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Asymptotic repetitive threshold of balanced sequences

L​’ubomíra Dvořáková FNSPE Czech Technical University in Prague, Czech Republic [email protected] Daniela Opočenská FNSPE Czech Technical University in Prague, Czech Republic [email protected]  and  Edita Pelantová FNSPE Czech Technical University in Prague, Czech Republic [email protected]
Abstract.

The critical exponent E(𝐮)E(\mathbf{u}) of an infinite sequence 𝐮\mathbf{u} over a finite alphabet expresses the maximal repetition of a factor in 𝐮\mathbf{u}. By the famous Dejean’s theorem, E(𝐮)1+1d1E(\mathbf{u})\geq 1+\frac{1}{d-1} for every dd-ary sequence 𝐮\mathbf{u}. We define the asymptotic critical exponent E(𝐮)E^{*}(\mathbf{u}) as the upper limit of the maximal repetition of factors of length nn. We show that for any d>1d>1 there exists a dd-ary sequence 𝐮\mathbf{u} having E(𝐮)E^{*}(\mathbf{u}) arbitrarily close to 11. Then we focus on the class of dd-ary balanced sequences. In this class, the values E(𝐮)E^{*}(\mathbf{u}) are bounded from below by a threshold strictly bigger than 1. We provide a method which enables us to find a dd-ary balanced sequence with the least asymptotic critical exponent for 2d102\leq d\leq 10.

2020 Mathematics Subject Classification:
68R15
The first and the third authors were supported by the Ministry of Education, Youth and Sports of the Czech Republic through the project CZ.02.1.01/0.0/0.0/16_019/0000778.
The second author was supported by Czech technical university in Prague, through the project SGS20/183/OHK4/3T/14.

1. Introduction

The concatenation of ee\in\mathbb{N} copies of a non-empty word uu is usually abbreviated as ueu^{e}. In 1972, Dejean extended this exponential notation to rational exponents. If uu is a non-empty word of length \ell and ee is a positive rational number of the form n/n/\ell, then ueu^{e} denotes the prefix of length nn of the infinite periodic sequence uuu=uωuuu\cdots=u^{\omega}. For instance, a Czech word starostastarosta (mayor) can be written in this formalism as (staro)8/5(staro)^{8/5}. The rational exponent ee describes the repetition rate of uu in the string ueu^{e}.

The critical exponent E(𝐮)E(\mathbf{u}) of an infinite sequence 𝐮=u0u1u2\mathbf{u}=u_{0}u_{1}u_{2}\cdots captures the maximal possible repetition rate of factors occurring in 𝐮\mathbf{u}, formally,

E(𝐮)=sup{e:ueis a factor of 𝐮for a non-empty wordu}.E(\mathbf{u})=\sup\{e\in\mathbb{Q}:\ u^{e}\ \text{is a factor of }\mathbf{u}\ \text{for a non-empty word}\ u\}\,.

The number E(𝐮)E(\mathbf{u}) can be rational or irrational. Krieger and Shallit [KrSh07] have shown that any positive real number larger than one is a critical exponent of some sequence over a finite alphabet. If the size of the alphabet is a fixed number d,d2d\in\mathbb{N},d\geq 2, then the critical exponent of any sequence 𝐮\mathbf{u} over this alphabet cannot be smaller than a threshold larger than one. This bound is denoted in [CuRa11] as RT(d)RT(d) and called the repetitive threshold, i.e.,

RT(d)=inf{E(𝐮):𝐮 is a sequence over a d-ary alphabet}.RT(d)=\inf\{E(\mathbf{u}):\mathbf{u}\text{\ is a sequence over a $d$-ary alphabet}\}\,.

For instance, if 𝐮\mathbf{u} is a binary sequence over {0,1}\{0,1\}, then 0000 or 1111 or 01010101 appears in 𝐮\mathbf{u}, and thus E(𝐮)2E(\mathbf{u})\geq 2. The existence of an infinite binary sequence 𝐮\mathbf{u} with E(𝐮)=2E(\mathbf{u})=2 was demonstrated by Thue in 1912. The binary sequence for which E(𝐮)=2E(\mathbf{u})=2 is nowadays known as the Thue-Morse sequence (or the Prouhet-Thue-Morse sequence). Therefore RT(2)=2RT(2)=2. Dejean [Dej72] proved that RT(3)=74RT(3)=\frac{7}{4} and this bound is attained. In the same paper she also conjectured:

  • RT(4)=7/5RT(4)=7/5;

  • RT(d)=1+1d1RT(d)=1+\frac{1}{d-1} for d5d\geq 5.

The conjecture had been proved step by step by many people [Pan84c, Mou92, MoCu07, Car07, CuRa11, Rao11].

Recently, Rampersad, Shallit and Vandome asked the same question for dd-ary balanced sequences. Let us recall that a sequence over a finite alphabet is balanced if, for any two of its factors uu and vv of the same length, the number of occurrences of each letter in uu and vv differs by at most 1. Binary balanced aperiodic sequences coincide with Sturmian sequences and the least critical exponent is equal to 2+1+522+\frac{1+\sqrt{5}}{2} and it is reached by the Fibonacci sequence [CaLu2000]. Let us denote the repetitive threshold of balanced sequences over a dd-ary alphabet by RTB(d)RTB(d), i.e.,

RTB(d)=inf{E(𝐯):𝐯 is a balanced sequence over a d-ary alphabet}.RTB(d)=\inf\{E(\mathbf{v}):\mathbf{v}\text{ is a balanced sequence over a }d\text{-ary alphabet}\}\,.

The following results have been proved so far:

  • RTB(3)=2+12RTB(3)=2+\frac{1}{\sqrt{2}} and RTB(4)=1+1+54RTB(4)=1+\frac{1+\sqrt{5}}{4} [RSV19];

  • RTB(d)=1+1d3RTB(d)=1+\frac{1}{d-3} for 5d105\leq d\leq 10 [Bar20, BaSh19, DDP21];

  • RTB(d)=1+1d2RTB(d)=1+\frac{1}{d-2} for d=11d=11 and all even numbers d12d\geq 12 [DOPS2022].

It remains as an open problem to prove the conjecture RTB(d)=1+1d2RTB(d)=1+\frac{1}{d-2} also for all odd numbers d13d\geq 13.

In this paper we focus on the asymptotic critical exponent E(𝐮)E^{*}(\mathbf{u}) of 𝐮\mathbf{u}. It is defined to be ++\infty if E(𝐮)=+E(\mathbf{u})=+\infty and

E(𝐮)=lim supn{e:ueis a factor of 𝐮for some uof lengthn},E^{*}(\mathbf{u})=\limsup_{n\to\infty}\{e\in\mathbb{Q}:\ u^{e}\ \text{is a factor of }\mathbf{u}\ \text{for some }u\ \text{of length}\ n\}\,,

otherwise. Obviously, E(𝐮)E(𝐮)E^{*}(\mathbf{u})\leq E(\mathbf{u}) and the equality holds true whenever E(𝐮)E(\mathbf{u}) is irrational. It is for instance the case of the Fibonacci sequence. Nevertheless, E(𝐮)E^{*}(\mathbf{u}) and E(𝐮)E(\mathbf{u}) can coincide even if E(𝐮)E(\mathbf{u}) is rational: it is the case of the Thue-Morse sequence.

While the value E(𝐮)E(\mathbf{u}) takes into account repetitions of all factors of 𝐮\mathbf{u}, E(𝐮)E^{*}(\mathbf{u}) considers only repetitions of factors of length tending to infinity. There is a huge literature devoted to questions situated between these two extremes. For example, Shur and Tunev [ShTu] construct a dd-ary sequence 𝐮\mathbf{u} whose all factors (except the trivial repetition of one letter in factors of the form a1a2ad1a1a_{1}a_{2}\cdots a_{d-1}a_{1}) have the exponent <1+1d1=RT(d)<1+\frac{1}{d-1}=RT(d). Other results of this flavour can be found in [BaCr, De, Sh].

We provide a simple construction showing that for any d,d2,d\in\mathbb{N},d\geq 2, and any ϵ>0\epsilon>0, there exists a dd-ary sequence 𝐮\mathbf{u} with E(𝐮)<1+ϵE^{*}(\mathbf{u})<1+\epsilon. Therefore, if we denote

RT(d)=inf{E(𝐮):𝐮 is a sequence over a d-ary alphabet},RT^{*}(d)=\inf\{E^{*}(\mathbf{u}):\mathbf{u}\text{\ is a sequence over a $d$-ary alphabet}\}\,,

we have RT(d)=1for all d2.RT^{*}(d)=1\quad\text{for all $d\geq 2$}\,. Then we restrict our study to balanced sequences and look for the threshold

RTB(d)=inf{E(𝐯):𝐯 is a balanced sequence over a d-ary alphabet}.RTB^{*}(d)=\inf\{E^{*}(\mathbf{v}):\mathbf{v}\text{\ is a balanced sequence over a $d$-ary alphabet}\}\,.

This threshold is bounded from below by 1+12d21+\frac{1}{2^{d-2}}, see Corollary 15. It is known that RTB(2)=RTB(2)=2+5+123.618RTB^{*}(2)=RTB(2)=2+\frac{\sqrt{5}+1}{2}\doteq 3.618 and it is reached by the Fibonacci sequence. We introduce a new tool – graphs of admissible tails – for computation of the exact value of RTB(d)RTB^{*}(d). Using it we obtain at once RTB(d)RTB^{*}(d) for d10d\leq 10, see Table 1. Let us point out that a similar result for RTB(d)RTB(d) had been done step by step by several research teams.

Comparing the results with the minimal critical exponent of balanced sequences, we can see that RTB(d)=RTB(d)RTB^{*}(d)=RTB(d) for d{2,3,4,5}d\in\{2,3,4,5\}, but RTB(d)<RTB(d)RTB^{*}(d)<RTB(d) for larger dd. Moreover, the precise values RTB(d)RTB^{*}(d) we have found for d10d\leq 10 suggest that RTB(d)<1+qdRTB^{*}(d)<1+{q^{d}} for some positive q<1q<1, whereas RTB(d)>RT(d)=1+1d1RTB(d)>RT(d)=1+\frac{1}{d-1}. Looking into Table 1 we can see that

RTB(6)<1+14RTB^{*}(6)<1+\frac{1}{4},   RTB(7)<1+17RTB^{*}(7)<1+\frac{1}{7},   RTB(8)<1+120RTB^{*}(8)<1+\frac{1}{20},   RTB(9)<1+130RTB^{*}(9)<1+\frac{1}{30},  RTB(10)<1+168RTB^{*}(10)<1+\frac{1}{68}.

The time and space complexity of our algorithm allowed us to determine RTB(d)RTB^{*}(d) only for d10d\leq 10. It seems that a new idea is needed to extend Table 1 for d11d\geq 11.

2. Preliminaries

An alphabet 𝒜\mathcal{A} is a finite set of symbols called letters. A word over 𝒜\mathcal{A} of length nn is a string u=u0u1un1u=u_{0}u_{1}\cdots u_{n-1}, where ui𝒜u_{i}\in\mathcal{A} for all i{0,1,,n1}i\in\{0,1,\ldots,n-1\}. The length of uu is denoted by |u||u|. The set of all finite words over 𝒜\mathcal{A} together with the operation of concatenation forms a monoid, denoted 𝒜\mathcal{A}^{*}. Its neutral element is the empty word ε\varepsilon and we denote 𝒜+=𝒜{ε}\mathcal{A}^{+}=\mathcal{A}^{*}\setminus\{\varepsilon\}. If u=xyzu=xyz for some x,y,z𝒜x,y,z\in\mathcal{A}^{*}, then xx is a prefix of uu, zz is a suffix of uu and yy is a factor of uu. To any word uu over 𝒜\mathcal{A} with cardinality #𝒜=d\#\mathcal{A}=d, we assign its Parikh vector V(u)d\vec{V}(u)\in\mathbb{N}^{d} defined as (V(u))a=|u|a(\vec{V}(u))_{a}=|u|_{a} for all a𝒜a\in\mathcal{A}, where |u|a|u|_{a} is the number of letters aa occurring in uu.

A sequence over 𝒜\mathcal{A} is an infinite string 𝐮=u0u1u2\mathbf{u}=u_{0}u_{1}u_{2}\cdots, where ui𝒜u_{i}\in\mathcal{A} for all ii\in\mathbb{N}. The notation 𝒜\mathcal{A}^{\mathbb{N}} stands for the set of all sequences over 𝒜\mathcal{A}. We always denote sequences by bold letters. The shift operator σ\sigma maps any sequence 𝐮=u0u1u2\mathbf{u}=u_{0}u_{1}u_{2}\cdots to the sequence σ(𝐮)=u1u2u3\sigma(\mathbf{u})=u_{1}u_{2}u_{3}\cdots.

A sequence 𝐮\mathbf{u} is eventually periodic if 𝐮=vwww=vwω\mathbf{u}=vwww\cdots=vw^{\omega} for some v𝒜v\in\mathcal{A}^{*} and w𝒜+w\in\mathcal{A}^{+}. It is periodic if 𝐮=wω\mathbf{u}=w^{\omega}. In both cases, the number |w||w| is a period of 𝐮\mathbf{u}. We write Per𝐮\mathrm{Per}\,\mathbf{u} for the minimal period of 𝐮\mathbf{u}. If 𝐮\mathbf{u} is not eventually periodic, then it is aperiodic. A factor of 𝐮=u0u1u2\mathbf{u}=u_{0}u_{1}u_{2}\cdots is a word yy such that y=uiui+1ui+2uj1y=u_{i}u_{i+1}u_{i+2}\cdots u_{j-1} for some i,ji,j\in\mathbb{N}, iji\leq j. The number ii is called an occurrence of the factor yy in 𝐮\mathbf{u}. In particular, if i=ji=j, the factor yy is the empty word ε\varepsilon and any index ii is its occurrence. If i=0i=0, the factor yy is a prefix of 𝐮\mathbf{u}. If each factor of 𝐮\mathbf{u} has infinitely many occurrences in 𝐮\mathbf{u}, the sequence 𝐮\mathbf{u} is recurrent. Moreover, if for each factor the distances between its consecutive occurrences are bounded, 𝐮\mathbf{u} is uniformly recurrent.

The language (𝐮)\mathcal{L}(\mathbf{u}) of a sequence 𝐮\mathbf{u} is the set of all its factors. A factor ww of 𝐮\mathbf{u} is right special if wa,wbwa,wb are in (𝐮)\mathcal{L}(\mathbf{u}) for at least two distinct letters a,b𝒜a,b\in\mathcal{A}. A left special factor is defined symmetrically. A factor is bispecial if it is both left and right special.

A sequence 𝐮𝒜\mathbf{u}\in\mathcal{A}^{\mathbb{N}} is balanced if for every letter a𝒜a\in\mathcal{A} and every pair of factors u,v(𝐮)u,v\in{\mathcal{L}}(\mathbf{u}) with |u|=|v||u|=|v|, we have |u|a|v|a1|u|_{a}-|v|_{a}\leq 1. Every recurrent balanced sequence over any alphabet is uniformly recurrent, see [DolceDP21]. An aperiodic balanced sequence over a binary alphabet is called Sturmian, [MoHe40]. There exist many equivalent definitions of Sturmian sequences, see for instance [BeSe02].

A morphism over 𝒜\mathcal{A} is a mapping ψ:𝒜𝒜\psi:\mathcal{A}^{*}\to\mathcal{A}^{*} such that ψ(uv)=ψ(u)ψ(v)\psi(uv)=\psi(u)\psi(v) for all u,v𝒜u,v\in\mathcal{A}^{*}. Morphisms can be naturally extended to 𝒜\mathcal{A}^{\mathbb{N}} by setting ψ(u0u1u2)=ψ(u0)ψ(u1)ψ(u2)\psi(u_{0}u_{1}u_{2}\cdots)=\psi(u_{0})\psi(u_{1})\psi(u_{2})\cdots\,. A fixed point of a morphism ψ\psi is a sequence 𝐮\mathbf{u} such that ψ(𝐮)=𝐮\psi(\mathbf{u})=\mathbf{u}.

Example 1.

The most famous Sturmian sequence is the Fibonacci sequence

𝐮f=𝚋𝚊𝚋𝚋𝚊𝚋𝚊𝚋𝚋𝚊𝚋𝚋𝚊𝚋𝚊𝚋𝚋𝚊𝚋𝚊𝚋𝚋,\mathbf{u}_{f}={\tt babbababbabbababbababb}\cdots\,,

defined as the fixed point of the morphism f:𝚋𝚋𝚊f:{\tt b}\mapsto{\tt ba}, 𝚊𝚋{\tt a}\mapsto{\tt b}. The critical exponent of 𝐮f\mathbf{u}_{f} is 2+1+522+\frac{1+\sqrt{5}}{2}. As shown by Carpi and de Luca [CaLu2000], it is the least critical exponent for Sturmian sequences, i.e., RTB(2)=E(𝐮f)RTB(2)=E(\mathbf{u}_{f}).

