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G.f. A(x) satisfies A(x) = 1 + x * A(x)^4 / (1 - x).
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%I #25 Sep 28 2025 09:46:40

%S 1,1,5,31,219,1678,13570,114014,985542,8708099,78298727,714105907,

%T 6590200215,61427125994,577456943614,5468604044500,52122539760992,

%U 499613409224137,4813105582181533,46576519080852235,452545041339982871,4413071971740021275,43177663974461532959

%N G.f. A(x) satisfies A(x) = 1 + x * A(x)^4 / (1 - x).

%H Seiichi Manyama, <a href="/A349331/b349331.txt">Table of n, a(n) for n = 0..985</a>

%F a(n) = Sum_{k=0..n} binomial(n-1,k-1) * binomial(4*k,k) / (3*k+1).

%F a(n) ~ 283^(n + 1/2) / (2^(7/2) * sqrt(Pi) * n^(3/2) * 3^(3*n + 3/2)). - _Vaclav Kotesovec_, Nov 15 2021

%F G.f.: 1 + Series_Reversion( x / (x+(1+x)^4) ). - _Seiichi Manyama_, Sep 28 2025

%p a:= n-> coeff(series(RootOf(1+x*A^4/(1-x)-A, A), x, n+1), x, n):

%p seq(a(n), n=0..22); # _Alois P. Heinz_, Nov 15 2021

%t nmax = 22; A[_] = 0; Do[A[x_] = 1 + x A[x]^4/(1 - x) + O[x]^(nmax + 1) // Normal, nmax + 1]; CoefficientList[A[x], x]

%t Table[Sum[Binomial[n - 1, k - 1] Binomial[4 k, k]/(3 k + 1), {k, 0, n}], {n, 0, 22}]

%o (PARI) {a(n) = my(A=[1]); for(m=1, n, A=concat(A, 0);

%o A[#A] = 1 + sum(k=1, m-1, (polcoeff(Ser(A)^4, k)) )); A[n+1]}

%o for(n=0, 30, print1(a(n), ", ")) \\ _Vaclav Kotesovec_, Nov 23 2024, after _Paul D. Hanna_

%Y Cf. A002212, A002293, A307678, A317133, A346646 (partial sums), A349332, A349333, A349334, A349335.

%K nonn

%O 0,3

%A _Ilya Gutkovskiy_, Nov 15 2021