I Read a Short Questions on Mathematics and get stuck in one challenging problem. any idea from my friends?
Suppose 2n points with arrangement $P_1,Q_1,P_2,Q_2,...,P_n,Q_n$ with equidistance is distributed on the circumference of one circle. We want to connect these points with $n$ Non-crossover Hypotenuse such that:
1) each Hypotenuse joins one point $P_i$ to one point $Q_i$ and
2) each of these $2n$ points, exactly place on one Hypotenuse.
if the number of ways to doing this equal to $a_n$, what is the value of $a_5$?
The recurrence similar to counting the number of triangulations of a convex polygon (for the Catalan numbers http://en.wikipedia.org/wiki/Catalan_number) seems to also work here. Begin with a circle that has $2n$ points ($n$ $P$'s and $n$ $Q$'s). View connecting $P_{i}$ to $Q_{j}$ as a cut of the circle into two sections, a right and a left. The number of $P$'s on the right will be the same as the number of $Q$'s on the right; likewise on the left. This is because of the interleaving of the $P$'s with the $Q$'s.
If $a_{n}$ is the number of ways of connecting up $2n$ pieces, then consider what happens when $P_{1}$ is connected to $Q_{i}$. The right side has $i-1$ $P$'s and $Q$'s, the left side has $n-i$ $P$'s and $Q$'s. (Here's the small jump...) It appears that the right side can be finished with in any of $a_{i-1}$ ways, and the left can be finished in any of $a_{n-i}$ ways. So what you have is:
$$a_{n}=\sum_{i=1}^{n} a_{i-1}a_{n-i}$$
We also have that $a_{1}=1$ and for sanity, $a_{0}$ should be 1 too. This can certainly be used to compute your $a_{5}$ as requested, it is very similar to the catalan recurrence (possibly shifted by 1?).
EDIT: So to determine $a_5$ we follow the process:
\begin{align*} a_{2} &= a_{0}a_{1} + a_{1}*a_{0} \\ &= 1(1) + 1(1) \\ &= 2 \\ a_{3} &= a_{0}a_{2} + a_{1}a_{1} + a_{2}a_{0} \\ &= 1(2) + 1(1) + 2(1) \\ &= 5 \\ a_{4} &= a_{0}a_{3} + a_{1}a_{2} + a_{2}a_{1} + a_{3}a_{0} \\ &= 1(5) + 1(2) + 2(1) + 5(1) \\ &= 14 \\ a_{5} &= a_{0}a_{4} + a_{1}a_{3} + a_{2}a_{2} + a_{3}a_{1} + a_{4}a_{0} \\ &= 1(14) + 1(5) + 2(2) + 5(1) + 14(1) \\ &= 42 \end{align*}