Consider a factor ww of a recurrent sequence 𝐮=u0u1u2\mathbf{u}=u_{0}u_{1}u_{2}\cdots. Let i<ji<j be two consecutive occurrences of ww in 𝐮\mathbf{u}. Then the word uiui+1uj1u_{i}u_{i+1}\cdots u_{j-1} is a return word to ww in 𝐮\mathbf{u}. The set of all return words to ww in 𝐮\mathbf{u} is denoted by 𝐮(w)\mathcal{R}_{\mathbf{u}}(w). If 𝐮\mathbf{u} is uniformly recurrent, the set 𝐮(w)\mathcal{R}_{\mathbf{u}}(w) is finite for each factor ww. If ww is a prefix of 𝐮\mathbf{u}, then 𝐮\mathbf{u} can be written as a concatenation 𝐮=rd0rd1rd2\mathbf{u}=r_{d_{0}}r_{d_{1}}r_{d_{2}}\cdots of return words to ww. The sequence 𝐝𝐮(w)=d0d1d2\mathbf{d}_{\mathbf{u}}(w)=d_{0}d_{1}d_{2}\cdots over the alphabet of cardinality #𝐮(w)\#\mathcal{R}_{\mathbf{u}}(w) is called the derived sequence of 𝐮\mathbf{u} to ww. If ww is not a prefix, then we construct the derived sequence in an analogous way starting from the first occurrence of ww in 𝐮\mathbf{u}. The concept of derived sequences was introduced by Durand [Dur98].

3. Asymptotic critical exponent and its relation to return words

In [DDP21] a handy formula for computation of the critical exponent and asymptotic critical exponent of uniformly recurrent sequences is deduced. It uses the notion of return words to a factor of a sequence.

Theorem 2 ([DDP21]).

Let 𝐮\mathbf{u} be a uniformly recurrent aperiodic sequence111Arseny M. Shur in private communication pointed out to us that the same formula remains valid for recurrent aperiodic sequences.. Let (un)(u_{n}) be a sequence of all bispecial factors ordered by their length. For every nn\in\mathbb{N}, let rnr_{n} be a shortest return word to unu_{n} in 𝐮\mathbf{u}. Then

E(𝐮)=1+supn{|un||rn|}andE(𝐮)=1+lim supn|un||rn|.\mathrm{E}(\mathbf{u})=1+\sup\limits_{n\in\mathbb{N}}\left\{\frac{|u_{n}|}{|r_{n}|}\right\}\qquad\text{and}\qquad\mathrm{E}^{*}(\mathbf{u})=1+\limsup\limits_{n\to\infty}\frac{|u_{n}|}{|r_{n}|}.
Theorem 3.

Let 𝒜\mathcal{A} be a finite alphabet of size at least 2. Then

inf{E(𝐮):𝐮𝒜,𝐮 uniformly recurrent}=1.\inf\{\,E^{*}(\mathbf{u})\ :\ \mathbf{u}\in\mathcal{A}^{\mathbb{N}},\ \mathbf{u}\text{\ uniformly recurrent}\,\}=1.

Consequently, RT(d)=1RT^{*}(d)=1 for every d2d\geq 2.

Proof.

It is enough to prove the statement for the alphabet 𝒜={0,1}\mathcal{A}=\{0,1\}. For every Fibonacci number222The Fibonacci sequence is defined recursively: F0=0,F1=1F_{0}=0,F_{1}=1, and Fn+2=Fn+1+FnF_{n+2}=F_{n+1}+F_{n} for every nn\in\mathbb{N}. FkF_{k}, with k7k\geq 7, we construct a binary sequence 𝐮=𝐮(k)\mathbf{u}=\mathbf{u}^{(k)} such that every bispecial factor uu of length at least 3(k+1)3(k+1) and every return word rr to uu in 𝐮\mathbf{u} satisfy |u||r|<2Fk3\frac{|u|}{|r|}<\frac{2}{F_{k}-3}. For the sequence 𝐮(k)\mathbf{u}^{(k)}, the second formula of Theorem 2 gives E(𝐮(k))1+2Fk3E^{*}\bigl{(}{\mathbf{u}^{(k)}}\bigr{)}\leq 1+\frac{2}{F_{k}-3}, which implies the theorem. Construction of 𝐮(k)\mathbf{u}^{(k)} follows.

Let 𝒟={0,1,,d1}\mathcal{D}=\{0,1,\ldots,d-1\}, where d=212Fk{Fk1,Fk}d=2\lfloor\frac{1}{2}F_{k}\rfloor\in\{F_{k}-1,F_{k}\}. By [DOPS2022] there exists a balanced (and hence uniformly recurrent) dd-ary sequence 𝐯\mathbf{v} having E(𝐯)=d1d2E(\mathbf{v})=\frac{d-1}{d-2}. Zeckendorf’s theorem [Ze72] says that every i𝒟i\in\mathcal{D} can be written in the form i=n=2k1cnFni=\sum_{n=2}^{k-1}c_{n}F_{n}, where ck1ck2c2c_{k-1}c_{k-2}\cdots c_{2} is a word over the alphabet {0,1}\{0,1\}, which does not contain two consecutive 11’s. We denote the (k2)(k-2)-tuple ck1ck2c2c_{k-1}c_{k-2}\cdots c_{2} representing ii by (i)Fib(i)_{Fib} and define a morphism ψ:𝒟{0,1}\psi:\mathcal{D}^{*}\mapsto\{0,1\}^{*} and the binary sequence 𝐮\mathbf{u} as follows:

ψ(i)=110(i)Fib for every i𝒟and𝐮=𝐮(k)=ψ(𝐯).\psi(i)=110(i)_{Fib}\ \text{ for every }i\in\mathcal{D}\quad\text{and}\quad\mathbf{u}=\mathbf{u}^{(k)}=\psi(\mathbf{v}).

The morphism ψ\psi is uniform since the image of any letter ii by ψ\psi has length |ψ(i)|=k+1|\psi(i)|=k+1. Moreover, 𝐮\mathbf{u} is uniformly recurrent as 𝐯\mathbf{v} is uniformly recurrent. ψ\psi is a coding since the factor 110110 occurs only as a prefix of each ψ(i)\psi(i). Hence any factor u(𝐮)u\in\mathcal{L}(\mathbf{u}) longer than k+2k+2 can be written uniquely in the form u=u(L)ψ(v)u(R)u=u^{(L)}\psi(v)u^{(R)}, where v(𝐯)v\in\mathcal{L}(\mathbf{v}), u(L)u^{(L)} is a proper suffix of ψ(i)\psi(i) and u(R)u^{(R)} is a proper prefix of ψ(j)\psi(j) for some i,j𝒟i,j\in\mathcal{D}. Obviously, if uu is a left special factor of 𝐮\mathbf{u}, i.e., 0u0u and 1u1u belong to the language of 𝐮\mathbf{u}, then vv is a left special factor of 𝐯\mathbf{v}. An analogous statement is true for right special factors of 𝐮\mathbf{u}.

Let uu be a bispecial factor of 𝐮\mathbf{u} with |u|3(k+1)|u|\geq 3(k+1) and rr be a return word to uu in 𝐮\mathbf{u}. Then there exist a bispecial factor v(𝐯)v\in\mathcal{L}(\mathbf{v}) of length |v|2|v|\geq 2 and s(𝐯)s\in\mathcal{L}(\mathbf{v}) such that u=u(L)ψ(v)u(R)u=u^{(L)}\psi(v)u^{(R)} and ru=u(L)ψ(sv)u(R)ru=u^{(L)}\psi(sv)u^{(R)}. Obviously, ss is a return word or concatenation of several return words to vv in 𝐯\mathbf{v} and |r|=|ψ(s)|=(k+1)|s||r|=|\psi(s)|=(k+1)|s|. As E(𝐯)=1+1d2E(\mathbf{v})=1+\frac{1}{d-2}, Theorem 2 implies |v||s|1d2<1Fk3\frac{|v|}{|s|}\leq\frac{1}{d-2}<\frac{1}{F_{k}-3}. Hence

|u||r|=|u(L)|+(k+1)|v|+|u(R)|(k+1)|s|2k+(k+1)|v|(k+1)|s|<2|v||s|<2Fk3,\frac{|u|}{|r|}=\frac{|u^{(L)}|+(k+1)|v|+|u^{(R)}|}{(k+1)|s|}\leq\frac{2k+(k+1)|v|}{(k+1)|s|}<\frac{2|v|}{|s|}<\frac{2}{F_{k}-3}\,,

as we wanted to show. ∎

4. Sturmian sequences

Sturmian sequences, i.e., aperiodic balanced sequences over a binary alphabet, are a principal tool in the study of balanced sequences over arbitrary alphabets. In the sequel, we will restrict our consideration to standard sequences. Let us recall that a Sturmian sequence 𝐮\mathbf{u} is called a standard sequence if both sequences 𝚊𝐮{\tt a}\mathbf{u} and 𝚋𝐮{\tt b}\mathbf{u} are Sturmian. For each Sturmian sequence there exists a unique standard sequence having the same language and thus the same critical exponent and the asymptotic critical exponent. Sturmian sequences have well defined frequencies of letters. Let us recall that the frequency of a letter aa in a sequence 𝐮\mathbf{u} is the limit ρa(𝐮)=limn|u0un1|an\rho_{a}(\mathbf{u})=\lim_{n\to\infty}\frac{|u_{0}\cdots u_{n-1}|_{a}}{n} if it exists.

We use the characterization of standard sequences by their directive sequences. To introduce them, we recall two morphisms

G={𝚊𝚊𝚋𝚊𝚋andD={𝚊𝚋𝚊𝚋𝚋.G=\ \left\{\,\begin{aligned} {\tt a}&\to{\tt a}\\ {\tt b}&\to{\tt ab}\,\end{aligned}\right.\quad\text{and}\quad D=\ \left\{\,\begin{aligned} {\tt a}&\to{\tt ba}\\ {\tt b}&\to{\tt b}\,\end{aligned}\right..
Proposition 4 ([JuPi02]).

For every standard sequence 𝐮\mathbf{u} there is a uniquely given directive sequence 𝚫=Δ0Δ1Δ2{G,D}\mathbf{\Delta}=\Delta_{0}\Delta_{1}\Delta_{2}\cdots\in\{G,D\}^{\mathbb{N}} of morphisms and a sequence (𝐮(n))(\mathbf{u}^{(n)}) of standard sequences such that

𝐮=Δ0Δ1Δn1(𝐮(n))for every n.\mathbf{u}={\Delta_{0}\Delta_{1}\cdots\Delta_{n-1}}\left(\mathbf{u}^{(n)}\right)\,\ \text{for every }\ n\in\mathbb{N}\,.

Both GG and DD occur in the sequence 𝚫\mathbf{\Delta} infinitely often.

If Δ0=D\Delta_{0}=D, then 𝚋\mathtt{b} is the most frequent letter in 𝐮\mathbf{u}. Otherwise, 𝚊\mathtt{a} is the most frequent letter in 𝐮\mathbf{u}. We adopt the convention that ρ𝚋(𝐮)>ρ𝚊(𝐮)\rho_{\tt b}(\mathbf{u})>\rho_{\tt a}(\mathbf{u}) and thus the directive sequence of 𝐮\mathbf{u} starts with DD. Let us write this sequence in the run-length encoded form 𝚫=Da1Ga2Da3Ga4\mathbf{\Delta}=D^{a_{1}}G^{a_{2}}D^{a_{3}}G^{a_{4}}\cdots, where all integers ana_{n} are positive. Then the number θ\theta having the continued fraction expansion θ=[0,a1,a2,a3,]\theta=[0,a_{1},a_{2},a_{3},\ldots] equals the ratio ρ𝚊(𝐮)ρ𝚋(𝐮)\frac{\rho_{\tt a}(\mathbf{u})}{\rho_{\tt b}(\mathbf{u})} (see [BeSe02]) and θ\theta is called the slope of 𝐮\mathbf{u}.

The convergents to the continued fraction of θ\theta, usually denoted pNqN\frac{p_{N}}{q_{N}}, have a close relation to return words in a Sturmian sequence. Recall that the sequences (pN)(p_{N}) and (qN)(q_{N}) both satisfy the recurrence relation

(1) XN+1=aN+1XN+XN1X_{N+1}=a_{N+1}X_{N}+X_{N-1}

with initial conditions p1=1p_{-1}=1, p0=0p_{0}=0 and q1=0q_{-1}=0, q0=1q_{0}=1. Two consecutive convergents satisfy pNqN1pN1qN=(1)N+1p_{N}q_{N-1}-p_{N-1}q_{N}=(-1)^{N+1} for every NN\in\mathbb{N}.

Vuillon [Vui01] showed that an infinite recurrent sequence 𝐮\mathbf{u} is Sturmian if and only if each of its factors has exactly two return words. Moreover, the derived sequence of a Sturmian sequence to any of its factors is also Sturmian.

All bispecial factors of any standard sequence 𝐮\mathbf{u} are its prefixes. So, one of the return words to a bispecial factor of 𝐮\mathbf{u} is a prefix of 𝐮\mathbf{u}.

Proposition 5 ([DvMePe20]).

Suppose that 𝐮\mathbf{u} is a standard sequence with slope θ=[0,a1,a2,a3,]\theta=[0,a_{1},a_{2},a_{3},\ldots] and zz is a bispecial factor of 𝐮\mathbf{u}. Let rr (resp., ss) denote the return word to zz which is (resp., is not) a prefix of 𝐮\mathbf{u}. Then

  1. (1)

    there exists a unique pair (N,m)2(N,m)\in\mathbb{N}^{2} with 0m<aN+10\leq m<a_{N+1} such that the Parikh vectors of rr, ss, and zz are respectively

    V(r)=(pNqN),V(s)=(mpN+pN1mqN+qN1),V(z)=V(r)+V(s)(11);\vec{V}(r)=\begin{pmatrix}p_{N}\\ q_{N}\end{pmatrix},\;\;\vec{V}(s)=\begin{pmatrix}m\,p_{N}+p_{N-1}\\ m\,q_{N}+q_{N-1}\end{pmatrix},\;\;\vec{V}(z)=\vec{V}(r)+\vec{V}(s)-\begin{pmatrix}1\\ 1\end{pmatrix};
  2. (2)

    the slope of the derived sequence 𝐝𝐮(z)\mathbf{d}_{\mathbf{u}}(z) is

    θ=[0,aN+1m,aN+2,aN+3,].{\theta}^{\prime}=[0,a_{N+1}-m,a_{N+2},a_{N+3},\ldots].

To express the length of bispecial factors and their return words in a Sturmian sequence with slope θ\theta, we use for each NN\in\mathbb{N} the notation

QN:=pN+qNQ_{N}:=p_{N}+q_{N}, where pNqN\frac{p_{N}}{q_{N}} is the NthN^{th} convergent to θ\theta,

i.e., the sequence (QN)(Q_{N}) satisfies (1) with initial conditions Q1=1=Q0Q_{-1}=1=Q_{0}.

Remark 6.

The formulae for computation of E(𝐮)E(\mathbf{u}) and E(𝐮)E^{*}(\mathbf{u}) for a Sturmian sequence 𝐮\mathbf{u} with slope θ=[0,a1,a2,a3,]\theta=[0,a_{1},a_{2},a_{3},\ldots] were provided by Daminik and Lenz in [DaLe2002], see also [CaLu2000].

E(𝐮)=2+supN{aN+1+QN12QN2}andE(𝐮)=2+lim supN{aN+1+QN1QN2}.E(\mathbf{u})=2+\sup_{N\in\mathbb{N}}\Bigl{\{}a_{N+1}+\tfrac{Q_{N-1}-2}{Q_{N-2}}\Bigr{\}}\quad\text{and}\quad E^{*}(\mathbf{u})=2+\limsup_{N\to\infty}\Bigl{\{}a_{N+1}+\tfrac{Q_{N-1}}{Q_{N-2}}\Bigr{\}}\,.

These formulae can be deduced easily using Proposition 5 and Theorem 2.

In the sequel, it will be necessary to recognize which vectors are Parikh vectors of some factors in a Sturmian sequence. The answer follows.

Lemma 7 ([DDP21]).

Let 𝐮\mathbf{u} be a Sturmian sequence with slope θ=ρ𝚊(𝐮)ρ𝚋(𝐮)<1\theta=\frac{\rho_{\tt a}(\mathbf{u})}{\rho_{\tt b}(\mathbf{u})}<1. Denote δ=θ1\delta=\theta^{-1}. Then 𝐮\mathbf{u} contains a factor ww such that |w|𝚋=k|w|_{\tt b}=k and |w|𝚊=|w|_{\tt a}=\ell if and only if

(2) |δk|<δ+1.|\ell\delta-k|<\delta+1.

5. Balanced sequences

In 2000 Hubert [Hubert00] characterized balanced sequences over alphabets of cardinality bigger than 2 in terms of Sturmian sequences, colourings, and constant gap sequences.

Definition 8.

Let 𝐮\mathbf{u} be a sequence over {𝚊,𝚋}\{\mathtt{a},\mathtt{b}\}, 𝐲\mathbf{y} and 𝐲\mathbf{y}^{\prime} be arbitrary sequences. The colouring of 𝐮\mathbf{u} by 𝐲\mathbf{y} and 𝐲\mathbf{y}^{\prime} is the sequence 𝐯=colour(𝐮,𝐲,𝐲)\mathbf{v}=\mathrm{colour}(\mathbf{u},\mathbf{y},\mathbf{y}^{\prime}) obtained from 𝐮\mathbf{u} by replacing the subsequence of all 𝚊\mathtt{a}’s with 𝐲\mathbf{y} and the subsequence of all 𝚋\mathtt{b}’s with 𝐲\mathbf{y}^{\prime}.

Definition 9.

A sequence 𝐲\mathbf{y} is a constant gap sequence if for each letter aa occurring in 𝐲\mathbf{y} the distance between any consecutive occurrences of aa in 𝐲\mathbf{y} is constant.

There is a rich literature on a notion equivalent to constant gap sequences, the so-called exact covering systems, see [Znam1969, PoSch2002, GoGrBrSh2019].

Example 10.

