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Expansion of (1/x) * Series_Reversion( x / ((1+x)^3+x^4) ).
2

%I #9 Jan 15 2024 09:03:32

%S 1,3,12,55,274,1443,7905,44593,257305,1511553,9010170,54361486,

%T 331336454,2037132958,12619056108,78682008194,493427982703,

%U 3110202012353,19693920616872,125214061831251,799059649687239,5116372686471627,32860439054510610

%N Expansion of (1/x) * Series_Reversion( x / ((1+x)^3+x^4) ).

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

%F a(n) = (1/(n+1)) * Sum_{k=0..floor(n/4)} binomial(n+1,k) * binomial(3*n-3*k+3,n-4*k).

%o (PARI) my(N=30, x='x+O('x^N)); Vec(serreverse(x/((1+x)^3+x^4))/x)

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

%Y Cf. A127902, A369126, A369158.

%K nonn

%O 0,2

%A _Seiichi Manyama_, Jan 15 2024