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A slow triangle map with a segment of indifferent fixed points and a complete tree of rational pairs

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Abstract

We study the two-dimensional continued fraction algorithm introduced in Garrity (J Number Theory 88(1):86–103, 2001) and the associated triangle map T, defined on a triangle \(\triangle \subseteq \mathbb {R}^2\). We introduce a slow version of the triangle map, the map S, which is ergodic with respect to the Lebesgue measure and preserves an infinite Lebesgue-absolutely continuous invariant measure. We discuss the properties that the two maps T and S share with the classical Gauss and Farey maps on the interval, including an analogue of the weak law of large numbers and of Khinchin’s weak law for the digits of the triangle sequence, the expansion associated to T. Finally, we confirm the role of the map S as a two-dimensional version of the Farey map by introducing a complete tree of rational pairs, constructed using the inverse branches of S, in the same way as the Farey tree is generated by the Farey map, and then, equivalently, generated by a generalised mediant operation.

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Notes

  1. We say that \(a_n\asymp b_n\) if and only if \(a_n =O(b_n)\) and \(b_n=O(a_n)\).

  2. The induced map \(R_A\) is an ergodic measure-preserving transformation of the probability space \((A,{\mathcal {B}}\cap A,\mu |_A)\). See, for instance, [1, Proposition 1.5.2 and 1.5.3].

  3. In [23] it is only assumed that V is measurable. We assume differentiability to simplify the approach to the system \((\bar{\triangle },\mu ,S)\).

  4. More precisely, \(\gamma \) is the only real root of the characteristic equation \(x^3-x^2-1=0\).

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Correspondence to Claudio Bonanno.

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Communicated by H. Bruin.

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We thank the referee for useful comments and for pointing out a mistake in the previous version. This research is part of the authors’ activity within the DinAmicI community, see www.dinamici.org and the Gruppo Nazionale di Fisica Matematica, INdAM, Italy. The authors are partially supported by the research project PRA_2017_22 “Dynamical systems in analysis, geometry, mathematical logic and celestial mechanics” of the University of Pisa. C. Bonanno is partially supported also by the research project PRIN 2017S35EHN_004 “Regular and stochastic behaviour in dynamical systems” of the Italian Ministry of Education and Research.

Appendices

Appendix A. Some results on the local inverses of V

In this appendix we prove some properties of the local inverses of the map V, needed for the argument of Sect. 3.1. We recall that V is the induced map of S on the set \(A=\{(x,y)\in \Gamma _0 \,:\, S(x,y)\in \Gamma _0\}\), and that each local inverse of V is a linear fractional map, as in (3.3). In general, a linear fractional map \(\psi \) of the form (3.3) with cofficients given by non-negative integers, can be expressed in projective coordinates by the \(3\times 3\) matrix

$$\begin{aligned} M_{\psi } :=\begin{pmatrix} r &{}\quad s &{}\quad t\\ r_1 &{}\quad s_1 &{}\quad t_1 \\ r_2 &{}\quad s_2 &{}\quad t_2 \end{pmatrix} \end{aligned}$$
(A.1)

by associating a point \((\frac{x}{z},\frac{y}{z})\in \mathbb {R}^2\) to a vector \(v=( z,\, x,\, y)^t\), so that \(\psi \left( \frac{x}{z},\frac{y}{z} \right) \) is associated to the vector \(M_\psi v\). For instance, the two inverse maps \(\phi _0\) and \(\phi _1\) have matrices

$$\begin{aligned} M_0 :=M_{\phi _0} = \begin{pmatrix} 1 &{}\quad 0 &{}\quad 1 \\ 1 &{}\quad 0 &{}\quad 0 \\ 0 &{}\quad 1 &{}\quad 0 \end{pmatrix} \qquad \text {and}\qquad M_1 :=M_{\phi _1} = \begin{pmatrix} 1 &{}\quad 0 &{}\quad 1 \\ 0 &{}\quad 1 &{}\quad 0 \\ 0 &{}\quad 0 &{}\quad 1 \end{pmatrix}. \end{aligned}$$

Note that the composition of linear fractional maps translates into the left multiplication of their matrices. As a consequence, since both \(M_0\) and \(M_1\) have unit determinant, every product involving these two matrices also has unit determinant. To every linear fractional map as above, we associate the vectors \(v_1(\psi ),\,v_2(\psi ),\,v(\psi ) \in \mathbb {R}^3\) corresponding to the rows of the associated matrix \(M_\psi \). In other words,

$$\begin{aligned} v_1(\psi ) = \begin{pmatrix} r_1 \\ s_1 \\ t_1 \end{pmatrix}, \qquad v_2(\psi ) = \begin{pmatrix} r_2 \\ s_2 \\ t_2 \end{pmatrix}, \qquad v(\psi ) = \begin{pmatrix} r \\ s \\ t \end{pmatrix}. \end{aligned}$$

In what follows, we use the notation \(\left\| \cdot \right\| \) for the Euclidean norm and \(\left\| \cdot \right\| _1\) for the 1-norm on \(\mathbb {R}^3\). In the following we use that

$$\begin{aligned} \Vert v\Vert \le \Vert v\Vert _1 \le \sqrt{3} \Vert v\Vert . \end{aligned}$$
(A.2)

Moreover, to each \(v\in \mathbb {R}^3{\setminus }\{0\}\) with non-negative components we associate the normalised vector \(P_v\) given by \(P_v :=\frac{v}{\Vert v\Vert _1}\).

Lemma A.1

For any \(v,w\in \mathbb {R}^3{\setminus }\{0\}\) with non-negative components, \( \Vert v \times w\Vert \le \sqrt{3} \Vert v\Vert \Vert w\Vert \Vert P_v - P_w\Vert . \)

Proof

Let \(\theta _{v,w}\in \left[ 0,\frac{\pi }{2}\right] \) be the angle between the two vectors v and w, so that \(\Vert v\times w\Vert = \Vert v\Vert \Vert w\Vert \sin \theta _{v,w}\), and between the two vectors \(P_v\) and \(P_w\). Let \(\lambda _{w,u}:=\Vert P_w -(P_w\cdot P_v) P_v\Vert \), the modulus of the component of \(P_w\) orthogonal to \(P_v\) (see Fig. 8). Then by simple geometric considerations we have

$$\begin{aligned} \sin \theta _{v,w} = \frac{\lambda _{w,v}}{\Vert P_w\Vert } = \frac{\lambda _{w,v}}{\Vert w\Vert }\Vert w\Vert _1 \overset{(\mathrm{A}.2)}{\le } \sqrt{3}\lambda _{w,v} \le \sqrt{3}\Vert P_v - P_w\Vert , \end{aligned}$$

\(\square \)

Fig. 8
figure 8

Graphical representation of vectors \(P_v\) and \(P_w\), along with the quantities involved in the proof of Lemma A.1

Lemma A.2

Let \(\Phi \) be an arbitrary composition of the maps \(\phi _0\) and \(\phi _1\). Then the matrix \(M_\Phi \) satisfies \(\Vert v(\Phi )\Vert _1\ge \Vert v_1(\Phi )\Vert _1\) and \(\Vert v(\Phi )\Vert _1\ge \Vert v_2(\Phi )\Vert _1\).

