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A120972 G.f. A(x) satisfies A(x/A(x)^3) = 1 + x ; thus A(x) = 1 + series_reversion(x/A(x)^3).

%I A120972 #23 Jun 04 2025 09:52:35
%S A120972 1,1,3,21,217,2814,42510,718647,13270944,263532276,5567092665,
%T A120972 124143735663,2905528740060,71058906460091,1809695198254281,
%U A120972 47861102278428198,1311488806252697283,37164457324943708739,1087356593493807164289,32801308084353988297404
%N A120972 G.f. A(x) satisfies A(x/A(x)^3) = 1 + x ; thus A(x) = 1 + series_reversion(x/A(x)^3).
%C A120972 More generally, if g.f. A(x) satisfies: A(x/A(x)^k) = 1 + x*A(x)^m, then
%C A120972 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 A120972 thus a(n) = [x^(n-1)] ((m+k)/(m+k*n))*A(x)^(m+k*n) for n>=1 with a(0)=1.
%F A120972 G.f. satisfies: A(x) = 1 + x*A(A(x) - 1)^3.
%F A120972 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 A120972 Let B(x) be the g.f. of A120973, then B(x) and g.f. A(x) are related by:
%F A120972 (a) B(x) = A(A(x)-1),
%F A120972 (b) B(x) = A(x*B(x)^3),
%F A120972 (c) A(x) = B(x/A(x)^3),
%F A120972 (d) A(x) = 1 + x*B(x)^3,
%F A120972 (e) B(x) = 1 + x*B(x)^3*B(A(x)-1)^3,
%F A120972 (f) A(B(x)-1) = B(A(x)-1) = B(x*B(x)^3).
%F A120972 From _Seiichi Manyama_, Jun 04 2025: (Start)
%F A120972 Let b(n,k) = [x^n] B(x)^k, where B(x) is the g.f. of A120973.
%F A120972 b(n,0) = 0^n; b(n,k) = k * Sum_{j=0..n} binomial(3*n+k,j)/(3*n+k) * b(n-j,3*j).
%F A120972 a(n) = b(n-1,3) for n > 0. (End)
%e A120972 G.f.: A(x) = 1 + x + 3*x^2 + 21*x^3 + 217*x^4 + 2814*x^5 + 42510*x^6 +...
%e A120972 Related expansions.
%e A120972 A(x)^3 = 1 + 3*x + 12*x^2 + 82*x^3 + 813*x^4 + 10212*x^5 + 150699*x^6 +...
%e A120972 A(A(x)-1) = 1 + x + 6*x^2 + 60*x^3 + 776*x^4 + 11802*x^5 + 201465*x^6 +...
%e A120972 A(A(x)-1)^3 = 1 + 3*x + 21*x^2 + 217*x^3 + 2814*x^4 + 42510*x^5 +...
%e A120972 x/A(x)^3 = x - 3*x^2 - 3*x^3 - 37*x^4 - 420*x^5 - 5823*x^6 -...
%e A120972 Series_Reversion(x/A(x)^3) = x + 3*x^2 + 21*x^3 + 217*x^4 + 2814*x^5 + 42510*x^6 +...
%e A120972 To illustrate the formula a(n) = [x^(n-1)] 3*A(x)^(3*n)/(3*n),
%e A120972 form a table of coefficients in A(x)^(3*n) as follows:
%e A120972   A^3:  [(1), 3,  12,   82,    813,   10212,   150699,   2503233, ...];
%e A120972   A^6:  [ 1, (6), 33,  236,   2262,   27270,   388906,   6289080, ...];
%e A120972   A^9:  [ 1,  9, (63), 489,   4671,   54684,   756012,  11904813, ...];
%e A120972   A^12: [ 1, 12, 102, (868),  8445,   97260,  1310040,  20112516, ...];
%e A120972   A^15: [ 1, 15, 150, 1400, (14070), 161343,  2130505,  31961175, ...];
%e A120972   A^18: [ 1, 18, 207, 2112,  22113, (255060), 3324003,  48876264, ...];
%e A120972   A^21: [ 1, 21, 273, 3031,  33222,  388563, (5030529), 72769014, ...]; ...
%e A120972 in which the main diagonal forms the initial terms of this sequence:
%e A120972 [3/3*(1), 3/6*(6), 3/9*(63), 3/12*(868), 3/15*(14070), 3/18*(255060), ...].
%t A120972 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 A120972 (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 A120972 (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 A120972 (PARI) /* This sequence is generated when k=3, m=0: A(x/A(x)^k) = 1 + x*A(x)^m */
%o A120972 {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 A120972 for(n=0,25,print1(a(n),", "))
%o A120972 (PARI) b(n, k) = if(k==0, 0^n, k*sum(j=0, n, binomial(3*n+k, j)/(3*n+k)*b(n-j, 3*j)));
%o A120972 a(n) = if(n==0, 1, b(n-1, 3)); \\ _Seiichi Manyama_, Jun 04 2025
%Y A120972 Cf. A120973; variants: A120970, A120974, A120976, A030266, A067145, A107096.
%Y A120972 Cf. A381603.
%K A120972 nonn
%O A120972 0,3
%A A120972 _Paul D. Hanna_, Jul 20 2006