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A214760 G.f. satisfies: A(x) = 1/A(-x*A(x)^3).

%I A214760 #8 Jan 06 2013 02:54:09
%S A214760 1,2,8,40,224,1328,8128,51664,344864,2419808,17690624,133255936,
%T A214760 1025049088,8010533504,63486137600,510583994560,4173502486400,
%U A214760 34713858911872,293845548560384,2528481737709056,22072195485742080,194999285532239872,1739475973516343296
%N A214760 G.f. satisfies: A(x) = 1/A(-x*A(x)^3).
%C A214760 Compare g.f. to: G(x) = 1/G(-x*G(x)^3) when G(x) = 1 + x*G(x)^2 (A000108).
%C A214760 An infinite number of functions G(x) satisfy (*) G(x) = 1/G(-x*G(x)^3); for example, (*) is satisfied by G(x) = C(m*x) = (1-sqrt(1-4*m*x))/(2*m*x) for all m, where C(x) is the Catalan function.
%H A214760 Paul D. Hanna, <a href="/A214760/b214760.txt">Table of n, a(n) for n = 0..200</a>
%F A214760 The g.f. of this sequence is the limit of the recurrence:
%F A214760 (*) G_{n+1}(x) = AGM(G_n(x), 1/G_n(-x*G_n(x)^3)) starting at G_0(x) = 1+2*x, where AGM(x,y) = AGM((x+y)/2,sqrt(x*y)) is the arithmetic-geometric mean of x and y.
%e A214760 G.f.: A(x) = 1 + 2*x + 8*x^2 + 40*x^3 + 224*x^4 + 1328*x^5 + 8128*x^6 +...
%e A214760 A(x)^2 = 1 + 4*x + 20*x^2 + 112*x^3 + 672*x^4 + 4192*x^5 + 26752*x^6 +...
%e A214760 A(x)^3 = 1 + 6*x + 36*x^2 + 224*x^3 + 1440*x^4 + 9456*x^5 + 62912*x^6 +...
%e A214760 To illustrate the recurrence for the g.f., start with G_0(x) = 1+2*x, then
%e A214760 1/G_0(-x*G_0(x)^3) = 1/(1-2*x-12*x^2-24*x^3-16*x^4), and so
%e A214760 G_1(x) = AGM(1+2*x, 1/(1-2*x-12*x^2-24*x^3-16*x^4)):
%e A214760 G_1(x) =  1 + 2*x + 8*x^2 + 40*x^3 + 192*x^4 + 976*x^5 + 4944*x^6 +...;
%e A214760 continuing in this way yields the g.f. as a limit.
%o A214760 (PARI) {a(n)=local(A=1+2*x); for(i=0, n, A=agm(A,1/subst(A, x, -x*A^3+x*O(x^n)))); polcoeff(A, n)}
%o A214760 for(n=0, 30, print1(a(n), ", "))
%Y A214760 Cf. A214763.
%K A214760 nonn
%O A214760 0,2
%A A214760 _Paul D. Hanna_, Jul 29 2012