Proof

We argue by induction on the length \(l\ge 1\) of \(\Phi \) as a composition of maps. If \(l=1\), then \(\Phi \) is either \(\phi _0\) or \(\phi _1\), and in both cases the thesis is true. For the inductive step, let \(l\ge 1\) and suppose that the thesis is true for a certain \(\Phi \) of length l. Let the matrix \(M_\Phi \) be as in (A.1). We have

$$\begin{aligned}&M_{\phi _0\circ \Phi } = M_0M_\Phi = \begin{pmatrix} r+r_2 &{}\quad s+s_2 &{}\quad t+t_2\\ r &{}\quad s &{}\quad t \\ r_1 &{}\quad s_1 &{}\quad t_1 \end{pmatrix} \quad \text {and} \\&\quad M_{\phi _1\circ \Phi } = M_1M_\Phi = \begin{pmatrix} r+r_2 &{}\quad s+s_2 &{}\quad t+t_2 \\ r_1 &{}\quad s_1 &{}\quad t_1 \\ r_2 &{}\quad s_2 &{}\quad t_2 \end{pmatrix}. \end{aligned}$$

For the first matrix it holds that

$$\begin{aligned} \Vert v(\phi _0\circ \Phi )\Vert _1=r+s+t+r_2+s_2+t_2\ge r+s+t = \Vert v_1(\phi _0\circ \Phi )\Vert _1 \end{aligned}$$

since \(r_2,\,s_2,\,t_2\ge 0\), and that

$$\begin{aligned} \Vert v(\phi _0\circ \Phi )\Vert _1\ge r+s+t\ge r_1+s_1+t_1=\Vert v_2(\phi _0\circ \Phi )\Vert _1 \end{aligned}$$

by the inductive assumption. Analogous estimates hold for \(M_{\phi _1\circ \Phi }\). \(\square \)

Lemma A.3

Let \(\Phi \) be an arbitrary composition of the maps \(\phi _0\) and \(\phi _1\).

(i):

If \(D\Phi \) denotes the Jacobian matrix of \(\Phi \), then

$$\begin{aligned} \begin{aligned}&\max \left\{ \sup _A\left( \left| (D\Phi )_{11} \right| + \left| (D\Phi )_{21}\right| \right) ,\, \sup _A\left( \left| (D\Phi )_{12} \right| + \left| (D\Phi )_{22}\right| \right) \right\} \\&\quad \le 27\sqrt{3} \left( \left\| P_{v(\Phi )}-P_{v_1(\Phi )}\right\| + \left\| P_{v(\Phi )}-P_{v_2(\Phi )}\right\| \right) . \end{aligned} \end{aligned}$$
(ii):

For \(k=0,1\), let \(D_k\) be the Jacobian matrix of \(\phi _k\circ \Phi \), then

$$\begin{aligned} \begin{aligned}&\max \left\{ \sup _A\left( \left| (D_k)_{11} \right| + \left| (D_k)_{21}\right| \right) ,\, \sup _A\left( \left| (D_k)_{12} \right| + \left| (D_k)_{22}\right| \right) \right\} \\&\quad \le 27\sqrt{3} \left( \left\| P_{v(\Phi )+v_2(\Phi )}-P_{v_1(\Phi )}\right\| + \left\| P_{v(\Phi )+v_2(\Phi )}-P_{v_2(\Phi )}\right\| \right) . \end{aligned} \end{aligned}$$

In particular the worst case is realised for \(k=0\).

(iii):

We have that

$$\begin{aligned}&\left\| P_{v(\Phi )+v_2(\Phi )}-P_{v_1(\Phi )}\right\| + \left\| P_{v(\Phi )+v_2(\Phi )}-P_{v_2(\Phi )}\right\| \\&\quad \le \left\| P_{v(\Phi )}-P_{v_1(\Phi )}\right\| + \left\| P_{v(\Phi )}-P_{v_2(\Phi )}\right\| . \end{aligned}$$

Proof

  1. (i)

    The map \(\Phi \) is a composition of \(\phi _0\) and \(\phi _1\), thus of the form (3.4). Since we are interested in the local inverses of V, we look at the supremum of the Jacobian matrix of \(\Phi \) on A. Hence we have

    $$\begin{aligned}&\sup _A\left( \left| (D\Phi )_{11} \right| + \left| (D\Phi )_{21}\right| \right) \\&\quad \le 9 \cdot \frac{\left| r_1s-rs_1\right| + \left| s_1t-st_1\right| + \left| r_2s-rs_2\right| + \left| s_2t-st_2\right| }{(r+s+t)^2} \\&\quad \le 9\cdot \frac{\left\| v(\Phi )\times v_1(\Phi )\right\| _1 + \left\| v(\Phi )\times v_2(\Phi )\right\| _1}{\left\| v(\Phi )\right\| ^2}\\&\quad \overset{(\mathrm{A}.2)}{\le } 9\sqrt{3}\cdot \frac{\left\| v(\Phi )\times v_1(\Phi )\right\| + \left\| v(\Phi )\times v_2(\Phi )\right\| }{\left\| v(\Phi )\right\| ^2}\\&\quad \overset{\text {Lem. A.1}}{\le } 27\cdot \frac{\left\| v_1(\Phi )\right\| \left\| P_{v(\Phi )}-P_{v_1(\Phi )}\right\| + \left\| v_2(\Phi )\right\| \left\| P_{v(\Phi )}-P_{v_2(\Phi )}\right\| }{\left\| v(\Phi )\right\| }\\&\quad \overset{(\mathrm{A}.2)}{\le }27\sqrt{3}\cdot \frac{\left\| v_1(\Phi )\right\| _1\left\| P_{v(\Phi )}-P_{v_1(\Phi )}\right\| + \left\| v_2(\Phi )\right\| _1\left\| P_{v(\Phi )}-P_{v_2(\Phi )}\right\| }{\left\| v(\Phi )\right\| _1}. \end{aligned}$$

    From Lemma A.2 we have \(\left\| v_1(\Phi )\right\| _1\le \left\| v(\Phi )\right\| _1\) and \(\left\| v_2(\Phi )\right\| _1\le \left\| v(\Phi )\right\| _1\), so that

    $$\begin{aligned} \sup _A\, \left( \left| (D\Phi )_{11} \right| + \left| (D\Phi )_{21}\right| \right) \le 27\sqrt{3}\cdot \left( \left\| P_{v(\Phi )}-P_{v_1(\Phi )}\right\| + \left\| P_{v(\Phi )}-P_{v_2(\Phi )}\right\| \right) \end{aligned}$$

    The same estimate holds for \(\sup _A\left( \left| (D\Phi )_{12} \right| + \left| (D\Phi )_{22}\right| \right) \) and thus (i) is proved.

