What does $(Z - \delta)^k [x,y]$ count?

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I am trying to understand my professor's lecture notes about zeta function and using it to count.


DEFINITIONS

The zeta function is defined as $Z[x,y] = \begin{cases} 1 & \text{if > } x \leq y \text{ in } P \\ 0 & \text{otherwise} \end{cases}$

$\delta : \text{Int}(P) \to \{0, 1\}$ such that $\delta[x,y] = > \begin{cases} 1 & \text{if } x = y \\ 0 & \text{otherwise} > \end{cases}$ is the identity of convolution product.


QUESTION

The question is the following:

What does $(Z - \delta)^k [x,y]$ count?


MY WORK

Now I know I can write it like this, where $*$ stands for convulation product:

$ (Z - \delta)^k [x,y] = (Z - \delta) * (Z - \delta) * \dots * (Z - \delta)[x,y]$

I think I can use the facts above, also from my professor's lecture notes:

  1. $(Z - \delta) \ast [x,y] = \begin{cases} 1 & \text{if } x < y \text{ (where } x \neq y\text{)} \\ 0 & \text{otherwise} \end{cases}$

and

  1. $(Z - \delta) \ast (Z - \delta) [x,y] = \sum_{x < z < y} (Z - \delta)[x,z] \cdot (Z - \delta)[z,y]$

I understand that $2$ is the definition of convolution product, and $1$ is how we subtract the identity function (?) from the zeta function I guess?

My professor also notes that $2.$ is equal to ($\sum_{x < z < y} 1$). That's where I'm totally lost, and I don't know how to manipulate this into the summation of $1$'s and then use it to count.