OFFSET
1,2
COMMENTS
The row length of this table is delta(2*l) = A055034(2*l), l >= 1.
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.
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.
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.
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).
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.
FORMULA
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.
EXAMPLE
The table a(l,m) (n = 2*l) starts (row length A055034(2*l)):
l, n\m 0 1 2 3 4 5 6 7 8 9 10 1
1, 2: 1
2, 4: 3 2
3, 6: 7 4
4, 8: -1 0 4 2
5, 10: -9 -4 8 4
6, 12: -1 0 8 4
7, 14: 15 8 -24 -12 8 4
8, 16: -1 0 16 8 -16 -8 4 2
9, 18: 7 4 -16 -8 8 4
10, 20: -1 0 24 12 -32 -16 8 4
11, 22: 23 12 -104 -52 128 64 -56 -28 8 4
12, 24: -1 0 32 16 -32 -16 8 4
13, 26: -25 -12 176 88 -320 -160 232 116 -72 -36 8 4
14, 28: -1 0 48 24 -160 -80 168 84 -64 -32 8 4
15, 30: -1 0 16 8 -24 -12 8 4
...
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.
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.
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.
CROSSREFS
KEYWORD
sign,tabf
AUTHOR
Wolfdieter Lang, Oct 09 2013
STATUS
approved