cp's OEIS Frontend

A230072 Coefficients of an algebraic number sqLhat(2*l) in the power basis of the number field Q(2*cos(Pi/2*l)), related to the square of all length in a regular (2*l)-gon inscribed in a circle of radius of 1 length unit.

%I A230072 #19 Feb 25 2025 00:29:55
%S A230072 1,3,2,7,4,-1,0,4,2,-9,-4,8,4,-1,0,8,4,15,8,-24,-12,8,4,-1,0,16,8,-16,
%T A230072 -8,4,2,7,4,-16,-8,8,4,-1,0,24,12,-32,-16,8,4,23,12,-104,-52,128,64,
%U A230072 -56,-28,8,4,-1,0,32,16,-32,-16,8,4,-25,-12,176,88,-320,-160,232,116,-72,-36,8,4
%N A230072 Coefficients of an algebraic number sqLhat(2*l) in the power basis of the number field Q(2*cos(Pi/2*l)), related to the square of all length in a regular (2*l)-gon inscribed in a circle of radius of 1 length unit.
%C A230072 The row length of this table is delta(2*l) = A055034(2*l), l >= 1.
%C A230072 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.
%C A230072 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.
%C A230072 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.
%C A230072 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).
%H A230072 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 A230072 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>.
%F A230072 a(l,m) = [rho(2*l)^m](sqLhat(2*l) (mod C(2*l,rho(2*l)))), l >= 1, m = 0, ..., delta(2*l)-1, with delta(2*l) = A055034(2*l) and the formula for sqLhat(2*l) is given in a comment above.
%e A230072 The table a(l,m) (n = 2*l) starts (row length A055034(2*l)):
%e A230072   l,   n\m  0   1    2   3    4    5    6    7    8   9 10  1
%e A230072   1,   2:   1
%e A230072   2,   4:   3   2
%e A230072   3,   6:   7   4
%e A230072   4,   8:  -1   0    4   2
%e A230072   5,  10:  -9  -4    8   4
%e A230072   6,  12:  -1   0    8   4
%e A230072   7,  14:  15   8  -24 -12    8    4
%e A230072   8,  16:  -1   0   16   8  -16   -8    4    2
%e A230072   9,  18:   7   4  -16  -8    8    4
%e A230072   10, 20:  -1   0   24  12  -32  -16    8    4
%e A230072   11, 22:  23  12 -104 -52  128   64  -56  -28   8    4
%e A230072   12, 24:  -1   0   32  16  -32  -16    8    4
%e A230072   13, 26: -25 -12  176  88 -320 -160  232  116  -72 -36  8  4
%e A230072   14, 28:  -1   0   48  24 -160  -80  168   84  -64 -32  8  4
%e A230072   15, 30:  -1   0   16   8  -24  -12    8    4
%e A230072   ...
%e A230072 l = 3, n=6: (hexagon) sqLhat(6) = 13 + 4*rho(6) - 2*rho(6)^2 = 7 + 4*sqrt(3), where rho(6) = sqrt(3) and s(6) = 1. C(6,x) = x^2 -3. sqLhat(6) is approximately 13.92820323, therefore Mustonen's L2^(10) is approximately 501.4153163.
%e A230072 l = 5, n=10: (decagon) sqLhat(10) = -9 - 4*rho(10) + 8*rho(10)^2 + 4*rho(10)^3 = 7 + 8*phi + 4*(-1 + (2+phi))*sqrt(2+phi) = 7 + 8*phi + 4*sqrt(7+11*phi), with the golden section phi = rho(5) = (1 + sqrt(5))/2. sqLhat(10) is approximately 39.86345818, therefore Mustonen's L2^(10) is about 3986.345818. Here rho(10) = sqrt(2+phi) and s(10) = phi - 1.
%e A230072 l=6, n = 12: (dodecagon) sqLhat(12) = -1 + 8*rho(12)^2 + 4*rho(12)^3 = 15 + 6*sqrt(6) + 10*sqrt(2) + 4*sqrt(2)*sqrt(6), approximately 57.69548054. rho(12) = sqrt(2+sqrt(3)) and s(12) = sqrt(2 - sqrt(3)). Therefore Mustonen's L2^(12) is approximately 8308.149198.
%Y A230072 Cf. A055034, A187360, A228780, A230073 (minimal polynomials).
%K A230072 sign,tabf
%O A230072 1,2
%A A230072 _Wolfdieter Lang_, Oct 09 2013