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a(n) = f(4*n+2)/(f(n-1)*f(n)*f(n+1)*f(n+2)), where f(k) = k!.
4

%I #14 Sep 22 2025 16:01:18

%S 60,12600,2522520,514594080,107550162720,22969641895200,

%T 4995897112206000,1103284402265073600,246784661070292144800,

%U 55803246694136969227200,12736017918577260592084800,2930174751896446667689440000,678879630375093886522676256000,158257286142440155623613107216000

%N a(n) = f(4*n+2)/(f(n-1)*f(n)*f(n+1)*f(n+2)), where f(k) = k!.

%C These are multinomial coefficients.

%H Clark Kimberling, <a href="/A248708/b248708.txt">Table of n, a(n) for n = 1..1000</a>

%F a(n) ~ 2^(8*n+7/2) / (Pi^(3/2) * n^(3/2)). - _Vaclav Kotesovec_, Oct 19 2014

%e a(2) = 10!/(1!*2!*3!*4!) = 12600.

%t Table[(4 n + 2)!/((n - 1)! n! (n + 1)! (n + 2)!), {n, 1, 20}]

%o (SageMath) [factorial(4*n + 2)/(factorial(n - 1)* factorial(n)*factorial(n + 1)*factorial(n + 2)) for n in range(1,14)] # _Stefano Spezia_, Aug 16 2024

%Y Cf. A000142, A001700, A248707, A248709, A248710.

%K nonn,easy

%O 1,1

%A _Clark Kimberling_, Oct 12 2014