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a(n) = Sum_{k=1..n-1} binomial (2*k, k)^n.
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%I #13 Jun 06 2021 09:04:49

%S 4,224,161312,1683907808,256213978094784,575112148876911852416,

%T 19248204431728945392010740480,9687459136669902998216039379883774976,

%U 73815961078227084527800998811241905249902260224,8562177846610881578580018959490439733543225146878872883200

%N a(n) = Sum_{k=1..n-1} binomial (2*k, k)^n.

%C 2^n divides a(n).

%C p divides a(p) for prime p of the form p = 6k + 1.

%F a(n) = Sum_{k=1..n-1} binomial(2*k, k)^n.

%F a(n) ~ exp(3/8) * 4^(n^2-n) / (Pi^(n/2) * n^(n/2)). - _Vaclav Kotesovec_, Mar 03 2014

%t Table[ Sum[ Binomial[2 k, k]^n, {k, 1, n - 1}], {n, 2, 13}]

%Y Cf. A211611, A238717.

%K nonn

%O 2,1

%A _Alexander Adamchuk_, Apr 17 2012