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a(n) = Product_{i=0..n, j=0..n} (i!^2 + j!^2).
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%I #12 Feb 17 2021 02:12:35

%S 2,16,80000,17272267776000000,

%T 277884245560378426290863196025651200000000,

%U 3337940951837185557810120427617693521487357301121536848574225250643001642844160000000000

%N a(n) = Product_{i=0..n, j=0..n} (i!^2 + j!^2).

%F a(n) ~ c * 2^(n*(n+3)) * Pi^(n*(n+2)) * n^((n+1)*(2*n+1)*(2*n+3)/3) / exp(2*n*(2*n+3)*(4*n+3)/9), where c = 401.488675138779168689540247334821476110398137334270208637438...

%t Table[Product[i!^2 + j!^2, {i, 0, n}, {j, 0, n}], {n, 0, 7}]

%t Clear[a]; a[n_] := a[n] = If[n == 0, 2, a[n-1] * Product[k!^2 + n!^2, {k, 0, n}]^2 / (2*n!^2)]; Table[a[n], {n, 0, 7}]

%o (Python)

%o from math import prod, factorial as f

%o def a(n): return prod(f(i)**2+f(j)**2 for i in range(n) for j in range(n))

%o print([a(n) for n in range(1, 7)]) # _Michael S. Branicky_, Feb 16 2021

%Y Cf. A306729, A324403, A325050.

%K nonn

%O 0,1

%A _Vaclav Kotesovec_, Mar 26 2019