%I #32 Oct 15 2024 15:42:30
%S 2,3,12,63,378,2457,16848,120042,880308,6602310,50417640,390736710,
%T 3065780340,24307258410,194458067280,1567818167445,12726994535730,
%U 103937122041795,853378475711580,7040372424620535,58334514375427290
%N Recursive sequence (n+1)*a(n) = 3*(3*n-2)*a(n-1).
%C This is built akin to (n+1)*C(n) = 2*(2*n-1)*C(n-1) for the Catalan numbers A000108.
%H Vincenzo Librandi, <a href="/A155579/b155579.txt">Table of n, a(n) for n = 0..1000</a>
%H Vladimir Kruchinin, D. V. Kruchinin, <a href="http://arxiv.org/abs/1103.2582">Composita and their properties </a>, arXiv:1103.2582 [math.CO], 2011-2013.
%F (n+1)*a(n) = 3*(3*n-2)*a(n-1).
%F From _Vladimir Kruchinin_, Sep 20 2010: (Start)
%F G.f.: A(x) = 1/3*(1-(1-9*x)^(2/3)).
%F a(n) = 3^(2*n-1)*sum(binomial(k,n-k)*2^(2*k-n)*(-1)^(n-k)*(if k=1 then (1/3) else 1/k*(1/3)^k*sum(binomial(i,k-1-i)*(-1/3)^(k-1-i)*binomial(i+k-1,k-1),i,1,k-1)),k,1,n),n>0. (End)
%F From _Vaclav Kotesovec_, Jul 20 2019: (Start)
%F a(n) = 2 * 3^(2*n) * Gamma(n + 1/3) / (Gamma(1/3) * Gamma(n+2)).
%F a(n) ~ 2 * 3^(2*n) / (Gamma(1/3) * n^(5/3)). (End)
%F a(n) = 3*A185047(n-1) for n >= 1. - _Peter Bala_, Oct 14 2024
%t a[0] = 1; a[n_] := a[n] = ((3*n - 2)/(n + 1))*a[n - 1];
%t Table[2*3^(n)*a[n], {n, 0, 30}]
%o (Maxima) a(n):=3^(2*n-1)*sum(binomial(k,n-k)*2^(2*k-n)*(-1)^(n-k)*(if k=1 then (1/3) else 1/k*(1/3)^k*sum(binomial(i,k-1-i)*(-1/3)^(k-1-i)*binomial(i+k-1,k-1),i,1,k-1)),k,1,n); /* _Vladimir Kruchinin_, Sep 20 2010 */
%Y Cf. A185047, A283150.
%K nonn,easy
%O 0,1
%A _Roger L. Bagula_, Jan 24 2009