OFFSET
0,3
COMMENTS
Sum_{n>=0} a(n)/(2*n)! = exp(1/3 + 2*sqrt(3)*Pi/27) = 2.08840341696864282...
LINKS
Renzo Sprugnoli, Sums of reciprocals of the central binomial coefficients, Integers: electronic journal of combinatorial number theory, 6 (2006) #A27, 1-18.
FORMULA
E.g.f.: exp(L(x)) = Sum_{n>=0} a(n)*x^(2*n)/(2*n)!,
where L(x) = x^2/(4-x^2) + 4*x*arctan(x/sqrt(4-x^2))/sqrt((4-x^2)^3)
from a formula given in the Sprugnoli link.
EXAMPLE
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)! +...
where
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) +...
In closed form,
log(A(x)) = x^2/(4-x^2) + 4*x*arctan(x/sqrt(4-x^2))/sqrt((4-x^2)^3).
PROG
(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)}
(PARI) /* Using formula for e.g.f. = exp(L(x)): */
{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)}
CROSSREFS
KEYWORD
nonn
AUTHOR
Paul D. Hanna, Jul 25 2011
STATUS
approved