login
G.f. satisfies A(x) = 1 + x/A(x)*(1 + 1/A(x)).
8

%I #30 Oct 21 2023 11:09:47

%S 1,2,-6,34,-238,1858,-15510,135490,-1223134,11320066,-106830502,

%T 1024144482,-9945711566,97634828354,-967298498358,9659274283650,

%U -97119829841854,982391779220482,-9990160542904134,102074758837531810,-1047391288012377774,10788532748880319298

%N G.f. satisfies A(x) = 1 + x/A(x)*(1 + 1/A(x)).

%F G.f.: A(x) = 1/B(-x) where B(x) is the g.f. of A027307.

%F a(n) = (-1)^(n-1) * (1/n) * Sum_{k=0..n} binomial(n,k) * binomial(2*n+k-2,n-1) = (-1)^(n-1) * A108424(n) for n > 0.

%F D-finite with recurrence n*(2*n-1)*a(n) +3*(6*n^2-10*n+3)*a(n-1) +(-46*n^2+227*n-279)*a(n-2) +2*(n-3)*(2*n-7)*a(n-3)=0. - _R. J. Mathar_, Jul 25 2023

%F a(n) ~ c*(-1)^(n-1)*4^n*2F1([-n, 2*n-1], [n], -1)*n^(-3/2), with c = 1/(4*sqrt(Pi)) = A087197/4. - _Stefano Spezia_, Oct 21 2023

%p A364394 := proc(n)

%p if n = 0 then

%p 1;

%p else

%p (-1)^(n-1)*add( binomial(n,k) * binomial(2*n+k-2,n-1),k=0..n)/n ;

%p end if;

%p end proc:

%p seq(A364394(n),n=0..80); # _R. J. Mathar_, Jul 25 2023

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

%Y Cf. A112478, A364396, A364398.

%Y Cf. A027307, A087197, A108424.

%K sign

%O 0,2

%A _Seiichi Manyama_, Jul 22 2023