Number of indecomposable modules with projective dimension equal to the global dimension in the Nakayama algebra corresponding to the Dyck path.

Return the Dyck path corresponding to the partition interpreted as a parallogram polyomino.
The Ferrers diagram of an integer partition can be interpreted as a parallogram polyomino, such that each part corresponds to a column.
This map returns the corresponding Dyck path.

The cut-out partition of a Dyck path.
The partition $\lambda$ associated to a Dyck path is defined to be the complementary partition inside the staircase partition $(n-1,\ldots,2,1)$ when cutting out $D$ considered as a path from $(0,0)$ to $(n,n)$.
In other words, $\lambda_{i}$ is the number of down-steps before the $(n+1-i)$-th up-step of $D$.
This map is a bijection between Dyck paths of size $n$ and partitions inside the staircase partition $(n-1,\ldots,2,1)$.

The zeta map on Dyck paths.
The zeta map $\zeta$ is a bijection on Dyck paths of semilength $n$.
It was defined in [1, Theorem 1], see also [2, Theorem 3.15] and sends the bistatistic (area, dinv) to the bistatistic (bounce, area). It is defined by sending a Dyck path $D$ with corresponding area sequence $a=(a_1,\ldots,a_n)$ to a Dyck path as follows:

• First, build an intermediate Dyck path consisting of $d_1$ north steps, followed by $d_1$ east steps, followed by $d_2$ north steps and $d_2$ east steps, and so on, where $d_i$ is the number of $i-1$'s within the sequence $a$.
  For example, given $a=(0,1,2,2,2,3,1,2)$, we build the path
  $$NE\ NNEE\ NNNNEEEE\ NE.$$
• Next, the rectangles between two consecutive peaks are filled. Observe that such the rectangle between the $k$th and the $(k+1)$st peak must be filled by $d_k$ east steps and $d_{k+1}$ north steps. In the above example, the rectangle between the second and the third peak must be filled by $2$ east and $4$ north steps, the $2$ being the number of $1$'s in $a$, and $4$ being the number of $2$'s. To fill such a rectangle, scan through the sequence a from left to right, and add east or north steps whenever you see a $k-1$ or $k$, respectively. So to fill the $2\times 4$ rectangle, we look for $1$'s and $2$'s in the sequence and see $122212$, so this rectangle gets filled with $ENNNEN$.
  The complete path we obtain in thus
  $$NENNENNNENEEENEE.$$