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%I #13 Jan 15 2018 08:52:38
%S 1,1,3,21,217,2814,42510,718647,13270944,263532276,5567092665,
%T 124143735663,2905528740060,71058906460091,1809695198254281,
%U 47861102278428198,1311488806252697283,37164457324943708739
%N G.f. satisfies: A(x/A(x)^3) = 1 + x ; thus A(x) = 1 + series_reversion(x/A(x)^3).
%C More generally, if g.f. A(x) satisfies: A(x/A(x)^k) = 1 + x*A(x)^m, then
%C A(x) = 1 + x*G(x)^(m+k) where G(x) = A(x*G(x)^k) and G(x/A(x)^k) = A(x);
%C thus a(n) = [x^(n-1)] ((m+k)/(m+k*n))*A(x)^(m+k*n) for n>=1 with a(0)=1.
%F G.f. satisfies: A(x) = 1 + x*A(A(x) - 1)^3.
%F a(n) = [x^(n-1)] A(x)^(3*n)/n for n>=1 with a(0)=1; i.e., a(n) equals the coefficient of x^(n-1) in A(x)^(3*n)/n for n>=1 (see comment).
%F Let B(x) be the g.f. of A120973, then B(x) and g.f. A(x) are related by:
%F (a) B(x) = A(A(x)-1),
%F (b) B(x) = A(x*B(x)^3),
%F (c) A(x) = B(x/A(x)^3),
%F (d) A(x) = 1 + x*B(x)^3,
%F (e) B(x) = 1 + x*B(x)^3*B(A(x)-1)^3,
%F (f) A(B(x)-1) = B(A(x)-1) = B(x*B(x)^3).
%e G.f.: A(x) = 1 + x + 3*x^2 + 21*x^3 + 217*x^4 + 2814*x^5 + 42510*x^6 +...
%e Related expansions.
%e A(x)^3 = 1 + 3*x + 12*x^2 + 82*x^3 + 813*x^4 + 10212*x^5 + 150699*x^6 +...
%e A(A(x)-1) = 1 + x + 6*x^2 + 60*x^3 + 776*x^4 + 11802*x^5 + 201465*x^6 +...
%e A(A(x)-1)^3 = 1 + 3*x + 21*x^2 + 217*x^3 + 2814*x^4 + 42510*x^5 +...
%e x/A(x)^3 = x - 3*x^2 - 3*x^3 - 37*x^4 - 420*x^5 - 5823*x^6 -...
%e Series_Reversion(x/A(x)^3) = x + 3*x^2 + 21*x^3 + 217*x^4 + 2814*x^5 + 42510*x^6 +...
%e To illustrate the formula a(n) = [x^(n-1)] 3*A(x)^(3*n)/(3*n),
%e form a table of coefficients in A(x)^(3*n) as follows:
%e A^3: [(1), 3, 12, 82, 813, 10212, 150699, 2503233, ...];
%e A^6: [1, (6), 33, 236, 2262, 27270, 388906, 6289080, ...];
%e A^9: [1, 9, (63), 489, 4671, 54684, 756012, 11904813, ...];
%e A^12: [1, 12, 102, (868), 8445, 97260, 1310040, 20112516, ...];
%e A^15: [1, 15, 150, 1400, (14070), 161343, 2130505, 31961175, ...];
%e A^18: [1, 18, 207, 2112, 22113, (255060), 3324003, 48876264, ...];
%e A^21: [1, 21, 273, 3031, 33222, 388563, (5030529), 72769014, ...]; ...
%e in which the main diagonal forms the initial terms of this sequence:
%e [3/3*(1), 3/6*(6), 3/9*(63), 3/12*(868), 3/15*(14070), 3/18*(255060), ...].
%t terms = 18; A[_] = 1; Do[A[x_] = 1 + x*A[A[x] - 1]^3 + O[x]^j // Normal, {j, terms}]; CoefficientList[A[x], x] (* _Jean-François Alcover_, Jan 15 2018 *)
%o (PARI) {a(n)=local(A=[1,1]);for(i=2,n,A=concat(A,0); A[ #A]=-Vec(subst(Ser(A),x,x/Ser(A)^3))[ #A]);A[n+1]}
%o (PARI) {a(n)=local(A=1+x+x*O(x^n));for(i=1,n,A=1+x*subst(A^3,x,A-1+x*O(x^n)));polcoeff(A,n)}
%o (PARI) /* This sequence is generated when k=3, m=0: A(x/A(x)^k) = 1 + x*A(x)^m */
%o {a(n, k=3, m=0)=local(A=sum(i=0, n-1, a(i, k, m)*x^i)); if(n==0, 1, polcoeff((m+k)/(m+k*n)*A^(m+k*n), n-1))}
%o for(n=0,25,print1(a(n),", "))
%Y Cf. A120973; variants: A120970, A120974, A120976, A030266, A067145, A107096.
%K nonn
%O 0,3
%A _Paul D. Hanna_, Jul 20 2006
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