cp's OEIS Frontend

The length of row n of this irregular array is the degree of the algebraic number rho(n):= 2*cos(Pi/n), given in A055034(n). See a Jul 19 2011 comment there.

The regular n-gon, inscribed in a circle of radius defining the length unit 1, has distinct line segments (chords) (V_0, V_j), j=1, ... , floor(n/2), with the n-gon vertices V_j, j=0, ... , n-1 distributed on the circle in the counterclockwise sense. The corresponding length ratios are denoted by L(n,j)/radius. The side length is s(n) = (V_0, V_1) = 2*sin(Pi/n), and for n >= 4 the first (the smallest) diagonal has length s(n)*rho(n), with rho(n) of degree delta(n) given above. s(2) = 2 is the ratio of the diameter of the circle. rho(2) = 0, but we use here rho(2)^0 = 1.

For n = 3: rho(3) = 1, s(3)^2 = 3. The algebraic number field Q(rho(n)) is the subject of the W. Lang link given below.

S2(n) := (sum(L(n,j)/radius, j=1, ... ,floor(n/2))^2 is seen below to be a number in the field Q(rho(n)) of degree delta(n), namely S2(n) = sum(a(n,k)*rho(n)^k, k=0..(delta(n)-1)). From the definition one has S2(n) = (s(n)*sum(S(j-1,rho(n)), j=1..floor(n/2)))^2, with the Chebyshev S-polynomials (see A049310). Due to s(n) = s(2*n)*rho(2*n), rho(2*n) = sqrt(2 + rho(n)) and an S-identity this becomes S2(n) = (s(2*n)*S(floor(n/2)-1, rho(2*n))*S(floor(n/2), rho(2*n)))^2. This can also be written as S2(n) = 4*(1 - T(2*floor(n/2), rho(2*n)/2))*(1 - T(2*(floor(n/2)+1), rho(2*n)/2))/(4-rho(2*n)^2), with Chebyshev's T-polynomials (see A053120). S2(n), written as a function of rho(n), has to be computed modulo the minimal polynomial C(n,rho(n)) of degree delta(n). These minimal polynomials are treated in A187360 (see the link to a Galois paper there, with its Table 2 and Section 3). The result is then the above given representation of S2(n) in the power basis of Q(rho(n)).

This computation was inspired by an email exchange with Seppo Mustonen. The author thanks him for sending the paper given as a link below. In this connection one should consider the even and odd n cases separately in order to find the square of the total length segments/radius in the regular n-gon, noticing that in the odd n case each distinct chord (side or diagonal) appears 2*(n/2) = n times, whereas in the even n case the longest diagonal of length 2 (in units of the radius) appears only n/2 times and the other chords appear n times.

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).