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a(n) = lcm({n! * binomial(n, k) for k = 0..n}).
1

%I #11 Feb 15 2023 09:41:38

%S 1,1,4,18,288,1200,43200,529200,11289600,91445760,9144576000,

%T 92207808000,13277924352000,160283515392000,2094371267788800,

%U 58904191906560000,15079473128079360000,242109318556385280000,78443419212268830720000,1415903716781452394496000

%N a(n) = lcm({n! * binomial(n, k) for k = 0..n}).

%F a(n) = n! * lcm({k for k = 1..n+1}) / (n+1) = n! * LCM(n + 1) / (n + 1).

%F a(n) / a(n-1) = n^2 if and only if n + 1 is prime, for n >= 1.

%p a := n -> ilcm(seq(n!*binomial(n, k), k=0..n)):

%p seq(a(n), n = 0..19);

%o (Python)

%o from math import factorial, lcm

%o def A360283(n): return factorial(n)*lcm(*(i for i in range(1,n+2)))//(n+1) # _Chai Wah Wu_, Feb 15 2023

%Y Cf. A002397, A003418 (LCM), A005722, A021012, A196347.

%K nonn

%O 0,3

%A _Peter Luschny_, Feb 14 2023