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

%I #11 Mar 23 2024 10:53:04

%S 1,2,5,13,34,87,212,471,858,740,-3674,-29291,-141951,-576379,-2111677,

%T -7161898,-22646026,-66408560,-176815194,-403468266,-641064024,

%U 337909918,9269952852,55908644837,256989808831,1033152002312,3792152422259,12903091079930,40749582818221

%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)} (-1)^k * 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, (-1)^k*binomial(n+1, k)*binomial(2*n-2*k+2, n-3*k))/(n+1);

%Y Cf. A001006, A371427.

%Y Cf. A369212.

%K sign

%O 0,2

%A _Seiichi Manyama_, Mar 23 2024