^(2*delta(n))*p(n, x/n^2).

The row length sequence of this table is delta(2*k), k >= 1, with the degree delta(n) = A055034(n) of the algebraic number rho(n):=2*cos(Pi/n).

The algebraic numbers S2(n) have been given in A228780 in the power basis of the degree delta(n) number field Q(rho(n)), with rho(n):=2*cos(Pi/n), n >= 2. Here the even case, n = 2*k, is considered. S2(n) is the square of the sum of the distinct length ratios side/radius and diagonal/radius with the radius of the circle in which a regular n-gon is inscribed. For two formulas for S2(n) in terms of powers of rho(n) see the comment section of A228780.

The minimal (monic) polynomial of S2(2*k) has degree delta(2*k) and is given by p(2*k,x) = Product_{j=1..delta(2*k)} (x - S2(2*k)^{(j-1)}) (mod C(2*k, rho(2*k))) = Sum_{m=0..delta(2*k)} a(k, m)*x^m, where S2(2*k)^{(0)} = S2(2*k) and S2(2*k)^{(j-1)} is the (j-1)-th conjugate of S2(2*L). For the conjugate of an algebraic number in Q(rho(n)) see a comment on A228781.

The motivation to look into this problem originated from emails by Seppo Mustonen, who found experimentally polynomials which had as one zero the square of the total length/radius of all chords (sides and diagonals) in the regular n-gon. See his paper given as a link below. The author thanks Seppo Mustonen for sending his paper.

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).

The length of row no. l of this table is delta(2*l) + 1 = A055034(2*l) + 1.

sqLhat(2*l), l >= 1, from A230072, an algebraic number (in fact integer) of degree delta(2*l) over the rationals, gives the square of the sum of the length ratios of all lines/R (also called chords/R) divided by (2*l)^2 in a regular (2*l)-gon inscribed in a circle of radius R. This number lives in the algebraic number field Q(rho(2*l)), with rho(2*l) = 2*cos(Pi/(2*l)) (see A187360 and the W. Lang link below for this number field).

The minimal polynomial for sqLhat(2*l) = Sum_{m=0..delta(2*l)} A230072(l,m)* rho(2*l)^m, called here psqLhat(l, x), is computed from the conjugates rho(2*l)^{(j)}, j = 0, ..., delta(2*l)-1, with rho(2*l)^{(0)} = rho(2*l), by calculating the conjugates sqLhat(2*l)^{(j)}, j = 0, ..., delta(2*l)-1, which are polynomials in rho(2*l) =: z, with the usual rules for conjugation. All results have to be taken modulo the minimal polynomial C(2*l, z) of rho(2*l) (see A187360 table 2 and section 3 for C(n,x)), in order to obtain finally elements written in the power basis of the field Q(rho(2*l)). The conjugates rho(2*l)^{(j)} are just the delta(2*l) roots of C(2*l, z). Therefore, psqLhat(l, x) = (Product_{j=0.. delta(2*l)-1} (x - substitute(rho(2*l) = z, sqLhat(2*l)^{(j)}))) mod C(2*l, z).

Thanks go to Seppo Mustonen, who asked me to look into this matter. I thank him for sending 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 = (2*l+1)^2*S2(2*l+1) (observing that all distinct line lengths occur precisely n times in the regular n-gon if n is odd). His polynomials given in his eq. (6) (here for the n even case) are in general not monic, and not irreducible. Instead one should consider the minimal (monic, irreducible and integer) polynomials PsqL(l, x) := (2*l)^(2*delta(2*l))* psqLhat(l, x/(2*l)^2), l >= 1 (for n = 2*l).

Mustonen's polynomials from his eq. (6) for the even n case coincide with PsqL(l, x) precisely for l = 2^k, k>=1, and for l = 1 (k=0) one has to take the negative. In all other cases the degrees do not fit (Mustonen's polynomials become reducible over the integers). His conjecture for the coefficients can then be rewritten as a conjecture for the present polynomials psqLhat(2^k, x), k >= 0 (see the formula section).

S. Mustonen also conjectured about the other zeros of his polynomials.