%I #17 Feb 15 2024 04:16:51
%S 1,1,2,8,29,105,414,1695,7046,29853,128644,561262,2474142,11006108,
%T 49343508,222715440,1011217425,4615519083,21165513228,97467424198,
%U 450541090701,2089777230606,9723511785608,45371996501895,212271904284993,995513843930049,4679212044797252
%N Expansion of (1/x) * Series_Reversion( x * (1-x) / (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(2*n-3*k,n-3*k).
%F a(n) = (1/(n+1)) * [x^n] ( 1/(1-x) * (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)/(1+x^3)^3)/x)
%o (PARI) a(n, s=3, t=3, u=1) = 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. A071969, A369266.
%Y Cf. A369269, A369270.
%K nonn
%O 0,3
%A _Seiichi Manyama_, Jan 18 2024