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

%I #10 Jan 15 2024 09:03:24

%S 1,2,5,14,43,142,495,1794,6686,25436,98311,384826,1522283,6075838,

%T 24437937,98956270,403080170,1650502292,6790018182,28050896964,

%U 116322826479,484029536374,2020386475025,8457397801150,35495812337114,149336478356692,629685490668799

%N Expansion of (1/x) * Series_Reversion( x / ((1+x)^2+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(2*n-2*k+2,n-4*k).

%o (PARI) my(N=30, x='x+O('x^N)); Vec(serreverse(x/((1+x)^2+x^4))/x)

%o (PARI) a(n) = sum(k=0, n\4, binomial(n+1, k)*binomial(2*n-2*k+2, n-4*k))/(n+1);

%Y Cf. A127902, A369126, A369159.

%K nonn

%O 0,2

%A _Seiichi Manyama_, Jan 15 2024