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%I #15 Jun 24 2023 16:57:54
%S 1,2,918,11592504000,86712397842439769400000,
%T 3472997049383321958747830928094241894400000,
%U 4152034082374349458781848863476555783741415883758270213129361920000000
%N a(n) = n!^(4*n) * Product_{k=1..n} binomial(n + 1/k^3, n).
%C In general, for m > 1, Product_{k=1..n} binomial(n + 1/k^m, n) ~ n^Zeta(m) / c(m), where c(m) = Product_{j>=1} Gamma(1 + 1/j^m)).
%C Equivalently, c(m) = -gamma * Zeta(m) + Sum_{k>=2} (-1)^k*Zeta(k)*Zeta(m*k)/k, where gamma is the Euler-Mascheroni constant A001620.
%F a(n) ~ n!^(4*n) * n^Zeta(3) / (Product_{j>=1} Gamma(1 + 1/j^3)).
%F a(n) ~ n^(4*n^2 + 2*n + Zeta(3)) * (2*Pi)^(2*n) / exp(4*n^2 - 1/3 - gamma*Zeta(3) + c), where c = A306778 = Sum_{k>=2} (-1)^k*Zeta(k)*Zeta(3*k)/k.
%p a:= n-> n!^(4*n)*mul(binomial(n+1/k^3, n), k=1..n):
%p seq(a(n), n=0..7); # _Alois P. Heinz_, Jun 24 2023
%t Table[n!^(4*n) * Product[Binomial[n + 1/j^3, n], {j, 1, n}], {n, 1, 8}]
%Y Cf. A306760, A324596, A306778.
%K nonn
%O 0,2
%A _Vaclav Kotesovec_, Mar 09 2019
%E a(0)=1 prepended by _Alois P. Heinz_, Jun 24 2023
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