Taylor Expansion of the Polynomial in Flajolet’s Fundamental Lemma

Question

I am currently looking at the proof of Flajolet’s Fundamental Lemma. Before I phrase the question, I need to review the definition of $(0,k)$-path and define its weight.

Define $(0,k)$-path as the following ($k\ge 0$):

1. It is a lattice path that start with $(0,0)\in \mathbb{N}^{2}$ ($0\in \mathbb{N}$).
2. For each step, it can go one unit, at either diagonally upper-right, diagonally down-right, or horizontally right direction.
3. It cannot go below $y=0$ or above $y=k$.
4. It ends at $(l,0)$, for some $l\ge 0$.

Let $P_{0}^{k}$ be the collection of all such $(0,k)$-paths. Let's define the weight on on each path as the following:

1. For each step diagonally upper-right or down-right, associate it with weight $1$.
2. For each step horizontally right at level $\not= k$, associate it with weight the indeterminant $z+1$. For step horizontally right at level $k$, associate it with weight $0$.
3. For the trivial path, the single point $(0,0)$ (i.e. $l=0$), define its weight as $1$.
4. Define the weight of the entire path, $\sigma$, as the product of the weight of all of its steps, which is a polynomial in $\mathbb{Z}[z]$. Denote it by $\omega(\sigma)$.

Let's consider the sum of the weight of all $(0,k)$-paths, i.e. define $$ p_{0}^{k}:=\sum_{\sigma\in P_{0}^{k}} \omega(\sigma) $$ which is a formal power series in $\mathbb{Z}[[z]]$. It can be proven that $$ p_{0}^{k}=\frac{1}{-z-\frac{1}{-z-\frac{1}{\frac{\vdots}{-z-1}}}} $$ which has $k$-layers. I notice that its Taylor expansion at $z=0$ always has integer coefficient, for all $k\ge 1$. For example $$ p_{0}^{2}=1+2x+3x^{2}+5x^{3}+8x^{4}+13x^{5}+21x^{6}+\cdots $$

My questions is this: I understand that the entire polynomial $p_{0}^{k}$ is counting the total weight of all $(0,k)$-paths, for all length $l\ge 0$. But what is the geometric intepretation of the coefficients of its Taylor expansion? What objects are they counting?

Initially, I guessed that, for $p_{0}^{k}$, the coefficient of $x^{l}$ is counting the weighted number of paths ends at $(l,0)$. However, it is not the case, but I feel that it is counting something like that.