  2. (ii)

    The matrices associated to the maps \(\phi _k\circ \Phi \) for \(k=0,1\) are

    $$\begin{aligned} M_0M_\Phi = \begin{pmatrix} v(\Phi )+v_2(\Phi )\\ v(\Phi )\\ v_1(\Phi ) \end{pmatrix} \quad \text {and}\quad M_1M_\psi = \begin{pmatrix} v(\Phi )+v_2(\Phi )\\ v_1(\Phi )\\ v_2(\Phi ) \end{pmatrix}. \end{aligned}$$

    Applying (i) to the map \(\phi _k\circ \Phi \) we have

    $$\begin{aligned}&\max \left\{ \sup _A\left( \left| (D_0)_{11} \right| + \left| (D_0)_{21}\right| \right) ,\, \sup _A \left( \left| (D_0)_{12} \right| + \left| (D_0)_{22}\right| \right) \right\} \\&\quad \le 27\sqrt{3} \left( \left\| P_{v(\Phi )+v_2(\Phi )}-P_{v(\Phi )}\right\| +\left\| P_{v(\Phi )+v_2(\Phi )}-P_{v_1(\Phi )}\right\| \right) \end{aligned}$$

    and

    $$\begin{aligned}&\max \left\{ \sup _A\left( \left| (D_1)_{11} \right| + \left| (D_1)_{21}\right| \right) ,\,\sup _A \left( \left| (D_1)_{12} \right| + \left| (D_1)_{22}\right| \right) \right\} \\&\quad \le 27\sqrt{3} \left( \left\| P_{v(\Phi )+v_2(\Phi )}-P_{v_1(\Phi )}\right\| + \left\| P_{v(\Phi )+v_2(\Phi )}-P_{v_2(\Phi )}\right\| \right) . \end{aligned}$$

    To finish the proof it suffices to show that

    $$\begin{aligned} \left\| P_{v(\Phi )+v_2(\Phi )}-P_{v(\Phi )}\right\| \le \left\| P_{v(\Phi )+v_2(\Phi )}-P_{v_2(\Phi )}\right\| . \end{aligned}$$
    (A.3)

    To this end, note that \(P_{v(\Phi )+v_2(\Phi )}\) is a convex combination of \(P_{v(\Phi )}\) and \(P_{v_2(\Phi )}\), in particular

    $$\begin{aligned} P_{v(\Phi )+v_2(\Phi )} = \frac{\left\| v(\Phi )\right\| _1}{\left\| v(\Phi )\right\| _1 + \left\| v_2(\Phi )\right\| _1} P_{v(\Phi )} + \frac{\left\| v_2(\Phi )\right\| _1}{\left\| v(\Phi )\right\| _1 + \left\| v_2(\Phi )\right\| _1}P_{v_2(\Phi )}, \end{aligned}$$

    and since from Lemma A.2 we have \(\left\| v_2(\Phi )\right\| _1 \le \left\| v(\Phi )\right\| _1\), (A.3) easily follows.

  3. (iii)

    The three points \(P_{v(\Phi )}\), \(P_{v_1(\Phi )}\) and \(P_{v_2(\Phi )}\) belong to the standard 2-symplex in \(\mathbb {R}^3\) and define a triangle \(\triangle _\Phi \) since they are linearly independent. Furthermore, (ii) implies that \(P_{v(\Phi )+v_2(\Phi )} = \lambda P_{v(\Phi )} + (1-\lambda ) P_{v_2(\Phi )}\) for some \(\lambda \in \left( \frac{1}{2},1\right) \). Also \(P_{v(\Phi )+v_2(\Phi )}\), \(P_{v_1(\Phi )}\) and \(P_{v_2(\Phi )}\) define a triangle \(\triangle _\Phi '\), which is a subtriangle of \(\triangle _\Phi \). In particular, \(\triangle _\Phi \) and \(\triangle _\Phi '\) have a common side and the non-common vertex \(P_{v(\Phi )+v_2(\Phi )}\) belongs to the side of \(\triangle _\Phi \) with vertices \(P_{v(\Phi )}\) and \(P_{v_2(\Phi )}\). The inequality to prove easily follows from this geometric interpretation, since perimeter of the subtriangle \(\triangle _\Phi '\) is less than or equal to the perimeter of \(\triangle _\Phi \). Besides this geometrical approach, an analytic estimate easily follows from the triangle inequality:

    $$\begin{aligned}&\left\| P_{v(\Phi )+v_2(\Phi )}-P_{v_1(\Phi )}\right\| + \left\| P_{v(\Phi )+v_2(\Phi )}-P_{v_2(\Phi )}\right\| \\&\quad = \left\| \lambda P_{v(\Phi )} + (1-\lambda ) P_{v_2(\Phi )}-P_{v_1(\Phi )}\right\| + \lambda \left\| P_{v(\Phi )}- P_{v_2(\Phi )}\right\| \\&\quad = \left\| -(1-\lambda )(P_{v(\Phi )}-P_{v_2(\Phi )})+P_{v(\Phi )}-P_{v_1(\Phi )}\right\| + \lambda \left\| P_{v(\Phi )}- P_{v_2(\Phi )}\right\| \\&\quad \le (1-\lambda )\left\| P_{v(\Phi )}-P_{v_2(\Phi )}\right\| +\left\| P_{v(\Phi )}-P_{v_1(\Phi )}\right\| + \lambda \left\| P_{v(\Phi )}- P_{v_2(\Phi )}\right\| \\&\quad =\left\| P_{v(\Phi )}-P_{v_1(\Phi )}\right\| + \left\| P_{v(\Phi )}-P_{v_2(\Phi )}\right\| . \end{aligned}$$

