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).

0, 54, 112500, 1111783050, 34170487129800, 2547904546473882750, 392729761077721859691900, 111952195111563752746582031250, 54383838324760862570403373592970000, 42287009764195722177648795731701042023750, 50090854432555768390858694036859976340444362500

Offset: 0

Author

Zhi-Wei Sun, Aug 14 2021

Keywords

• nonn

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

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}.

Examples

			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.
		

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)

Crossrefs

Cf. A000182, A346934.

Programs

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. *)

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

Formula

a(n) ~ 2^(4*n+3) * n^(4*n+2) / exp(2*n-1). - Vaclav Kotesovec, Aug 19 2021

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