The periodic sequences 𝐲=(34)ω\mathbf{y}=(\texttt{34})^{\omega} and 𝐲=(0102)ω\mathbf{y}^{\prime}=(\texttt{0102})^{\omega} are constant gap sequences over binary and ternary alphabet, respectively. The sequence 𝐯=colour(𝐮f,𝐲,𝐲)\mathbf{v}=\mathrm{colour}(\mathbf{u}_{f},\mathbf{y},\mathbf{y}^{\prime}), where 𝐮f\mathbf{u}_{f} is the Fibonacci sequence defined in Example 1, looks as follows:

𝐮f\displaystyle\mathbf{u}_{f} =𝚋𝚊𝚋𝚋𝚊𝚋𝚊𝚋𝚋𝚊𝚋𝚋𝚊𝚋𝚊𝚋𝚋𝚊𝚋𝚊𝚋𝚋𝚊𝚋𝚋𝚊𝚋𝚊𝚋𝚋𝚊\displaystyle=\mathtt{babbababbabbababbababbabbababba}\cdots
𝐯\displaystyle\mathbf{v} =𝟶𝟹𝟷𝟶𝟺𝟸𝟹𝟶𝟷𝟺𝟶𝟸𝟹𝟶𝟺𝟷𝟶𝟹𝟸𝟺𝟶𝟷𝟹𝟶𝟸𝟺𝟶𝟹𝟷𝟶𝟺\displaystyle=\mathtt{{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}0}{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}3}{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}10}{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}4}{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}2}{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}3}{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}01}{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}4}{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}02}{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}3}{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}0}{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}4}{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}10}{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}3}{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}2}{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}4}{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}01}{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}3}{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}02}{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}4}{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}0}{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}3}{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}10}{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}4}}\cdots
Theorem 11 ([Hubert00]).

A recurrent aperiodic sequence 𝐯\mathbf{v} is balanced if and only if 𝐯=colour(𝐮,𝐲,𝐲)\mathbf{v}=\mathrm{colour}(\mathbf{u},\mathbf{y},\mathbf{y}^{\prime}) for some Sturmian sequence 𝐮\mathbf{u} and constant gap sequences 𝐲,𝐲{\mathbf{y}},{\mathbf{y}^{\prime}} over two disjoint alphabets.

Let 𝒜,\mathcal{A},\mathcal{B} be two disjoint alphabets. The “discolouration map” π\pi is defined for any word or sequence over 𝒜\mathcal{A}\cup\mathcal{B}; it replaces all letters from 𝒜\mathcal{A} by 𝚊\mathtt{a} and all letters from \mathcal{B} by 𝚋\mathtt{b}. If 𝐯=colour(𝐮,𝐲,𝐲)\mathbf{v}=\mathrm{colour}(\mathbf{u},\mathbf{y},\mathbf{y}^{\prime}), where 𝐲𝒜\mathbf{y}\in\mathcal{A}^{\mathbb{N}}, 𝐲\mathbf{y}^{\prime}\in\mathcal{B}^{\mathbb{N}}, then π(𝐯)=𝐮\pi(\mathbf{v})=\mathbf{u} and π(v)(𝐮)\pi(v)\in\mathcal{L}(\mathbf{u}) for every v(𝐯)v\in\mathcal{L}(\mathbf{v}).

Corollary 12 ([DDP21]).

Let 𝐯=colour(𝐮,𝐲,𝐲)\mathbf{v}=\mathrm{colour}(\mathbf{u},\mathbf{y},\mathbf{y}^{\prime}) and u(𝐮)u\in\mathcal{L}(\mathbf{u}). For any i,ji,j\in\mathbb{N}, the word vv obtained by colouring uu with shifted constant gap sequences σi(𝐲)\sigma^{i}(\mathbf{y}) and σj(𝐲)\sigma^{j}(\mathbf{y}^{\prime}) is in (𝐯)\mathcal{L}(\mathbf{v}). In particular, if a Sturmian sequence 𝐮~\widetilde{\mathbf{u}} has the same language as 𝐮\mathbf{u}, then, E(𝐯)=E(𝐯~)E(\mathbf{v})=E(\widetilde{\mathbf{v}}) and E(𝐯)=E(𝐯~)E^{*}(\mathbf{v})=E^{*}(\widetilde{\mathbf{v}}).

Example 13.

Let 𝐮f,𝐯,𝐲\mathbf{u}_{f},\mathbf{v},\mathbf{y} and 𝐲\mathbf{y}^{\prime} be as in Example 10. Let u=𝚋𝚊𝚋u={\tt bab}. The reader is invited to check that all words 𝟶𝟹𝟷,𝟷𝟹𝟶,𝟶𝟹𝟸,𝟸𝟹𝟶,𝟶𝟺𝟷,𝟷𝟺𝟶,𝟶𝟺𝟸,𝟸𝟺𝟶{\tt 031},{\tt 130},{\tt 032},{\tt 230},{\tt 041},{\tt 140},{\tt 042},{\tt 240} are factors of 𝐯\mathbf{v}.

As a consequence of the previous corollary when studying the asymptotic critical exponent of balanced sequences, we can limit our consideration to colourings of standard Sturmian sequences.

Proposition 14 ([DDP21]).

Let 𝐯=colour(𝐮,𝐲,𝐲)\mathbf{v}=\mathrm{colour}(\mathbf{u},\mathbf{y},\mathbf{y}^{\prime}) and θ=[0,a1,a2,a3,]\theta=[0,a_{1},a_{2},a_{3},\ldots]. One has:

  1. (1)

    E(𝐯)E(𝐯)1+1Per𝐲Per𝐲.E(\mathbf{v})\geq E^{*}(\mathbf{v})\geq 1+\frac{1}{\mathrm{Per}\,{\mathbf{y}}\cdot\mathrm{Per}\,{\mathbf{y}^{\prime}}}\,.

  2. (2)

    E(𝐯)E^{*}(\mathbf{v}) depends on Per𝐲\mathrm{Per}\,{\mathbf{y}} and Per𝐲\mathrm{Per}\,{\mathbf{y}^{\prime}}, not on the structure of 𝐲\mathbf{y} and 𝐲\mathbf{y}^{\prime}.

  3. (3)

    E(𝐯)E^{*}(\mathbf{v}) is finite \Leftrightarrow E(𝐮)E^{*}(\mathbf{u}) is finite \Leftrightarrow (an)(a_{n}) is a bounded sequence.

Hoffman et al. proved that any dd-ary constant gap sequence satisfies Per𝐲2d1\mathrm{Per}\,\mathbf{y}\leq 2^{d-1} (see Theorem 2 in [Hof2011] based on Corollary 2 from [Si85]). Therefore, the first item of Proposition 14 leads to the following corollary.

Corollary 15.

RTB(d)1+12d2RTB^{*}(d)\geq 1+\frac{1}{2^{d-2}} for every positive integer dd.

The main task we solve in the paper is the following: for given PP and PP^{\prime} find a balanced sequence 𝐯\mathbf{v} such that its asymptotic critical exponent E(𝐯)E^{*}(\mathbf{v}) has the least value among all balanced sequences which arise as colouring of Sturmian sequences by two constant gap sequences 𝐲\mathbf{y} and 𝐲\mathbf{y}^{\prime} with P=Per𝐲P=\mathrm{Per}\,\mathbf{y} and P=Per𝐲P^{\prime}=\mathrm{Per}\,\mathbf{y}^{\prime}.

Having a method for solving this task, we are able to determine RTB(d)RTB^{*}(d) for a fixed dd. We apply the method to all pairs of P,PP,P^{\prime} such that P=Per𝐲P=\mathrm{Per}\,\mathbf{y} for some constant gap sequence 𝐲\mathbf{y} over d𝚊d_{\tt a}-ary alphabet and P=Per𝐲P^{\prime}=\mathrm{Per}\,\mathbf{y}^{\prime} for some constant gap sequence 𝐲\mathbf{y}^{\prime} over d𝚋d_{\tt b}-ary alphabet, where d𝚊+d𝚋=dd_{\tt a}+d_{\tt b}=d.

For a fixed dd there are only finitely many pairs P,PP,P^{\prime} with the described property. It seems that to find the periods of all constant gap sequences over a dd-ary alphabet is a difficult problem. Nevertheless, it is known (see [Schnabel2015]) that constant gap sequences over dd letters with d12d\leq 12 may be obtained by interlacing.

Definition 16.

Let 𝒜(0),𝒜(1),,𝒜(k1)\mathcal{A}^{(0)},\mathcal{A}^{(1)},\ldots,\mathcal{A}^{(k-1)} be mutually disjoint alphabets and let 𝐲(i)=y0(i)y1(i)y2(i)\mathbf{y}^{(i)}=y_{0}^{(i)}y_{1}^{(i)}y_{2}^{(i)}\cdots be a constant gap sequence over the alphabet 𝒜(i)\mathcal{A}^{(i)} for every i{0,1,,k1}i\in\{0,1,\ldots,k-1\}. The interlacing of 𝐲(0),𝐲(1),,𝐲(k1)\mathbf{y}^{(0)},\mathbf{y}^{(1)},\ldots,\mathbf{y}^{(k-1)} is the sequence 𝐲=y0y1y2\mathbf{y}=y_{0}y_{1}y_{2}\cdots, where ykn+j=yn(j)y_{kn+j}=y^{(j)}_{n} for every nn\in\mathbb{N} and j{0,1,,k1}j\in\{0,1,\ldots,k-1\}.

In other words, the interlacing of 𝐲(0),𝐲(1),,𝐲(k1)\mathbf{y}^{(0)},\mathbf{y}^{(1)},\ldots,\mathbf{y}^{(k-1)} is a sequence obtained by listing step by step the first letters of 𝐲(0),𝐲(1),,𝐲(k1)\mathbf{y}^{(0)},\mathbf{y}^{(1)},\ldots,\mathbf{y}^{(k-1)}, then the second letters, the third letters etc.

The interlacing 𝐲\mathbf{y} of 𝐲(0),𝐲(1),,𝐲(k1)\mathbf{y}^{(0)},\mathbf{y}^{(1)},\ldots,\mathbf{y}^{(k-1)} satisfies

Per𝐲=klcm{Per𝐲(0),Per𝐲(1),,Per𝐲(k1)}.\mathrm{Per}\,{\mathbf{y}}=k\cdot\mathrm{lcm}\{\mathrm{Per}{\,\mathbf{y}^{(0)}},\mathrm{Per}{\,\mathbf{y}^{(1)}},\ldots,\mathrm{Per}{\,\mathbf{y}^{(k-1)}}\}.
Example 17.

The interlacing of 𝐲(0)=(𝟶𝟷𝟶𝟸)ω\mathbf{y}^{(0)}=(\tt 0102)^{\omega} and 𝐲(1)=(𝟹𝟺)ω\mathbf{y}^{(1)}=(\tt 34)^{\omega} equals (𝟶𝟹𝟷𝟺𝟶𝟹𝟸𝟺)ω(\tt 03140324)^{\omega}. The reader may easily verify that it is again a constant gap sequence and its period equals 2lcm{Per𝐲(0),Per𝐲(1)}=82\cdot\mathrm{lcm}\{\mathrm{Per}{\,\mathbf{y}^{(0)}},\mathrm{Per}{\,\mathbf{y}^{(1)}}\}=8.

Remark 18.

Using the fact that all constant gap sequences with at most 12 letters are obtained by interlacing, it is not difficult to verify that the periods of constant gap sequences over alphabet of size dd are as follows:

dperiod112233,444,6,855,6,8,9,12,1666,8,10,12,16,18,24,3277,8,9,10,12,15,16,18,20,24,27,32,36,48,6488,10,12,14,16,18,20,24,30,32,36,40,48,54,64,72,96,12899,10,12,14,15,16,18,20,21,24,25,27,28,30,32,36,40,45,48,54,60,64,72,80,81,96,108,128,144,192,256\begin{array}[]{l|l}d&\text{period}\\ \hline\cr 1&1\\ 2&2\\ 3&3,4\\ 4&4,6,8\\ 5&5,6,8,9,12,16\\ 6&6,8,10,12,16,18,24,32\\ 7&7,8,9,10,12,15,16,18,20,24,27,32,36,48,64\\ 8&8,10,12,14,16,18,20,24,30,32,36,40,48,54,64,72,96,128\\ 9&9,10,12,14,15,16,18,20,21,24,25,27,28,30,32,36,40,45,\\ &48,54,60,64,72,80,81,96,108,128,144,192,256\end{array}

6. Formula for the asymptotic critical exponent of balanced sequences

In this section, θ=[0,a1,a2,a3,]\theta=[0,a_{1},a_{2},a_{3},\ldots] denotes the continued fraction of the slope of a standard Sturmian sequence 𝐮\mathbf{u}, P=Per𝐲P=\mathrm{Per}\,\mathbf{y} and P=Per𝐲P^{\prime}=\mathrm{Per}\,\mathbf{y}^{\prime} are periods of two constant gap sequences and 𝐯=colour(𝐮,𝐲,𝐲)\mathbf{v}=\mathrm{colour}(\mathbf{u},\mathbf{y},\mathbf{y}^{\prime}). The computation of the asymptotic critical exponent of 𝐯\mathbf{v} is based on the knowledge of bispecial factors of the Sturmian sequence 𝐮\mathbf{u} and their return words. To provide an explicit formula let us first fix some notation.

Convention: Given positive c,dc,d\in\mathbb{N}.

  • If a1=b1modc and a2=b2modda_{1}=b_{1}\bmod c\text{ and }a_{2}=b_{2}\bmod d, we write

    (a1a2)=(b1b2)mod(cd).\left(\begin{smallmatrix}a_{1}\\ a_{2}\end{smallmatrix}\right)=\left(\begin{smallmatrix}b_{1}\\ b_{2}\end{smallmatrix}\right)\bmod\left(\begin{smallmatrix}c\\ d\end{smallmatrix}\right).
  • If (a1ia2i)=(b1ib2i)mod(cd)\left(\begin{smallmatrix}a_{1i}\\ a_{2i}\end{smallmatrix}\right)=\left(\begin{smallmatrix}b_{1i}\\ b_{2i}\end{smallmatrix}\right)\bmod\left(\begin{smallmatrix}c\\ d\end{smallmatrix}\right) for i=1,2i=1,2, we write

    (a11a12a21a22)=(b11b12b21b22)mod(cd).\left(\begin{smallmatrix}a_{11}&a_{12}\\ a_{21}&a_{22}\end{smallmatrix}\right)=\left(\begin{smallmatrix}b_{11}&b_{12}\\ b_{21}&b_{22}\end{smallmatrix}\right)\bmod\left(\begin{smallmatrix}c\\ d\end{smallmatrix}\right).

A formula for computation of E(𝐯)E^{*}(\mathbf{v}) is deduced in [DDP21]. To keep this paper self-contained, we provide a sketch of its proof. It is based on the following simple observation.

Observation 19.

Let vv be a factor of 𝐯\mathbf{v} and u=π(v)u=\pi(v), where |u|𝚊P,|u|𝚋P|u|_{\tt a}\geq P,\ |u|_{\tt b}\geq P^{\prime}. Then

  1. (1)

    vv is bispecial in 𝐯\mathbf{v} if and only if uu is bispecial in 𝐮\mathbf{u};

  2. (2)

    if ii\in\mathbb{N} is an occurrence of vv in 𝐯\mathbf{v}, then ii is an occurrence of uu in 𝐮\mathbf{u}.

Denote AN=(pN1pNqN1qN),δN=[aN+1,aN+2,]A_{N}=\left(\begin{smallmatrix}p_{N-1}&p_{N}\\ q_{N-1}&q_{N}\end{smallmatrix}\right),\ \ \delta_{N}=[a_{N+1},a_{N+2},\ldots].

Let (N,m)(N,m) be a pair associated in Proposition 5 to a bispecial factor of 𝐮\mathbf{u}. We assign to (N,m)(N,m) the sets:

𝒮1(N,m)={(k):AN(10m1)(k)=(00)mod(PP)};\mathcal{S}_{1}(N,m)=\bigl{\{}\left(\begin{smallmatrix}\ell\\ k\end{smallmatrix}\right):A_{N}\left(\begin{smallmatrix}1&0\\ m&1\end{smallmatrix}\right)\left(\begin{smallmatrix}\ell\\ k\end{smallmatrix}\right)=\left(\begin{smallmatrix}0\\ 0\end{smallmatrix}\right)\mod\left(\begin{smallmatrix}P\\ P^{\prime}\end{smallmatrix}\right)\bigr{\}}\,;

𝒮2(N,m)={(k):|(δNm)k|<δNm+1 and k+>0};\mathcal{S}_{2}(N,m)=\{\left(\begin{smallmatrix}\ell\\ k\end{smallmatrix}\right):|\ell(\delta_{N}-m)-k|<\delta_{N}-m+1\text{ and }k+\ell>0\}\,;

𝒮(N,m)=𝒮1(N,m)𝒮2(N,m).\mathcal{S}(N,m)=\mathcal{S}_{1}(N,m)\cap\mathcal{S}_{2}(N,m)\,.

Proposition 20.

Let 𝐮\mathbf{u} be a Sturmian sequence with slope θ=[0,a1,a2,a3,]\theta=[0,a_{1},a_{2},a_{3},\ldots] and 𝐲\mathbf{y} and 𝐲\mathbf{y}^{\prime} be two constant gap sequences. Put

(3) ΦN:=max{1+m+QN1QNk+m+QN1QN:(k)𝒮(N,m)and 0m<aN+1}.\Phi_{N}:=\max\left\{\frac{1+m+\frac{Q_{N-1}}{Q_{N}}}{k+\ell m+\ell\frac{Q_{N-1}}{Q_{N}}}\ :\ \left(\begin{smallmatrix}\ell\\ k\end{smallmatrix}\right)\in\mathcal{S}(N,m)\ \text{and}\ 0\leq m<a_{N+1}\right\}\,.

