A193441 E.g.f.: exp( Sum_{n>=1} n!^2*x^(2*n)/(2*n)! ) = Sum_{n>=0} a(n)*x^(2*n)/(2*n)!.
1, 1, 7, 111, 3089, 131985, 7977687, 645443295, 67165412385, 8722553971041, 1380689271177255, 261365482010524815, 58252017195624969009, 15086874107373899187825, 4490370671139664566269175, 1521257907398602231501780095, 581762614758928225569542394945
Offset: 0
Keywords
Examples
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).
Links
- Renzo Sprugnoli, Sums of reciprocals of the central binomial coefficients, Integers: electronic journal of combinatorial number theory, 6 (2006) #A27, 1-18.
Programs
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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)}
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.
Comments