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

1, 2, 8, 40, 224, 1328, 8128, 51664, 344864, 2419808, 17690624, 133255936, 1025049088, 8010533504, 63486137600, 510583994560, 4173502486400, 34713858911872, 293845548560384, 2528481737709056, 22072195485742080, 194999285532239872, 1739475973516343296

Offset: 0

Author

Paul D. Hanna, Jul 29 2012

Keywords

• nonn

Comments

Compare g.f. to: G(x) = 1/G(-x*G(x)^3) when G(x) = 1 + x*G(x)^2 (A000108).

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.

Examples

			G.f.: A(x) = 1 + 2*x + 8*x^2 + 40*x^3 + 224*x^4 + 1328*x^5 + 8128*x^6 +...
A(x)^2 = 1 + 4*x + 20*x^2 + 112*x^3 + 672*x^4 + 4192*x^5 + 26752*x^6 +...
A(x)^3 = 1 + 6*x + 36*x^2 + 224*x^3 + 1440*x^4 + 9456*x^5 + 62912*x^6 +...
To illustrate the recurrence for the g.f., start with G_0(x) = 1+2*x, then
1/G_0(-x*G_0(x)^3) = 1/(1-2*x-12*x^2-24*x^3-16*x^4), and so
G_1(x) = AGM(1+2*x, 1/(1-2*x-12*x^2-24*x^3-16*x^4)):
G_1(x) =  1 + 2*x + 8*x^2 + 40*x^3 + 192*x^4 + 976*x^5 + 4944*x^6 +...;
continuing in this way yields the g.f. as a limit.
		

Links

• Paul D. Hanna, Table of n, a(n) for n = 0..200

Crossrefs

Cf. A214763.

Programs

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)}
for(n=0, 30, print1(a(n), ", "))

Formula

The g.f. of this sequence is the limit of the recurrence:

(*) 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.