The sequence x(n) goes 2, 6, 20, 70, 252, 924, 3432
The OIES gives the formula x(n) = (2n)! / (n!)^2 from which it is clear that every term in the denominator will reduce to 1, leaving an integer.
Proof of the formula:
Show it works for n=1. (2*1)! / (1!)^2 = 2
Assume it works for x(n) and show it works for x(n+1)
x(n+1) = 2(2(n+1)1)*x(n)/(n+1)
=[2(2n+1)*(2n)!] / [(n+1)*(n!)^2]
multiply by (n+1)/(n+1)
=[(2n+2)(2n+1)(2n)!] / [(n+1)^2*(n!)^2]
=(2n+2)! / ((n+1)!)^2

Posted by Jer
on 20101004 15:01:41 