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.



Fibonacci-Motzkin paths and J-type continued fractions
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 $$\cfrac{1}{(1 - z\gamma\_0) - \cfrac{(1 - z\gamma\_0) z^2 \beta\_1}{(1 - z \gamma\_1) + z^2 \beta\_1 - \cfrac{(1 - z \gamma\_1) z^2 \beta\_2}{(1 - z \gamma\_2) + z^2 \beta\_2 - \ddots}}}$$