Lemma A.4

Let \(\psi _{i_1,\,\dots ,\,i_n} : A\rightarrow C_{i_1,\,\dots ,\,i_n}\) be a local inverse of \(V^n\). Then

$$\begin{aligned} \left\| P_{v(\psi _{i_1,\,\dots ,\,i_n})}-P_{v_1(\psi _{i_1,\,\dots ,\,i_n})}\right\| + \left\| P_{v(\psi _{i_1,\,\dots ,\,i_n})}-P_{v_2(\psi _{i_1,\,\dots ,\,i_n})}\right\| \le \tilde{d}(n) \end{aligned}$$

where

$$\begin{aligned} \tilde{d}(n):=\left\| P_{v(\phi _0^{n})}-P_{v_1(\phi _0^{n})}\right\| + \left\| P_{v(\phi _0^{n})}-P_{v_2(\phi _0^{n})}\right\| . \end{aligned}$$

Proof

As outlined in Sect. 3.1, \(\psi _{i_1,\,\ldots ,\,i_n} = \psi _{i_1}\circ \dots \circ \psi _{i_n}\) and, for \(h=1,\,\ldots ,\,n\), \(\psi _{i_h}=\phi _0\circ \Phi _{i_h}\), where \(\Phi _{i_h}\) is empty or a composition of the maps \(\phi _0\) and \(\phi _1\) beginning with \(\phi _0\). In Proposition A.3 (iii) we proved that the estimation function introduced in (i) is decreasing with respect to the number of compositions of the maps \(\phi _0\) or \(\phi _1\). The inequality of this lemma follows, since \(\phi _0^n\) contains the least possible number of compositions of the maps \(\phi _0\) and \(\phi _1\) compatible with the definition of the local inverses, and by Proposition A.3-(ii) realises the worst case. \(\square \)

Lemma A.5

Let \((\tilde{d}(n))_{n\ge 0}\) be the real sequence introduced in Lemma A.4. Then \(\lim _{n\rightarrow +\infty } \tilde{d}(n) = 0\).

Proof

Arguing by induction on \(n\ge 0\), it is easy to prove that

$$\begin{aligned} M_{\phi _0^n} = M_0^n = \begin{pmatrix} f_{n+4} &{}\quad f_{n+2} &{}\quad f_{n+3}\\ f_{n+3} &{}\quad f_{n+1} &{}\quad f_{n+2}\\ f_{n+2} &{}\quad f_{n} &{}\quad f_{n+1} \end{pmatrix}, \end{aligned}$$

where \((f_n)_{n\ge 0}\) is recursively defined to be \(f_0=0\), \(f_1=1\), \(f_2=0\), and \( f_{n+3} = f_{n+2}+f_n\) for all \(n\ge 1\). The sequence \((\nu _n)_{n\ge 0}\), \(\nu _n:=f_{n+4}\), is also referred to as the Narayana’s cows sequence. It is known that this sequence has a ratio limit, i.e. there exists \(\lim _{n\rightarrow \infty }\frac{\nu _{n+1}}{\nu _n} =:\gamma <+\infty \) [26]Footnote 4. Note that for \(n\ge 4\) and for \(r\ge 1\) we have \(\frac{f_{n+r}}{f_n} = \prod _{j=0}^{r-1} \frac{f_{n+1+j}}{f_{n+j}}\), so that

$$\begin{aligned} \lim _{n\rightarrow +\infty } \frac{f_{n+r}}{f_n} = \gamma ^r. \end{aligned}$$
(A.4)

For \(n\ge 0\) we have

$$\begin{aligned} P_{v_1(\phi _0^{n})}&= \frac{1}{f_{n+1}+f_{n+2}+f_{n+3}} \begin{pmatrix}f_{n+3}\\ f_{n+1}\\ f_{n+2}\end{pmatrix} = \frac{1}{f_{n+5}} \begin{pmatrix}f_{n+3}\\ f_{n+1}\\ f_{n+2}\end{pmatrix}\\ P_{v_2(\phi _0^{n})}&= \frac{1}{f_{n}+f_{n+1}+f_{n+2}} \begin{pmatrix}f_{n+2}\\ f_{n}\\ f_{n+1}\end{pmatrix} = \frac{1}{f_{n+4}} \begin{pmatrix}f_{n+2}\\ f_{n}\\ f_{n+1}\end{pmatrix},\\ P_{v(\phi _0^{n})}&= \frac{1}{f_{n+2}+f_{n+3}+f_{n+4}} \begin{pmatrix}f_{n+4}\\ f_{n+2}\\ f_{n+3}\end{pmatrix} = \frac{1}{f_{n+6}} \begin{pmatrix}f_{n+4}\\ f_{n+2}\\ f_{n+3}\end{pmatrix} \end{aligned}$$

so that

$$\begin{aligned} \tilde{d}(n)\le & {} \left\| P_{v(\phi _0^{n})}-P_{v_1(\phi _0^{n})}\right\| _1 + \left\| P_{v(\phi _0^{n})}-P_{v_2(\phi _0^{n})}\right\| _1 \\= & {} \sum _{k=0}^2 \left( \left| \frac{f_{n+k+2}}{f_{n+6}}-\frac{f_{n+k+1}}{f_{n+5}}\right| + \left| \frac{f_{n+k+2}}{f_{n+6}}-\frac{f_{n+k}}{f_{n+4}}\right| \right) . \end{aligned}$$

Using (A.4), for each \(k=0,\,1,\,2\) we have \(\frac{f_{n+k+2}}{f_{n+6}}-\frac{f_{n+k+1}}{f_{n+5}}\rightarrow \gamma ^{4-k}-\gamma ^{4-k}=0\) and analogously \(\frac{f_{n+k+2}}{f_{n+6}}-\frac{f_{n+k}}{f_{n+4}} \rightarrow 0\) as \(n\rightarrow +\infty \). This proves that \(\lim _{n\rightarrow +\infty } \tilde{d}(n)= 0\). \(\square \)

Proof of Proposition 3.9

It follows directly from Lemma A.3-(i), A.4 and A.5 with \(d(n)=27 \sqrt{3}\, \tilde{d}(n)\). \(\square \)

