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%I A228783 #11 Oct 19 2013 02:03:42 %S A228783 2,0,1,1,0,-3,0,1,-1,1,0,4,0,-1,-1,-1,1,0,-7,0,14,0,-7,0,1,2,1,-1,0,8, %T A228783 0,-18,0,8,0,-1,1,2,-3,-1,1,0,-8,0,6,0,-1,0,0,-1,3,3,-4,-1,1,0,12,0, %U A228783 -67,0,96,0,-52,0,12,0,-1,-2,3,1,-1,0,-15,0,140,0,-378,0,450,0,-275,0,90,0,-15,0,1 %N A228783 Table of coefficients of the algebraic number s(2*l) = 2*sin(Pi/2*l) as a polynomial in powers of rho(2*l) = 2*cos(Pi/(2*l)) if l is even and of rho(l) = 2*cos(Pi/l) if l is odd (reduced version). %C A228783 In the regular (2*l)-gon inscribed in a circle of radius R the length ratio side/R is s(2*l) = 2*sin(Pi/(2*l)). This can be written as a polynomial in the length ratio (smallest diagonal)/side given by rho(2*l) = 2*cos(Pi/(2*l)). (For the 2-gon there is no such diagonal and rho(2) = 0). This leads, in a first step, to the triangle A127672 (see the Oct 05 2013 comment there referring also to the bisections signed A111125 and A127677). Because the minimal polynomial of the algebraic number rho(2*l) of degree delta(2*l) = A055034(2*l), called C(2*l,x) (with coefficients given in A187360) one can eliminate all powers rho(2*l)^k with k >= delta(2*l) by using C(2*l,rho(2*l)) = 0. Furthermore, because for odd l only even powers of rho(2*l) appear, but rho(2*l)^2 = 2 + rho(l), one will obtain a reduced table for s(2*l) with powers rho(2*l)^(2*k+1), k= 0, ..., (delta(2*l)-2)/2 if l is even, and with powers rho(l)^m, m=0, ... , delta(l)-1 if l is odd. %C A228783 This leads to a reduction of the triangle A127672, when applied for the s(2*l) computation, giving the present table with row length delta(4*L) = A055034(4*L) = phi(8*L)/2 if l =2*L, if L >= 1, and phi(2*L+1)/2 = A055035(4*L+2), if l = 2*L + 1, L >= 1, where phi(n) = A000010(n) (Euler totient). %C A228783 This table gives the coefficients of s(2*l) in the power basis of the algebraic number field Q(rho(2*l)) of degree delta(2*l) = A055034(2*l) if l is even, and in Q(rho(l)) of degree delta(2*l)/2 if l is odd. s(2) and s(6) are rational integers of degree 1. %C A228783 Thanks go to Seppo Mustonen whose question about the square of the sum of all length in a regular n-gon, led me to this computation. %C A228783 If l = 2*L+1, L >= 0, that is n == 2 (mod 4), one can obtain s(2*l) more directly in powers of rho(l) by s(2*l) = R(l-1, rho(l)) (mod C(l,rho(l))), with the monic (except for l=1) Chebyshev T-polynomials, called R, in A127672, and the C polynomials from A187360. - _Wolfdieter Lang_, Oct 10 2013 %F A228783 a(2*L,m) = [x^m](s(4*L,x)(mod C(4*L,x))), with s(4*L,x) = sum((-1)^(L-1-s)*A111125(L-1,s)*x^(2*s+1),s=0..L-1), L >= 1, m =0, ..., delta(4*L)-1, and %F A228783 a(2*L+1,m) = [x^m](s(4*L+2,x)(mod C(2*L+1,x))), with s(4*L+2,x) = sum(A127677(L,s)*(2+x)^(L-s)),s=0..L) (with s(2,x) = 2 for L = 0), L >= 0, m = 0, ..., delta(4*L+2)/2, with delta(n) = A055034(2*l). %e A228783 The table a(l,m), with n = 2*l, begins: %e A228783 n, l \m 0 1 2 3 4 5 6 7 8 9 10 11 ... %e A228783 2 1: 2 %e A228783 4 2: 0 1 %e A228783 6 3: 1 %e A228783 8 4: 0 -3 0 1 %e A228783 10 5: -1 1 %e A228783 12 6: 0 4 0 -1 %e A228783 14 7: -1 -1 1 %e A228783 16 8: 0 -7 0 14 0 -7 0 1 %e A228783 18 9: 2 1 -1 %e A228783 20 10: 0 8 0 -18 0 8 0 -1 %e A228783 22 11: 1 2 -3 -1 1 %e A228783 24 12: 0 -8 0 6 0 -1 0 0 %e A228783 26 13: -1 3 3 -4 -1 1 %e A228783 28 14: 0 12 0 -67 0 96 0 -52 0 12 0 -1 %e A228783 30 15: -2 3 1 -1 %e A228783 ... %e A228783 n = 8, l = 4: s(8) = -3*rho(8) + rho(8)^3 = -3*sqrt(2 + sqrt(2)) + (sqrt(2 + sqrt(2)))^3 = (sqrt(2) - 1)*sqrt(2 + sqrt(2)). %e A228783 n = 10, l = 5: s(10) = -1 + rho(5), where rho(5) = tau = (1 + sqrt(5))/2, the golden section. %Y A228783 Cf. A127672, A111125, A127677, A055034, A187360, A228785 (odd n case), A228786 (minimal polynomials). %K A228783 sign,tabf %O A228783 1,1 %A A228783 _Wolfdieter Lang_, Oct 06 2013