%I #25 Apr 15 2021 23:40:04
%S -4,1,4,-12,1,36,-24,1,16,-96,136,-40,1,16,-96,136,-56,1,16,-320,456,
%T -80,1,3136,-12544,14896,-7168,1484,-112,1,256,-7168,41216,-73472,
%U 53344,-17472,2576,-144,1,64,-1152,5424,-6080,2124,-168,1,256,-13312,62720,-104192,76384,-26048,3920,-208,1
%N Irregular triangle read by rows: coefficients of minimal polynomial of a certain algebraic number S2(2*k) from Q(2*cos(Pi/(2*k))) related to the regular (2*k)-gon.
%C 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).
%C 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.
%C 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.
%C 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.
%H Wolfdieter Lang, <a href="http://arxiv.org/abs/1210.1018">The field Q(2cos(pi/n)), its Galois group and length ratios in the regular n-gon</a>, arXiv:1210.1018 [math.GR], 2012-2017.
%H Seppo Mustonen, <a href="http://www.survo.fi/papers/Roots2013.pdf"> Lengths of edges and diagonals and sums of them in regular polygons as roots of algebraic equations</a>.
%H Seppo Mustonen, <a href="/A108716/a108716.pdf">Lengths of edges and diagonals and sums of them in regular polygons as roots of algebraic equations</a> [Local copy]
%F a(k,m) = [x^m] p(2*k, x), with the minimal polynomial p(2*k, x) of S2(2*k) given in the power basis in A228780.
%F p(2*k, x) is given in a comment above in terms of the S2(2*k) and its conjugates S2(2*k)^{(j-1)}, j = 2, ..., delta(2*k), where delta(2*k) = A055034(2*k).
%e The irregular triangle a(k, m) begins:
%e n k / m 0 1 2 3 4 5 6 7 8
%e 2 1: -4 1
%e 4 2: 4 -12 1
%e 6 3: 36 -24 1
%e 8 4: 16 -96 136 -40 1
%e 10 5: 16 -96 136 -56 1
%e 12 6: 16 -320 456 -80 1
%e 14 7: 3136 -12544 14896 -7168 1484 -112 1
%e 16 8: 256 -7168 41216 -73472 53344 -17472 2576 -144 1
%e ...
%e n = 18, k = 9: 64, -1152, 5424, -6080, 2124, -168, 1;
%e n = 20, k = 10: 256, -13312, 62720, -104192, 76384, -26048, 3920, -208, 1.
%e n = 6, k = 3: p(6), x) = (x - S2(6))*(x - S2(6)^{(1)}),
%e with S2(6) = 12 + 6*rho(6), where rho(6) = sqrt(3). C(6, x) = x^2 - 3 = (x - rho(6))*(x - (-rho(6))), hence rho(6)^{(1)} - -rho(6) and S2(6)^{(1)} = 12 - 6*rho(6). Thus p(6, x) = 144 - 36*rho(6)^2 - 24*x + x^2, reduced with C(6, rho(6)) = 0, i.e., rho(6)^2 = 3; this becomes finally 36 - 24*x + x^2.
%Y Cf. A055034, A187360, A228780, A228781 (odd case).
%K sign,tabf
%O 1,1
%A _Wolfdieter Lang_, Oct 01 2013
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