Number of ways to pair off $2n$ points such that no chords intersect
First off, there is a missing case for $n = 3$ : the one where you pair $(2,3), (1,4), (6,5)$.
Now the recursion to count the number of pairing would go like this :
Let us take a fixed point $O$ among one of the $2n$ points on the circle. A line going from this point would have to divide the circle into two regions, each containing an even number of points. There are thus $n$ possible choices for a chord leaving from $O$ (make a drawing).
Choosing one such cord will lead to unique divisions of the circle (two different cords leaving from $O$ will give different divisions). Reciprocally, any division of the circle would match with one of these cords. So we have correctly divided our problem into a set of disjoint subproblems.
Then, to get a recursive formula, we need to count how many points are on each side of the cord, for each possible cord : the regions have $(2k, 2(n-k-1))$ points, for $0\leq k \leq n-1$.
So finally, a recursive formula for $h(n)$ would be :
$h(n) = \sum_{k=0}^{n-1} h(k) \times h(n-k-1)$.