count pairs of non-crossing partitions that differ by one transposition 
Problem:
$\sigma$ and $\delta$ are non-crossing partitions of $(1,2,\cdots,n)$. They only differ only by a single transposition. In another words, $\sigma \delta^{-1}$ is a single transposition.
Let the number of pairs of $\sigma, \delta$ to be $a_n$. Find
  an analytic expression for $a_n$.

Example:
When $n = 3$, we have 12 pairs:
\begin{equation}
\begin{aligned}
\sigma &= \left\lbrace
          \begin{aligned}
            &(123)&  \\
            & I&  \\
          \end{aligned} \right. 
\delta = \left\lbrace
         \begin{aligned}
          (12)  \\
          (13)  \\
          (23)
         \end{aligned} \right. 
\end{aligned}
+ \text{interchange} 
\end{equation}
Numerics suggests a surprisingly simple result
\begin{equation}
a_n = 2{2n \choose n - 2 }
\end{equation}
which makes me believe there is a simple proof. 
non-crossing partition
Let $\sigma$ be a permutation element which composes of cycles. We can draw it on a circle by connecting points within a circle. $\sigma$ is called a non-crossing partition if the resulting diagram has no intersection. See wikipedia for more about its precise definition and properties. 
For example, $\sigma = (13)2(456)$ is drew on the circle below, and it is a non-crossing partition. 

Attempt:
Since non-crossing partition has orders, we can count the case of $\sigma > \delta$ and multiply by 2. This means that $\delta$ is a finer partition than $\sigma$, i.e. it can only break cycles inside $\sigma$. 
Suppose $\sigma$ has $k$ cycles $c_1, c_2, \cdots, c_k$. $\delta$ can be constructed by breaking any one of the cycles into two. Suppose cycle 1 has $|c_1| = m$ elements. Without loss of generality, let it be $(123\cdots m)$. There are ${m \choose 2 }$ ways for $\delta$ to break this cycle into two cycles, i.e. taking $1 \le i < j \le m$, $\delta = (ij) \sigma$. There are 
\begin{equation}
\sum_{i=1}^k  { |c_i|  \choose 2 }
\end{equation}
ways to construct $\delta$ out of this $\sigma$. Hence
\begin{equation}
a_n = \sum_{\substack{\sigma \in \text{NC}\\ \sigma = \{c_1, c_2, \cdots, c_k \}}} \sum_{i=1}^k  { |c_i|  \choose 2 }
\end{equation}
I do not find a way to construct a recursive relation or generating function for $a_n $. It may be related to the free probability theory because it is a summation over the non-crossing partition lattice. But from what I know, the moment-cumulant relation in free probability is a summation over the product on the functions on each cycle, not a summation. 
Edit 11/10/2017:
According to the article 1 and cited Kreweras's paper, there is a simple formula of counting the non-crossing partition according to cycle types. 
Let $(r_1, r_2, \cdots, r_n )$ be a set of non-negative integers that satisfy
\begin{equation}
\sum_{m=1}^n m r_m = n 
\end{equation}
We can then use $r_m$ to denote the number m-cycles. The number of non-crossing partitions that has $r_m$ m-cycles ($m = 1, \cdots, n$) is
\begin{equation}
\frac{n!}{(n - \sum_m r_m +1 )!} \prod_m \frac{1}{r_m!} = \frac{1}{n+1} { n+1 \choose r_1, r_2, \cdots, r_n } 
\end{equation}
Then 
\begin{equation}
a_n = \sum_{\{r_i\}} \frac{1}{n+1} { n+1 \choose r_1, r_2, \cdots, r_n } \sum_{m=1}^{n} r_m { m \choose 2 } 
\end{equation}
where the summation is performed over the constraint $\sum_{m=1}^n m r_m = n$. 
1 Liaw, S.C.; Yeh, H.G.; Hwang, F.K.; Chang, G.J., A simple and direct derivation for the number of noncrossing partitions, Proc. Am. Math. Soc. 126, No.6, 1579-1581 (1998). ZBL0892.05007. 
 A: First, apply the bijection from noncrossing partitions of size $n$ to Dyck paths of order $n$, where for each vertex in order we write

*

*${↗↘}$ if it’s not connected,

*${↗↗}$ if it’s connected only to following vertices,

*${↘↗}$ if it’s connected to a preceding vertex and a following vertex,

*${↘↘}$ if it’s connected only to preceding vertices.

(Your example partition $(13)2(456)$ is written $↗↗↗↘↘↘↗↗↘↗↘↘$.)
Under this bijection, a single-transposition pair $σ, δ$ corresponds precisely to a pair of Dyck paths where we swap a matched $↗, ↘$ pair of nonzero height, or a $↘, ↗$ pair that would become matched; that is,
$$a_L{↗}b{\color{red}↗}c{\color{red}↘}d{↘}a_R \\
a_L{↗}b{\color{red}↘}c{\color{red}↗}d{↘}a_R$$
where $a_La_R, b, c, d$ are four Dyck paths, or the opposite. Breaking a cycle swaps a $↗$ at an even position with a $↘$ at an odd position, while joining two cycles swaps a $↗$ at an odd position with a $↘$ at an even position.
We can count these pairs as follows. Consider any staircase path $e$ (not a Dyck path) with $n + 2$ $↗$s and $n - 2$ $↘$s. Split it into $e_L{↗}e_R$ at the last $↗$ of minimum height. Then $e_R{↘↘↘}e_L$ is a Dyck path. We can let $a_R = e_L$, decompose $e_R$ uniquely as $a_L{↗}b{↗}c{↗}d$ where $b, c, d$ are Dyck paths, then plug $a_L, b, c, d, a_R$ into the above schema and its opposite. This construction gives each valid pair exactly once, so there are $2\binom{2n}{n - 2}$ of them.
