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%I #51 Feb 26 2020 08:14:28
%S 1,-2,-4,4,48,-160,-32,2176,6912,0,-273408,41984,19456,-37027840,
%T -141705216,0,3833856,-34359869440,0,1625620480000,11045440585728,
%U -47710208,-520279482695680,7719016726528,909115392000,-207717914210467840,0,0,100736516652659638272,-721057900040447590400
%N Determinants of the equally spaced angles sines to tangents matrices.
%H O. D. Biesel, D. V. Ingerman, J. A. Morrow, and W. T. Shore, <a href="http://www.math.washington.edu/~reu/papers/2008/william/layered.pdf">Layered networks, the discrete Laplacian, and a continued fraction identity</a>, Mathematics REU 2008, University of Washington.
%H Zhi-Wei Sun, <a href="https://arxiv.org/abs/1901.04837">On some determinants involving the tangent function</a>, arXiv:1901.04837 [math.NT], 2019.
%H Wikibooks, <a href="https://en.wikibooks.org/wiki/On_2D_Inverse_Problems/The_case_of_the_unit_disc">On 2D Inverse Problems/The case of the unit disc</a>
%F Let n be a natural number. a(n) = det(T(n)), where T(n) is the n X n matrix with entries 1,0 and -1, such that 2*T(n)*s(n) = t(n), where s(n) and t(n) are vectors of length n, given by s(n) = sin(k*Pi/(2n+1)) and t(n) = tan(k*Pi/(2n+1)), for k=1..n.
%F Existence of the matrix T(n) is proved for prime 2n+1, in which case the entries of T(n) are 1 and -1. Computer checked for small 2n+1...
%F Examples:
%F 2*sin(Pi/3) = tan(Pi/3),
%F 2*(-sin(Pi/5) + sin(2*Pi/5)) = tan(Pi/5),
%F 2*(sin(Pi/5) + sin(2*Pi/5)) = tan(2*Pi/5),
%F 2*(sin(Pi/7) + sin(2*Pi/7) - sin(3*Pi/7)) = tan(Pi/7),
%F 2*(sin(Pi/7) + sin(2*Pi/7) + sin(3*Pi/7)) = tan(3*Pi/7),
%F ...
%e a(1) = det([1]) = 1,
%e a(2) = det([-1 1], [1 1]) = -2,
%e a(3) = det([1 1 -1], [1 -1 1], [1 1 1]) = -4,
%e a(4) = det([-1 1 1 -1], [-1 1 -1 1], [0 0 1 0], [1 1 1 1]) = 4.
%e ...
%o (SageMath)
%o def binary_trig(n):
%o N=2*n+1
%o print(N,"th root of unity")
%o T=matrix(ZZ,n,n)
%o for ll in range(n):
%o l=ll*2+1
%o for kk in range(n):
%o k=kk+1
%o T[min(l*n%N,N-l*n%N)-1,min(k*l%N,N-k*l%N)-1]=sign(RDF(sin(k*l*pi/N)))
%o s=matrix(RDF,n,1)
%o for k in range(n):
%o s[k,0]=sin((k+1)*pi/N)
%o for k in range(n):
%o if (T*s)[k,0]<0:
%o #if prod(T[k])==0:
%o T[k]=-T[k]
%o return det(T)
%Y Related to A007318 by the continued fraction.
%K sign
%O 1,2
%A _David V. Ingerman_, Oct 15 2016
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