%I #17 Feb 15 2024 04:30:32
%S 1,3,15,94,657,4902,38236,308025,2542965,21401780,182934144,
%T 1583745114,13858675065,122379042879,1089156646584,9759520978270,
%U 87975115569873,797233088237190,7258632128721117,66367727370376632,609132332475784548,5610015849998778144
%N Expansion of (1/x) * Series_Reversion( x * (1-x)^3 * (1-x^3)^3 ).
%H Peter 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+k+2,k) * binomial(4*n-3*k+2,n-3*k).
%F a(n) = (1/(n+1)) * [x^n] 1/( (1-x)^3 * (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)^3*(1-x^3)^3)/x)
%o (PARI) a(n, s=3, t=3, u=3) = sum(k=0, n\s, binomial(t*(n+1)+k-1, k)*binomial((u+1)*(n+1)-s*k-2, n-s*k))/(n+1);
%Y Cf. A369299, A369300.
%Y Cf. A369270.
%K nonn
%O 0,2
%A _Seiichi Manyama_, Jan 18 2024