OFFSET
1,3
COMMENTS
Conjecture: For any odd prime p, we have a(p) == 1/2 (mod p).
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).
LINKS
Zhi-Wei Sun, Arithmetic properties of some permanents, arXiv:2108.07723 [math.GM], 2021.
EXAMPLE
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.
MATHEMATICA
a[n_]:=a[n]=Permanent[Table[Abs[j-k+1], {j, 1, n}, {k, 1, n}]]
Table[a[n], {n, 1, 22}]
PROG
(PARI) a(n) = matpermanent(matrix(n, n, j, k, abs(j-k+1))); \\ Michel Marcus, Sep 13 2021
(Python)
from sympy import Matrix
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
CROSSREFS
KEYWORD
nonn
AUTHOR
Zhi-Wei Sun, Sep 12 2021
STATUS
approved