

A228785


Table of coefficients of the algebraic number s(2*l+1) = 2*sin(Pi/(2*l+1)) as a polynomial in odd powers of rho(2*(2*l+1)) = 2*cos(Pi/(2*(2*l+1))) (reduced version).


6



1, 3, 1, 5, 5, 1, 4, 5, 1, 9, 30, 27, 9, 1, 11, 55, 77, 44, 11, 1, 4, 13, 7, 1, 15, 140, 378, 450, 275, 90, 15, 1, 17, 204, 714, 1122, 935, 442, 119, 17, 1, 4, 25, 26, 9, 1, 0, 21, 385, 2079, 5148, 7007, 5733, 2940, 952, 189, 21, 1, 8, 126, 539, 967, 870, 429, 118, 17, 1, 0
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OFFSET

1,2


COMMENTS

In the regular (2*l+1)gon, l >= 1, inscribed in a circle of radius R the length ratio side/R is s(2*l+1) = 2*sin(Pi/(2*l+1)). This can be written as a polynomial in the length ratio (smallest diagonal)/side in the (2*(2*l+1))gon given by rho(2*(2*l+1)) = 2*cos(Pi/(2*(2*l+1))). This leads, in a first step, to the signed triangle A111125. Because of the minimal polynomial of the algebraic number rho(2*(2*l+1)) of degree delta(2*(2*l+1)) = A055034(2*(2*l+1)), called C(2*(2*l+1),x) (with coefficients given in A187360), one can eliminate all powers rho(2*(2*l+1))^k with k >= delta(2*(2*l+1)) by using C(2*(2*l+1),rho(2*(2*l+1))) = 0. This leads to the present table expressing s(2*(l+1)) in terms of odd powers of rho(2*(2*l+1)) with maximal exponent delta(2*(2*l+1))1.
This table gives the coefficients of s(2*l+1), related to the (2*l+1)gon, in the power basis of the algebraic number field Q(rho(2*(2*l+1))) of degree delta(2*(2*l+1)), related to rho from the (2*(2*l+1))gon, provided one inserts zeros for the even powers, starting each row with a zero and filling zeros at the end in order to obtain the row length delta(2*(2*l+1)). Note that some trailing zeros in the present table (e.g., row l = 10) have been given such that the row length for the s(2*l+1) coefficients in the power basis Q(rho(2*(2*l+1))) becomes just twice the one of this table.
Thanks go to Seppo Mustonen for telling me about his findings regarding the square of the sum of all length in the regular ngon, which led me to consider this entry (even though for odd n this is not needed because only s(2*l+1)^2 = 4  rho(2*l+1)^2 enters).


LINKS

Table of n, a(n) for n=1..68.


FORMULA

a(l,m) = [x^(2*m+1)](s(2*l+1,x)(mod C(2*(2l+1),x))), with s(2*l+1,x) = sum((1)^(l1s)* A111125(l1,s)*x^(2*s+1), s=0..l1), l >= 1, m=0, ..., (delta(2*(2*l+1))/2  1), with delta(n) = A055034(n).
Rows 9,15,21,27 are coefficients of polynomials in reciprocal powers of u for rows n=2,4,6,8 generated by the o.g.f. (u4)/(uux+x^2) of A267633. These polynomials in u occur in a moving average of the polynomials of A140882 interlaced with these polynomials.  Tom Copeland, Jan 16 2016


EXAMPLE

The table a(l,m), with n = 2*l+1, begins:
n, l \m 0 1 2 3 4 5 6 7 8 9 10
3, 1: 1
5, 2: 3 1
7, 3: 5 5 1
9, 4: 4 5 1
11, 5: 9 30 27 9 1
13, 6: 11 55 77 44 11 1
15, 7: 4 13 7 1
17, 8: 15 140 378 450 275 90 15 1
19, 9: 17 204 714 1122 935 442 119 17 1
21, 10: 4 25 26 9 1 0
23, 11: 21 385 2079 5148 7007 5733 2940 952 189 21 1
25, 12: 8 126 539 967 870 429 118 17 1 0
27, 13: 4 41 70 43 11 1 0 0 0
...
n = 29 l = 14: 27, 819, 7371, 30888, 72930, 107406, 104652, 69768, 32319, 10395, 2277, 324, 27, 1.
n = 5, l=2: s(5) = 3*rho(10) + rho(10)^3 = (tau  1)*sqrt(2 + tau), approximately 1.175570504, where tau = (1 + sqrt(5))/2 (golden section).
n = 17, l = 8: s(17) = 15*x + 140*x^3  378*x^5 + 450*x^7  275*x^9 + 90*x^11  15*x^13 + 1*x^15, with x = rho(34) = 2*cos(Pi/34). s(17) is approximately 0.3674990356. With the length row l = 8 the degree of the algebraic number s(17) = 2*sin(Pi/17) is therefore 2*8 = 16. See A228787 for the decimal expansion of s(17) and A228788 for the one of rho(34).


CROSSREFS

Cf. A055034, A187360, A228783 (even n case), A228786 (minimal polynomials).
Cf. A140882, A267633.
Sequence in context: A096975 A145174 A351965 * A135184 A210224 A131304
Adjacent sequences: A228782 A228783 A228784 * A228786 A228787 A228788


KEYWORD

sign,tabf


AUTHOR

Wolfdieter Lang, Oct 07 2013


STATUS

approved



