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

%I #11 Apr 29 2024 09:06:55

%S 1,1,1,3,9,21,54,161,470,1347,4007,12199,37141,113802,352905,1101969,

%T 3455220,10891968,34515825,109814395,350616323,1123368287,3610647348,

%U 11637410625,37605280548,121812321775,395455199269,1286446544052,4192913001804,13690359696969

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

%o (PARI) my(N=40, x='x+O('x^N)); Vec(serreverse(x*(1+x)/(1+x+x^3)^2)/x)

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

%Y Cf. A369483, A369484, A369485.

%Y Cf. A372371.

%K nonn

%O 0,4

%A _Seiichi Manyama_, Apr 29 2024