OFFSET
1,5
COMMENTS
The relevant trigonometric identity (used in the D. H. Lehmer and I. Niven references, given in A181871) is 2*sin(2*Pi/n) = 2*cos(2*Pi*(1/n -1/4)) = 2*cos(Pi*abs(n-4)/(2*n)) = 2*cos(Pi*p(n)/q(n)), with gcd(p(n), q(n)) = 1 (fraction p(n)/q(n) in lowest terms). One finds p(n) = A106609(n-4), n >=4, with p(1) = 3 , p(2) = 1 = p(3), and q(n) = A225975(n), n >= 1. See the comments on these two A-numbers. Therefore, 2*sin(2*Pi/n) = R(p(n), rho(q(n))), with rho(k) = 2*cos(Pi/k), and the R-polynomials (monic version of Chebyshev's T-polynomials) are given in A127672. It may happen that p(n), the degree of R, is >= delta(q(n)), the degree of the algebraic number rho(q(n)). Here delta(k) = A055034(k) is the degree of the minimal polynomial C(k, x) of rho(k) found under A187360. In this case one can reduce all rho(q(n)) powers >= delta(q(n)) with the help of the equation C(q(n), rho(q(n))) = 0. Thus the final result is 2*sin(2*Pi/n) = R(p(n), x) (mod C(q(n), x)) with x = rho(q(n)). Because R is an integer polynomial this shows that 2*sin(2*Pi/n) is an integer in the algebraic number field Q(rho(q(n))) of degree delta(q(n)).
The power basis of Q(rho(q(n))) is <1, rho(q(n)), ..., rho(q(n))^(delta(q(n))-1)>. Therefore the length of row n of this table is delta(q(n)).
The values n for which mod C(q(n), x) is in operation for the given formula for 2*sin(2*Pi/n) are those for which delta(q(n)) - p(n) <= 0, that is n = 1, 2, 12, 15, 18, 20, 21, 24, 25, 27, 28, 30,...
For the minimal polynomials of 2*sin(2*Pi/n) see the coefficient table A231188.
FORMULA
a(n,m) = [x^m] (R(p(n), x) (mod C(q(n), x)), n >= 1, m = 0, 1, ..., delta(q(n)) - 1, where the R and C polynomials are found in A187360 and A127672, respectively. p(n) = A106609(n-4), n >=4, with p(1) = 3 , p(2) = 1 = p(3), and q(n) = A225975(n). Powers of x = rho(q(n)) = 2*cos(Pi/q(n)) appear in the table in increasing order.
EXAMPLE
[0], [0], [0, 1], [2], [0, 1, 0, 0], [0, 1], [0, -3, 0, 1, 0, 0], [0, 1], [0, 5, 0, -5, 0, 1], ...
The table a(n,m) begins (the trailing zeros are needed to have the correct degree for Q(rho(q(n)))):
n\m 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 ...
1: 0
2: 0
3: 0 1
4: 2
5: 0 1 0 0
6: 0 1
7: 0 -3 0 1 0 0
8: 0 1
9: 0 5 0 -5 0 1
10: 0 -3 0 1
11: 0 -7 0 14 0 -7 0 1 0 0
12: 1
13: 0 9 0 -30 0 27 0 -9 0 1 0 0
14: 0 5 0 -5 0 1
15: 0 -7 0 22 0 -13 0 2
16: 0 -3 0 1
17: 0 13 0 -91 0 182 0 -156 0 65 0 -13 0 1 0 0
18: 0 -4 0 5 0 -1
19: 0 -15 0 140 0 -378 0 450 0 -275 0 90 0 -15 0 1 0 0
20: -1 1
...
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n=1: 2*sin(2*Pi/1) = 0. rho(q(1)) = rho(2) = 2*cos(Pi/2) = 0 and p(1) = 3. R(3, x) = -3*x + x^3 and C(2, x) = x. Therefore R(3, x) (mod C(2, x)) = 0. The degree of C(2, x) is delta(2) = A055034(2) = 1. Here one should use 1 for the undefined rho(q(1))^0 in order to obtain a(1, 0) = 0.
n=2: 2*sin(2*Pi/2) = 0; rho(q(2)) = rho(2) = 0; p(2) = 1, R(1, x) = x , C(2, x) = x and delta(2) = 1. Therefore R(1, x) (mod C(1, x)) = 0. Again, rho(2)^0 is put to 1 here, and a(2, 0) = 0.
n=5: 2*sin(2*Pi/5) = R(1, rho(10)) (mod C(10, rho(10)) =1* rho(10) (the degree of C(10,x) is delta(10) = 4, therefore the mod prescription is not needed). Therefore, a(5, 0) =0, a(5,1) =1, a(n, m) = 0 for m=2, 3.
n =11: 2*sin(2*Pi/11) = R(7, x) (mod(C(22, x)) with x = rho(22), because p(11) = 7 and q(11) = 22. The degree of C(22, x) is delta(22) = 10, therefore the mod restriction is not needed and R(7, x) = -7*x + 14*x^3 - 7*x^5 + x^7. The coefficients produce the row [0, -7, 0, 14, 0, -7, 0, 1, 0, 0] with the two trailing zeros needed to obtain the correct row length, namely delta(q(11)) = 10.
CROSSREFS
KEYWORD
sign,tabf,easy
AUTHOR
Wolfdieter Lang, Dec 04 2013
STATUS
approved