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%I #23 Oct 13 2014 04:06:12
%S 1,1,7,111,3089,131985,7977687,645443295,67165412385,8722553971041,
%T 1380689271177255,261365482010524815,58252017195624969009,
%U 15086874107373899187825,4490370671139664566269175,1521257907398602231501780095,581762614758928225569542394945
%N E.g.f.: exp( Sum_{n>=1} n!^2*x^(2*n)/(2*n)! ) = Sum_{n>=0} a(n)*x^(2*n)/(2*n)!.
%C Sum_{n>=0} a(n)/(2*n)! = exp(1/3 + 2*sqrt(3)*Pi/27) = 2.08840341696864282...
%H Renzo Sprugnoli, <a href="http://www.emis.de/journals/INTEGERS/papers/g27/g27.Abstract.html">Sums of reciprocals of the central binomial coefficients</a>, Integers: electronic journal of combinatorial number theory, 6 (2006) #A27, 1-18.
%F E.g.f.: exp(L(x)) = Sum_{n>=0} a(n)*x^(2*n)/(2*n)!,
%F where L(x) = x^2/(4-x^2) + 4*x*arctan(x/sqrt(4-x^2))/sqrt((4-x^2)^3)
%F from a formula given in the Sprugnoli link.
%e E.g.f.: A(x) = 1 + x^2/2! + 7*x^4/4! + 111*x^6/6! + 3089*x^8/8! + 131985*x^10/10! + 7977687*x^12/12! +...+ a(n)*x^(2*n)/(2*n)! +...
%e where
%e log(A(x)) = x^2/2 + x^4/6 + x^6/20 + x^8/70 + x^10/252 + x^12/924 + x^14/3432 + x^16/12870 +...+ x^(2*n)/A000984(n) +...
%e In closed form,
%e log(A(x)) = x^2/(4-x^2) + 4*x*arctan(x/sqrt(4-x^2))/sqrt((4-x^2)^3).
%o (PARI) {a(n)=(2*n)!*polcoeff(exp(sum(m=1,n,x^(2*m)/binomial(2*m,m))+O(x^(2*n+1))),2*n)}
%o (PARI) /* Using formula for e.g.f. = exp(L(x)): */
%o {a(n)=local(Ox=O(x^(2*n+1)), L=x^2/(4-x^2 +Ox) + 4*x*atan(x/sqrt(4-x^2 +Ox))/sqrt((4-x^2 +Ox)^3)); (2*n)!*polcoeff(exp(L), 2*n)}
%Y Cf. A193442, A193443, A193444, A000984 (C(2*n,n)).
%K nonn
%O 0,3
%A _Paul D. Hanna_, Jul 25 2011
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