A200002 G.f.: exp( Sum_{n>=1} C(2*n,n)^n/2^n * x^n/n ).

1, 1, 5, 338, 375502, 6351970709, 1620698781098852, 6259260939361008796229, 367534769386519350929158503892, 329474737492618783473185792974307067503, 4525697838840190793599072589249813785373031191426, 955617474162634862818320009634143510233705849191099879550608

Offset: 0

Author

Paul D. Hanna, Dec 05 2011

Keywords

• nonn

Comments

Compare to the g.f. C(x) = 1 + x*C(x)^2 of the Catalan numbers (A000108), where C(x) = exp( Sum_{n>=1} C(2*n,n)/2 * x^n/n ).

Examples

			G.f.: A(x) = 1 + x + 5*x^2 + 338*x^3 + 375502*x^4 + 6351970709*x^5 +...
where
log(A(x)) = x + 3^2*x^2/2 + 10^3*x^3/3 + 35^4*x^4/4 + 126^5*x^5/5 + 462^6*x^6/6 + 1716^7*x^7/7 +...+ A001700(n+1)^n*x^n/n +...
		

Crossrefs

Cf. A224732, A201556, A001700.

Programs

Mathematica

nmax = 10; b = ConstantArray[0, nmax+1]; b[[1]] = 1; Do[b[[n+1]] = 1/n*Sum[Binomial[2*k,k]^k/2^k * b[[n-k+1]], {k, 1, n}], {n, 1, nmax}]; b  (* Vaclav Kotesovec, Mar 06 2014 *)

PARI

{a(n)=polcoeff(exp(sum(m=1,n,binomial(2*m,m)^m/2^m*x^m/m)+x*O(x^n)),n)}

Formula

Conjecture: Limit n->infinity a(n)^(1/n^2) = 4. - Vaclav Kotesovec, Mar 06 2014