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G.f. satisfies A(x) = 1 + x/A(x)^3*(1 + 1/A(x)).
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%I #30 Oct 21 2023 11:08:52

%S 1,2,-14,162,-2270,35234,-582958,10076354,-179802046,3287029698,

%T -61246957902,1158889656930,-22207636788894,430106644358242,

%U -8405699952109166,165557885912786818,-3282954949273886590,65487784219460233602,-1313225110482709157518

%N G.f. satisfies A(x) = 1 + x/A(x)^3*(1 + 1/A(x)).

%F a(n) = (-1)^(n-1) * (1/n) * Sum_{k=0..n} binomial(n,k) * binomial(4*n+k-2,n-1) for n > 0.

%F D-finite with recurrence 2*n*(462919*n -714364)*(4*n-3) *(2*n-1)*(4*n-1)*a(n) +(625365036*n^5 -2723245780*n^4 +4202103460*n^3 -2471353250*n^2 +81675089*n +289227120)*a(n-1) +(-484851248*n^5 +5501638270*n^4 -25122933600*n^3 +57439557800*n^2 -65490996232*n +29691239955)*a(n-2) +(2*n-5)*(652184*n -1103659)*(4*n-13) *(n-3)*(4*n-11)*a(n-3)=0. - _R. J. Mathar_, Jul 25 2023

%F a(n) ~ c*(-1)^(n-1)*256^n*27^(-n)*2F1([1-n, 4*n], [3*n], -1)*n^(-3/2), with c = sqrt(3/(32*Pi)). - _Stefano Spezia_, Oct 21 2023

%p A364398 := proc(n)

%p if n = 0 then

%p 1;

%p else

%p (-1)^(n-1)*add( binomial(n,k) * binomial(4*n+k-2,n-1),k=0..n)/n ;

%p end if;

%p end proc:

%p seq(A364398(n),n=0..70); # _R. J. Mathar_, Jul 25 2023

%t nmax = 18; A[_] = 1; Do[A[x_] = 1+x/A[x]^3*(1+1/A[x]) + O[x]^(nmax+1) // Normal, {nmax}]; CoefficientList[A[x], x] (* _Jean-François Alcover_, Oct 21 2023 *)

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

%Y Cf. A112478, A364394, A364396.

%Y Cf. A364399, A364400.

%Y Cf. A260332.

%K sign

%O 0,2

%A _Seiichi Manyama_, Jul 22 2023