Then the asymptotic critical exponent of the balanced sequence 𝐯=colour(𝐮,𝐲,𝐲)\mathbf{v}=\mathrm{colour}(\mathbf{u},\mathbf{y},\mathbf{y}^{\prime}) equals E(𝐯)=1+lim supNΦNE^{*}(\mathbf{v})=1+\limsup\limits_{N\to\infty}\Phi_{N}.

Proof.

Let vv be a bispecial factor of 𝐯\mathbf{v} satisfying assumptions of Observation 19 and i<ji<j be two consecutive occurrences of vv in 𝐯\mathbf{v} and w=vivi+1vj1w=v_{i}v_{i+1}\cdots v_{j-1}. By Item 1 of Observation 19 the factor u=π(v)u=\pi(v) is bispecial in 𝐮\mathbf{u}. Let (N,m)(N,m) be the pair associated by Proposition 5 with uu and rr and ss be two return words to uu in 𝐮\mathbf{u}. By Item 2 of Observation 19 the projection π(w)\pi(w) is concatenated from \ell return words ss and kk return words rr for some ,k,k+>0.\ell,k\in\mathbb{N},k+\ell>0. Obviously, the vector (k)\left(\begin{smallmatrix}\ell\\ k\end{smallmatrix}\right) is the Parikh vector of the derived sequence d𝐮(u)d_{\mathbf{u}}(u). By Proposition 5 the slope θ\theta^{\prime} of the derived sequence is [0,aN+1m,aN+2,aN+3,][0,a_{N+1}-m,a_{N+2},a_{N+3},\ldots]. The inverse of θ\theta^{\prime} is δNm=[aN+1m,aN+2,aN+3,]\delta_{N}-m=[a_{N+1}-m,a_{N+2},a_{N+3},\ldots]. By Lemma 7 the pair ,k\ell,k satisfies the inequality |(δNm)k|<δNm+1|\ell(\delta_{N}-m)-k|<\delta_{N}-m+1. Hence (k)\left(\begin{smallmatrix}\ell\\ k\end{smallmatrix}\right) belongs to 𝒮2(N,m)\mathcal{S}_{2}(N,m).

Since ii and jj are occurrences of vv in 𝐯\mathbf{v}, the number of letters 𝚊{\tt a}, resp. 𝚋{\tt b} in π(w)\pi(w) is a multiple of PP, resp. PP^{\prime}. Using the corresponding Parikh vectors we have V(π(w))=V(s)+kV(r)=(00)mod(PP)\vec{V}(\pi(w))=\ell\vec{V}(s)+k\vec{V}(r)=\left(\begin{smallmatrix}0\\ 0\end{smallmatrix}\right)\bmod\left(\begin{smallmatrix}P\\ P^{\prime}\end{smallmatrix}\right). By Proposition 5, V(s)+kV(r)=AN(10m1)(k)\ell\vec{V}(s)+k\vec{V}(r)=A_{N}\left(\begin{smallmatrix}1&0\\ m&1\end{smallmatrix}\right)\left(\begin{smallmatrix}\ell\\ k\end{smallmatrix}\right), hence (k)\left(\begin{smallmatrix}\ell\\ k\end{smallmatrix}\right) belongs to 𝒮1(N,m)\mathcal{S}_{1}(N,m) and thus to 𝒮(N,m)\mathcal{S}(N,m).

Let us evaluate the ratio |v||w|\frac{|v|}{|w|}. We abbreviate xN=QN1QNx_{N}=\frac{Q_{N-1}}{Q_{N}}. By Proposition 5

|v||w|=|r|+|s|2k|r|+|s|=(1+m)QN+QN12(k+m)QN+QN1=1+m+xNk+m+xN2|w|.\frac{|v|}{|w|}=\frac{|r|+|s|-2}{k|r|+\ell|s|}=\frac{(1+m)Q_{N}+Q_{N-1}-2}{(k+\ell m)Q_{N}+\ell Q_{N-1}}=\frac{1+m+x_{N}}{k+\ell m+\ell x_{N}}-\frac{2}{|w|}.

The previous equality is valid for each factor ww occurring between two occurrences of vv in 𝐯\mathbf{v}. If ww is a shortest return word to vv, then

|v||w|=max{1+m+xNk+m+xN:(k)𝒮(N,m)}2|w|.\frac{|v|}{|w|}=\max\left\{\frac{1+m+x_{N}}{k+\ell m+\ell x_{N}}:\left(\begin{smallmatrix}\ell\\ k\end{smallmatrix}\right)\in\mathcal{S}(N,m)\right\}-\frac{2}{|w|}\,.

Hence ΦN\Phi_{N} expresses (up to the subtracted fraction 2|w|\frac{2}{|w|}) the maximal value of the ratio |v|/|w||v|/|w| among all possible pairs (N,m)(N,m) with a fixed NN. Since the length |w||w| tends to infinity with growing NN, Theorem 2 concludes the proof. ∎

If the slope of a Sturmian sequence is quadratic irrational, then the asymptotic critical exponent can be computed explicitly. The algorithm is explained in details in [DDP21]. We implemented it and throughout this paper we use our computer program which for a given eventually periodic continued fraction θ\theta and a pair P=Per𝐲P=\mathrm{Per}\,\mathbf{y}, P=Per𝐲P^{\prime}=\mathrm{Per}\,\mathbf{y}^{\prime} finds the exact value of E(𝐯)E^{*}(\mathbf{v}).

7. Colouring with linked parameters

In this section we study how to restrict without loss of generality the periods P,PP,P^{\prime} of constant gap sequences when searching for balanced sequences with the minimal asymptotic critical exponent. Let us recall that in the whole paper we work without loss of generality with Sturmian sequences 𝐮\mathbf{u} having the slope θ=ρ𝚊(𝐮)ρ𝚋(𝐮)(0,1)\theta=\frac{\rho_{\tt a}({\mathbf{u}})}{\rho_{\tt b}({\mathbf{u}})}\in(0,1), i.e., the letter 𝚋\tt b is the most frequent letter in 𝐮{\mathbf{u}}.

Proposition 21.

Let 𝐲\mathbf{y} and 𝐲\mathbf{y}^{\prime} be two constant gap sequences and 𝐮{𝚊,𝚋}\mathbf{u}\in\{\tt a,\tt b\}^{\mathbb{N}} be a Sturmian sequence with 𝚋\tt b being the most frequent letter in 𝐮{\mathbf{u}}. Then there exists a Sturmian sequence 𝐮~{𝚊,𝚋}\widetilde{\mathbf{u}}\in\{\tt a,\tt b\}^{\mathbb{N}} with 𝚋\tt b being the most frequent letter in 𝐮~\widetilde{\mathbf{u}} such that

E(colour(𝐮~,𝐲,𝐲))=E(colour(𝐮,𝐲,𝐲)).E^{*}\bigl{(}\mathrm{colour}(\widetilde{\mathbf{u}},\mathbf{y}^{\prime},{\mathbf{y}})\bigr{)}=E^{*}\bigl{(}\mathrm{colour}({\mathbf{u}},\mathbf{y},{\mathbf{y}^{\prime}})\bigr{)}\,.
Proof.

Let θ=[0,a1,a2,a3,]\theta=[0,a_{1},a_{2},a_{3},\ldots] be the slope of 𝐮\mathbf{u} and pNqN\frac{p_{N}}{q_{N}} be the NthN^{th} convergent to θ\theta. Define the slope θ~=[0,b1,b2,]\widetilde{\theta}=[0,b_{1},b_{2},\ldots] of the Sturmian sequence 𝐮~\widetilde{\mathbf{u}} by b1=P=Per𝐲b_{1}=P=\mathrm{Per}\,{\mathbf{y}} and bN+1=aNb_{N+1}=a_{N} for each NN\in\mathbb{N}. For the slope θ\theta, we use the notation 𝒮(N,m)\mathcal{S}(N,m) and ΦN\Phi_{N} as introduced in Section 6 and QN=pN+qNQ_{N}=p_{N}+q_{N}. Analogously, for the slope θ~\widetilde{\theta}, we use the notation 𝒮~(N,m)\widetilde{\mathcal{S}}(N,m), Φ~N\widetilde{\Phi}_{N} and Q~N\widetilde{Q}_{N}.

Using the recurrent relation (1), we get p~N+1=qN,q~N+1=pN+qNP\widetilde{p}_{N+1}=q_{N},\quad\widetilde{q}_{N+1}=p_{N}+q_{N}P and Q~N+1=QN+qNP\widetilde{Q}_{N+1}=Q_{N}+q_{N}P for each NN\in\mathbb{N}. It follows then immediately for each NN\in\mathbb{N} and m<aN+1=bN+2m<a_{N+1}=b_{N+2} that

  1. (1)

    𝒮~(N+1,m)=𝒮(N,m)\widetilde{\mathcal{S}}(N+1,m)={\mathcal{S}}(N,m);

  2. (2)

    |Q~NQ~N+1QN1QN|PQN2|\frac{\widetilde{Q}_{N}}{\widetilde{Q}_{N+1}}-\tfrac{Q_{N-1}}{Q_{N}}|\leq\tfrac{P}{Q_{N}^{2}}.

Items 1 and 2 imply that limN(Φ~N+1ΦN)=0\lim\limits_{N\to\infty}\bigl{(}\widetilde{\Phi}_{N+1}-\Phi_{N}\bigr{)}=0. Hence by Proposition 20

E(colour(𝐮~,𝐲,𝐲))=1+lim supNΦ~N=1+lim supNΦN=E(colour(𝐮,𝐲,𝐲)).E^{*}\bigl{(}\mathrm{colour}(\widetilde{\mathbf{u}},\mathbf{y}^{\prime},{\mathbf{y}})\bigr{)}=1+\limsup\limits_{N\to\infty}\widetilde{\Phi}_{N}=1+\limsup\limits_{N\to\infty}{\Phi}_{N}=E^{*}\bigl{(}\mathrm{colour}({\mathbf{u}},\mathbf{y},{\mathbf{y}^{\prime}})\bigr{)}.

In the statement of Proposition 21, the asymptotic critical exponent cannot be replaced by the critical exponent. The following example demonstrates this fact.

Example 22.

Baranwal and Shallit show in [BaSh19] that the minimal critical exponent of 55-ary balanced sequences equals 32\frac{3}{2} and it is reached by the sequence 𝐯=colour(𝐮,𝐲,𝐲)\mathbf{v}=\mathrm{colour}(\mathbf{u},\mathbf{y},\mathbf{y}^{\prime}), where 𝐮{𝚊,𝚋}\mathbf{u}\in\{\tt a,\tt b\}^{\mathbb{N}} is a Sturmian sequence with slope [0,1,2¯][0,1,\overline{2}] and 𝚊\tt a’s are coloured by 𝐲=(𝟶𝟷)ω\mathbf{y}=(\tt 01)^{\omega} and 𝚋\tt b’s are coloured by 𝐲=(𝟸𝟹𝟸𝟺)ω\mathbf{y}^{\prime}=(\tt 2324)^{\omega}.

Now we colour 𝚊\tt a’s by 𝐲=(𝟸𝟹𝟸𝟺)ω\mathbf{y}^{\prime}=(\tt 2324)^{\omega} and 𝚋\tt b’s by 𝐲=(𝟶𝟷)ω\mathbf{y}=(\tt 01)^{\omega}. We will explain that for each sequence 𝐯~=colour(𝐮~,𝐲,𝐲)\widetilde{\mathbf{v}}=\mathrm{colour}(\widetilde{\mathbf{u}},\mathbf{y}^{\prime},{\mathbf{y}}), where 𝐮~\widetilde{\mathbf{u}} is a Sturmian sequence with slope [0,a1,a2,a3,][0,a_{1},a_{2},a_{3},\ldots], we have E(𝐯~)>32E(\widetilde{\mathbf{v}})>\frac{3}{2}.

  • If a12a_{1}\geq 2, then by Proposition 4 and the definition of slope, Da1(𝚊𝚊)=𝚋a1𝚊𝚋a1𝚊D^{a_{1}}({\tt aa})={\tt b}^{a_{1}}{\tt a}{\tt b}^{a_{1}}{\tt a} is a factor of 𝐮~\widetilde{\mathbf{u}}. If a1=1a_{1}=1 and a2=1a_{2}=1, then DG(𝚋𝚋𝚊)=𝚋𝚊𝚋𝚋𝚊𝚋𝚋𝚊DG({\tt bba})=\tt babbabba is a factor of 𝐮~\widetilde{\mathbf{u}}. In both cases, 𝚋𝚋𝚊𝚋𝚋(𝐮~){\tt bbabb}\in\mathcal{L}(\widetilde{\mathbf{u}}). Hence by Corollary 12 the factor 𝟶𝟷𝟸𝟶𝟷(𝐯~){\tt 01201}\in\mathcal{L}(\widetilde{\mathbf{v}}) and thus E(𝐯~)53>32E(\widetilde{\mathbf{v}})\geq\frac{5}{3}>\frac{3}{2}.

  • If a1=1a_{1}=1 and a22a_{2}\geq 2, then DGa2(𝚊𝚋)=𝚋𝚊(𝚋𝚊)a2𝚋DG^{a_{2}}({\tt ab})={\tt ba}({\tt ba})^{a_{2}}{\tt b} is a factor of 𝐮~\widetilde{\mathbf{u}}, in particular, 𝚋𝚊𝚋𝚊𝚋𝚊𝚋(𝐮~){\tt bababab}\in\mathcal{L}(\widetilde{\mathbf{u}}). Hence 𝟶𝟸𝟷𝟹𝟶𝟸𝟷(𝐯~){\tt 0213021}\in\mathcal{L}(\widetilde{\mathbf{v}}) and thus E(𝐯~)74>32E(\widetilde{\mathbf{v}})\geq\frac{7}{4}>\frac{3}{2}.

Lemma 23.

Let 𝐯=colour(𝐮,𝐲,𝐲)\mathbf{v}=\mathrm{colour}(\mathbf{u},\mathbf{y},{\mathbf{y}}^{\prime}) and 𝐯^=colour(𝐮,𝐲^,𝐲^)\widehat{\mathbf{v}}=\mathrm{colour}(\mathbf{u},\hat{\mathbf{y}},{\hat{\mathbf{y}}}^{\prime}). If Per𝐲^\mathrm{Per}\,\hat{\mathbf{y}} is divisible by Per𝐲\mathrm{Per}\,\mathbf{y} and Per𝐲^\mathrm{Per}\,{\hat{\mathbf{y}}}^{\prime} is divisible by Per𝐲\mathrm{Per}\,{\mathbf{y}}^{\prime}, then E(𝐯)E(𝐯^)E^{*}(\mathbf{v})\geq E^{*}(\widehat{\mathbf{v}}).

Proof.

Let us denote by 𝒮(N,m){\mathcal{S}}(N,m) the set corresponding to 𝐯\mathbf{v} defined in Section 6 and similarly 𝒮^(N,m)\hat{{\mathcal{S}}}(N,m) for 𝐯^\widehat{\mathbf{v}}. Since we colour the same Sturmian sequence 𝐮\mathbf{u}, we have 𝒮^(N,m)𝒮(N,m)\hat{{\mathcal{S}}}(N,m)\subset{\mathcal{S}}(N,m). Applying Proposition 20 we obtain E(𝐯)E(𝐯^)E^{*}(\mathbf{v})\geq E^{*}(\widehat{\mathbf{v}}). ∎

8. Equivalence on unimodular matrices

By Proposition 14, the asymptotic critical exponent depends only on the periods of constant gap sequences 𝐲\mathbf{y} and 𝐲\mathbf{y}^{\prime}, not on their structure. In the sequel we use for a fixed pair P=Per𝐲P=\mathrm{Per}\,\mathbf{y}\, and P=Per𝐲P^{\prime}=\mathrm{Per}\,\mathbf{y}^{\prime}\, the notation H,L,Y,YH,L,Y,Y^{\prime} introduced by the following relations:

(4) H=gcd(P,P),P=HY,P=HYand L=lcm(P,P).H=\gcd(P,\ P^{\prime}\,),\ \ P=HY\,,\ \ P^{\prime}=HY^{\prime}\ \ \text{and }\ \ L=\mathrm{lcm}(P,P^{\prime}\,).

Obviously, YY and YY^{\prime} are coprime. We always consider L>1L>1 since for Per𝐲=Per𝐲=1\mathrm{Per}\,\mathbf{y}=\mathrm{Per}\,\mathbf{y}^{\prime}=1, the sequences 𝐯=colour(𝐮,𝐲,𝐲)\mathbf{v}=\mathrm{colour}(\mathbf{u},\mathbf{y},\mathbf{y}^{\prime}) and 𝐮\mathbf{u} are the same and the minimal asymptotic critical exponent for Sturmian sequences is known.

In this section we will study the form of the set

𝒮1(N,m)={(k):AN(10m1)(k)=(00)mod(PP)},\mathcal{S}_{1}(N,m)=\bigl{\{}\left(\begin{smallmatrix}\ell\\ k\end{smallmatrix}\right):A_{N}\left(\begin{smallmatrix}1&0\\ m&1\end{smallmatrix}\right)\left(\begin{smallmatrix}\ell\\ k\end{smallmatrix}\right)=\left(\begin{smallmatrix}0\\ 0\end{smallmatrix}\right)\mod\left(\begin{smallmatrix}P\\ P^{\prime}\end{smallmatrix}\right)\bigr{\}}\,,

which plays an essential role in the definition of ΦN\Phi_{N} (see Proposition 20), and consequently in the computation of the asymptotic critical exponent of balanced sequences. Note that the determinant of AN(10m1)A_{N}\left(\begin{smallmatrix}1&0\\ m&1\end{smallmatrix}\right) equals ±1\pm 1, i.e., the matrix is unimodular. A solution (k)\left(\begin{smallmatrix}\ell\\ k\end{smallmatrix}\right) depends only on entries of the matrix counted modP\bmod\,P in the first row and modP\bmod\,P^{\prime} in the second row. Hence it is possible to group matrices into classes of the same behaviour with respect to the form of 𝒮1(N,m)\mathcal{S}_{1}(N,m). The following lemma prepares such grouping.

