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a(n) = 8^n * binomial(n * 3/2, n).
2

%I #19 Jan 31 2024 07:21:20

%S 1,12,192,3360,61440,1153152,22020096,425677824,8304721920,

%T 163176499200,3224446697472,64012657213440,1275708366127104,

%U 25506581874278400,511404848311173120,10278423735852072960,207016682596362878976,4177272328882468945920,84430333294202899660800

%N a(n) = 8^n * binomial(n * 3/2, n).

%H Paolo Xausa, <a href="/A358367/b358367.txt">Table of n, a(n) for n = 0..750</a>

%F a(n) ~ c*2^(2*n)*3^(3*n/2)/sqrt(n) where c = sqrt(3/(2*Pi)). - _Stefano Spezia_, Nov 14 2022

%p seq(8^n * binomial(n*3/2, n), n = 0..18);

%t A358367[n_] := 8^n*Binomial[3/2*n, n];

%t Array[A358367, 20, 0] (* _Paolo Xausa_, Jan 31 2024 *)

%o (Python)

%o from sympy import binomial, S

%o def A358367(n): return (1<<n*3)*binomial(3*S.Half*n,n) # _Chai Wah Wu_, Nov 14 2022

%o (PARI) a(n) = 8^n * binomial(n * 3/2, n); \\ _Michel Marcus_, Nov 15 2022

%K nonn

%O 0,2

%A _Peter Luschny_, Nov 14 2022