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Permanent of the n X n matrix with (j,k)-entry |j-k+1| (j,k = 1..n).
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%I #19 Jul 29 2023 06:36:11

%S 1,1,8,80,1568,42308,1632544,82377984,5340729728,429905599744,

%T 42164608801024,4944386388782080,683353973472423936,

%U 109907353260811403520,20352830852731108528128,4299139435513999926820864,1027450150728092835655335936,275824741022588671077713641472

%N Permanent of the n X n matrix with (j,k)-entry |j-k+1| (j,k = 1..n).

%C Conjecture: For any odd prime p, we have a(p) == 1/2 (mod p).

%C It is easy to show that for any integer n > 1 the determinant of the n X n matrix with (j,k)-entry |j-k+1| (j,k=1..n) has the value 2^(n-2).

%H Zhi-Wei Sun, <a href="https://arxiv.org/abs/2108.07723">Arithmetic properties of some permanents</a>, arXiv:2108.07723 [math.GM], 2021.

%e a(2) = 1 since the permanent of the matrix [|1-1+1|,|1-2+1|; |2-1+1|,|2-2+1|] = [1,0;2,1] has the value 1.

%t a[n_]:=a[n]=Permanent[Table[Abs[j-k+1],{j,1,n},{k,1,n}]]

%t Table[a[n],{n,1,22}]

%o (PARI) a(n) = matpermanent(matrix(n, n, j, k, abs(j-k+1))); \\ _Michel Marcus_, Sep 13 2021

%o (Python)

%o from sympy import Matrix

%o def A347768(n): return Matrix(n,n,[abs(j-k+1) for j in range(n) for k in range(n)]).per() # _Chai Wah Wu_, Sep 14 2021

%Y Cf. A085807.

%K nonn

%O 1,3

%A _Zhi-Wei Sun_, Sep 12 2021