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a(n) = floor((3n)!/(n!^3 a(n-1)), where a(0)=1.
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%I #4 Jul 15 2024 23:34:02

%S 1,6,15,112,309,2449,7004,56977,166128,1371673,4046880,33736358,

%T 100328902,842011587,2518460195,21241145637,63807550099,540215987658,

%U 1628274148825,13827134974963,41789656029594,355743023641567,1077551461258952,9191485155260759

%N a(n) = floor((3n)!/(n!^3 a(n-1)), where a(0)=1.

%t a[0] = 1; a[n_] := a[n] = Floor[(3 n)!/( (n!)^3*a[n - 1])];

%t Table[a[n], {n, 0, 30}]

%Y Cf. A372989.

%K nonn

%O 0,2

%A _Clark Kimberling_, Jul 15 2024