OFFSET
0,2
COMMENTS
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.
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}.
LINKS
Vaclav Kotesovec, Table of n, a(n) for n = 0..100
Zhi-Wei Sun, On some determinants involving the tangent function, arXiv:1901.04837 [math.NT], 2019-2021.
Zhi-Wei Sun, Arithmetic properties of some permanents, arXiv:2108.07723 [math.GM], 2021. (Cf. Section 2)
FORMULA
a(n) ~ 2^(4*n+3) * n^(4*n+2) / exp(2*n-1). - Vaclav Kotesovec, Aug 19 2021
EXAMPLE
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.
MATHEMATICA
a[n_]:=Det[Table[Tan[Pi*(j-k)/(2n+1)]^2, {j, 0, 2n}, {k, 0, 2n}]]
(* 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. *)
PROG
(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
CROSSREFS
KEYWORD
nonn
AUTHOR
Zhi-Wei Sun, Aug 14 2021
STATUS
approved