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a(n) = (1/(4n+2))*M(3n; n,n,n).
2

%I #22 Aug 05 2024 06:28:40

%S 1,9,120,1925,34398,659736,13302432,278397405,5996669250,132166590270,

%T 2967978162240,67694635250424,1564409223571600,36561597688116000,

%U 862822254602816640,20535537339485077005,492426552811873991850,11886753074132473787250,288645723487776840570000

%N a(n) = (1/(4n+2))*M(3n; n,n,n).

%C Number of standard Young tableaux of shape (n,n,{1}^n). - _Alois P. Heinz_, Apr 05 2013

%H Alois P. Heinz, <a href="/A024487/b024487.txt">Table of n, a(n) for n = 1..250</a>

%F a(n) ~ 3^(3*n+1/2) / (8*Pi*n^2). - _Vaclav Kotesovec_, Sep 06 2014

%F a(n) = A006480(n) / (4*n + 2) if n>0. - _Michael Somos_, Oct 25 2014

%F D-finite with recurrence: n^2*(2*n+1)*a(n) -3*(3*n-1)*(2*n-1)*(3*n-2)*a(n-1)=0. - _R. J. Mathar_, Apr 27 2020

%e G.f. = x + 9*x^2 + 120*x^3 + 1925*x^4 + 34398*x^5 + 659736*x^6 + ...

%p with(combinat):

%p a:= n-> multinomial(3*n, n$3)/(4*n+2):

%p seq(a(n), n=1..30); # _Alois P. Heinz_, Apr 05 2013

%t a[ n_] := If[ n < 1, 0, (3 n)! / (n!^3 (4 n + 2))]; (* _Michael Somos_, Oct 25 2014 *)

%o (PARI) {a(n) = if( n<1, 0, (3*n)! / (n!^3 * (4*n + 2)))}; /* _Michael Somos_, Oct 25 2014 */

%Y Cf. A006480.

%K nonn,easy

%O 1,2

%A _Clark Kimberling_

%E Corrected and extended by _Alois P. Heinz_, Apr 05 2013