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Expansion of (1/x) * Series_Reversion( x / (1+x+x^3/(1+x)) ).
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%I #13 Jan 27 2024 10:31:47

%S 1,1,1,2,4,7,14,30,62,131,287,629,1385,3096,6967,15735,35782,81823,

%T 187781,432689,1000919,2322584,5405094,12614260,29512587,69205602,

%U 162634994,382961435,903431963,2134945637,5053385429,11979405642,28438444486,67601886687

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

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

%o (PARI) my(N=40, x='x+O('x^N)); Vec(serreverse(x/(1+x+x^3/(1+x)))/x)

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

%Y Cf. A126042.

%K nonn

%O 0,4

%A _Seiichi Manyama_, Jan 26 2024