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Expansion of (1/x) * Series_Reversion( x / ((1+x)^2 * (1+x^3)^2) ).
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%I #21 Jan 23 2024 11:14:12

%S 1,2,5,16,62,264,1170,5310,24599,116090,556569,2703098,13268900,

%T 65721840,328050639,1648535856,8333536002,42348587700,216211838178,

%U 1108514508608,5704874555112,29460504457692,152612723209700,792833380805160,4129639139612133

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

%o (PARI) my(N=30, x='x+O('x^N)); Vec(serreverse(x/((1+x)^2*(1+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(u*(n+1), n-s*k))/(n+1);

%Y Cf. A369442, A369443.

%K nonn

%O 0,2

%A _Seiichi Manyama_, Jan 23 2024