Lemma 24.

Let A2×2A\in\mathbb{Z}^{2\times 2} be unimodular and ,k\ell,k\in\mathbb{Z}. Then A(k)=(00)mod(PP)A\left(\begin{smallmatrix}\ell\\ k\end{smallmatrix}\right)=\left(\begin{smallmatrix}0\\ 0\end{smallmatrix}\right)\ {\rm mod}\ \left(\begin{smallmatrix}P\\ P^{\prime}\end{smallmatrix}\right) if and only if there exist λ,κ\lambda,\kappa\in\mathbb{Z} such that

(k)=H(λκ)andA(λκ)=(00)mod(YY).\left(\begin{smallmatrix}\ell\\ k\end{smallmatrix}\right)=H\left(\begin{smallmatrix}\lambda\\ \kappa\end{smallmatrix}\right)\ \ \ \text{and}\ \ \ A\left(\begin{smallmatrix}\lambda\\ \kappa\end{smallmatrix}\right)=\left(\begin{smallmatrix}0\\ 0\end{smallmatrix}\right)\ {\rm mod}\left(\begin{smallmatrix}Y\\ Y^{\prime}\end{smallmatrix}\right).
Proof.

()(\Rightarrow) We have A(k)=(aPbP)=H(aYbY)A\left(\begin{smallmatrix}\ell\\ k\end{smallmatrix}\right)=\left(\begin{smallmatrix}a\,P\\ b\,P^{\prime}\end{smallmatrix}\right)=H\left(\begin{smallmatrix}aY\hskip 2.84544pt\\ bY^{\prime}\end{smallmatrix}\right) for some a,ba,b\in\mathbb{Z}. Since AA is unimodular, it is invertible in \mathbb{Z}, hence

(λκ):=1H(k)=A1(aYbY)2.\left(\begin{smallmatrix}\lambda\\ \kappa\end{smallmatrix}\right):=\tfrac{1}{H}\left(\begin{smallmatrix}\ell\\ k\end{smallmatrix}\right)=A^{-1}\left(\begin{smallmatrix}aY\hskip 2.84544pt\\ bY^{\prime}\end{smallmatrix}\right)\in\mathbb{Z}^{2}.

Moreover, A(k)=H(aYbY)A\left(\begin{smallmatrix}\ell\\ k\end{smallmatrix}\right)=H\left(\begin{smallmatrix}aY\hskip 2.84544pt\\ bY^{\prime}\end{smallmatrix}\right) implies A(λκ)=(aYbY)=(00)mod(YY).A\left(\begin{smallmatrix}\lambda\\ \kappa\end{smallmatrix}\right)=\left(\begin{smallmatrix}aY\hskip 2.84544pt\\ bY^{\prime}\end{smallmatrix}\right)=\left(\begin{smallmatrix}0\\ 0\end{smallmatrix}\right)\ {\rm mod}\left(\begin{smallmatrix}Y\\ Y^{\prime}\end{smallmatrix}\right).

()(\Leftarrow) The reasoning is analogous. ∎

Remark 25.

To solve the equation A(λκ)=(00)mod(YY)A\left(\begin{smallmatrix}\lambda\\ \kappa\end{smallmatrix}\right)=\left(\begin{smallmatrix}0\\ 0\end{smallmatrix}\right)\ {\rm mod}\ \left(\begin{smallmatrix}Y\\ Y^{\prime}\end{smallmatrix}\right), where AA is a unimodular matrix, we just need to know the remainders of division by YY, resp. YY^{\prime}, of the first row, resp. of the second row of AA. Therefore, we consider instead of a matrix A=(a11a12a21a22)A=\left(\begin{smallmatrix}a_{11}&a_{12}\\ a_{21}&a_{22}\end{smallmatrix}\right) the so-called

(Y,Y)-name of A defined as MA=(a11modYa12modYa21modYa22modY).\text{$(Y,Y^{\prime})$-name of $A$ defined as }\quad M_{A}=\left(\begin{smallmatrix}a_{11}\bmod Y\ &\ a_{12}\bmod Y\\ a_{21}\bmod Y^{\prime}\ &\ a_{22}\bmod Y^{\prime}\end{smallmatrix}\right).

Obviously, A(λκ)=(00)mod(YY)A\left(\begin{smallmatrix}\lambda\\ \kappa\end{smallmatrix}\right)=\left(\begin{smallmatrix}0\\ 0\end{smallmatrix}\right)\ {\rm mod}\left(\begin{smallmatrix}Y\\ Y^{\prime}\end{smallmatrix}\right) if and only if MA(λκ)=(00)mod(YY)M_{A}\left(\begin{smallmatrix}\lambda\\ \kappa\end{smallmatrix}\right)=\left(\begin{smallmatrix}0\\ 0\end{smallmatrix}\right)\ {\rm mod}\left(\begin{smallmatrix}Y\\ Y^{\prime}\end{smallmatrix}\right). Hence, matrices with the same (Y,Y)(Y,Y^{\prime})-name have the same solutions λ,κ\lambda,\kappa. By Lemma 24, the values HH, mm and the (Y,Y)(Y,Y^{\prime})-name of ANA_{N} capture all pieces of information we need to decide whether (k)\left(\begin{smallmatrix}\ell\\ k\end{smallmatrix}\right) belongs to the set 𝒮1(N,m)\mathcal{S}_{1}(N,m).

Nevertheless, matrices with distinct (Y,Y)(Y,Y^{\prime})-names can have the same solutions as well. The following example illustrates it.

Example 26.

Let Y=3Y=3 and Y=4Y^{\prime}=4. Consider the unimodular matrices A=(53121130)A=\left(\begin{smallmatrix}5&31\\ {21}&130\end{smallmatrix}\right) and B=(111334)B=\left(\begin{smallmatrix}1&11\\ 3&34\end{smallmatrix}\right). Their (3,4)(3,4)-names are MA=(2112)M_{A}=\left(\begin{smallmatrix}2&1\\ {1}&2\end{smallmatrix}\right) and MB=(1232)M_{B}=\left(\begin{smallmatrix}1&2\\ {3}&2\end{smallmatrix}\right). By the previous remark, a pair λ,κ\lambda,\kappa solves A(λκ)=(00)mod(34)A\left(\begin{smallmatrix}\lambda\\ \kappa\end{smallmatrix}\right)=\left(\begin{smallmatrix}0\\ 0\end{smallmatrix}\right)\ {\rm mod}\left(\begin{smallmatrix}3\\ 4\end{smallmatrix}\right) if and only if

2λ+κ=0mod3λ+2κ=0mod42(2λ+κ)=0mod33(λ+2κ)=0mod4λ+2κ=0mod33λ+2κ=0mod4,\begin{smallmatrix}2\lambda+\kappa=0\mod 3\\ \lambda+2\kappa=0\mod 4\end{smallmatrix}\quad\Longleftrightarrow\quad\begin{smallmatrix}2(2\lambda+\kappa)=0\mod 3\\ 3(\lambda+2\kappa)=0\mod 4\end{smallmatrix}\quad\Longleftrightarrow\quad\begin{smallmatrix}\lambda+2\kappa=0\mod 3\\ 3\lambda+2\kappa=0\mod 4\end{smallmatrix}\,,

which is equivalent to B(λκ)=(00)mod(34)B\left(\begin{smallmatrix}\lambda\\ \kappa\end{smallmatrix}\right)=\left(\begin{smallmatrix}0\\ 0\end{smallmatrix}\right)\ {\rm mod}\left(\begin{smallmatrix}3\\ 4\end{smallmatrix}\right).

To group unimodular matrices into classes with the same pairs λ,κ\lambda,\kappa of solutions, we introduce an equivalence.

Definition 27.

Let AA and BB be unimodular matrices in 2×2\mathbb{Z}^{2\times 2}. We say that AA is equivalent to BB, and write ABA\equiv B, if there exist cc\in\mathbb{Z} coprime with YY and cc^{\prime}\in\mathbb{Z} coprime with YY^{\prime} such that

(5) (c00c)A=Bmod(YY).\left(\begin{smallmatrix}c&0\\ 0&c^{\prime}\end{smallmatrix}\right)A=B\ \ {\rm mod}\left(\begin{smallmatrix}Y\\ Y^{\prime}\end{smallmatrix}\right)\,.

The equivalence class containing a matrix AA will be denoted [A][A]_{\equiv}.

Remark 28.

The relation \equiv is an equivalence. Evidently AAA\equiv A. For a fixed zz\in\mathbb{Z}, the set of elements coprime with zz is closed under inverse modulo zz and multiplication modulo zz, which implies that \equiv is symmetric and transitive. Obviously, if AA and BB satisfy (5), then the (Y,Y)(Y,Y^{\prime})-names MAM_{A} and MBM_{B} satisfy (c00c)MA=MBmod(YY)\left(\begin{smallmatrix}c&0\\ 0&c^{\prime}\end{smallmatrix}\right)M_{A}=M_{B}\ \ {\rm mod}\left(\begin{smallmatrix}Y\\ Y^{\prime}\end{smallmatrix}\right).

Lemma 29.

Let k,k,\ell\in\mathbb{N} and let AA and BB be unimodular matrices in 2×2\mathbb{Z}^{2\times 2} such that ABA\equiv B. Then

  1. (1)

    A(k)=(00)mod(PP) if and only if B(k)=(00)mod(PP).A\left(\begin{smallmatrix}\ell\\ k\end{smallmatrix}\right)=\left(\begin{smallmatrix}0\\ 0\end{smallmatrix}\right)\ {\rm mod}\ \left(\begin{smallmatrix}P\\ P^{\prime}\end{smallmatrix}\right)\text{\ \ if and only if \ \ }B\left(\begin{smallmatrix}\ell\\ k\end{smallmatrix}\right)=\left(\begin{smallmatrix}0\\ 0\end{smallmatrix}\right)\ {\rm mod}\ \left(\begin{smallmatrix}P\\ P^{\prime}\end{smallmatrix}\right).

  2. (2)

    ACBCAC\equiv BC   for any unimodular matrix C2×2C\in\mathbb{Z}^{2\times 2}.

Proof.

Let AA and BB satisfy (5) and λ,κ\lambda,\kappa\in\mathbb{N}. Then

(6) (c00c)A(λκ)=B(λκ)mod(YY).\left(\begin{smallmatrix}c&0\\ 0&c^{\prime}\end{smallmatrix}\right)A\left(\begin{smallmatrix}\lambda\\ \kappa\end{smallmatrix}\right)=B\left(\begin{smallmatrix}\lambda\\ \kappa\end{smallmatrix}\right)\ \ {\rm mod}\left(\begin{smallmatrix}Y\\ Y^{\prime}\end{smallmatrix}\right)\,.

Let us point out a simple fact: If cc is coprime with YY, then cx=0modYcx=0\mod Y if and only if x=0modYx=0\mod Y for every xx\in\mathbb{Z} and analogously for the coprime values cc^{\prime} and YY^{\prime}. Hence

(7) (c00c)A(λκ)=(00)mod(YY) if and only if A(λκ)=(00)mod(YY).\left(\begin{smallmatrix}c&0\\ 0&c^{\prime}\end{smallmatrix}\right)A\left(\begin{smallmatrix}\lambda\\ \kappa\end{smallmatrix}\right)=\left(\begin{smallmatrix}0\\ 0\end{smallmatrix}\right)\ {\rm mod}\ \left(\begin{smallmatrix}Y\\ Y^{\prime}\end{smallmatrix}\right)\text{\ \ if and only if \ \ }A\left(\begin{smallmatrix}\lambda\\ \kappa\end{smallmatrix}\right)=\left(\begin{smallmatrix}0\\ 0\end{smallmatrix}\right)\ {\rm mod}\ \left(\begin{smallmatrix}Y\\ Y^{\prime}\end{smallmatrix}\right).

Equations (6) and (7) imply

(8) A(λκ)=(00)mod(YY) if and only if B(λκ)=(00)mod(YY).A\left(\begin{smallmatrix}\lambda\\ \kappa\end{smallmatrix}\right)=\left(\begin{smallmatrix}0\\ 0\end{smallmatrix}\right)\ {\rm mod}\ \left(\begin{smallmatrix}Y\\ Y^{\prime}\end{smallmatrix}\right)\text{\ \ if and only if \ \ }B\left(\begin{smallmatrix}\lambda\\ \kappa\end{smallmatrix}\right)=\left(\begin{smallmatrix}0\\ 0\end{smallmatrix}\right)\ {\rm mod}\ \left(\begin{smallmatrix}Y\\ Y^{\prime}\end{smallmatrix}\right).

Lemma 24 together with Equation (8) prove Item 1.

Item 2 is a direct consequence of Equation (6) in which (λκ)\left(\begin{smallmatrix}\lambda\\ \kappa\end{smallmatrix}\right) represents the first column of the matrix CC and then its second column. ∎

9. A lower bound on the asymptotic critical exponent for fixed periods of constant gap sequences

We associate to θ=[0,a1,a2,a3,]\theta=[0,a_{1},a_{2},a_{3},\ldots] a sequence of classes of equivalent matrices ([AN])\bigl{(}[A_{N}]_{\equiv}\bigr{)} with representatives AN=(pN1pNqN1qN)A_{N}=\left(\begin{smallmatrix}p_{N-1}&p_{N}\\ q_{N-1}&q_{N}\end{smallmatrix}\right) and a sequence of (δN)(\delta_{N}) with δN=[aN+1,aN+2,]\delta_{N}=[a_{N+1},a_{N+2},\ldots]. Let us stress that ANA_{N} depends only of the first NN coefficients of the continued fraction of θ\theta, whereas δN\delta_{N} depends only on the remaining coefficients of θ\theta. Representatives of two consecutive classes satisfy

(9) AN+1=AN(011aN+1).A_{N+1}=A_{N}\left(\begin{smallmatrix}0&1\\ 1&a_{N+1}\end{smallmatrix}\right)\,.
Definition 30.

Let A2×2A\in\mathbb{Z}^{2\times 2} be unimodular and β>0\beta>0. We say that δ>1\delta>1 is (1+β)(1+\beta)-forcing for the class [A][A]_{\equiv} if there exist m,k,,+k>0m,k,\ell\in\mathbb{N},\ell+k>0, such that

𝔓1\mathfrak{P}1:

A(10m1)(k)=(00)mod(PP);A\begin{pmatrix}1&0\\ m&1\end{pmatrix}\begin{pmatrix}\ell\\ k\end{pmatrix}=\begin{pmatrix}0\\ 0\end{pmatrix}\mod\begin{pmatrix}P\\ P^{\prime}\end{pmatrix}\,;

𝔓2\mathfrak{P}2:

m+1<δm+1<\delta   and    |(δm)k|<δm+1;|\ell(\delta-m)-k|<\delta-m+1\,;

𝔓3\mathfrak{P}3:
  • •:

    if k=k=\ell, then 1k>β\frac{1}{k}>\beta;

  • •:

    if k>k>\ell, then 1+mk+mβ\frac{1+m}{k+m\ell}\geq\beta;

  • •:

    if k<k<\ell, then 2+mk+(m+1)β\frac{2+m}{k+(m+1)\ell}\geq\beta.

The set of (1+β)(1+\beta)-forcing δ\delta’s for the class [A][A]_{\equiv} is denoted (β,A)\mathcal{F}(\beta,A).

Note that the definition is correct because it does not depend on the choice of the representative AA from the class of equivalence. Indeed, only 𝔓1\mathfrak{P}1 depends on AA and by Lemma 29 the solutions ,k\ell,k are the same for each matrix in [A][A]_{\equiv}.

Theorem 31.

Let 𝐯=colour(𝐮,𝐲,𝐲)\mathbf{v}=\mathrm{colour}(\mathbf{u},\mathbf{y},\mathbf{y}^{\prime}), where 𝐮\mathbf{u} is a Sturmian sequence with slope θ=[0,a1,a2,a3,]\theta=[0,a_{1},a_{2},a_{3},\ldots], and let β\beta be a fixed positive number.

Assume that there exist infinitely many NN\in\mathbb{N} such that δN\delta_{N} is (1+β)(1+\beta)-forcing for the class [AN][A_{N}]_{\equiv}. Then E(𝐯)>1+βE^{*}(\mathbf{v})>1+\beta.

Proof.

First assume that the sequence (an)(a_{n}) of coefficients in the continued fraction expansion of θ\theta is unbounded, then by Proposition 14 E(𝐯)=+E^{*}(\mathbf{v})=+\infty. In the sequel, assume that (an)(a_{n}) is bounded, say by KK\in\mathbb{N}.

