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%I #48 Dec 23 2024 22:49:05
%S 3,65,1995,149121,13399155,2141638785,387394882875,107278565018625,
%T 32598647029023075,13887464253877202625,6371929420307657080875,
%U 3868428575833744110890625,2498109253610200166983921875,2048940276336573213783855290625,1771722889715428355094671454046875,1887395844638188493284600904244890625
%N a(n) = (-1)^n*permanent[tan(Pi*(j+k)/(2n+1))]_{1<=j,k<=2n}.
%C 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.
%C Conjecture 1: a(n)/(2n+1) is always a positive integer congruent to 1 modulo 4.
%C Conjecture 2: We have (n+1)*a(n) == 0 (mod (2n+1)!!) for all n > 0.
%H Zhi-Wei Sun, <a href="http://arxiv.org/abs/2108.07723">Arithmetic properties of some permanents</a>, arXiv:2108.07723 [math.GM], 2021.
%e 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.
%t a[n_]:=a[n]=(-1)^n*Permanent[Table[Tan[Pi*(j+k)/(2n+1)],{j,1,2n},{k,1,2n}]]
%t (* 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. *)
%o (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
%Y Cf. A346934, A347061.
%K nonn,changed
%O 1,1
%A _Zhi-Wei Sun_, Aug 22 2021
%E a(7)-a(12) from _Michel Marcus_, Aug 22 2021
%E a(13)-a(16) from _Vaclav Kotesovec_, Aug 22 2021