A278847 a(n) = permanent M_n where M_n is the n X n matrix m(i,j) = i^2 + j^2.

1, 2, 41, 3176, 620964, 246796680, 174252885732, 199381727959680, 345875291854507584, 864860593764292790400, 2996169331694350840741440, 13929521390709644084719495680, 84659009841182126038701730464000, 658043094413184868424932006273344000

Offset: 0

Author

Vaclav Kotesovec, Nov 29 2016

Keywords

• nonn

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.

Crossrefs

Cf. A005249, A085750, A085807, A204249, A278845, A278925, A278926, A346934, A346949.

Programs

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}]}]

PARI

a(n)={matpermanent(matrix(n, n, i, j, i^2 + j^2))} \\ Andrew Howroyd, Aug 21 2018

Formula

a(n) ~ c * d^n * (n!)^3 / n, where d = 3.809076776112918119... and c = 1.07739642254738...