Let NN\in\mathbb{N} be such that δN\delta_{N} is (1+β)(1+\beta)-forcing for the class [AN][A_{N}]_{\equiv}. We point out that 𝔓1\mathfrak{P}1 and 𝔓2\mathfrak{P}2 of Definition 30 together mean that m<aN+1m<a_{N+1} and (k)\left(\begin{matrix}\ell\\ k\end{matrix}\right) belongs to 𝒮(N,m)\mathcal{S}(N,m) – the set used in the definition of ΦN\Phi_{N} in Proposition 20. Hence ΦN\Phi_{N} can be rewritten as

(10) ΦN=max{1+m+xNk+m+xN:m,k,satisfy 𝔓1 and 𝔓2 for A=AN and δ=δN},\Phi_{N}=\max\Bigl{\{}\tfrac{1+m+x_{N}}{k+\ell m+\ell x_{N}}:m,k,\ell\ \text{satisfy }\mathfrak{P}1\text{ and }\mathfrak{P}2\text{ for }A=A_{N}\text{ and }\delta=\delta_{N}\Bigr{\}},

where we abbreviate notation putting xN=QN1QN[0,1]x_{N}=\frac{Q_{N-1}}{Q_{N}}\in[0,1]. Using 𝔓3\mathfrak{P}3, we get

(11) 1+m+xNk+m+xNmin{1+m+xk+m+x:x[0,1]}={1k>βifk=;1+mk+mβifk>;2+mk+(m+1)βifk<.\tfrac{1+m+x_{N}}{k+\ell m+\ell x_{N}}\geq\min\Bigl{\{}\tfrac{1+m+x}{k+\ell m+\ell x}:x\in[0,1]\Bigr{\}}=\left\{\begin{array}[]{cl}\frac{1}{k}>\beta&\text{if}\ k=\ell\,;\\ &\\ \frac{1+m}{k+\ell m}\geq\beta&\text{if}\ k>\ell\,;\\ &\\ \frac{2+m}{k+\ell(m+1)}\geq\beta&\text{if}\ k\ <\ell\,.\end{array}\right.

Hence ΦNβ\Phi_{N}\geq\beta for infinitely many NN. It implies E(𝐯)=1+lim supNΦN1+βE^{*}(\mathbf{v})=1+\limsup\limits_{N\to\infty}\Phi_{N}\geq 1+\beta.

To prove the strict inequality E(𝐯)>1+βE^{*}(\mathbf{v})>1+\beta, we need to show more, namely, that there exists a positive number, say μ>0\mu>0, such that ΦNμ+β\Phi_{N}\geq\mu+\beta for each NN, where δN\delta_{N} is (1+β)(1+\beta)-forcing. Existence of such μ\mu follows from the three facts:

1) xN[1K+1,K+1K+2][0,1]x_{N}\in\bigl{[}\frac{1}{K+1},\frac{K+1}{K+2}\bigr{]}\subset[0,1]. Indeed, the recurrence relation QN=aNQN1+QN2Q_{N}=a_{N}Q_{N-1}+Q_{N-2} and the inequalities 1anK1\leq a_{n}\leq K imply on one hand

xN=QN1QN=QN1aNQN1+QN2QN1KQN1+QN1=1K+1x_{N}=\frac{Q_{N-1}}{Q_{N}}=\frac{Q_{N-1}}{a_{N}Q_{N-1}+Q_{N-2}}\geq\frac{Q_{N-1}}{KQ_{N-1}+Q_{N-1}}=\frac{1}{K+1}

and on the other hand

xN=1aN+QN2QN111+1K+1=K+1K+2.x_{N}=\frac{1}{a_{N}+\frac{Q_{N-2}}{Q_{N-1}}}\leq\frac{1}{1+\frac{1}{K+1}}=\frac{K+1}{K+2}\,.

2) If kk\neq\ell, the function fm,k,(x)=1+m+xk+m+xf_{m,k,\ell}(x)=\tfrac{1+m+x}{k+\ell m+\ell x} is strictly monotonous and thus the minimum of fm,k,f_{m,k,\ell} on the interval [1K+1,K+1K+2][\frac{1}{K+1},\frac{K+1}{K+2}] is strictly bigger than the minimum on [0,1][0,1]. If k=k=\ell, the strict inequality is required by 𝔓3\mathfrak{P}3.

3) There are only finitely many triplets m,k,m,k,\ell\in\mathbb{N} satisfying 𝔓2\mathfrak{P}2, 𝔓3\mathfrak{P}3 and m<Km<K. ∎

Example 32.

Using the previous proposition we show that P=Per𝐲=1P=\mathrm{Per}\,\mathbf{y}=1 and P=Per𝐲=3P^{\prime}=\mathrm{Per}\,\mathbf{y}^{\prime}=3 implies E(𝐯)2E^{*}(\mathbf{v})\geq 2 for every 𝐯=colour(𝐮,𝐲,𝐲)\mathbf{v}=\mathrm{colour}(\mathbf{u},\mathbf{y},\mathbf{y}^{\prime}). In particular, we show that for any β(0,1)\beta\in(0,1) every sequence of (δN)(\delta_{N}) contains infinitely many (1+β)(1+\beta)-forcing δN\delta_{N}.

There are four classes of equivalent matrices with (Y,Y)(Y,Y^{\prime})-names:

M1=(0011),M2=(0012),M3=(0010),M4=(0001)M_{1}=\left(\begin{smallmatrix}0&0\\ 1&1\end{smallmatrix}\right),M_{2}=\left(\begin{smallmatrix}0&0\\ 1&2\end{smallmatrix}\right),M_{3}=\left(\begin{smallmatrix}0&0\\ 1&0\end{smallmatrix}\right),M_{4}=\left(\begin{smallmatrix}0&0\\ 0&1\end{smallmatrix}\right).

If [A][A]_{\equiv} has the name

  • M1M_{1}, then every δ>2\delta>2 is (1+β)(1+\beta)-forcing for [A][A]_{\equiv}, as δ\delta and AA satisfy Properties 𝔓1\mathfrak{P}1, 𝔓2\mathfrak{P}2 and 𝔓3\mathfrak{P}3 with the triplet m=k==1m=k=\ell=1.

  • M2M_{2}, then every δ>1\delta>1 is (1+β)(1+\beta)-forcing for [A][A]_{\equiv}, the corresponding triplet is m=0m=0, k==1k=\ell=1.

  • M3M_{3}, then every δ>1\delta>1 is 22-forcing for [A][A]_{\equiv}, the corresponding triplet is m=0m=0, k=1,=0k=1,\ell=0.

  • M4M_{4}, then every δ>1\delta>1 is 22-forcing for [A][A]_{\equiv}, the corresponding triplet is m=0m=0, k=0,=1k=0,\ell=1.

Therefore, if E(𝐯)E^{*}(\mathbf{v}) is smaller than 22 for some 𝐯\mathbf{v}, then necessarily δN<2\delta_{N}<2 and M1M_{1} is the name of [AN][A_{N}]_{\equiv} for all N>N0N>N_{0}. In particular, aN+1=δN=1a_{N+1}=\lfloor\delta_{N}\rfloor=1 for all N>N0N>N_{0}. However, if [AN][A_{N}]_{\equiv} has the name M1M_{1}, then the class [AN+1][A_{N+1}]_{\equiv} containing the matrix AN+1=AN(0111)A_{N+1}=A_{N}\left(\begin{smallmatrix}0&1\\ 1&1\end{smallmatrix}\right) has the name M3M_{3} – a contradiction.

Example 33.

Now we apply Theorem 31 to balanced sequences obtained by colouring with two constant gap sequences both having the period 2, i.e., P=P=2P=P^{\prime}=2. Using notation (4), we have H=2,Y=1,Y=1H=2,Y=1,Y^{\prime}=1. According to Definition 27 all integer unimodular matrices belong to the same class of equivalence. By Lemma 24, the triplet m=1m=1, k=2k=2 and =0\ell=0 satisfies Property 𝔓1\mathfrak{P}1 for any unimodular matrix AA.

If we fix β=1\beta=1, then every δ>2\delta>2 with the same triplet satisfies 𝔓2\mathfrak{P}2 and 𝔓3\mathfrak{P}3. In other words, δ>2\delta>2 is 22-forcing. Therefore, the only candidates for 𝐯\mathbf{v} with E(𝐯)<2E^{*}(\mathbf{v})<2 are colourings of Sturmian sequences with slope θ=[0,w,1¯]\theta=[0,w,\overline{1}], where ww is any finite preperiod. For every such θ\theta, the formula (3) in Proposition 20 gives ΦN=1+QN1QN2\Phi_{N}=\frac{1+\frac{Q_{N-1}}{Q_{N}}}{2} for sufficiently large NN, where (QN)(Q_{N}) fulfils the recurrence relation QN+1=QN+QN1Q_{N+1}=Q_{N}+Q_{N-1}. Consequently, E(𝐯)=1+limΦN=1+5+141.809E^{*}(\mathbf{v})=1+\lim\Phi_{N}=1+\frac{\sqrt{5}+1}{4}\doteq 1.809 is the minimum value of the asymptotic critical exponent for P=P=2P=P^{\prime}=2 and it is attained if and only if 𝐯\mathbf{v} is a colouring of a Sturmian sequence with slope θ=[0,w,1¯]\theta=[0,w,\overline{1}] for a finite preperiod ww.

10. Admissible tails of continued fractions

In the previous section, we have associated with a continued fraction θ=[0,a1,a2,a3,]\theta=[0,a_{1},a_{2},a_{3},\ldots] a sequence of classes of equivalent matrices ([AN])\bigl{(}[A_{N}]_{\equiv}\bigr{)}. Since the number of classes is finite, there exists N0N_{0} such that any class of equivalence either occurs in ([AN])N>N0\bigl{(}[A_{N}]_{\equiv}\bigr{)}_{N>N_{0}} infinitely many times or does not occur at all in it. To find a balanced sequence 𝐯=colour(𝐮,𝐲,𝐲)\mathbf{v}=\mathrm{colour}(\mathbf{u},\mathbf{y},\mathbf{y}^{\prime}) with E(𝐯)1+βE^{*}(\mathbf{v})\leq 1+\beta, we should at least guarantee that δN\delta_{N} is not (1+β)(1+\beta)-forcing for the class [AN][A_{N}]_{\equiv} for each N>N0N>N_{0}. Formally, δN(β,AN)\delta_{N}\notin\mathcal{F}(\beta,A_{N}).

Remark 34.

In Definition 30 only Property 𝔓2\mathfrak{P}2 depends on δ\delta. Therefore, for each triplet m,k,m,k,\ell satisfying 𝔓1\mathfrak{P}1 and 𝔓3\mathfrak{P}3, we add to (β,A)\mathcal{F}(\beta,A) the interval of δ\delta’s satisfying 𝔓2\mathfrak{P}2, i.e., the interval

{(m+k1,+)(m+1,+) if =0;(m+k12,+)(m+1,+) if =1;(m+k1+1,m+k+11)(m+1,+) if 2.\begin{cases*}\ (m+k-1,+\infty)\cap(m+1,+\infty)&\quad if $\ell=0$;\\ \ (m+\frac{k-1}{2},+\infty)\cap(m+1,+\infty)&\quad if $\ell=1$;\\ \ (m+\frac{k-1}{\ell+1},m+\frac{k+1}{\ell-1})\cap(m+1,+\infty)&\quad if $\ell\geq 2$.\end{cases*}

The set (β,A)\mathcal{F}(\beta,A) is a union of several open intervals. Boundaries of these intervals are rational. Rational δ\delta’s have finite continued fractions and do not occur as tails of the continued fraction expansion of slopes of Sturmian sequences.

Definition 35.

Let β>0\beta>0 and A2×2A\in\mathbb{Z}^{2\times 2} be unimodular. We denote

𝒟(β,A)={δ>1:δ is NOT in the closure of (β,A)}.\mathcal{D}(\beta,A)=\{\delta>1:\delta\text{ is NOT in the closure of }\ \mathcal{F}(\beta,A)\}.

Using the notation of 𝒟(β,A)\mathcal{D}(\beta,A), Theorem 31 can be rephrased as the following corollary.

Corollary 36.

Let θ=[0,a1,a2,a3,]\theta=[0,a_{1},a_{2},a_{3},\ldots] be the slope of a Sturmian sequence 𝐮\mathbf{u} and β>0\beta>0. If E(𝐯)1+βE^{*}(\mathbf{v})\leq 1+\beta, then there exists N0N_{0} such that for every N>N0N>N_{0} the set 𝒟(β,AN)\mathcal{D}(\beta,A_{N}) contains δN\delta_{N}.

Lemma 37.

Let β>0\beta>0 and L=lcm(P,P)>1L=\mathrm{lcm}(P,P^{\prime}\,)>1. Then the set 𝒟(β,A)\mathcal{D}(\beta,A) is a subset of (1,L(1+β)2)(1,\lceil L(1+\beta)\rceil-2). In particular 𝒟(β,A)\mathcal{D}(\beta,A) is bounded for each equivalence class [A][A]_{\equiv}.

Proof.

Note that the pair k=Lk=L and =0\ell=0 fulfills 𝔓1\mathfrak{P}1 of Definition 30 independently on AA and mm\in\mathbb{N}. If we take m=Lβ1m=\lceil L\beta\rceil-1, then 1+mLβ\frac{1+m}{L}\geq\beta and 𝔓3\mathfrak{P}3 of Definition 30 is satisfied as well. By Remark 34, the set (m+k1,+)=(L(1+β)2,+)(m+k-1,+\infty)=(\lceil L(1+\beta)\rceil-2,+\infty) belongs to (β,A)\mathcal{F}(\beta,A). ∎

To compute easily 𝒟(β,A)\mathcal{D}(\beta,A), we collect several practical observations on triplets m,k,m,k,\ell\in\mathbb{N} that may influence the form of 𝒟(β,A)\mathcal{D}(\beta,A). We will apply these rules in examples worked out in hand. They follow immediately from Definitions 30 and 35, from Remark 34 and Lemmas 24 and 37.

Remark 38.

Let AA be a unimodular matrix, MM denotes its (Y,Y)(Y,Y^{\prime})-name. The following statements hold true.

  1. (1)

    The number of triplets m,k,m,k,\ell that may affect 𝒟(β,A)\mathcal{D}(\beta,A) is bounded. In particular, by Definition 35 and Lemma 37 it holds (1,m+1)𝒟(β,A)(1,L(1+β)2)(1,m+1)\subset\mathcal{D}(\beta,A)\subset(1,\lceil L(1+\beta)\rceil-2). It together with Property 𝔓3\mathfrak{P}3 forces m,k,m,k,\ell to satisfy 0m<L(1+β)2andk+m1β(m+2).0\leq m<\lceil L(1+\beta)\rceil-2\ \text{and}\ k+\ell m\leq\frac{1}{\beta}(m+2). The number of such triplets is finite.

    Each triplet adds to the set (β,A)\mathcal{F}(\beta,A) an interval as described in Remark 34. In fact, only few of the triplets add really new elements to (β,A)\mathcal{F}(\beta,A), or equivalently, erase some elements of 𝒟(β,A)\mathcal{D}(\beta,A). Hence, when we present in examples the set 𝒟(β,A)\mathcal{D}(\beta,A), we list only the triplets which determine the set. It means that no other triplet reduces the set 𝒟(β,A)\mathcal{D}(\beta,A) any more.

  2. (2)

    𝔓2\mathfrak{P}2 is satisfied whenever m+1<δm+1<\delta and

    (k){(10),(01),(11),(02),(12),(13)}.\left(\begin{smallmatrix}\ell\\ k\end{smallmatrix}\right)\in\{\left(\begin{smallmatrix}1\\ 0\end{smallmatrix}\right),\left(\begin{smallmatrix}0\\ 1\end{smallmatrix}\right),\left(\begin{smallmatrix}1\\ 1\end{smallmatrix}\right),\left(\begin{smallmatrix}0\\ 2\end{smallmatrix}\right),\left(\begin{smallmatrix}1\\ 2\end{smallmatrix}\right),\left(\begin{smallmatrix}1\\ 3\end{smallmatrix}\right)\}.
  3. (3)

    (k)\left(\begin{smallmatrix}\ell\\ k\end{smallmatrix}\right) with k+2\ell\geq k+2 does not fulfil 𝔓2\mathfrak{P}2 for any mm\in\mathbb{N} and δ>m+1\delta>m+1.

  4. (4)

    If H2H\geq 2, then only kk\geq\ell may satisfy 𝔓1\mathfrak{P}1 and 𝔓2\mathfrak{P}2 for some mm\in\mathbb{N} and δ>m+1\delta>m+1.

  5. (5)

    Let β<1\beta<1 and 𝔓1\mathfrak{P}1 be fulfilled for some mm\in\mathbb{N} and (k){(10),(01),(11)}\left(\begin{smallmatrix}\ell\\ k\end{smallmatrix}\right)\in\{\left(\begin{smallmatrix}1\\ 0\end{smallmatrix}\right),\left(\begin{smallmatrix}0\\ 1\end{smallmatrix}\right),\left(\begin{smallmatrix}1\\ 1\end{smallmatrix}\right)\}. Then 𝔓3\mathfrak{P}3 is fulfilled, too. Moreover, 𝔓2\mathfrak{P}2 is satisfied for all δ>m+1\delta>m+1.

    Consequently,

    𝒟(β,A)=\mathcal{D}(\beta,A)=\emptyset for m=0m=0    and     𝒟(β,A)(1,m+1)\mathcal{D}(\beta,A)\subset(1,m+1) for m1m\geq 1.

  6. (6)

    If a triplet m,k,m,k,\ell satisfies Properties 𝔓1\mathfrak{P}1 and 𝔓2\mathfrak{P}2, then the triplet m,Tk,Tm,Tk,T\ell with T,T2T\in\mathbb{N},T\geq 2, need not be taken into account when constructing 𝒟(β,A)\mathcal{D}(\beta,A).

