A277445 - OEIS

%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