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%I #15 Sep 22 2019 11:48:28
%S 6,300,47040,14968800,7991343360,6422134118400,7240779786240000,
%T 10899907851216384000,21115899915689779200000,
%U 51167604130438090014720000,151615591667542267763097600000,539306547534817468755148800000000,2267795648217238975260881584128000000
%N a(n) = (2/3)*(2*n+1)*(2*n-1)!*binomial(3*n,2*n).
%H Robert Israel, <a href="/A081321/b081321.txt">Table of n, a(n) for n = 1..194</a>
%H M. Bousquet and C. Lamathe, <a href="http://dx.doi.org/10.1016/j.disc.2004.11.015">Enumeration of solid trees according to edge number and edge degree distribution</a>, Discr. Math., 298 (2005), 115-141.
%F (27*n^4+27*n^3+6*n^2)*a(n)+(108*n^4+648*n^3+1373*n^2+1233*n+400)*a(n+1)+(-4*n^2-12*n-8)*a(n+2) = 0. - _Robert Israel_, Mar 29 2017
%F a(n) = (2n+1)(3n-1)!/n!. - _Jean-François Alcover_, Sep 22 2019
%p seq(2/3*(2*n+1)*(2*n-1)!*binomial(3*n,2*n),n=1..30); # _Robert Israel_, Mar 29 2017
%t a[n_] := (2n+1)(3n-1)!/n!;
%t Array[a, 30] (* _Jean-François Alcover_, Sep 22 2019 *)
%K nonn
%O 1,1
%A _N. J. A. Sloane_, May 22 2003
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