A063902 a(n+1) = sum{j = 0,...n}[C(2n,2j)a(j)a(n-j)] with a(0) = 1.

1, 1, 2, 10, 80, 1000, 17600, 418000, 12848000, 496672000, 23576960000, 1348404640000, 91442700800000, 7255463564800000, 665885747225600000, 69994087116448000000, 8354181454767104000000, 1123646013779238400000000, 169165728883243642880000000

Offset: 0

Author

Henry Bottomley, Aug 30 2001

Keywords

• nonn

Comments

Define e.g.f. A(x) = Sum_{n>=0} a(n)*x^(2*n)/(2*n)!. Then A(x) = P(z + x/sqrt(6)) where P is the Weierstrass P-function with g_2 = 0, g_3 = 4 and z = 2.10327313773966...*i satisfies P(z) = 1, P'(z) = 0. - Michael Somos, Apr 18 2022

Examples

			a(3) = 1*a(0)a(2) + 6*a(1)a(1) + 1*a(2)a(0) = 2+6+2 = 10.
E.g.f.: A(x) = 1 + x^2/2! + 2*x^4/4! + 10*x^6/6! + 80*x^8/8! + ...
G.f. = 1 + x + 2*x^2 + 10*x^3 + 80*x^4 + 1000*x^8 + 17600*x^9 + ...
		

Links

• Vaclav Kotesovec, Table of n, a(n) for n = 0..270
• Markus Kuba, Alois Panholzer, Combinatorial families of multilabelled increasing trees and hook-length formulas, arXiv:1411.4587 [math.CO], 2014.

Crossrefs

a(n+1) = sum[C(n, j)a(j)a(n-j)] would give factorials A000142, a(n+1) = sum[C(2n, 2j)a(j)a(n-j)]/a(n) would give Catalan numbers A000108, a(n+1) = sum[C(n, j)a(j)a(n-j)]/a(n) would give central binomials A001405.

Cf. A007558, A258895.

Programs

Mathematica

Clear[a]; a[n_]:=a[n]=If[n<2,1,Sum[a[j]*a[n-1-j]*Binomial[2*n-2,2*j],{j,0,n-1}]]; Table[a[n], {n,0,30}] (* Vaclav Kotesovec, Jun 14 2015 *)
a[ n_] := If[n < 0, 0, Module[{x, y}, Nest[Dt[#] /. {Dt[x] -> y, Dt[y] -> x^2} &, x, 2*n] /. {x -> 1, y -> 0}]]; (* Michael Somos, Apr 18 2022 *)

PARI

/* E.g.f. A(x) = exp( Integral 1/A(x) * Integral A(x)^2 dx dx ) */
{a(n) = local(A=1+x); for(i=1,n, A = exp( intformal( 1/A * intformal( A^2 + x*O(x^n)) ) ) ); n!*polcoeff(A,n)}
for(n=0,20,print1(a(2*n),", ")) \\ Paul D. Hanna, Jun 02 2015

PARI

/* By definition: */
{a(n) = if(n==0,1,sum(k=0,n-1, binomial(2*n-2,2*k)*a(k)*a(n-k-1)))}
for(n=0,20,print1(a(n),", ")) \\ Paul D. Hanna, Jun 02 2015

PARI

{a(n) = if(n<0, 0, my(x='x, y = 1 + O(x^2)); for(i=1, n, y = 1 + intformal(intformal(y^2))); polcoeff(y, 2*n)*(2*n)!)}; /* Michael Somos, Apr 18 2022 */

Formula

E.g.f. satisfies: A(x) = exp( Integral 1/A(x) * Integral A(x)^2 dx dx ), where A(x) = Sum_{n>=0} a(n)*x^(1*n)/(2*n)! and the constant of integration is zero. - Paul D. Hanna, Jun 02 2015

From Vaclav Kotesovec, Jun 14 2015: (Start)

a(n) ~ c * d^n * n!^2 * sqrt(n), where d = A258895 = 32*Pi / (Gamma(1/6) * Gamma(1/3))^2 = 2^(17/3) * Pi^2 / (3*Gamma(1/3)^6) = 0.452104299183420528841..., c = 1.53043521765866544548745... = 2^(20/3) * Pi^(3/2) / Gamma(1/3)^6 = 192*sqrt(Pi) / (Gamma(1/3)*Gamma(1/6))^2.

a(n) ~ 3 * 2^(5*n+7) * Pi^(n+3/2) * n^(2*n+3/2) / (exp(2*n) * Gamma(1/6)^(2*n+2) * Gamma(1/3)^(2*n+2)).

a(n) ~ 2^((17*n+23)/3) * Pi^(2*n+5/2) * n^(2*n+3/2) / (3^n * exp(2*n) * Gamma(1/3)^(6*n+6)).

(End)

Define e.g.f. A(x) = Sum_{n>=0} a(n)*x^(2*n)/(2*n)!. Then 3*A'(x)^2 = 2*A(x)^3 - 2. - Michael Somos, Jan 07 2022

Extensions

More terms from Vaclav Kotesovec, Jun 14 2015