OFFSET
0,2
COMMENTS
From Zhi-Wei Sun, Aug 19 2021: (Start)
I have proved that a(n) == (-1)^(n-1)*2*n! (mod 2n+1) whenever 2n+1 is prime.
Conjecture 1: If 2n+1 is composite, then a(n) == 0 (mod 2n+1).
Conjecture 2: If p = 4n+1 is prime, then the sum of those Product_{j=1..2n}(j^2-f(j)^2)^{-1} with f over all the derangements of {1,...,2n} is congruent to 1/(n!)^2 modulo p. (End)
LINKS
Vaclav Kotesovec, Table of n, a(n) for n = 0..36
Zhi-Wei Sun, Arithmetic properties of some permanents, arXiv:2108.07723 [math.GM], 2021.
FORMULA
a(n) ~ c * d^n * (n!)^3 / n, where d = 3.809076776112918119... and c = 1.07739642254738...
MAPLE
with(LinearAlgebra):
a:= n-> `if`(n=0, 1, Permanent(Matrix(n, (i, j)-> i^2+j^2))):
seq(a(n), n=0..16); # after Alois P. Heinz
MATHEMATICA
Flatten[{1, Table[Permanent[Table[i^2+j^2, {i, 1, n}, {j, 1, n}]], {n, 1, 15}]}]
PROG
(PARI) a(n)={matpermanent(matrix(n, n, i, j, i^2 + j^2))} \\ Andrew Howroyd, Aug 21 2018
CROSSREFS
KEYWORD
nonn
AUTHOR
Vaclav Kotesovec, Nov 29 2016
STATUS
approved