login

Year-end appeal: Please make a donation to the OEIS Foundation to support ongoing development and maintenance of the OEIS. We are now in our 61st year, we have over 378,000 sequences, and we’ve reached 11,000 citations (which often say “discovered thanks to the OEIS”).

a(n) = (-1)^n*permanent[tan(Pi*(j+k)/(2n+1))]_{1<=j,k<=2n}.
0

%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