%I #18 Apr 12 2018 14:27:38
%S 1,2,1,0,1,-1,1,-1,0,1,0,-1,1,-2,0,1,0,0,-1,1,0,0,-1,0,1,-1,-1,0,1,2,
%T 0,-2,0,1,0,0,0,0,-1,1,-1,0,-1,0,1,0,0,0,0,0,-1,1,-2,0,2,0,-2,0,1,-1,
%U -2,-1,1,1
%N Array of conversion coefficients for the minimal polynomials C of 2 cos(Pi/n) in terms of Chebyshev's S-polynomials.
%C The row length sequence is 1, 1 + A055034(n), n >= 1.
%C For the minimal polynomial C(n, x) of the algebraic number rho(n) := 2*cos(Pi/n) (the length ratio of the smallest diagonal and the side of a regular n-gon) see the coefficient array in A187360. The coefficient triangle of Chebyshev's S-polynomials is given in A049310.
%C The conversion is C(n, x) = sum(T(n, m)*S(m, x), m=0..delta(n)), for n >= 0 with C(0, x) := 1 (undefined product), delta(0) = 0 and delta(n) = A055034(n), n >= 1.
%C Originally Ahmet Zahid KÜÇÜK observed the structure for prime n. The precise formula for odd primes prime(n) = A000040(n), n >= 2, is C(prime(n), x) = S((prime(n)-1)/2, x) - S((prime(n)-3)/2, x).
%C This is equivalent to C(prime(n),x) = (-1)^((p(n)-1)/2)*S(prime(n)-1,I*sqrt(x-2)), with I^2 = -1.
%C Proof: The known identity S(n, x) - S(n-1, x) = (-1)^n*S(2*n, I*sqrt(x-2)) (from bisection). The degrees of the monic polynomials of both sides match, as do the known zeros.
%C The row sums give 1, 3, 1, 0, 0, 0, -1, 0, 0, -1, 1, 0, -1, 0, -1, -2, 0, 0, -1, 0 ...
%C The alternating row sums give 1, 1, -1, -2, 0, 2, -1, -2, 0, -1, 1, -2, -1, 2, -1, 0, 0, 2, -1, -2, ...
%C For the reverse problem, the factorization of S polynomials into C polynomials see a Apr 12 2018 comment in A049310. - _Wolfdieter Lang_, Apr 12 2018
%F The conversion is C(n, x) = sum(T(n, m)*S(m, x), m = 0..delta(n)), that is
%F T(n, m) = [S(m, x)] C(n, x), n >= 0, m = 0, ..., delta(n), with C(0, x) := 1, delta(0) = 0 and delta(n) = A055034(n), n >= 1. For the C and S polynomials see A187360 and A049310, respectively.
%F For n >= 2: T(prime(n), (prime(n) -1)/2) = +1, T(prime(n), (prime(n) -3)/2) = -1 and T(prime(n), m) = 0 otherwise.
%e The array T(n, m) begins:
%e n\m 0 1 2 3 4 5 6 ...
%e 0: 1
%e 1: 2 1
%e 2: 0 1
%e 3: -1 1
%e 4: -1 0 1
%e 5: 0 -1 1
%e 6: -2 0 1
%e 7: 0 0 -1 1
%e 8: 0 0 -1 0 1
%e 9: -1 -1 0 1
%e 10: 2 0 -2 0 1
%e 11: 0 0 0 0 -1 1
%e 12: -1 0 -1 0 1
%e 13: 0 0 0 0 0 -1 1
%e 14: -2 0 2 0 -2 0 1
%e 15: -1 -2 -1 1 1
%e ...
%e n=0: C(0, x) = 1 = 1*S(0, x),
%e n=1: C(1, x) = 2 + x = 2*S(0, x) + 1*S(1, x),
%e n=2: C(2, x) = x = 0*S(0, x) + 1*S(1, x),
%e n=3: C(3, x) = -1 + x = -1*S(0, x) + 1*S(1, x),
%e n=4: C(4, x) = -2 + x^2 = -1*S(0, x) + 0 + 1*S(2, x) = -1 + (-1 + x^2), ...
%Y Cf. A187360, A055034, A049310.
%K sign,easy,tabf
%O 0,2
%A Ahmet Zahid KÜÇÜK and _Wolfdieter Lang_, Mar 11 2015