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a(n) = n!^(3*n) * Product_{k=1..n} binomial(n + 1/k^2, n).
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%I #8 Jun 24 2023 16:56:20

%S 1,2,270,74692800,419731620267960000,

%T 252716802910471719823692648960000,

%U 59736659298524125157504488525739821430187940800000000,16079377413231597423103950774423398920424350187193326745026311068057600000000000

%N a(n) = n!^(3*n) * Product_{k=1..n} binomial(n + 1/k^2, n).

%F a(n) ~ n!^(3*n) * n^(Pi^2/6) / A303670.

%F a(n) ~ n^(3*n*(2*n+1)/2 + Pi^2/6) * (2*Pi)^(3*n/2) / exp(3*n^2 - 1/4 - 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.

%p a:= n-> n!^(3*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!^(3*n) * Product[Binomial[n + 1/k^2, n], {k, 1, n}], {n, 1, 8}]

%Y Cf. A303670, A306760, A306774, A324589, A324597.

%K nonn

%O 0,2

%A _Vaclav Kotesovec_, Mar 09 2019

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