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Expansion of (1/x) * Series_Reversion( x / ((1+x)^2 * (1+x+x^3)^2) ).
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%I #11 Jan 24 2024 05:56:58

%S 1,4,22,142,1007,7590,59683,484112,4021061,34029532,292373296,

%T 2543542676,22360917140,198341377680,1772860026933,15952960500612,

%U 144397901220980,1313835276189792,12009823111155481,110240431974732436,1015727265740887873

%N Expansion of (1/x) * Series_Reversion( x / ((1+x)^2 * (1+x+x^3)^2) ).

%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(2*n+2,k) * binomial(4*n-k+4,n-3*k).

%o (PARI) my(N=30, x='x+O('x^N)); Vec(serreverse(x/((1+x)^2*(1+x+x^3)^2))/x)

%o (PARI) a(n, s=3, t=2, u=2) = sum(k=0, n\s, binomial(t*(n+1), k)*binomial((t+u)*(n+1)-k, n-s*k))/(n+1);

%Y Cf. A369483, A369484.

%K nonn

%O 0,2

%A _Seiichi Manyama_, Jan 23 2024