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a(n) = binomial(2*n, n) * n^n.
2

%I #17 Nov 29 2023 17:22:17

%S 1,2,24,540,17920,787500,43110144,2826399576,215922769920,

%T 18836384175180,1847560000000000,201267982422458952,

%U 24110526523754151936,3150082833623386551800,445775933018180704665600,67925014299030761718750000,11087976122055526032070410240

%N a(n) = binomial(2*n, n) * n^n.

%H Michael De Vlieger, <a href="/A367271/b367271.txt">Table of n, a(n) for n = 0..322</a>

%F a(n) = A059297(2*n, n).

%F a(n) ~ 2^(2*n)*n^n/sqrt(n*Pi). - _Stefano Spezia_, Nov 29 2023

%p seq(binomial(2*n, n) * n^n, n = 0..16);

%t A367271[n_]:=If[n==0,1,Binomial[2n,n]n^n];

%t Array[A367271,20,0] (* _Paolo Xausa_, Nov 29 2023 *)

%Y Cf. A059297, A293270.

%K nonn

%O 0,2

%A _Peter Luschny_, Nov 11 2023