OFFSET
2,1
COMMENTS
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.
LINKS
Wolfdieter Lang, The field Q(2cos(pi/n)), its Galois group and length ratios in the regular n-gon, arXiv:1210.1018 [math.GR], 2012-2017.
Seppo Mustonen, Lengths of edges and diagonals and sums of them in regular polygons as roots of algebraic equations.
Seppo Mustonen, Lengths of edges and diagonals and sums of them in regular polygons as roots of algebraic equations [Local copy]
FORMULA
a(n,k) = [rho^k] (S2(n) modulo C(n,rho(n)), with S2(n) the square of the sum of the distinct length/radius ratios in the regular n-gon, with rho(n) = 2*cos(Pi/n) given above in a comment, and C(n,x) the minimal polynomial of rho(n) given in A187360 (see Table 2 and section 3 of the paper given in the W. Lang link below).
EXAMPLE
The table a(n,k) begins:
n\k 0 1 2 3 4 5 ...
2: 4
3: 3
4: 6 4
5: 3 4
6: 12 6
7: -1 4 4
8: -2 -4 6 4
9: 3 8 4
10: -16 -8 12 6
11: 3 -4 -8 4 4
12: 0 4 10 4
13: 3 8 -8 -12 4 4
14: 28 14 -40 -20 12 6
15: -1 8 2 4
...
n=5: S2(5) = (4-rho(5)^2)*(sum(S(j-1,rho(5)), j=1..2))^2 =
4 + 8*rho(5) + 3*rho(5)^2 - 2*rho(5)^3 - rho(5)^4, reduced with C(5,x) =x^2 -x -1, with x = rho(5), using C(5,rho(5)=0, to eliminate all powers of rho(5) starting with power 2.
This leads to S2(5) = 3*1 + 4*rho(5). rho(5) = phi, the golden section.
The exact or approximate real values for S2(n) are, for n = 2, ..., 15: 4, 3, 11.65685426, 9.472135960, 22.39230484, 19.19566936, 36.32882142, 32.16343753, 53.49096128, 48.37415020, 73.88698896, 67.82742928, 97.52047276, 90.52313112.
CROSSREFS
KEYWORD
sign,tabf
AUTHOR
Wolfdieter Lang, Oct 01 2013
STATUS
approved