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G.f. A(x) satisfies: A(x) = 1 / (1 + x) + x * A(x)^3.
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%I #13 Nov 17 2021 07:40:58

%S 1,0,1,2,7,23,82,300,1129,4334,16914,66899,267586,1080516,4398850,

%T 18035084,74402361,308624282,1286428765,5385578256,22635057148,

%U 95471113565,403983783772,1714494024947,7295949019114,31124885587680,133085594104222,570266646942488

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

%C Inverse binomial transform of A200753.

%H Seiichi Manyama, <a href="/A346627/b346627.txt">Table of n, a(n) for n = 0..1000</a>

%F G.f.: Sum_{k>=0} ( binomial(3*k,k) / (2*k + 1) ) * x^k / (1 + x)^(2*k+1).

%F a(n) = (-1)^n + Sum_{i=0..n-1} Sum_{j=0..n-i-1} a(i) * a(j) * a(n-i-j-1).

%F a(n) = Sum_{k=0..n} (-1)^(n-k) * binomial(n+k,n-k) * binomial(3*k,k) / (2*k + 1).

%F a(n) ~ sqrt(198 + 38*sqrt(33)) * (19 + 3*sqrt(33))^n / (9 * sqrt(Pi) * n^(3/2) * 2^(3*n + 3)). - _Vaclav Kotesovec_, Jul 30 2021

%t nmax = 27; A[_] = 0; Do[A[x_] = 1/(1 + x) + x A[x]^3 + O[x]^(nmax + 1) // Normal, nmax + 1]; CoefficientList[A[x], x]

%t a[n_] := a[n] = (-1)^n + Sum[Sum[a[i] a[j] a[n - i - j - 1], {j, 0, n - i - 1}], {i, 0, n - 1}]; Table[a[n], {n, 0, 27}]

%t Table[Sum[(-1)^(n - k) Binomial[n + k, n - k] Binomial[3 k, k]/(2 k + 1), {k, 0, n}], {n, 0, 27}]

%Y Cf. A001764, A005043, A199475, A200753, A346628, A349299, A349300, A349301, A349302, A349303.

%K nonn

%O 0,4

%A _Ilya Gutkovskiy_, Jul 25 2021