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a(n) = f(5*n)/(f(n-2)*f(n-1)*f(n)*f(n+1)*f(n+2)), where f(k) = k!.
4

%I #13 Aug 16 2024 19:07:16

%S 12600,37837800,97772875200,247365374256000,629483036137956000,

%T 1621828071329658192000,4234824783966213204768000,

%U 11198994141198650820008976000,29959571750765218953790679280000,80980722442318386832096206093840000,220917676017677910841226480887103040000

%N a(n) = f(5*n)/(f(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="/A248709/b248709.txt">Table of n, a(n) for n = 2..200</a>

%F a(n) ~ 5^(5*n+1/2) / (4*Pi^2*n^2). - _Vaclav Kotesovec_, Oct 19 2014

%e a(3) = 15!/(1!*2!*3!*4!*5!) = 37837800.

%t Table[(5 n)!/((n - 2)! (n - 1)! n! (n + 1)! (n + 2)!), {n, 2, 20}]

%o (Sage) [factorial(5*n)/(factorial(n - 2)*factorial(n - 1)*factorial(n)*factorial(n + 1)*factorial(n + 2)) for n in range(2,14)] # _Stefano Spezia_, Aug 16 2024

%Y Cf. A000142, A001700, A248707, A248708, A248710.

%K nonn,easy

%O 2,1

%A _Clark Kimberling_, Oct 12 2014