The smallest index i such that the i-th simple module has projective dimension equal to the global dimension of the corresponding Nakayama algebra.

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
  $$NENNENNNENEEENEE.$$