%I #7 Mar 14 2015 11:41:32

%S 1,1,8,101,1639,30665,630225,13836981,319062453,7640441894,

%T 188534274850,4767113222750,122998902095908,3228067183537455,

%U 85960229675478804,2317956019913480326,63193008693741620771,1739473925024629613227,48292271242981605779173

%N G.f.: exp( Sum_{n>=1} A000984(n)*A000172(n)/4 * x^n/n ), which involves central binomial coefficients (A000984) and Franel numbers (A000172).

%C Sum_{k=0..n} C(n,k)^2 = A000984(n) defines central binomial coefficients.

%C Sum_{k=0..n} C(n,k)^3 = A000172(n) defines Franel numbers.

%C Compare to the g.f. of the Catalan numbers (A000108): exp(Sum_{n>=1} A000984(n)/2*x^n/n) and to the g.f. of A166991: exp(Sum_{n>=1} A000172(n)/2*x^n/n).

%F Convolution 4th power yields A199813.

%e G.f.: A(x) = 1 + x + 8*x^2 + 101*x^3 + 1639*x^4 + 30665*x^5 +...

%e where

%e log(A(x)) = 1*1*x + 3*5*x^2/2 + 10*28*x^3/3 + 35*173*x^4/4 + 126*1126*x^5/5 + 462*7592*x^6/6 +...+ A000984(n)/2*A000172(n)/2*x^n/n +...

%o (PARI) {a(n)=polcoeff(exp(sum(m=1, n, binomial(2*m, m)/2*sum(k=0, m, binomial(m, k)^3)/2*x^m/m)+x*O(x^n)), n)}

%Y Cf. A199813, A166991, A000108, A181418, A000984, A000172.

%K nonn

%O 0,3

%A _Paul D. Hanna_, Nov 11 2011