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Expansion of A(x) = 1 + x + x*A(x) + x^2*A(x)^2 + x^3*A(x)^3.
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%I #25 May 16 2018 04:11:52

%S 1,2,3,8,24,73,230,751,2512,8549,29520,103192,364441,1298336,4660273,

%T 16837743,61187249,223489715,820040293,3021286338,11172619317,

%U 41454901924,154285693649,575826677260,2154643706124,8081453618583,30377809192246,114421431917805

%N Expansion of A(x) = 1 + x + x*A(x) + x^2*A(x)^2 + x^3*A(x)^3.

%F a(n) = Sum_{i=0..n} Sum_{k=0..n-i} Sum_{j=0..k} C(i,j)*C(j,k-j)*C(k+1,n-k-i)* C(k+i,i)/(k+1).

%p a:= n-> coeff(series(RootOf((A*x)^3+(A*x)^2+

%p A*(x-1)+x+1, A), x, n+1), x, n):

%p seq(a(n), n=0..30); # _Alois P. Heinz_, May 14 2018

%t Array[Sum[Sum[Sum[Binomial[i, j] Binomial[j, k - j], {j, 0, k}] Binomial[k + 1, # - k - i] Binomial[k + i, i]/(k + 1), {k, 0, # - i}], {i, 0, # + 1}] &, 31, 0] (* _Michael De Vlieger_, May 10 2018 *)

%t n = 27; A = Sum[a[k] x^k, {k, 0, n}] + x O[x]^n; Table[a[k], {k, 0, n}] /.

%t Solve[LogicalExpand[1 + x + x A + x^2 A^2 + x^3 A^3 == A]] // Flatten (* _Peter Luschny_, May 14 2018 *)

%o (Maxima)

%o a(n):=sum(sum(((sum(binomial(i,j)*binomial(j,k-j),j,0,k))*binomial(k+1,n-k-i)* binomial(k+i,i))/(k+1),k,0,n-i),i,0,n);

%K nonn

%O 0,2

%A _Vladimir Kruchinin_, May 07 2018