This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.
%I A347061 #26 Dec 24 2024 07:17:39
%S A347061 0,54,112500,1111783050,34170487129800,2547904546473882750,
%T A347061 392729761077721859691900,111952195111563752746582031250,
%U A347061 54383838324760862570403373592970000,42287009764195722177648795731701042023750,50090854432555768390858694036859976340444362500
%N A347061 Determinant of the (2n+1) X (2n+1) matrix with the (j,k)-entry (tan(Pi*(j-k)/(2n+1)))^2 (j,k = 0..2n).
%C A347061 The author has proved that a(n) = (2n)*(2n+1)^(2n-1)*((2n+1)!!)^2 for all n >= 0. He also showed that for any positive odd integer n, the determinant of the matrix [(sec(Pi*(j-k)/n))^2]_{0<=j,k<n} has the value n^(n-1)*(n!!)^2.
%C A347061 It is easy to see that det[(tan(Pi*(j+k)/(2n+1)))^2]_{0<=j,k<=2n} = (-1)^n*a(n) and det[(sec(Pi*(j+k)/(2n+1)))^2]_{0<=j,k<=2n} = (-1)^n*det[(sec(Pi*(j-k)/(2n+1)))^2]_{0<=j,k<=2n}.
%H A347061 Vaclav Kotesovec, <a href="/A347061/b347061.txt">Table of n, a(n) for n = 0..100</a>
%H A347061 Zhi-Wei Sun, <a href="http://arxiv.org/abs/1901.04837">On some determinants involving the tangent function</a>, arXiv:1901.04837 [math.NT], 2019-2021.
%H A347061 Zhi-Wei Sun, <a href="http://arxiv.org/abs/2108.07723">Arithmetic properties of some permanents</a>, arXiv:2108.07723 [math.GM], 2021. (Cf. Section 2)
%F A347061 a(n) ~ 2^(4*n+3) * n^(4*n+2) / exp(2*n-1). - _Vaclav Kotesovec_, Aug 19 2021
%e A347061 a(1) = 54 since the determinant of the 3 X 3 matrix [(tan(Pi*(j-k)/(2*1+1)))^2]_{j,k=0,1,2} = [(tan(Pi/3*(j-k)))^2]_{j,k=0,1,2} has the value 54.
%t A347061 a[n_]:=Det[Table[Tan[Pi*(j-k)/(2n+1)]^2,{j,0,2n},{k,0,2n}]]
%t A347061 (* Actually Mathematica could not yield the exact value of a(n) for a general n. Instead, we find an approximate value of a(n) via Mathematica, such as N[a[2],10] = 112500.0000. *)
%o A347061 (PARI) a(n) = round(matdet(matrix(2*n+1, 2*n+1, j, k, j--; k--; tan(Pi*(j-k)/(2*n+1))^2))); \\ _Michel Marcus_, Aug 15 2021
%Y A347061 Cf. A000182, A346934.
%K A347061 nonn
%O A347061 0,2
%A A347061 _Zhi-Wei Sun_, Aug 14 2021