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a(n) = Sum_{k=0..n} binomial(2*n,k) * binomial(3*n-k-1,n-k).
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%I #11 Sep 17 2024 10:45:39

%S 1,4,32,292,2816,28004,284000,2919620,30316544,317222212,3339504032,

%T 35329425124,375282559232,4000059761572,42760427177696,

%U 458259268924292,4921911787962368,52965710906750084,570951048018417440,6164049197776406180,66639047280436354816

%N a(n) = Sum_{k=0..n} binomial(2*n,k) * binomial(3*n-k-1,n-k).

%F a(n) = [x^n] ( (1+x)^2/(1-x)^2 )^n.

%F The g.f. exp( Sum_{k>=1} a(k) * x^k/k ) has integer coefficients and equals (1/x) * Series_Reversion( x*(1-x)^2/(1+x)^2 ).

%F a(n) = 2 * A103885(n) for n >= 1. - _Peter Bala_, Sep 16 2024

%o (PARI) a(n) = sum(k=0, n, binomial(2*n, k)*binomial(3*n-k-1, n-k));

%Y Cf. A370098, A370102.

%Y Cf. A032349, A103885.

%K nonn,easy

%O 0,2

%A _Seiichi Manyama_, Feb 10 2024