Appendix B. The wandering rate of the set A

The set A is defined in (3.2) and it is the triangle with vertices \(Q_1=(\frac{1}{2}, \frac{1}{2})\), \(Q_2= (\frac{2}{3}, \frac{1}{3})\) and \(Q_3=(1,1)\), with the sides \(Q_1Q_2\) and \(Q_2Q_3\) not included. We consider the wandering rate \(w_n(A)\) for \(n\ge 1\), which is defined to be

$$\begin{aligned} w_n(A) :=\sum _{k=0}^{n-1} \mu (A \cap \{\varphi >k\}), \end{aligned}$$

where \(\varphi \) is the first-return time function in A. Extending the function \(\varphi \) to all \(\bar{\triangle }\) by

$$\begin{aligned} \varphi (x,y) :=\min \left\{ n\ge 1\, :\, S^n(x,y)\in A\right\} \end{aligned}$$

we obtain the hitting time function of A, which is well-defined and finite \(\mu \)-almost everywhere since the system \((\bar{\triangle }, \mu , S)\) is conservative and ergodic. We now recall that, for \(k\ge 1\),

$$\begin{aligned} \mu (A \cap \{ \varphi >k\}) = \mu (A^\complement \cap \{ \varphi =k\}), \end{aligned}$$

where \(A^\complement :=\bar{\triangle }{\setminus } A\) [27, Lemma 1]. We thus study the diverging sequence \(\sum _{k=1}^n \mu (A^\complement \cap \{ \varphi =k\})\). The first step is to study the structure of \(A^\complement \cap \{ \varphi =k\}\) for \(k\ge 1\), the set of points in \(A^\complement \) which hit A for the first time after exactly k iterations of the map S. This set can be expressed in terms of the local inverse of S as follows. Let

$$\begin{aligned}&\Omega _k :=\left\{ \omega \in \{0,1\}^k \,:\, \pi _{\omega _i \omega _{i+1}}=1\ \forall i=0,\,\ldots ,\,k-2,\ \omega _{k-1}=1\right\} , \\&\quad \text {where } \Pi = (\pi _{ij})_{i,j=0,1} = \begin{pmatrix} 0 &{}\quad 1\\ 1 &{}\quad 1\end{pmatrix}. \end{aligned}$$

In this way, \(\Omega _k\) is the set of binary words of length k, which all end with a “1”, and in which the string “00” never appears. Then

$$\begin{aligned} A^\complement \cap \{\varphi =k\} = \bigcup _{\omega \in \Omega _k} \phi _\omega (A) = \bigcup _{\omega \in \Omega _k} \phi _{\omega _0} \circ \phi _{\omega _1} \circ \dots \circ \phi _{\omega _{k-2}} \circ \phi _1 (A). \end{aligned}$$

Indeed, a point in \(\phi _{\omega _0} \circ \phi _{\omega _1} \circ \dots \circ \phi _{\omega _{k-2}} \circ \phi _1 (A)\) has symbolic code given by \(\omega _0 \omega _1\dots \omega _{k-2} 1 00\) with the word “00” not appearing in the first k symbols. This is equivalent to saying that such a point does not visit A in the first \(k-1\) iterations, hence the point is in \(A^\complement \cap \{ \varphi =k\}\). The converse also obviously holds. Note that, in case \(k=1\), we have \(A^\complement \cap \{\varphi =1\}=\phi _1(A)\).

We first obtain an estimate from above for the wandering rate. In what follows, we write \(a_n \lesssim b_n\) if and only if \(a_n=O(b_n)\).

Proposition B.1

The wandering rate \(w_n(A)\) satisfies \(w_n(A) \lesssim \log ^2 n\).

Proof

Using the properties of the map S and its local inverses, one immediately verifies that

$$\begin{aligned} \bigcup _{k=1}^n \left( A^\complement \cap \{ \varphi =k\}\right) \subseteq \bigcup _{k=0}^{n-1} \triangle _k, \end{aligned}$$

where \(\{\triangle _k\}_{k\ge 0}\) is the partition represented in Fig. 1. Hence \(w_n(A) \le \sum _{k=0}^{n-1} \mu (\triangle _k)\). Using now the dynamical system defined in Sect. 2.2 on the strip \(\Sigma \), we have \(\mu (\triangle _k) = \rho (\Sigma _k)\) for all \(k\ge 0\), so that

$$\begin{aligned} w_n(A)\le & {} \sum _{k=0}^{n-1} \mu (\triangle _k) = \sum _{k=0}^{n-1} \rho (\Sigma _k) = \sum _{k=0}^{n-1} \int _k^{k+1} \left( \int _0^1 \frac{1}{1+uv} du\right) dv \\= & {} \int _0^n \frac{\log (1+v)}{v} dv \lesssim \log ^2 n. \end{aligned}$$

\(\square \)

To obtain an estimate from below, we use the matrix representation of the local inverses defined in Appendix A.

Lemma B.2

For a map \(\psi = \phi _{\omega _0} \circ \phi _{\omega _1} \circ \dots \circ \phi _{\omega _{k-2}} \circ \phi _1\) with matrix representation (A.1) it holds that

$$\begin{aligned}&\frac{m(A)}{(r_1+s_1+t_1)(r_2+s_2+t_2)(r+s+t)} \le \mu (\psi (A)) \\&\quad \le \frac{27 m(A)}{(r_1+s_1+t_1)(r_2+s_2+t_2)(r+s+t)} \end{aligned}$$

Proof

By definition of \(\mu \), denoting \(\psi (x,y) = (\psi _1(x,y),\psi _2(x,y))\),

$$\begin{aligned} \mu (\psi (A)) = \iint _{\psi (A)} \frac{1}{xy} dx dy = \iint _A \frac{1}{\psi _1(x,y) \psi _2(x,y)} \left| J\psi (x,y) \right| dxdy. \end{aligned}$$

Moreover by Proposition 3.10, we have

$$\begin{aligned}&\psi _1(x,y) = \frac{r_1 +s_1 x +t_1y}{r +s x +ty} , \quad \psi _2(x,y) = \frac{r_2 +s_2 x +t_2y}{r +s x +ty} , \\&\quad J\psi (x,y) = \frac{1}{(r +s x +ty)^3} \end{aligned}$$

hence

$$\begin{aligned} \mu (\psi (A)) = \iint _A \frac{1}{(r_1 +s_1 x +t_1y)(r_2 +s_2 x +t_2y)(r +s x +ty)} dxdy \end{aligned}$$

Since for \((x,y)\in A\) we can use \(\frac{1}{2}\le x\le 1\) and \(\frac{1}{3}\le y \le 1\), the proof is complete. \(\square \)