    Indeed, if m,Tk,Tm,Tk,T\ell satisfies 𝔓1\mathfrak{P}1 and 𝔓2\mathfrak{P}2, then

    (m+Tk1,+)(m+k1,+)(m+Tk12,+)(m+k12,+)(m+Tk1T+1,m+Tk+1T1)(m+k1+1,m+k+11)for2.\begin{array}[]{l}(m+Tk-1,+\infty)\subset(m+k-1,+\infty)\\[3.00003pt] (m+\frac{Tk-1}{2},+\infty)\subset(m+\frac{k-1}{2},+\infty)\\[3.00003pt] (m+\frac{Tk-1}{T\ell+1},m+\frac{Tk+1}{T\ell-1})\subset(m+\frac{k-1}{\ell+1},m+\frac{k+1}{\ell-1})\ \text{for}\ \ell\geq 2.\end{array}

    The statement follows by Remark 34.

Example 39.

Let P=2P=2, P=4P^{\prime}=4 and β=12\beta=\frac{1}{2}. In this case H=2H=2, L=4L=4, Y=1Y=1, Y=2Y^{\prime}=2 and there are three classes of equivalent matrices with (Y,Y)(Y,Y^{\prime})-names:

M1=(0010),M2=(0001),M3=(0011).M_{1}=\left(\begin{smallmatrix}0&0\\ 1&0\end{smallmatrix}\right),M_{2}=\left(\begin{smallmatrix}0&0\\ 0&1\end{smallmatrix}\right),M_{3}=\left(\begin{smallmatrix}0&0\\ 1&1\end{smallmatrix}\right).

To find 𝒟(β,Mi)\mathcal{D}(\beta,M_{i}) we will apply Remark 38.

  • M1=(0010)M_{1}=\left(\begin{smallmatrix}0&0\\ 1&0\end{smallmatrix}\right)
    Consider the triplet m=0,k=2,=0m=0,k=2,\ell=0. 𝔓1\mathfrak{P}1 is satisfied by Lemma 24. 𝔓2\mathfrak{P}2 is satisfied for δ>1\delta>1, by Item 2. As 2=k>=02=k>\ell=0, 𝔓3\mathfrak{P}3 says 12=1+mk+mβ=12\frac{1}{2}=\frac{1+m}{k+m\ell}\geq\beta=\frac{1}{2}.

    Hence, any δ>1\delta>1 is (1+12)(1+\frac{1}{2})-forcing and thus 𝒟(β,M1)=\mathcal{D}(\beta,M_{1})=\emptyset.

  • M2=(0001)M_{2}=\left(\begin{smallmatrix}0&0\\ 0&1\end{smallmatrix}\right)
    Due to Lemma 24 we consider only triplets with k=2κk=2\kappa , =2λ\ell=2\lambda, where M2(λκ)=(00)mod(12)M_{2}\left(\begin{smallmatrix}\lambda\\ \kappa\end{smallmatrix}\right)=\left(\begin{smallmatrix}0\\ 0\end{smallmatrix}\right)\bmod\left(\begin{smallmatrix}1\\ 2\end{smallmatrix}\right) forces κ\kappa to be even. Item 4 gives the restriction κλ\kappa\geq\lambda. Hence by Items 1 and 6, we need to work only with (k){(04),(24),(44)}\left(\begin{smallmatrix}\ell\\ k\end{smallmatrix}\right)\in\{\left(\begin{smallmatrix}0\\ 4\end{smallmatrix}\right),\left(\begin{smallmatrix}2\\ 4\end{smallmatrix}\right),\left(\begin{smallmatrix}4\\ 4\end{smallmatrix}\right)\} and m<4m<4. The inequality 1+mk+m12\frac{1+m}{k+m\ell}\geq\frac{1}{2} (in case k>k>\ell) and 1k>12\frac{1}{k}>\frac{1}{2} (in case k=k=\ell) required by 𝔓3\mathfrak{P}3 is fulfilled only if k=4,=0k=4,\ell=0 and 1m<41\leq m<4.

    By Remark 34, the set of (1+12)(1+\frac{1}{2})-forcing δ\delta’s is (4,+)(4,+\infty) and thus 𝒟(β,M2)=(1,4)\mathcal{D}(\beta,M_{2})=(1,4).

  • M3=(0011)M_{3}=\left(\begin{smallmatrix}0&0\\ 1&1\end{smallmatrix}\right)
    By Lemma 24 we need to check only the values k=2κk=2\kappa, =2λ\ell=2\lambda, where λ+κ\lambda+\kappa is even. Item 4 gives κλ\kappa\geq\lambda. Items 1 and 6 restrict (k){(04),(22)}\left(\begin{smallmatrix}\ell\\ k\end{smallmatrix}\right)\in\{\left(\begin{smallmatrix}0\\ 4\end{smallmatrix}\right),\left(\begin{smallmatrix}2\\ 2\end{smallmatrix}\right)\} and m<4m<4. The inequality in 𝔓3\mathfrak{P}3 is fulfilled only for k=4,=0k=4,\ell=0 and 1m<41\leq m<4.

    Analogously to the previous case, 𝒟(β,M3)=(1,4)\mathcal{D}(\beta,M_{3})=(1,4).

Note an interesting fact. If we replace the value β=12\beta=\frac{1}{2} by some β<12\beta_{-}<\frac{1}{2}, the triplets m=0,k=2,=2m=0,k=2,\ell=2 and m=2,k=2,=2m=2,k=2,\ell=2 extend the set of (1+β)(1+\beta_{-})-forcing δ\delta’s and 𝒟(β,M3)=\mathcal{D}(\beta_{-},M_{3})=\emptyset.

11. Graphs of admissible tails

In this section we create the main tool that enables us to find balanced sequences with a small value of EE^{*}. More precisely, for given P=Per𝐲P=\mathrm{Per}\,\mathbf{y} and P=Per𝐲P^{\prime}=\mathrm{Per}\,\mathbf{y}^{\prime} and β>0\beta>0, we will be able to identify candidates for suffixes (we call them “admissible tails”) of the continued fraction expansion of the slope θ\theta of a Sturmian sequence 𝐮\mathbf{u} whose colouring 𝐯=colour(𝐮,𝐲,𝐲)\mathbf{v}=\mathrm{colour}(\mathbf{u},\mathbf{y},{\mathbf{y}^{\prime}}) satisfies E(𝐯)1+βE^{*}(\mathbf{v})\leq 1+\beta.

Definition 40.

Given β>0\beta>0 and P,PP,\ P^{\prime}\in\mathbb{N}. An oriented graph (V,E)(V,E) is called the graph of (1+β)(1+\beta)-admissible tails, and denoted Γβ\Gamma_{\beta}, if

  • the set of vertices VV consists of classes of equivalence \equiv;

  • a pair ([A],[B])\bigl{(}[A]_{\equiv},[B]_{\equiv}\bigr{)} labeled by aa belongs to the set EE of oriented edges if a=δa=\lfloor\delta\rfloor for some δ𝒟(β,A)\delta\in\mathcal{D}(\beta,A) and B[A(011a)]B\in[A\left(\begin{smallmatrix}0&1\\ 1&a\end{smallmatrix}\right)]_{\equiv}.

The graph Γβ\Gamma_{\beta} is finite: it has a finite number of vertices as the number of classes of equivalence is finite and a finite number of edges as by Lemma 37 the set 𝒟(β,A){\mathcal{D}}(\beta,A) is bounded for each class [A][A]_{\equiv}. Moreover, if β1>β2\beta_{1}>\beta_{2}, then 𝒟(β2,A)𝒟(β1,A)\mathcal{D}(\beta_{2},A)\subset\mathcal{D}(\beta_{1},A) for every unimodular matrix AA. Hence Γβ2\Gamma_{\beta_{2}} is a subgraph of Γβ1\Gamma_{\beta_{1}}.

Let us recall that an oriented infinite path in a graph (V,E)(V,E) is a sequence v0,e0,v1,e1,v2,e2,v_{0},e_{0},v_{1},e_{1},v_{2},e_{2},\ldots such that viV,eiEv_{i}\in V,e_{i}\in E and (vi,vi+1)=ei(v_{i},v_{i+1})=e_{i} for every ii\in\mathbb{N}.

The graph terminology allows us to rephrase Corollary 36 into the following theorem.

Theorem 41.

Let β>0\beta>0 and 𝐯=colour(𝐮,𝐲,𝐲)\mathbf{v}=\mathrm{colour}(\mathbf{u},\mathbf{y},\mathbf{y}^{\prime}), where 𝐮\mathbf{u} is a Sturmian sequence with slope θ=[0,a1,a2,a3,]\theta=[0,a_{1},a_{2},a_{3},\ldots]. Assume that E(𝐯)1+βE^{*}(\mathbf{v})\leq 1+\beta.

Then there exists an infinite oriented path v0,e0,v1,e1,v2,e2,v_{0},e_{0},v_{1},e_{1},v_{2},e_{2},\ldots in Γβ\Gamma_{\beta} and N0N_{0}\in\mathbb{N} such that for every NN\in\mathbb{N}

  1. (1)

    [AN+N0]=vN[A_{N+N_{0}}]_{\equiv}=v_{N};

  2. (2)

    aN+N0+1a_{N+N_{0}+1} is the label of the edge eNe_{N}.

In the graph Γβ\Gamma_{\beta} of admissible tails we are interested in infinite paths on which vertices occur infinitely many times. Therefore it is enough to consider strongly connected components, i.e., subgraphs with an oriented path from each vertex to each vertex. In particular, if a component has only one vertex, it has to have a loop.

Corollary 42.

Let β>0\beta>0 and P,PP,\ P^{\prime}\in\mathbb{N}. If Γβ\Gamma_{\beta} contains no oriented cycle, then E(𝐯)>1+βE^{*}(\mathbf{v})>1+\beta for every colouring 𝐯\mathbf{v} of a Sturmian sequence by constant gap sequences of periods PP and PP^{\prime}.

Example 43.

Let us construct the graph of (1+β)(1+\beta)-admissible tails Γβ\Gamma_{\beta} for parameters P=2,P=4P=2,P^{\prime}=4 and β=12\beta=\frac{1}{2}. By Example 39, the graph has three vertices with (Y,Y)(Y,Y^{\prime})-names

M1=(0010),M2=(0001),M3=(0011)M_{1}=\left(\begin{smallmatrix}0&0\\ 1&0\end{smallmatrix}\right),M_{2}=\left(\begin{smallmatrix}0&0\\ 0&1\end{smallmatrix}\right),M_{3}=\left(\begin{smallmatrix}0&0\\ 1&1\end{smallmatrix}\right)

and the corresponding sets

𝒟(β,M1)=,𝒟(β,M2)=(1,4),𝒟(β,M3)=(1,4).\mathcal{D}(\beta,M_{1})=\emptyset,\quad\mathcal{D}(\beta,M_{2})=(1,4),\quad\mathcal{D}(\beta,M_{3})=(1,4).

The graph Γβ\Gamma_{\beta} and its only strongly connected component are depicted in Figure 1.

By Theorem 41 the suffix 2¯\overline{2} is the only candidate for the suffix of θ\theta associated with a Sturmian sequence 𝐮\mathbf{u} such that 𝐯=colour(𝐮,𝐲,𝐲)\mathbf{v}=\mathrm{colour}(\mathbf{u},\mathbf{y},\mathbf{y}^{\prime}) with Per𝐲=2,Per𝐲=4\mathrm{Per}\,\mathbf{y}=2,\ \mathrm{Per}\,\mathbf{y}^{\prime}=4 has E(𝐯)32E^{*}(\mathbf{v})\leq\frac{3}{2}. And indeed, E(𝐯)=32E^{*}(\mathbf{v})=\frac{3}{2} for θ=[0,1,2¯]\theta=[0,1,\overline{2}]. The reader is invited to check that ΦN=12\Phi_{N}=\frac{1}{2} in Proposition 20 (the maximum is attained for m=0m=0, k==2k=\ell=2).

As we have noticed in Example 39, if we choose instead of β=12\beta=\frac{1}{2} the value β<12\beta_{-}<\tfrac{1}{2}, then D(β,M3)=D(\beta_{-},M_{3})=\emptyset and the corresponding graph contains no oriented cycle. In other words, we can see immediately that there is no 𝐯=colour(𝐮,𝐲,𝐲)\mathbf{v}=\mathrm{colour}(\mathbf{u},\mathbf{y},\mathbf{y}^{\prime}) with E(𝐯)<32E^{*}(\mathbf{v})<\frac{3}{2} for Per𝐲=2,Per𝐲=4\mathrm{Per}\,\mathbf{y}=2,\ \mathrm{Per}\,\mathbf{y}^{\prime}=4.

(0010)\left(\begin{array}[]{cc}0&0\\ 1&0\\ \end{array}\right)(0001)\left(\begin{array}[]{cc}0&0\\ 0&1\\ \end{array}\right)(0011)\left(\begin{array}[]{cc}0&0\\ 1&1\\ \end{array}\right)132132
(0011)\left(\begin{array}[]{cc}0&0\\ 1&1\\ \end{array}\right)2
Figure 1. (a)  The graph of admissible tails for β=12\beta=\frac{1}{2} and P=2,P=4P=2,\ P^{\prime}=4 and (b)  its unique strongly connected component.
Example 44.

The authors of [RSV19] show that the least critical exponent E(𝐯)E(\mathbf{v}) on 4 letters – in our notation RTB(4)RTB(4) – equals 1+5+141+\frac{\sqrt{5}+1}{4} and it is reached for the colouring of the Fibonacci sequence by constant gap sequences with P=P=2P=P^{\prime}=2. Let us deduce that RTB(4)=RTB(4)RTB^{*}(4)=RTB(4).

Thanks to Proposition 21, Examples 32 and 33, and Remark 18, it remains to inspect only the case P=1P=1 and P=4P^{\prime}=4.

There are six classes of equivalent matrices with (Y,Y)(Y,Y^{\prime})-names

M1=(0010),M2=(0001),M3=(0011),M4=(0021),M5=(0012),M6=(0013).M_{1}=\left(\begin{smallmatrix}0&0\\ 1&0\end{smallmatrix}\right),M_{2}=\left(\begin{smallmatrix}0&0\\ 0&1\end{smallmatrix}\right),M_{3}=\left(\begin{smallmatrix}0&0\\ 1&1\end{smallmatrix}\right),M_{4}=\left(\begin{smallmatrix}0&0\\ 2&1\end{smallmatrix}\right),M_{5}=\left(\begin{smallmatrix}0&0\\ 1&2\end{smallmatrix}\right),M_{6}=\left(\begin{smallmatrix}0&0\\ 1&3\end{smallmatrix}\right).

Let us write down for each MiM_{i} the triplets m,k,m,k,\ell that influence the set 𝒟(β,Mi)\mathcal{D}(\beta,M_{i}) (the other triplets satisfying Property 𝔓1\mathfrak{P}1 and 𝔓3\mathfrak{P}3 do not reduce 𝒟(β,Mi)\mathcal{D}(\beta,M_{i}) any more):

M1=(0010)m=0,k=1,=0𝒟(β,M1)=M2=(0001)m=0,k=0,=1𝒟(β,M2)=M3=(0011)m=2,k=1,=1𝒟(β,M3)=(1,3)M4=(0021)m=1,k=1,=1𝒟(β,M4)=(1,2)M5=(0012)m=1,k=2,=0𝒟(β,M5)=(1,2)M6=(0013)m=0,k=1,=1𝒟(β,M6)=\begin{array}[]{rcll}M_{1}=\left(\begin{smallmatrix}0&0\\ 1&0\end{smallmatrix}\right)&m=0,k=1,\ell=0&\mathcal{D}(\beta,M_{1})=\emptyset\\ M_{2}=\left(\begin{smallmatrix}0&0\\ 0&1\end{smallmatrix}\right)&m=0,k=0,\ell=1&\mathcal{D}(\beta,M_{2})=\emptyset\\ M_{3}=\left(\begin{smallmatrix}0&0\\ 1&1\end{smallmatrix}\right)&m=2,k=1,\ell=1&\mathcal{D}(\beta,M_{3})=(1,3)\\ M_{4}=\left(\begin{smallmatrix}0&0\\ 2&1\end{smallmatrix}\right)&m=1,k=1,\ell=1&\mathcal{D}(\beta,M_{4})=(1,2)\\ M_{5}=\left(\begin{smallmatrix}0&0\\ 1&2\end{smallmatrix}\right)&m=1,k=2,\ell=0&\mathcal{D}(\beta,M_{5})=(1,2)\\ M_{6}=\left(\begin{smallmatrix}0&0\\ 1&3\end{smallmatrix}\right)&m=0,k=1,\ell=1&\mathcal{D}(\beta,M_{6})=\emptyset\end{array}
(0011)\left(\begin{array}[]{cc}0&0\\ 1&1\\ \end{array}\right)(0012)\left(\begin{array}[]{cc}0&0\\ 1&2\\ \end{array}\right)(0021)\left(\begin{array}[]{cc}0&0\\ 2&1\\ \end{array}\right)(0013)\left(\begin{array}[]{cc}0&0\\ 1&3\\ \end{array}\right)(0001)\left(\begin{array}[]{cc}0&0\\ 0&1\\ \end{array}\right)(0010)\left(\begin{array}[]{cc}0&0\\ 1&0\\ \end{array}\right)1211
Figure 2. The graph of admissible tails for β=5+14\beta=\frac{\sqrt{5}+1}{4} and P=1,P=4P=1,\ P^{\prime}=4.

In the graph Γβ\Gamma_{\beta} with parameters P=1,P=4P=1,\ P^{\prime}=4 and β=5+14\beta=\frac{\sqrt{5}+1}{4} (see Figure 2) there is no strongly connected component. This completes the proof that RTB(4)=1+5+14RTB^{*}(4)=1+\frac{\sqrt{5}+1}{4}.

12. Asymptotic repetitive threshold for binary to quinary balanced sequences

Using graphs of admissible tails we are able to list the least asymptotic critical exponent of dd-ary balanced sequences for 2d52\leq d\leq 5.

