Fibonacci-Motzkin paths and J-type continued fractions

Question

Recall that a Motzkin path is a piece-wise linear planar path connecting points in the integer lattice quadrant $\Bbb{Z}_{\geq 0} \times \Bbb{Z}_{\geq 0}$ beginning at the origin $(0,0)$ and ending at $(n,0)$ for some $n \in \Bbb{Z}_{>0}$ whose steps are either

\begin{equation} \begin{array}{ll} \nearrow & = (1,1) \\ \searrow & = (1,-1) \\ \rightarrow & = (1,0) \end{array} \end{equation}

Alternatively we may view a Motzkin path as an $n$-tuple $\underline{\pi} = \big(\pi_1, \dots, \pi_n \big)$ where $\pi_i \in \{ \nearrow \, , \searrow \, , \rightarrow \}$ for each $1 \leq i \leq n$ and where $\pi_1 + \cdots + \pi_n = (n,0)$. The set of Motzkin paths $\underline{\pi}$ terminating at $(n,0)$ will be denoted $\mathcal{M}_n$. Among the Motzkin paths $\mathcal{M}_n$ is the subset of what I'll call Fibonacci-Motzkin paths: These are paths that weakly increase until a threshold is reached, after which they strictly decrease. More specifically $\underline{\pi} = (\pi_1, \dots, \pi_n) \in \mathcal{M}_n$ is a Fibonacci-Motzkin path if there exists a threshold $ 0 \leq k \leq \lfloor {1 \over 2} n \rfloor $ such that $\pi_i \in \{ \nearrow, \, \rightarrow \}$ for all $1 \leq i \leq n - k$ and $\pi_i = \searrow$ for all $n-k < i \leq n$. The threshold $k$ equals the number of $\nearrow$ steps taken in the initial ascent of the path. A moment's reflection should convince the reader that the cardinality of the set $\mathcal{F}_n$ of all Fibonacci-Motzkin paths is indeed the $n$-th Fibonacci number, thus justifying the choice of terminology.

Received wisdom tells us to introduce two infinite families of generic parameters $\beta_1, \beta_2, \beta_3, \dots$ and $\gamma_0, \gamma_1, \gamma_2, \dots$ and then assign a weight

\begin{equation} \mathrm{w}(\underline{\pi}) \, := \ \beta_1 \cdots \beta_k \cdot \gamma_0^{m_0} \cdots \gamma_k^{m_k} \end{equation}

to a Motzkin path $\underline{\pi} \in \mathcal{M}_n$ where

\begin{equation} \begin{array}{ll} k &\text{$=$ number of $\nearrow$ steps taken by $\underline{\pi}$} \\ m_\ell &\text{$=$ number of $\rightarrow$ steps taken by $\underline{\pi}$ at height $\ell$} \end{array} \end{equation}

The corresponding generating function of all Motzkin paths, given by

\begin{equation} M(z) := \ 1 \ + \ \sum_{n \, \geq \, 1} z^n \sum_{\underline{\pi} \, \in \, \mathcal{M}_n} \, \mathrm{w}(\underline{\pi}) \end{equation}

is then seen to coincide with the formal expansion of the $J$-type continued fraction

\begin{equation} \ {1 \over {1 - z \gamma_0 \ - \ {\displaystyle z^2 \beta_1 \over {\displaystyle 1 - z \gamma_1 \ - \ {z^2 \beta_2 \over {\displaystyle 1 - z \gamma_2 \ - \ {z^2 \beta_3 \over {\ddots}}}}}}}} \end{equation}

Let us introduce a Fibonacci analogue of the generating function $M(z)$ namely

\begin{equation} F(z) := \ 1 \ + \ \sum_{n \, \geq \, 1} z^n \sum_{\underline{\pi} \, \in \, \mathcal{F}_n} \, \mathrm{w}(\underline{\pi}) \end{equation}

Question: Is there some kind of continued fraction whose expansion is $F(z)$?

thanks, ines.

Answer 1

Grouping the terms of $F(z)$ by the height reached, we get $$F(z) = \frac{1}{(1 - z\gamma_0)} + \frac{z^2 \beta_1}{(1 - z\gamma_0) (1 - z\gamma_1)} + \frac{z^4 \beta_1 \beta_2}{(1 - z\gamma_0) (1 - z\gamma_1) (1 - z\gamma_2)} + \cdots \\ $$ This has the form of Euler's continued fraction $$a_0 + a_0a_1 + a_0a_1a_2 + \cdots = \cfrac{a_0}{1 - \cfrac{a_1}{1 + a_1 - \cfrac{a_2}{1 + a_2 - \ddots}}}$$ with $$a_0 = \frac{1}{1 - z\gamma_0} \\ a_1 = \frac{z^2 \beta_1}{1 - z \gamma_1} \\ a_2 = \frac{z^2 \beta_2}{1 - z \gamma_2} \\ \vdots $$ It is perhaps more natural to drop the denominators to the next level: i.e. instead of $$\cfrac{\frac{1}{1 - z\gamma_0}}{1 - \cfrac{\frac{z^2 \beta_1}{(1 - z \gamma_1)}}{1 + \frac{z^2 \beta_1}{(1 - z \gamma_1)} - \cfrac{\frac{z^2 \beta_2}{(1 - z \gamma_2)}}{1 + \frac{z^2 \beta_2}{(1 - z \gamma_2)} - \ddots}}}$$ we could write $$\cfrac{1}{(1 - z\gamma_0) - \cfrac{(1 - z\gamma_0) z^2 \beta_1}{(1 - z \gamma_1) + z^2 \beta_1 - \cfrac{(1 - z \gamma_1) z^2 \beta_2}{(1 - z \gamma_2) + z^2 \beta_2 - \ddots}}}$$