We are then led to study the terms

$$\begin{aligned} t_{\omega _0 \omega _1\dots \omega _{k-2} 1} :=\frac{1}{(r_1+s_1+t_1)(r_2+s_2+t_2)(r+s+t)} \end{aligned}$$
(B.1)

for the maps \(\psi = \phi _{\omega _0} \circ \phi _{\omega _1} \circ \dots \circ \phi _{\omega _{k-2}} \circ \phi _1\) with \(\omega \in \Omega _k\). We shall also write simply \(t_{\psi }\) to shorten the notation. Thus we consider the sequence

$$\begin{aligned} \tau _n :=\sum _{k=1}^n \sum _{\omega \in \Omega _k} t_{\omega _0 \omega _1\dots \omega _{k-2} 1}, \end{aligned}$$

which by Lemma B.2 satisfies

$$\begin{aligned} m(A)\sum _{\omega \in \Omega _k} t_{\omega _0 \omega _1\dots \omega _{k-2} 1} \le \mu (A^\complement \cap \{ \varphi =k\}) \le 27m(A)\sum _{\omega \in \Omega _k} t_{\omega _0 \omega _1\dots \omega _{k-2} 1}. \end{aligned}$$
(B.2)

and then

$$\begin{aligned} m(A)\, \tau _n \le w_n(A) \le 27\, m(A)\, \tau _n\, . \end{aligned}$$
(B.3)

Moreover, given a linear fractional map \(\psi \) with matrix representation (A.1), if we introduce the vector

$$\begin{aligned} V_\psi :=\begin{pmatrix} r + s + t \\ r_1 + s_1 + t_1 \\ r_2 + s_2 + t_2 \end{pmatrix}, \end{aligned}$$

the term \(t_\psi \) in (B.1) is the inverse of the product of the components of \(V_\psi \). We also use the notation \(t_{V_\psi }\) for \(t_\psi \).

We now define a tree \({\mathcal {V}}\) of vectors, in such a way that the k-th level of \({\mathcal {V}}\) is associated to the set \(A^\complement \cap \{\varphi =k\}\). We first make a small modification in order to simplify the argument. For each \(k\ge 1\), we consider the subsets

$$\begin{aligned} \Phi _k :=A^\complement \cap \{ \varphi =k\} \cap \Gamma _1, \end{aligned}$$

so that

$$\begin{aligned} \Phi _k = \bigcup _{\omega \in \Omega _k} \phi _1 \circ \phi _{\omega _1} \circ \dots \circ \phi _{\omega _{k-2}} \circ \phi _1 (A) \end{aligned}$$

Obviously \(\Phi _1 = A^\complement \cap \{ \varphi =1\}=\{\phi _1(A)\}\), whereas for example

$$\begin{aligned} \Phi _2 = \left\{ \phi _1 \circ \phi _1 (A)\right\} \subsetneq A^\complement \cap \{ \varphi =2\} = \left\{ \phi _0 \circ \phi _1 (A),\, \phi _1 \circ \phi _1 (A)\right\} . \end{aligned}$$

We are now ready to introduce the levels of our tree \({\mathcal {V}}\). For each \(k\ge 1\) we define

$$\begin{aligned} L_k :=\left\{ V_\psi \, :\, \psi = \phi _1 \circ \phi _{\omega _1} \circ \dots \circ \phi _{\omega _{k-2}} \circ \phi _1\right\} \quad \text {and}\quad \lambda _k :=\sum _{V \in L_k} t_V, \end{aligned}$$

where \(t_V\) is the inverse of the product of the components of the vector V. The k-th row of \({\mathcal {V}}\) is the set \(L_k\). We have then associated two objects to each set \(A^\complement \cap \{\varphi =1\}\): the list of vectors \(L_k\) and the quantity \(\lambda _k\). For instance, corresponding to \(\Phi _1\) we obtain

$$\begin{aligned} V_1 :=V_{\phi _1} = \begin{pmatrix} 2 \\ 1 \\ 1 \end{pmatrix} \end{aligned}$$

and \(\lambda _1 = t_1=\frac{1}{2}\). The vector \(V_1\) is the root of our tree \({\mathcal {V}}\). Then

$$\begin{aligned} L_2 = \left\{ V_{\phi _1 \circ \phi _1} = \begin{pmatrix} 3 \\ 1 \\ 1 \end{pmatrix}\right\} \quad \text {and}\quad L_3 = \left\{ V_{\phi _1 \circ \phi _1 \circ \phi _1} = \begin{pmatrix} 4 \\ 1 \\ 1 \end{pmatrix} , V_{\phi _1 \circ \phi _0 \circ \phi _1} = \begin{pmatrix} 4 \\ 2 \\ 1 \end{pmatrix}\right\} , \end{aligned}$$

as follows by writing the matrix representation of the involved maps. Furthermore, for the first rows, one easily finds \(\lambda _2 = t_{V_{\phi _1 \circ \phi _1} } = \frac{1}{3}\), \(\lambda _3 = t_{V_{\phi _1 \circ \phi _1 \circ \phi _1} } + t_{V_{\phi _1 \circ \phi _0 \circ \phi _1} } = \frac{1}{4} + \frac{1}{8}\), and so on.

Moreover the tree \({\mathcal {V}}\) can be generated from the root vector \(V_1\) by the following algorithm, without using the maps \(\psi \). Let us consider the matrices

$$\begin{aligned} M_1 = \begin{pmatrix} 1 &{}\quad 0 &{}\quad 1 \\ 0 &{}\quad 1 &{}\quad 0 \\ 0 &{}\quad 0 &{}\quad 1 \end{pmatrix} \quad \text {and} \quad M_{10}= \begin{pmatrix} 1 &{}\quad 1 &{}\quad 1 \\ 1 &{}\quad 0 &{}\quad 0 \\ 0 &{}\quad 1 &{}\quad 0 \end{pmatrix}, \end{aligned}$$

that are the matrix representations of the maps \(\phi _1\) and \(\phi _1\circ \phi _0\) respectively. Let them act on the vectors of the tree to generate new vectors. When we apply \(M_1\) to a vector \(V\in L_k\), we obtain a vector in \(L_{k+1}\), and when we apply \(M_{10}\) we obtain a vector in \(L_{k+2}\). Hence, vectors in the k-th row of \({\mathcal {V}}\) are generated by applying \(M_1\) to all vectors in the \((k-1)\)-th row and \(M_{10}\) to all vectors in the \((k-2)\)-th row. Applying this algorithm starting from \(L_1=\left\{ V_1\right\} \), we immediately obtain for the first rows

