Recurrence relation for 2nCn

Question

Does there exist an easy to compute (by hand) recurrence relation for the central column of pascals triangle?

I'm trying to avoid factorials.

A recurrence for {1, 2, 6, 20, 70, 252... } Thanks in advance, Ben

Answer 1

This is sequence $A000984$ in $OEIS$. You will find a lot of informations about it and, in particular, $$n a_n+2(1-2n)a_{n-1}=0 \qquad \text{with} \qquad a_0=1$$

Edit

As I wrote in comments, in the $OEIS$ page, Gerry Martens gave in 2011 the nice $$a_n=\frac{ (-4)^n \sqrt{\pi }}{\Gamma \left(\frac{1}{2}-n\right) \Gamma (n+1)}$$ If you want an approximation which is quite accurate (using Stirling approximation and Padé approximants for making the series more compact) $$\log(a_n)\simeq 2\log(2)n-\log(\sqrt{\pi n})-\frac{5 n \left(5208 n^2+4121\right)}{6 \left(34720 n^4+28920 n^2+771\right)}$$ which is equivalent to an $O\left(\frac{1}{n^9}\right)$ series expansion.

For example, for $a_{10}$ the result is $184756.00000013$.