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a(n) = n!^(4*n) * Product_{k=1..n} binomial(n + 1/k^2, n).
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%I #10 Jun 24 2023 16:06:28

%S 1,2,1080,16133644800,139256878046022696960000,

%T 6288402750181849898716908922601472000000000,

%U 8322157105451357856813375261666887975745751468393837363200000000000000

%N a(n) = n!^(4*n) * Product_{k=1..n} binomial(n + 1/k^2, n).

%F a(n) ~ n!^(4*n) * n^(Pi^2/6) / A303670.

%F a(n) ~ n^(4*n^2 + 2*n + Pi^2/6) * (2*Pi)^(2*n) / exp(4*n^2 - 1/3 - gamma*Pi^2/6 + c), where gamma is the Euler-Mascheroni constant A001620 and c = A306774 = Sum_{k>=2} (-1)^k * Zeta(k) * Zeta(2*k) / k.

%F a(n) = n!^n * A324596(n).

%p a:= n-> n!^(4*n)*mul(binomial(n+1/k^2, n), k=1..n):

%p seq(a(n), n=0..7); # _Alois P. Heinz_, Jun 24 2023

%t Table[n!^(4*n) * Product[Binomial[1/k^2 + n, n], {k, 1, n}], {n, 1, 8}]

%Y Cf. A303670, A306760, A306774, A324589.

%K nonn

%O 0,2

%A _Vaclav Kotesovec_, Mar 09 2019

%E a(0)=1 prepended by _Alois P. Heinz_, Jun 24 2023