$$\begin{aligned} L_2= & {} \left\{ M_1 V_1 = \begin{pmatrix} 1 &{}\quad 0 &{}\quad 1 \\ 0 &{}\quad 1 &{}\quad 0 \\ 0 &{}\quad 0 &{}\quad 1 \end{pmatrix}\begin{pmatrix} 2 \\ 1 \\ 1 \end{pmatrix} = \begin{pmatrix} 3 \\ 1 \\ 1 \end{pmatrix}\right\} \\ L_3= & {} \left\{ M_1(M_1 V_1) = \begin{pmatrix} 1 &{}\quad 0 &{}\quad 1 \\ 0 &{}\quad 1 &{}\quad 0 \\ 0 &{}\quad 0 &{}\quad 1 \end{pmatrix}\begin{pmatrix} 3 \\ 1 \\ 1 \end{pmatrix} = \begin{pmatrix} 4 \\ 1 \\ 1 \end{pmatrix},\ M_{10}V_1 \right. \\= & {} \left. \begin{pmatrix} 1 &{}\quad 1 &{}\quad 1 \\ 1 &{}\quad 0 &{}\quad 0 \\ 0 &{}\quad 1 &{}\quad 0 \end{pmatrix}\begin{pmatrix} 2 \\ 1 \\ 1 \end{pmatrix} = \begin{pmatrix} 4 \\ 2 \\ 1 \end{pmatrix}\right\} , \end{aligned}$$

as above.

Lemma B.3

For \(n\ge 1\) define

$$\begin{aligned} \tilde{\tau }_n :=\sum _{k=1}^n \lambda _k = \sum _{k=1}^n \sum _{V \in L_k} t_V. \end{aligned}$$

Then \(\tilde{\tau }_n< \tau _n < \tilde{\tau }_n + \frac{\mu (\Gamma _0)}{m(A)}\) and \(\tilde{\tau }_n \gtrsim \log ^2 n\).

Proof

The difference between \(\tilde{\tau }_n\) and \(\tau _n\) is that for each \(k=1,\,\ldots ,\,n\), in \(\tilde{\tau }_n\) we are not considering the terms \(t_{V_\psi }\) for the maps \(\psi = \phi _{\omega _0} \circ \phi _{\omega _1} \circ \dots \circ \phi _{\omega _{k-2}} \circ \phi _1\) with \(\omega _0=0\). Recalling that for such maps \(t_{V_\psi } \le \frac{\mu (\psi (A))}{m(A)}\) by Lemma B.2, that \(\psi (A) \subseteq \Gamma _0\) if \(\omega _0=0\), and that the sets \(\psi (A)\) are disjoint for different maps \(\psi \) by definition, for all \(n\ge 1\) we have that

$$\begin{aligned} \tilde{\tau }_n< \tau _n < \tilde{\tau }_n + \frac{\mu (\Gamma _0)}{m(A)}. \end{aligned}$$

We prove by induction that each row \(L_k\) with \(k\ge 2\) contains the vectors

$$\begin{aligned} \begin{pmatrix} k+1 \\ j \\ 1 \end{pmatrix}\quad j=1,\,\dots ,\,k-1. \end{aligned}$$

By the definition of \(\lambda _k\), this implies that \(\tilde{\tau }_n :=\sum _{k=1}^n \lambda _k \ge \sum _{k=1}^n\, \frac{1}{k+1}\, \sum _{j=1}^{k-1}\, \frac{1}{j} \gtrsim \log ^2 n\). For \(k=2\), the row \(L_2\) contains only the vector \(M_1 V_1\), and the base case is proved. Let us assume that the statement is true for \(r=2,\,\dots ,\,k\), then using the algorithm to construct \({\mathcal {V}}\), we have that \(L_{k+1}\) contains the vectors

$$\begin{aligned} M_1 \begin{pmatrix} k+1 \\ j \\ 1 \end{pmatrix} = \begin{pmatrix} k+2 \\ j \\ 1 \end{pmatrix}\quad j=1,\,\dots ,\,k-1 \quad \text {and}\quad M_{10} \begin{pmatrix} k \\ 1 \\ 1 \end{pmatrix} = \begin{pmatrix} k+2 \\ k \\ 1 \end{pmatrix} . \end{aligned}$$

Hence the statement is true for \(L_{k+1}\). \(\square \)

Proposition B.4

The wandering rate \(w_n(A)\) satisfies \(w_n(A) \gtrsim \log ^2 n\).

Proof

It follows from (B.3) and Lemma B.3. \(\square \)

Finally we discuss the property of regular variation for \(w_n(A)\). The first remark is that if \(w_n(A)\) is regularly varying then it is slowly varying. By (3.1), if \(w_n(A)\) is regularly varying then there exists \(\alpha \in \mathbb {R}\) such that

$$\begin{aligned} \lim _{n\rightarrow \infty }\, \frac{w_{cn}}{w_n} = c^\alpha \end{aligned}$$

for all \(c\in \mathbb {N}\). However by Propositions B.1 and B.4, there exist two constants \(k_1,k_2\) with \(0<k_1< 1 < k_2\) such that

$$\begin{aligned} k_1\, \frac{\log ^2 (cn)}{\log ^2 (n)} \le \frac{w_{cn}}{w_n} \le k_2\, \frac{\log ^2 (cn)}{\log ^2 (n)} \end{aligned}$$

and passing to the limit we obtain \(k_1 \le c^\alpha \le k_2\) for all \(c\in \mathbb {N}\). Hence \(\alpha =0\).

A second remark is that we have a sufficient condition on the sequence \(\lambda _k\) from Lemma B.3 for \(w_n(A)\) being slowly varying. Since \(\tilde{\tau }_n(A) \gtrsim \log ^2 n\), it is immediate that

$$\begin{aligned} \liminf _{k\rightarrow \infty } \, \frac{k\, \lambda _k}{\log ^2 k} = 0 \end{aligned}$$

To have that \(w_n(A)\) is slowly varying it is enough that also the limsup vanishes.

Lemma B.5

If \(\lambda _k = o(\frac{\log ^2 k}{k})\) then \(w_n(A)\) is slowly varying.

Proof

From (B.2) we obtain that if \(\tau _n\) is slowly varying the same holds for \(w_n(A)\). Indeed

$$\begin{aligned} w_{2n}(A) - w_n(A)= & {} \sum _{k=n+1}^{2n} \mu (A^\complement \cap \{ \varphi =k\}) \\\le & {} 27m(A) \sum _{k=n+1}^{2n} \sum _{\omega \in \Omega _k} t_{\omega _0 \omega _1\dots \omega _{k-2} 1} = 27m(A) (\tau _{2n} - \tau _n) \end{aligned}$$

and

$$\begin{aligned} w_n(A) = \sum _{k=1}^n \mu (A^\complement \cap \{ \varphi =k\}) \ge m(A) \sum _{k=1}^n \sum _{\omega \in \Omega _k} t_{\omega _0 \omega _1\dots \omega _{k-2} 1} = m(A) \tau _n. \end{aligned}$$

In conclusion

$$\begin{aligned} 1\le \frac{w_{2n}(A)}{w_n(A)} = 1+ \frac{w_{2n}(A) - w_n(A)}{w_n(A)} \le 1 + 27 \frac{\tau _{2n}-\tau _n}{\tau _n} . \end{aligned}$$

If \((\tau _n)_{n\ge 1}\) is slowly varying the term \(\frac{\tau _{2n}-\tau _n}{\tau _n}\) is vanishing, and the result follows.

