A sequence {x_n} is defined as:
x₁ = 2, and:
n*x_n = 2(2n-1)*x_n-1, whenever n ≥ 2

Prove that x_n is always an integer for every positive integer value of n.

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 2010-10-04 15:01:41