A346162 a(n) = (-1)^n*permanent[tan(Pi*(j+k)/(2n+1))]_{1<=j,k<=2n}.

3, 65, 1995, 149121, 13399155, 2141638785, 387394882875, 107278565018625, 32598647029023075, 13887464253877202625, 6371929420307657080875, 3868428575833744110890625, 2498109253610200166983921875, 2048940276336573213783855290625, 1771722889715428355094671454046875, 1887395844638188493284600904244890625

Offset: 1

Author

Zhi-Wei Sun, Aug 22 2021

Keywords

• nonn

Comments

The author has proved that a(n) is always an integer, and that a((p-1)/2) == -2*p (mod p^2) for any odd prime p.

Conjecture 1: a(n)/(2n+1) is always a positive integer congruent to 1 modulo 4.

Conjecture 2: We have (n+1)*a(n) == 0 (mod (2n+1)!!) for all n > 0.

Examples

			a(1) = 3 since the permanent of the matrix [tan(Pi*(1+1)/3),tan(Pi*(1+2)/3); tan(Pi*(2+1)/3),tan(Pi*(2+2)/3)] = [-sqrt(3),0; 0,sqrt(3)] is -3.
		

Links

• Zhi-Wei Sun, Arithmetic properties of some permanents, arXiv:2108.07723 [math.GM], 2021.

Crossrefs

Cf. A346934, A347061.

Programs

Mathematica

a[n_]:=a[n]=(-1)^n*Permanent[Table[Tan[Pi*(j+k)/(2n+1)],{j,1,2n},{k,1,2n}]]
(* Actually Mathematica could not yield the exact value of a(n) for a general n > 0. Instead, we find an approximate value of a(n) via Mathematica, such as N[a[3],10] = 1995.000000. *)

PARI

a(n) = (-1)^n*round(matpermanent(matrix(2*n, 2*n, j, k, tan(Pi*(j+k)/(2*n+1))))); \\ Michel Marcus, Aug 22 2021

Extensions

a(7)-a(12) from Michel Marcus, Aug 22 2021

a(13)-a(16) from Vaclav Kotesovec, Aug 22 2021