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

%I A346934 #28 Aug 19 2021 03:29:31
%S A346934 -1,52,-18660,24446016,-85000104000,647188836814080,
%T A346934 -9486416237249952000,244072502056661870592000,
%U A346934 -10282514440038927957603532800,671904022157076034864609763328000,-65203712913305114275839483698454528000
%N A346934 Permanent of the 2n X 2n matrix with the (i,j)-entry i-j (i,j=1..2n).
%C A346934 The author has proved that a((p-1)/2) == 2 (mod p) for any odd prime p.
%C A346934 Conjecture 1: (-1)^n*a(n) > 0 for all n > 0. Also, a(n) == 0 (mod 2n+1) if 2n+1 is composite.
%C A346934 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.
%C A346934 It is easy to see that per[i+j]_{1<=i,j<n} = per[i+(n-j)]_{1<=i,j<n} == per[i-j]_{1<=i,j<n} (mod n). Thus A204249(2n) == a(n) (mod 2n+1).
%C A346934 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).
%C A346934 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.
%H A346934 Vaclav Kotesovec, <a href="/A346934/b346934.txt">Table of n, a(n) for n = 1..19</a>
%H A346934 Zhi-Wei Sun, <a href="http://arxiv.org/abs/2108.07723">Arithmetic properties of some permanents</a>, arXiv:2108.07723 [math.GM], 2021.
%e A346934 a(1) is the permanent of the matrix [1-1,1-2;2-1,2-2] = [0,-1;1,0], which equals -1.
%t A346934 a[n_]:=a[n]=Permanent[Table[i-j,{i,1,2n},{j,1,2n}]];
%t A346934 Table[a[n],{n,1,11}]
%o A346934 (PARI) a(n) = matpermanent(matrix(2*n, 2*n, i, j, i-j)); \\ _Michel Marcus_, Aug 08 2021
%Y A346934 Cf. A000166, A204249, A346949.
%K A346934 sign
%O A346934 1,2
%A A346934 _Zhi-Wei Sun_, Aug 08 2021