A346934 Permanent of the 2n X 2n matrix with the (i,j)-entry i-j (i,j=1..2n).

-1, 52, -18660, 24446016, -85000104000, 647188836814080, -9486416237249952000, 244072502056661870592000, -10282514440038927957603532800, 671904022157076034864609763328000, -65203712913305114275839483698454528000

Offset: 1

Author

Zhi-Wei Sun, Aug 08 2021

Keywords

• sign

Comments

The author has proved that a((p-1)/2) == 2 (mod p) for any odd prime p.

Conjecture 1: (-1)^n*a(n) > 0 for all n > 0. Also, a(n) == 0 (mod 2n+1) if 2n+1 is composite.

For any permutation f of {1,...,2n+1}, clearly Product_{j=1..2n+1} (j-f(j)) = -Product_{k=1..2n+1} (k-f^{-1}(k)). Thus the permanent of the matrix [i-j]_{1<=i,j<=2n+1} vanishes.

It is easy to see that per[i+j]{1<=i,j<n} = per[i+(n-j)]{1<=i,jA204249(2n) == a(n) (mod 2n+1).

Let D(2n) be the set of all derangements of {1,...,2n}. Clearly, a(n) is the sum of those Product_{j=1..2n}(j-f(j)) with f in the set D(2n).

Conjecture 2: For any odd prime p, the sum of those 1/Product_{j=1..p-1}(j-f(j)) with f in the set D(p-1) is congruent to (-1)^((p-1)/2) modulo p.

Examples

			a(1) is the permanent of the matrix [1-1,1-2;2-1,2-2] = [0,-1;1,0], which equals -1.
		

Links

• Vaclav Kotesovec, Table of n, a(n) for n = 1..19
• Zhi-Wei Sun, Arithmetic properties of some permanents, arXiv:2108.07723 [math.GM], 2021.

Crossrefs

Cf. A000166, A204249, A346949.

Programs

Mathematica

a[n_]:=a[n]=Permanent[Table[i-j,{i,1,2n},{j,1,2n}]];
Table[a[n],{n,1,11}]

PARI

a(n) = matpermanent(matrix(2*n, 2*n, i, j, i-j)); \\ Michel Marcus, Aug 08 2021