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

%I #5 Jan 16 2024 08:32:41

%S 1,2,5,15,50,177,652,2473,9594,37892,151846,615859,2523217,10427471,

%T 43415259,181941198,766841846,3248517320,13823977350,59067577266,

%U 253315964424,1089998388418,4704475230340,20361365646315,88351705071583,384280788724692,1675063399090659

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

%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(n+1,k) * binomial(2*n-2*k+2,n-3*k).

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

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

%Y Cf. A002212, A071356, A369158.

%Y Cf. A071879, A192132.

%K nonn

%O 0,2

%A _Seiichi Manyama_, Jan 16 2024