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%I #37 Aug 19 2021 03:28:54
%S 1,2,41,3176,620964,246796680,174252885732,199381727959680,
%T 345875291854507584,864860593764292790400,2996169331694350840741440,
%U 13929521390709644084719495680,84659009841182126038701730464000,658043094413184868424932006273344000
%N a(n) = permanent M_n where M_n is the n X n matrix m(i,j) = i^2 + j^2.
%C From _Zhi-Wei Sun_, Aug 19 2021: (Start)
%C I have proved that a(n) == (-1)^(n-1)*2*n! (mod 2n+1) whenever 2n+1 is prime.
%C Conjecture 1: If 2n+1 is composite, then a(n) == 0 (mod 2n+1).
%C 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)
%H Vaclav Kotesovec, <a href="/A278847/b278847.txt">Table of n, a(n) for n = 0..36</a>
%H Zhi-Wei Sun, <a href="http://arxiv.org/abs/2108.07723">Arithmetic properties of some permanents</a>, arXiv:2108.07723 [math.GM], 2021.
%F a(n) ~ c * d^n * (n!)^3 / n, where d = 3.809076776112918119... and c = 1.07739642254738...
%p with(LinearAlgebra):
%p a:= n-> `if`(n=0, 1, Permanent(Matrix(n, (i, j)-> i^2+j^2))):
%p seq(a(n), n=0..16); # after _Alois P. Heinz_
%t Flatten[{1, Table[Permanent[Table[i^2+j^2, {i, 1, n}, {j, 1, n}]], {n, 1, 15}]}]
%o (PARI) a(n)={matpermanent(matrix(n, n, i, j, i^2 + j^2))} \\ _Andrew Howroyd_, Aug 21 2018
%Y Cf. A005249, A085750, A085807, A204249, A278845, A278925, A278926, A346934, A346949.
%K nonn
%O 0,2
%A _Vaclav Kotesovec_, Nov 29 2016
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