cp's OEIS Frontend

A231189 Coefficients of the algebraic number 2*sin(2*Pi/n) in the power basis of Q(2*cos(Pi/q(n))), with q(n) = A225975(n), n >= 1.

%I A231189 #7 Dec 07 2013 05:07:59
%S A231189 0,0,0,1,2,0,1,0,0,0,1,0,-3,0,1,0,0,0,1,0,5,0,-5,0,1,0,-3,0,1,0,-7,0,
%T A231189 14,0,-7,0,1,0,0,1,0,9,0,-30,0,27,0,-9,0,1,0,0,0,5,0,-5,0,1,0,-7,0,22,
%U A231189 0,-13,0,2,0,-3,0,1,0,13,0,-91,0,182,0,-156,0,65,0,-13,0,1,0,0,0,-4,0,5,0,-1,0,-15,0,140,0,-378,0,450,0,-275,0,90,0,-15,0,1,0,0,-1,1
%N A231189 Coefficients of the algebraic number 2*sin(2*Pi/n) in the power basis of Q(2*cos(Pi/q(n))), with q(n) = A225975(n), n >= 1.
%C A231189 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)).
%C A231189 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)).
%C A231189 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,...
%C A231189 For the minimal polynomials of 2*sin(2*Pi/n) see the coefficient table A231188.
%F A231189 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.
%e A231189 [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], ...
%e A231189 The table a(n,m) begins (the trailing zeros are needed to have the correct degree for Q(rho(q(n)))):
%e A231189 n\m  0   1  2   3  4    5  6    7  8    9  10  11 12  13 14 15 16 17 ...
%e A231189 1:   0
%e A231189 2:   0
%e A231189 3:   0   1
%e A231189 4:   2
%e A231189 5:   0   1  0   0
%e A231189 6:   0   1
%e A231189 7:   0  -3  0   1  0    0
%e A231189 8:   0   1
%e A231189 9:   0   5  0  -5  0    1
%e A231189 10:  0  -3  0   1
%e A231189 11:  0  -7  0  14  0   -7  0    1  0    0
%e A231189 12:  1
%e A231189 13:  0   9  0 -30  0   27  0   -9  0    1   0   0
%e A231189 14:  0   5  0  -5  0    1
%e A231189 15:  0  -7  0  22  0  -13  0    2
%e A231189 16:  0  -3  0   1
%e A231189 17:  0  13  0 -91  0  182  0 -156  0   65   0 -13  0   1  0  0
%e A231189 18:  0  -4  0   5  0   -1
%e A231189 19:  0 -15  0 140  0 -378  0  450  0 -275   0  90  0 -15  0  1  0  0
%e A231189 20: -1   1
%e A231189 ...
%e A231189 --------------------------------------------------------------------------
%e A231189 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.
%e A231189 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.
%e A231189 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.
%e A231189 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.
%Y A231189 Cf. A055034 (for delta), A106609 (for p), A225975 (for q), A127672 (for R), A187360 (for C), A181871, A231188.
%K A231189 sign,tabf,easy
%O A231189 1,5
%A A231189 _Wolfdieter Lang_, Dec 04 2013