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a(n) = Sum_{k=0..n-1} ( 1 + a(k) * a(n-k-1) ) for n >= 1, a(0) = 1.
1

%I #22 Feb 16 2024 01:31:15

%S 1,2,6,19,66,249,996,4148,17784,77939,347516,1571304,7187288,33196887,

%T 154611392,725284721,3423760262,16251813715,77523741208,371428985796,

%U 1786623827240,8624669381161,41769772877288,202893913979291,988224403828490,4825331506973445

%N a(n) = Sum_{k=0..n-1} ( 1 + a(k) * a(n-k-1) ) for n >= 1, a(0) = 1.

%H Seiichi Manyama, <a href="/A305118/b305118.txt">Table of n, a(n) for n = 0..1000</a>

%F G.f. A(x) satisfies: A(x) = 1 + x * (1/(1 - x)^2 + A(x)^2). - _Ilya Gutkovskiy_, Jun 30 2020

%F From _Vaclav Kotesovec_, Jun 30 2020: (Start)

%F G.f.: (1 - sqrt(1 - 4*x*(1 + x/(1 - x)^2))) / (2*x).

%F a(n) ~ sqrt(1/r + 2/(1 - r)^3) / (2*sqrt(Pi) * n^(3/2) * r^n), where r = 0.19288682865259090392018... is the real root of the equation -1 + 6*r - 5*r^2 + 4*r^3 = 0. (End)

%t CoefficientList[Series[(1 - Sqrt[1 - 4*x*(1 + x/(1 - x)^2)]) / (2*x), {x, 0, 20}], x] (* _Vaclav Kotesovec_, Jun 30 2020 *)

%o (PARI)

%o seq(N) = {

%o my(a = vector(N)); a[1] = 1;

%o for (n=2, N, a[n] = sum(k=1, n-1, 1 + a[k]*a[n-k])); a;

%o };

%o seq(32)

%Y Cf. A007317, A000699, A088716.

%K nonn

%O 0,2

%A _Joerg Arndt_, May 26 2018