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G.f.: exp( Sum_{n>=1} C(2*n,n)^n/2^n * x^n/n ).
4

%I #18 Mar 06 2014 12:19:09

%S 1,1,5,338,375502,6351970709,1620698781098852,6259260939361008796229,

%T 367534769386519350929158503892,

%U 329474737492618783473185792974307067503,4525697838840190793599072589249813785373031191426,955617474162634862818320009634143510233705849191099879550608

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

%C 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 ).

%F Conjecture: Limit n->infinity a(n)^(1/n^2) = 4. - _Vaclav Kotesovec_, Mar 06 2014

%e G.f.: A(x) = 1 + x + 5*x^2 + 338*x^3 + 375502*x^4 + 6351970709*x^5 +...

%e where

%e 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 +...

%t 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 *)

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

%Y Cf. A224732, A201556, A001700.

%K nonn

%O 0,3

%A _Paul D. Hanna_, Dec 05 2011