%I #4 Jul 31 2014 09:48:44
%S 1,1,3,8,25,80,271,952,3443,12758,48212,185283,722227,2849955,
%T 11366379,45757142,185726603,759401542,3125472832,12939604503,
%U 53856950922,225250407802,946253665230,3991221520996,16897320866269,71782331694315
%N G.f.: exp( Sum_{n>=0} [ Sum_{k=0..n} C(n,k)^2*x^k ]^2* x^n/n ).
%C Compare the g.f. of this sequence to the g.f.s:
%C . exp( Sum_{n>=0} [Sum_{k=0..n} C(n,k)^2*x^k]*x^n/n ) = (G(x)-1)/x where G(x) = g.f. of A004148.
%C . exp( Sum_{n>=0} [Sum_{k=0..n} C(n,k)*x^k]^2*x^n/n ) = 1/(1-x*(1+x)^2).
%H Vaclav Kotesovec, <a href="/A180718/b180718.txt">Table of n, a(n) for n = 0..900</a>
%e G.f.: A(x) = 1 + x + 3*x^2 + 8*x^3 + 25*x^4 + 80*x^5 + 271*x^6 +...
%e The logarithm (A180719) begins:
%e log(A(x)) = x + 5*x^2/2 + 16*x^3/3 + 61*x^4/4 + 226*x^5/5 + 884*x^6/6 + 3543*x^7/7 + 14429*x^8/8 +...
%e which equals the sum of the series:
%e log(A(x)) = (1 + x)^2*x
%e + (1 + 4*x + x^2)^2*x^2/2
%e + (1 + 9*x + 9*x^2 + x^3)^2*x^3/3
%e + (1 + 16*x + 36*x^2 + 16*x^3 + x^4)^2*x^4/4
%e + (1 + 25*x + 100*x^2 + 100*x^3 + 25*x^4 + x^5)^2*x^5/5
%e + (1 + 36*x + 225*x^2 + 400*x^3 + 225*x^4 + 36*x^5 + x^6)^2*x^6/6 +...
%o (PARI) {a(n)=polcoeff(exp(sum(m=1,n,sum(k=0,m,binomial(m,k)^2*x^k)^2*x^m/m)+x*O(x^n)),n)}
%Y Cf. A180719, A004148.
%K nonn
%O 0,3
%A _Paul D. Hanna_, Sep 24 2010