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a(n) = (n^2)!/(n!)^2.
10

%I #14 Oct 14 2023 23:33:08

%S 1,1,6,10080,36324288000,1077167364120207360000,

%T 717579719887926731226850787328000000,

%U 23946596436219275985459662514223331478629410406400000000

%N a(n) = (n^2)!/(n!)^2.

%C Based on an observation of _Hugo Pfoertner_, _W. Edwin Clark_ conjectured and Xiang-dong Hou proved that (n^2)!/(n!)^2 gives the number of distinct determinants of the generic n X n matrix whose entries are n^2 different indeterminates under all (n^2)! permutations of the entries.

%C Using J. T. Schwarz's Sparse Zeros Lemma this implies that for any positive integer n there is an n X n matrix A with positive integer entries such that the set of determinant values obtained from A by permuting the elements of A is (n^2)!/(n!)^2.

%C Moreover, for any entries, no larger number of determinants can be obtained. In fact, by the Sparse Zeros Lemma one can select the entries of A from any sufficiently large subset of real numbers.

%H Vincenzo Librandi, <a href="/A088021/b088021.txt">Table of n, a(n) for n = 0..21</a>

%F a(n) = A088020(n)/A001044(n).

%o (Magma) [Factorial(n^2)/Factorial(n)^2: n in [0..10]]; // _Vincenzo Librandi_, May 31 2011

%Y Cf. A001044, A088020.

%K nonn

%O 0,3

%A _Hugo Pfoertner_, Sep 18 2003