Moreover from Lemma B.3 it is immediate that if \(\tilde{\tau }_n\) is slowly varying then the same is true for \(\tau _n\). We are thus reduced to study \(\tilde{\tau }_n\). We first claim that it is enough to show that (3.1) holds with \(\alpha =0\) only for \(c=2\) (see for example [3, Proposition 1.10.1]). Indeed, for \(1<c<2\) we write

$$\begin{aligned} 1 \le \frac{\tilde{\tau }_{\lfloor cn \rfloor }}{\tilde{\tau }_n} \le \frac{\tilde{\tau }_{2n}}{\tilde{\tau }_n}. \end{aligned}$$

For \(c>2\), let \(k\ge 1\) such that \(c\le 2^k\), then we write

$$\begin{aligned} 1\le \frac{\tilde{\tau }_{\lfloor cn \rfloor }}{\tilde{\tau }_n} \le \frac{\tilde{\tau }_{2^k n}}{\tilde{\tau }_{2^{k-1}n}} \frac{\tilde{\tau }_{2^{k-1} n}}{\tilde{\tau }_{2^{k-2}n}} \cdots \frac{\tilde{\tau }_{2 n}}{\tilde{\tau }_{n}} \end{aligned}$$

and for all \(j=1,\,\dots ,\,k \) we have \(\frac{\tilde{\tau }_{2^j n}}{\tilde{\tau }_{2^{j-1}n}} \rightarrow 1\), because it is a subsequence of \(\frac{\tilde{\tau }_{2n}}{\tilde{\tau }_{n}}\). Hence (3.1) follows again with \(\alpha =0\) for \(c>2\). We can proceed analogously for the case \(0<c<1\), which completes the proof of the claim.

Moreover we follow the proof of [3, Theorem 1.5.4] to show that (3.1) holds with \(\alpha =0\) for \(c=2\). Let \(\alpha >0\), then by definition the sequence \(\phi (n):=n^\alpha \, \tilde{\tau }_n\) is non-decreasing. We also show that the sequence \(\psi (n):=n^{-\alpha }\, \tilde{\tau }_n\) is eventually non-increasing. Indeed

$$\begin{aligned} \psi (n)-\psi (n+1) = \psi (n)\, \left( 1- \frac{\tilde{\tau }_{n+1}}{\tilde{\tau }_n}\, \frac{n^\alpha }{(n+1)^\alpha } \right) = \psi (n)\, \left( 1- \frac{1+\frac{\lambda _{n+1}}{\tilde{\tau }_n}}{(1+\frac{1}{n})^\alpha } \right) \end{aligned}$$

and \(\lambda _{n+1} = o(\frac{\log ^2 (n+1)}{n+1})\) together with \(\tilde{\tau }_n \gtrsim \log ^2 n\) implies

$$\begin{aligned} 1+\frac{\lambda _{n+1}}{\tilde{\tau }_n} = o \left( \frac{1}{n} \right) . \end{aligned}$$

Hence we have that

$$\begin{aligned} 1+\frac{\lambda _{n+1}}{\tilde{\tau }_n} < \left( 1+\frac{1}{n}\right) ^\alpha = 1+\alpha \frac{1}{n} + o \left( \frac{1}{n} \right) \end{aligned}$$

for n big enough. It follows that \(\psi (n)-\psi (n+1)\ge 0\) eventually. For n big enough we can then write

$$\begin{aligned} 2^{-\alpha } = \frac{\tilde{\tau }_{2n}}{\tilde{\tau }_n}\, \frac{\phi (n)}{\phi (2n)} \le \frac{\tilde{\tau }_{2n}}{\tilde{\tau }_n} \le \frac{\tilde{\tau }_{2n}}{\tilde{\tau }_n}\, \frac{\psi (n)}{\psi (2n)} = 2^\alpha \end{aligned}$$

hence

$$\begin{aligned} 2^{-\alpha } \le \liminf _{n\rightarrow \infty }\, \frac{\tilde{\tau }_{2n}}{\tilde{\tau }_n}\le \limsup _{n\rightarrow \infty }\, \frac{\tilde{\tau }_{2n}}{\tilde{\tau }_n}\le 2^\alpha . \end{aligned}$$

Since the previous argument can be repeated for all \(\alpha >0\) it follows that

$$\begin{aligned} \lim _{n\rightarrow \infty }\, \frac{\tilde{\tau }_{2n}}{\tilde{\tau }_n} = 1\, . \end{aligned}$$

\(\square \)

Finally, we recall from [1] and [27] that for the pointwise dual ergodic system \((\bar{\triangle }, \mu ,S)\) the renormalising sequence \(a_n(S)\) is defined in terms of the wandering rate of a subset on which the induced map is \(\psi \)-mixing (see Proposition 3.6), and \(a_n(S)\) is asymptotically independent on the chosen subset with this property. In particular this implies that if \(\Gamma _0\) satisfies Proposition 3.6 then we have

$$\begin{aligned} a_n(S) \asymp \frac{n}{w_n(\Gamma _0)} \end{aligned}$$

where

$$\begin{aligned} w_n(\Gamma _0) = \sum _{k=0}^{n-1}\, \mu (\triangle _k) = \int _0^n\, \frac{\log (1+v)}{v}\, dv \end{aligned}$$

as shown in Proposition B.1. Since it is not difficult to show that \(w_n(\Gamma _0)\) is slowly varying, then we would have

$$\begin{aligned} a_n(S) \sim \frac{n}{\int _0^n\, \frac{\log (1+v)}{v}\, dv} \end{aligned}$$

as discussed in Remark 3.11. Unfortunately it is not known whether \(\Gamma _0\) is a good set to which apply Proposition 3.6.

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Bonanno, C., Del Vigna, A. & Munday, S. A slow triangle map with a segment of indifferent fixed points and a complete tree of rational pairs. Monatsh Math 194, 1–40 (2021). https://doi.org/10.1007/s00605-020-01500-w

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