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1
vote
Accepted
Series reversion using something like continued fraction
We assume $F(0) \neq 0$, since otherwise we don't satisfy the assumptions for the series reversion. Let $G = G(0)$ be the fixpoint of the recurrence given:
$$G(x) = F\left(\frac{x}{G(x)}\right)$$
Mult …
3
votes
A special class of weighted Motzkin paths
It's going to be useful to say that horizontal steps at height $0$ have weight $u$ so that we can derive a recurrence: therefore I shall consider $c_n(t, u)$ and look to specialise it to $u=1$ later.
…
3
votes
Accepted
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\ …