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

%I #8 Jan 25 2024 07:49:32

%S 1,6,47,420,4059,41316,436345,4737018,52535950,592667532,6779699073,

%T 78458218746,916886214115,10805128064100,128260666769895,

%U 1532180536574580,18405744106135914,222204347510440092,2694506677864591810,32804976554127379680,400837173223351237295

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

%H <a href="/index/Res#revert">Index entries for reversions of series</a>

%F a(n) = (1/(n+1)) * Sum_{k=0..n} binomial(2*n+2,k) * binomial(4*n-2*k+4,n-k).

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

%o (PARI) a(n) = sum(k=0, n, binomial(2*n+2, k)*binomial(4*n-2*k+4, n-k))/(n+1);

%Y Cf. A369503, A369504.

%Y Cf. A003645, A369505.

%K nonn

%O 0,2

%A _Seiichi Manyama_, Jan 25 2024