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A346934
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Permanent of the 2n X 2n matrix with the (i,j)-entry i-j (i,j=1..2n).
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5
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-1, 52, -18660, 24446016, -85000104000, 647188836814080, -9486416237249952000, 244072502056661870592000, -10282514440038927957603532800, 671904022157076034864609763328000, -65203712913305114275839483698454528000
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OFFSET
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1,2
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COMMENTS
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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,j<n} == per[i-j]_{1<=i,j<n} (mod n). Thus A204249(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.
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LINKS
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EXAMPLE
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a(1) is the permanent of the matrix [1-1,1-2;2-1,2-2] = [0,-1;1,0], which equals -1.
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MATHEMATICA
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a[n_]:=a[n]=Permanent[Table[i-j, {i, 1, 2n}, {j, 1, 2n}]];
Table[a[n], {n, 1, 11}]
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PROG
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(PARI) a(n) = matpermanent(matrix(2*n, 2*n, i, j, i-j)); \\ Michel Marcus, Aug 08 2021
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CROSSREFS
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KEYWORD
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sign
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AUTHOR
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STATUS
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approved
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