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%I #15 Jun 06 2021 09:04:54
%S 1,9,642,540982,5496576970,698491214560174,1147342896257677900291,
%T 25005346993500437111980892595,7381619397278667883874693730628586499,
%U 30009934325456999669083059570156145437948880627,1703283943023520710008632777768663744247664926649672215939
%N a(n) = Sum_{k=1..n-1} C(k)^n, where C(k) is a Catalan number.
%C The C(k) are the Catalan numbers, C(k) = A000108(k) = (2k)!/(k!*(k+1)!) = C(2*k,k)/(k+1).
%C p divides a(p) for prime p of the form p = 6k + 1 (A002476).
%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/CatalanNumber.html">Catalan Number</a>
%F a(n) = Sum_{k=1..n-1} binomial(2*k, k)/(k+1)^n.
%F a(n) ~ exp(3/8) * 4^(n^2-n) / (Pi^(n/2) * n^(3*n/2)). - _Vaclav Kotesovec_, Mar 03 2014
%t Table[ Sum[ (Binomial[2 k, k]/(k + 1))^n, {k, 1, n - 1}], {n, 2, 13}]
%Y Cf. A000108, A002476, A211610, A238717.
%K nonn
%O 2,2
%A _Alexander Adamchuk_, Apr 17 2012