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

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

%S 1,4,22,142,1005,7546,59033,475962,3927204,33001024,281449964,

%T 2429922400,21196031340,186521336460,1653830553417,14761130834428,

%U 132516050272100,1195778542160992,10839917478886459,98671228898404032,901509955793840923

%N Expansion of (1/x) * Series_Reversion( x / ((1+x)^2+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-2*k+4,n-3*k).

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

%o (PARI) a(n) = sum(k=0, n\3, binomial(2*n+2, k)*binomial(4*n-2*k+4, n-3*k))/(n+1);

%Y Cf. A369502, A369503.

%Y Cf. A369212.

%K nonn

%O 0,2

%A _Seiichi Manyama_, Jan 25 2024