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%I #9 Apr 12 2023 10:04:47
%S 1,1,2,8,49,380,3400,33469,352763,3914105,45203847,539095203,
%T 6600723606,82616454685,1053503618516,13650703465841,179351890161617,
%U 2385294488375623,32066177447127597,435218601202213040
%N G.f.: A(x) = exp( Sum_{n>=1,k>=0} CATALAN(n,k)^2*x^(n+k)/n ), where CATALAN(n,k) = n*C(n+2*k-1,k)/(n+k) is the coefficient of x^k in C(x)^n and C(x) is the g.f. of the Catalan numbers.
%C Compare the g.f. of this sequence to the g.f. of the Catalan numbers:
%C C(x) = exp( Sum_{n>=1,k>=0} C(n+k-1,k)^2*x^(n+k)/n ).
%e G.f.: A(x) = 1 + x + 2*x^2 + 8*x^3 + 49*x^4 + 380*x^5 + 3400*x^6 +...
%e The logarithm of the g.f. (A183069) begins:
%e log(A(x)) = x + 3*x^2/2 + 19*x^3/3 + 163*x^4/4 + 1626*x^5/5 +...
%e and equals the series:
%e log(A(x)) = (1 + x + 2^2*x^2 + 5^2*x^3 + 14^2*x^4 +...)*x
%e + (1 + 2^2*x + 5^2*x^2 + 14^2*x^3 + 42^2*x^4 +...)*x^2/2
%e + (1 + 3^2*x + 9^2*x^2 + 28^2*x^3 + 90^2*x^4 +...)*x^3/3
%e + (1 + 4^2*x + 14^2*x^2 + 48^2*x^3 + 165^2*x^4 +...)*x^4/4
%e + (1 + 5^2*x + 20^2*x^2 + 75^2*x^3 + 275^2*x^4 +...)*x^5/5 +...
%e which consists of the squares of coefficients in powers of C(x),
%e where C(x) = 1 + x*C(x)^2 is g.f. of the Catalan numbers (A000108).
%e ...
%e Compare the above series for log(A(x)) to log(C(x)):
%e log(C(x)) = (1 + x + x^2 + x^3 + x^4 + x^5 +...)*x
%e + (1 + 2^2*x + 3^2*x^2 + 4^2*x^3 + 5^2*x^4 +...)*x^2/2
%e + (1 + 3^2*x + 6^2*x^2 + 10^2*x^3 + 15^2*x^4 +...)*x^3/3
%e + (1 + 4^2*x + 10^2*x^2 + 20^2*x^3 + 35^2*x^4 +...)*x^4/4
%e + (1 + 5^2*x + 15^2*x^2 + 35^2*x^3 + 70^2*x^4 +...)*x^5/5 +...
%e which consists of the squares of coefficients in powers of 1/(1-x).
%o (PARI) {a(n)=polcoeff(exp(sum(m=1, n, sum(k=0, n, (m*binomial(m+2*k-1,k)/(m+k))^2*x^k)*x^m/m)+x*O(x^n)), n)}
%Y Cf. A183069 (log), A000108, A009766.
%K nonn
%O 0,3
%A _Paul D. Hanna_, Dec 23 2010
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