Does singularity confinement imply a fixed pattern of irreducible factors?

Consider a rational map $$f \colon (x_1,\ldots,x_n) \mapsto (P_1(x_1,\ldots,x_n),\ldots,P_n(x_1,\ldots,x_n))$$, where the $$P_i$$ are rational functions. Via iteration this map defines a discrete dynamical system on $$\mathbb{k}^n$$, where $$\mathbb{k}$$ is some algebraically closed field (take $$\mathbb{k}=\mathbb{C}$$ if you want). This system is said to have singularity confinement if all singularities of the system are "confined" in the sense that if there is $$\mathbf{x}\in\mathbb{k}^n$$ for which $$f(\mathbf{x})$$ is not defined (because of division-by-zero issues), then there is always some $$k\geq 0$$ for which $$f^k(\mathbf{x})$$ is defined.

In fact, for simplicity, in this question we can just consider one-dimensional'' dynamics $$x_{n+1}=P(x_1,\ldots,x_n)$$ for a rational function $$P$$. So singularity confinement here means that $$x_{n+1}$$ might not be defined for a certain choice of $$x_1,\ldots,x_n\in\mathbb{k}^n$$, but then there must be some $$k\geq 1$$ so that $$x_{n+k}$$ is defined.

One reason that singularity confinement might hold is because of Laurentification, as explained for instance in Section 4 of https://arxiv.org/abs/1709.00578. This means that there is a sequence of irreducible, coprime Laurent polynomials $$y_1,y_2,y_3,\ldots,$$ in $$x_1,\ldots,x_n$$ so that $$x_{n+k}=Q(y_{(k-1)\ell+1},\ldots,y_{k\ell})$$ for all $$k\geq N$$ for some fixed rational function $$Q$$ and $$\ell, N\geq 1$$.

Is there any example of a rational mapping system which has singularity confinement, but not because of Laurentification? In other words, this would mean that the irreducible factors appearing in $$x_{n+k}$$ do not eventually form any fixed pattern, but these factors still manage to always "diappear" in a finite amount of time.

• How about $x_3=x_2+\frac{a}{{x_2}^2}-x_1$? It has singularity confinement according to arxiv.org/pdf/solv-int/9711014.pdf. I've coded up for $x_{10}$ using $a=2$ and I don't think it is laurentificable, maybe because it's not positive. – Wenze 'Sylvester' Zhang Mar 19 at 23:15
• @Wenze'Sylvester'Zhang: really interesting example! I'll accept it as an answer if you provide more details (about why it has singularity confinement and what the pattern of irreducible factors looks like) – Sam Hopkins Mar 19 at 23:24