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A120970 G.f. A(x) satisfies A(x/A(x)^2) = 1 + x ; thus A(x) = 1 + Series_Reversion(x/A(x)^2).

%I A120970 #41 Jun 04 2025 09:52:39
%S A120970 1,1,2,9,60,504,4946,54430,655362,8496454,117311198,1711459903,
%T A120970 26228829200,420370445830,7021029571856,121859518887327,
%U A120970 2192820745899978,40831103986939664,785429260324068156,15585831041632684997,318649154587152781210,6704504768568697046504
%N A120970 G.f. A(x) satisfies A(x/A(x)^2) = 1 + x ; thus A(x) = 1 + Series_Reversion(x/A(x)^2).
%C A120970 From _Paul D. Hanna_, Nov 16 2008: (Start)
%C A120970 More generally, if g.f. A(x) satisfies: A(x/A(x)^k) = 1 + x*A(x)^m, then
%C A120970 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 A120970 thus a(n) = [x^(n-1)] ((m+k)/(m+k*n))*A(x)^(m+k*n) for n>=1 with a(0)=1. (End)
%H A120970 Vaclav Kotesovec, <a href="/A120970/b120970.txt">Table of n, a(n) for n = 0..350</a>
%F A120970 G.f. satisfies: A(x) = 1 + x*A(A(x) - 1)^2.
%F A120970 Let B(x) be the g.f. of A120971, then B(x) and g.f. A(x) are related by:
%F A120970 (a) B(x) = A(A(x)-1),
%F A120970 (b) B(x) = A(x*B(x)^2),
%F A120970 (c) A(x) = B(x/A(x)^2),
%F A120970 (d) A(x) = 1 + x*B(x)^2,
%F A120970 (e) B(x) = 1 + x*B(x)^2*B(A(x)-1)^2,
%F A120970 (f) A(B(x)-1) = B(A(x)-1) = B(x*B(x)^2).
%F A120970 a(n) = [x^(n-1)] (1/n)*A(x)^(2n) for n>=1 with a(0)=1; i.e., a(n) equals 1/n times the coefficient of x^(n-1) in A(x)^(2n) for n>=1. [_Paul D. Hanna_, Nov 16 2008]
%F A120970 From _Seiichi Manyama_, Jun 04 2025: (Start)
%F A120970 Let b(n,k) = [x^n] B(x)^k, where B(x) is the g.f. of A120971.
%F A120970 b(n,0) = 0^n; b(n,k) = k * Sum_{j=0..n} binomial(2*n+k,j)/(2*n+k) * b(n-j,2*j).
%F A120970 a(n) = b(n-1,2) for n > 0. (End)
%e A120970 G.f.: A(x) = 1 + x + 2*x^2 + 9*x^3 + 60*x^4 + 504*x^5 + 4946*x^6 + ...
%e A120970 Related expansions.
%e A120970 A(x)^2 = 1 + 2*x + 5*x^2 + 22*x^3 + 142*x^4 + 1164*x^5 + 11221*x^6 + ...
%e A120970 A(A(x)-1) = 1 + x + 4*x^2 + 26*x^3 + 218*x^4 + 2151*x^5 + 23854*x^6 + ...
%e A120970 A(A(x)-1)^2 = 1 + 2*x + 9*x^2 + 60*x^3 + 504*x^4 + 4946*x^5 + ...
%e A120970 x/A(x)^2 = x - 2*x^2 - x^3 - 10*x^4 - 73*x^5 - 662*x^6 - 6842*x^7 - ...
%e A120970 Series_Reversion(x/A(x)^2) = x + 2*x^2 + 9*x^3 + 60*x^4 + 504*x^5 + 4946*x^6 + ...
%e A120970 To illustrate the formula a(n) = [x^(n-1)] 2*A(x)^(2*n)/(2*n),
%e A120970 form a table of coefficients in A(x)^(2*n) as follows:
%e A120970   A^2:  [(1), 2,   5,   22,   142,   1164,   11221,   121848, ...];
%e A120970   A^4:  [ 1, (4), 14,   64,   397,   3116,   29002,   306468, ...];
%e A120970   A^6:  [ 1,  6, (27), 134,   825,   6270,   56492,   580902, ...];
%e A120970   A^8:  [ 1,  8,  44, (240), 1502,  11200,   98144,   983016, ...];
%e A120970   A^10: [ 1, 10,  65,  390, (2520), 18672,  160115,  1565260, ...];
%e A120970   A^12: [ 1, 12,  90,  592,  3987, (29676), 250730,  2399388, ...];
%e A120970   A^14: [ 1, 14, 119,  854,  6027,  45458, (381010), 3582266, ...]; ...
%e A120970 in which the main diagonal forms the initial terms of this sequence:
%e A120970 [2/2*(1), 2/4*(4), 2/6*(27), 2/8*(240), 2/10*(2520), 2/12*(29676), ...].
%t A120970 terms = 21; A[_] = 1; Do[A[x_] = 1 + x*A[A[x] - 1]^2 + O[x]^j // Normal, {j, terms}]; CoefficientList[A[x], x] (* _Jean-François Alcover_, Jan 15 2018 *)
%o A120970 (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)^2))[ #A]);A[n+1]}
%o A120970 for(n=0,30,print1(a(n),", "))
%o A120970 (PARI) /* This sequence is generated when k=2, m=0: A(x/A(x)^k) = 1 + x*A(x)^m */ {a(n,k=2,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))} \\ _Paul D. Hanna_, Nov 16 2008
%o A120970 for(n=0,30,print1(a(n),", "))
%o A120970 (PARI) b(n, k) = if(k==0, 0^n, k*sum(j=0, n, binomial(2*n+k, j)/(2*n+k)*b(n-j, 2*j)));
%o A120970 a(n) = if(n==0, 1, b(n-1, 2)); \\ _Seiichi Manyama_, Jun 04 2025
%Y A120970 Cf. A120971; variants: A120972, A120974, A120976, A030266, A067145, A107096.
%Y A120970 Cf. related variants: A145347, A145348, A147664, A145349, A145350. - _Paul D. Hanna_, Nov 16 2008
%Y A120970 Cf. A381602.
%K A120970 nonn
%O A120970 0,3
%A A120970 _Paul D. Hanna_, Jul 20 2006