%I #16 Feb 15 2024 04:22:06
%S 1,2,7,33,173,962,5586,33498,205846,1289386,8202247,52845855,
%T 344129832,2261377872,14976646685,99863119809,669860309538,
%U 4517037850220,30603008068997,208211448723097,1421986458302466,9745007758311114,66993247112160800
%N Expansion of (1/x) * Series_Reversion( x * (1-x)^2 / (1+x^3)^3 ).
%H P. Bala, <a href="/A251592/a251592.pdf">Fractional iteration of a series inversion operator</a>
%H <a href="/index/Res#revert">Index entries for reversions of series</a>
%F a(n) = (1/(n+1)) * Sum_{k=0..floor(n/3)} binomial(3*n+3,k) * binomial(3*n-3*k+1,n-3*k).
%F a(n) = (1/(n+1)) * [x^n] ( 1/(1-x)^2 * (1+x^3)^3 )^(n+1). - _Seiichi Manyama_, Feb 14 2024
%o (PARI) my(N=30, x='x+O('x^N)); Vec(serreverse(x*(1-x)^2/(1+x^3)^3)/x)
%o (PARI) a(n, s=3, t=3, u=2) = sum(k=0, n\s, binomial(t*(n+1), k)*binomial((u+1)*(n+1)-s*k-2, n-s*k))/(n+1);
%Y Cf. A369268, A369270.
%Y Cf. A369265, A369267.
%K nonn
%O 0,2
%A _Seiichi Manyama_, Jan 18 2024