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

%I #7 Mar 02 2024 10:37:41

%S 1,2,5,16,61,256,1133,5191,24403,117066,570835,2821026,14097839,

%T 71121660,361718339,1852640518,9547375955,49469352300,257564997407,

%U 1346840074300,7070283106575,37246786128714,196849114734855,1043398553112059,5545408681615257

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

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

%F a(n) = Sum_{k=0..n} binomial(n,k) * b(k), where g.f. B(x) = Sum_{k>=0} b(k)*x^k satisfies B(x) = (1/x) * Series_Reversion( x*(1-x-x^3) ).

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

%Y Cf. A218225, A370798.

%Y Cf. A049140.

%K nonn

%O 0,2

%A _Seiichi Manyama_, Mar 02 2024