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a(n) = Sum_{i+j+k=n, 0<=i<=n, 0<=j<=n, 0<=k<=n} (n+i+j)!/((i+j)! * j! * k!).
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%I #13 Oct 30 2021 05:35:10

%S 1,5,43,495,7281,133173,2945755,76769759,2306295265,78492222693,

%T 2985018589323,125449316558415,5773653823774929,288808141870191765,

%U 15601413322486382523,905170780889312826303,56136189828704013001665

%N a(n) = Sum_{i+j+k=n, 0<=i<=n, 0<=j<=n, 0<=k<=n} (n+i+j)!/((i+j)! * j! * k!).

%H Seiichi Manyama, <a href="/A092471/b092471.txt">Table of n, a(n) for n = 0..200</a>

%F From _Vaclav Kotesovec_, Oct 30 2021: (Start)

%F a(n) ~ 2^(2*n + 1/2) * n^n / exp(n - 3/2).

%F Recurrence: n*(2*n - 5)*a(n) = (2*n - 5)*(2*n - 1)*(2*n + 3)*a(n-1) - (2*n - 3)*(24*n^2 - 72*n + 29)*a(n-2) + (2*n - 9)*(2*n - 5)*(2*n - 1)*a(n-3) + (n-3)*(2*n - 1)*a(n-4). (End)

%o (PARI) a(n)=sum(i=0,n,sum(j=0,n,sum(k=0,n,if(i+j+k-n,0,(n+i+j)!/(i+j)!/j!/k!))))

%K nonn

%O 0,2

%A _Benoit Cloitre_, Mar 25 2004