  • It is known that RTB(2)=RTB(2)=2+5+123.618RTB^{*}(2)=RTB(2)=2+\frac{\sqrt{5}+1}{2}\doteq 3.618 and it is reached for the Fibonacci sequence.

  • RTB(3)=2+12RTB^{*}(3)=2+\frac{1}{\sqrt{2}}:   It suffices to consider P=1,P=2P=1,P^{\prime}=2 and β=1+12\beta=1+\frac{1}{\sqrt{2}}. There are three classes of equivalent matrices with (Y,Y)(Y,Y^{\prime})-names:

    M1=(0010),M2=(0001),M3=(0011).M_{1}=\left(\begin{smallmatrix}0&0\\ 1&0\end{smallmatrix}\right),M_{2}=\left(\begin{smallmatrix}0&0\\ 0&1\end{smallmatrix}\right),M_{3}=\left(\begin{smallmatrix}0&0\\ 1&1\end{smallmatrix}\right).

    Let us write down for each MiM_{i} the triplets m,k,m,k,\ell that influence the set 𝒟(β,Mi)\mathcal{D}(\beta,M_{i}):

    M1=(0010)m=1,k=1,=0𝒟(β,M1)=(1,2)M2=(0001)m=0,k=0,=1𝒟(β,M2)=M3=(0011)m=3,k=2,=0𝒟(β,M3)=(1,4)\begin{array}[]{rcll}M_{1}=\left(\begin{smallmatrix}0&0\\ 1&0\end{smallmatrix}\right)&m=1,k=1,\ell=0&\mathcal{D}(\beta,M_{1})=(1,2)\\ M_{2}=\left(\begin{smallmatrix}0&0\\ 0&1\end{smallmatrix}\right)&m=0,k=0,\ell=1&\mathcal{D}(\beta,M_{2})=\emptyset\\ M_{3}=\left(\begin{smallmatrix}0&0\\ 1&1\end{smallmatrix}\right)&m=3,k=2,\ell=0&\mathcal{D}(\beta,M_{3})=(1,4)\end{array}

    There is a unique strongly connected component in the graph Γβ\Gamma_{\beta}, which is the same as the one depicted in Figure 1. By Theorem 41 the suffix 2¯\overline{2} is the only candidate for the suffix of the continued fraction θ\theta associated with a Sturmian sequence 𝐮\mathbf{u} such that 𝐯=colour(𝐮,𝐲,𝐲)\mathbf{v}=\mathrm{colour}(\mathbf{u},\mathbf{y},\mathbf{y}^{\prime}) with Per𝐲=1,Per𝐲=2\mathrm{Per}\,\mathbf{y}=1,\ \mathrm{Per}\,\mathbf{y}^{\prime}=2 has E(𝐯)2+12E^{*}(\mathbf{v})\leq 2+\frac{1}{\sqrt{2}}. And indeed, E(𝐯)=2+12E^{*}(\mathbf{v})=2+\frac{1}{\sqrt{2}} for θ=[0,1,2¯]\theta=[0,1,\overline{2}]. The reader is invited to check that ΦN=2+QN1QN1+QN1QN\Phi_{N}=\frac{2+\frac{Q_{N-1}}{Q_{N}}}{1+\frac{Q_{N-1}}{Q_{N}}} (the maximum in (3) is attained for m=1,k=0,=1m=1,\ k=0,\ \ell=1) and QN+1=2QN+QN1Q_{N+1}=2Q_{N}+Q_{N-1} for N1N\geq 1. Using Proposition 20 we have E(𝐯)=1+limNΦN=2+12E^{*}(\mathbf{v})=1+\lim_{N\to\infty}\Phi_{N}=2+\frac{1}{\sqrt{2}}.

  • RTB(4)=1+5+14RTB^{*}(4)=1+\frac{\sqrt{5}+1}{4}:   It was shown in Example 44.

  • RTB(5)=32RTB^{*}(5)=\frac{3}{2}:   We set β=12\beta=\frac{1}{2}. By Remark 18, Proposition 21 and Lemma 23 we have to inspect the pairs (P,P){(2,3),(2,4),(1,6),(1,8)}(P,P^{\prime})\in\{(2,3),(2,4),(1,6),(1,8)\}. In all cases besides (2,4)(2,4) there are no strongly connected components in the graph Γβ\Gamma_{\beta}. The case of P=2,P=4P=2,\ P^{\prime}=4 was examined in Example 43, where it was shown that the unique admissible tail is 2¯\overline{2} and E(𝐯)=32E^{*}(\mathbf{v})=\frac{3}{2} for θ=[0,1,2¯]\theta=[0,1,\overline{2}].

The method used to determine RTB(d)RTB^{*}(d) for d5d\leq 5 does not work for d=6d=6. Let us explain why. To find RTB(6)RTB^{*}(6) we have to inspect the pairs (P,P){(3,3),(3,4),(2,6),(2,8),(1,5),(1,9),(1,12),(1,16)}(P,P^{\prime})\in\{(3,3),(3,4),(2,6),(2,8),(1,5),(1,9),(1,12),(1,16)\}. In the sequel, we will show that RTB(6)=1+3655801.2398RTB^{*}(6)=1+\frac{3\sqrt{65}-5}{80}\doteq 1.2398. If we construct the graph Γβ\Gamma_{\beta} with the optimal value β=365580\beta=\frac{3\sqrt{65}-5}{80}, we find out that in all cases besides (P,P)=(1,16)(P,P^{\prime})=(1,16) there are no strongly connected components in Γβ\Gamma_{\beta}.

Hence we focus on the case (P,P)=(1,16)(P,P^{\prime})=(1,16). For this pair, the strongly connected component of Γβ\Gamma_{\beta} is depicted in Figure 3 (we will show later that the bold cycle corresponds to the unique (1+β)(1+\beta)-admissible tail).

(0016)\left(\begin{array}[]{cc}0&0\\ 1&6\\ \end{array}\right)(0025)\left(\begin{array}[]{cc}0&0\\ 2&5\\ \end{array}\right)(00112)\left(\begin{array}[]{cc}0&0\\ 1&12\\ \end{array}\right)(0043)\left(\begin{array}[]{cc}0&0\\ 4&3\\ \end{array}\right)(00113)\left(\begin{array}[]{cc}0&0\\ 1&13\\ \end{array}\right)(00111)\left(\begin{array}[]{cc}0&0\\ 1&11\\ \end{array}\right)(0014)\left(\begin{array}[]{cc}0&0\\ 1&4\\ \end{array}\right)(0041)\left(\begin{array}[]{cc}0&0\\ 4&1\\ \end{array}\right)131211211212
Figure 3. The strongly connected component of the graph of admissible tails for β=365580\beta=\frac{3\sqrt{65}-5}{80} and P=1,P=16P=1,\ P^{\prime}=16.

Although 1,1,3¯\overline{1,1,3} is the label of an infinite path in Γβ\Gamma_{\beta}, the asymptotic critical exponent E(𝐯)1.41>1+βE^{*}(\mathbf{v})\geq 1.41>1+\beta for any colouring of a Sturmian sequence with slope θ\theta having eventually periodic continued fraction with the period 1,1,3¯\overline{1,1,3}. In other words, the implication in Theorem 41 cannot be reversed. Our next goal is to reduce the graph Γβ\Gamma_{\beta} in order to exclude paths corresponding to the asymptotic critical exponent >1+β>1+\beta.

13. Reduction of the graphs

We have seen that the existence of an infinite path in the graph of admissible tails does not guarantee a small value of EE^{*}. There may be even uncountably many paths in the graph Γβ\Gamma_{\beta} corresponding to E(𝐯)>1+βE^{*}(\mathbf{v})>1+\beta. However, we will show how to reduce the graph so that the number of unsuitable paths is diminished. This method enables us to find RTB(d)RTB^{*}(d) for 6d106\leq d\leq 10.

We describe two reasons why the implication in Theorem  41 cannot be reversed.

  1. (1)

    Let d1,d2,d3,d_{1},d_{2},d_{3},\ldots be labels of edges in an infinite path v0,e0,v1,e1,v2,e2,v_{0},e_{0},v_{1},e_{1},v_{2},e_{2},\ldots in Γβ\Gamma_{\beta} and all vertices of the path occurs in it infinitely many times. If the path corresponds to an asymptotic critical exponent 1+β\leq 1+\beta, then for each NN\in\mathbb{N}, the continued fraction [dN+1,dN+2,][d_{N+1},d_{N+2},\ldots] belongs to 𝒟(β,A)\mathcal{D}(\beta,A), where AvNA\in v_{N}. But the construction of the graph requires only dN+1=δd_{N+1}=\lfloor\delta\rfloor for some δ𝒟(β,A)\delta\in\mathcal{D}(\beta,A).

    Hence, if it is not possible to extend the edge starting in vNv_{N} and labeled by dN+1d_{N+1} to an infinite path whose labels make the continued fraction expansion of some δ𝒟(β,A)\delta\in\mathcal{D}(\beta,A), then this edge may be erased without influencing the validity of Theorem 41.

    We call this process forward reduction.

  2. (2)

    The terms used to describe Property 𝔓3\mathfrak{P}3 in Definition 30 represent the minimal value of the function fm,k,(x)=1+m+xk+m+xf_{m,k,\ell}(x)=\frac{1+m+x}{k+\ell m+\ell x} for x[0,1]x\in[0,1], see (11). The actual value of xx we use in evaluation of ΦN\Phi_{N} in (10) is xN=QN1/QN=[0,aN,aN1,aN2,,a1]x_{N}=Q_{N-1}/Q_{N}=[0,a_{N},a_{N-1},a_{N-2},\ldots,a_{1}]. It may happen that fm,k,(xN)>βf_{m,k,\ell}(x_{N})>\beta even if the minimum of fm,k,f_{m,k,\ell} on [0,1][0,1] is smaller than β\beta.

    To construct Γβ\Gamma_{\beta} we use only the fact that xN[0,1]x_{N}\in[0,1]. As xN=[0,aN,aN1,aN2,,a1]x_{N}=[0,a_{N},a_{N-1},a_{N-2},\ldots,a_{1}], the value xNx_{N} is given by the labels of edges of a path of length NN ending in vN=[A]v_{N}=[A]_{\equiv}. Assume that the structure of Γβ\Gamma_{\beta} enables to deduce that all paths entering the vertex [A][A]_{\equiv} have xNJ[0,1]x_{N}\in J\subset[0,1]. If for some m,k,m,k,\ell satisfying Property 𝔓1\mathfrak{P}1 the following inequality holds true

    min{fm,k,(x):xJ}>β,\min\{f_{m,k,\ell}(x):x\in J\}\ >\ \beta\,,\ \ \

    then δ\delta satisfying Property 𝔓2\mathfrak{P}2 is (1+β)(1+\beta)-forcing, i.e., all such δ\delta’s should be deleted from the set 𝒟(β,A)\mathcal{D}(\beta,A). The new set 𝒟(β,A)\mathcal{D}^{\prime}(\beta,A) may cause that some edges starting in the vertex [A][A]_{\equiv} are deleted as well.

    We call this process backward reduction.

We always apply the forward and backward reductions together with searching for strongly connected components as long as the graph changes.

Let us illustrate both kinds of reductions on the following example.

Example 45.

Let us construct Γβ\Gamma_{\beta} for P=3,P=4,β=13P=3,\ P^{\prime}=4,\ \beta=\tfrac{1}{3}. Then H=1H=1 and L=12L=12. There are 24 classes of equivalence with the following (Y,Y)(Y,Y^{\prime})-names:

(0101),(1010),(1012),(1121),(1213),(0111),(0110),(1110),(1210),(0113),(1013),(1113),\left(\begin{smallmatrix}0&1\\ 0&1\end{smallmatrix}\right),\left(\begin{smallmatrix}1&0\\ 1&0\end{smallmatrix}\right),\left(\begin{smallmatrix}1&0\\ 1&2\end{smallmatrix}\right),\left(\begin{smallmatrix}1&1\\ 2&1\end{smallmatrix}\right),\left(\begin{smallmatrix}1&2\\ 1&3\end{smallmatrix}\right),\left(\begin{smallmatrix}0&1\\ 1&1\end{smallmatrix}\right),\left(\begin{smallmatrix}0&1\\ 1&0\end{smallmatrix}\right),\left(\begin{smallmatrix}1&1\\ 1&0\end{smallmatrix}\right),\left(\begin{smallmatrix}1&2\\ 1&0\end{smallmatrix}\right),\left(\begin{smallmatrix}0&1\\ 1&3\end{smallmatrix}\right),\left(\begin{smallmatrix}1&0\\ 1&3\end{smallmatrix}\right),\left(\begin{smallmatrix}1&1\\ 1&3\end{smallmatrix}\right),
(1001),(1021),(1011),(1201),(1101),(0112),(1112),(1212),(1211),(0121),(1221),(1111).\left(\begin{smallmatrix}1&0\\ 0&1\end{smallmatrix}\right),\left(\begin{smallmatrix}1&0\\ 2&1\end{smallmatrix}\right),\left(\begin{smallmatrix}1&0\\ 1&1\end{smallmatrix}\right),\left(\begin{smallmatrix}1&2\\ 0&1\end{smallmatrix}\right),\left(\begin{smallmatrix}1&1\\ 0&1\end{smallmatrix}\right),\left(\begin{smallmatrix}0&1\\ 1&2\end{smallmatrix}\right),\left(\begin{smallmatrix}1&1\\ 1&2\end{smallmatrix}\right),\left(\begin{smallmatrix}1&2\\ 1&2\end{smallmatrix}\right),\left(\begin{smallmatrix}1&2\\ 1&1\end{smallmatrix}\right),\left(\begin{smallmatrix}0&1\\ 2&1\end{smallmatrix}\right),\left(\begin{smallmatrix}1&2\\ 2&1\end{smallmatrix}\right),\left(\begin{smallmatrix}1&1\\ 1&1\end{smallmatrix}\right).

By Lemma 37, 𝒟(β,Mi)(1,14)\mathcal{D}(\beta,M_{i})\subset(1,14). We write down for each class the triplets m,k,m,k,\ell that influence the set 𝒟(β,Mi)\mathcal{D}(\beta,M_{i}):

list of triplets(m,k,)M1=(0101)𝒟(β,M1)=(0,0,1)M2=(1010)𝒟(β,M2)=(0,1,0)M3=(1012)𝒟(β,M3)=(0,2,0)M4=(1121)𝒟(β,M4)=(0,2,1)M5=(1213)𝒟(β,M5)=(0,1,1)M6=(0111)𝒟(β,M6)=(0,3,1)M7=(0110)𝒟(β,M7)=(1,2)(0,3,0)M8=(1110)𝒟(β,M8)=(1,2)(0,3,0)M9=(1210)𝒟(β,M9)=(1,2)(0,3,0)M10=(0113)𝒟(β,M10)=(1,2)(1,4,2),(2,7,1)M11=(1013)𝒟(β,M11)=(1,4)(1,4,0)M12=(1113)𝒟(β,M12)=(1,52)(1,4,1)M13=(1001)𝒟(β,M13)=(1,3)(72,4)(1,4,0),(2,2,3)M14=(1021)𝒟(β,M14)=(32,4)(0,2,3),(1,4,0)M15=(1011)𝒟(β,M15)=(1,2)(52,4)(1,2,3),(1,4,0)M16=(1201)𝒟(β,M16)=(1,2)(1,3,1)M17=(1101)𝒟(β,M17)=(1,2)(4,92)(1,2,2),(2,6,1)M18=(0112)𝒟(β,M18)=(0,3,2),(1,6,0),(2,5,2)M19=(1112)𝒟(β,M19)=(2,3)(0,1,2),(1,6,0),(2,3,2)M20=(1212)𝒟(β,M20)=(1,2)(1,3,2),(1,6,0),(3,5,2)M21=(1211)𝒟(β,M21)=(3,4)(0,2,2),(2,5,1)M22=(0121)𝒟(β,M22)=(1,3)(1,5,1)M23=(1221)𝒟(β,M23)=(1,3)(2,4,2),(3,7,1)M24=(1111)𝒟(β,M24)=(1,4)(3,4,2),(3,8,1)\begin{array}[]{lll}&&\textbf{list of triplets}\ (m,k,\ell)\\ &&\\ M_{1}=\left(\begin{smallmatrix}0&1\\ 0&1\end{smallmatrix}\right)&\mathcal{D}(\beta,M_{1})=\emptyset&\quad\ \ (0,0,1)\\ M_{2}=\left(\begin{smallmatrix}1&0\\ 1&0\end{smallmatrix}\right)&\mathcal{D}(\beta,M_{2})=\emptyset&\quad\ \ (0,1,0)\\ M_{3}=\left(\begin{smallmatrix}1&0\\ 1&2\end{smallmatrix}\right)&\mathcal{D}(\beta,M_{3})=\emptyset&\quad\ \ (0,2,0)\\ M_{4}=\left(\begin{smallmatrix}1&1\\ 2&1\end{smallmatrix}\right)&\mathcal{D}(\beta,M_{4})=\emptyset&\quad\ \ (0,2,1)\\ M_{5}=\left(\begin{smallmatrix}1&2\\ 1&3\end{smallmatrix}\right)&\mathcal{D}(\beta,M_{5})=\emptyset&\quad\ \ (0,1,1)\\ M_{6}=\left(\begin{smallmatrix}0&1\\ 1&1\end{smallmatrix}\right)&\mathcal{D}(\beta,M_{6})=\emptyset&\quad\ \ (0,3,1)\\ M_{7}=\left(\begin{smallmatrix}0&1\\ 1&0\end{smallmatrix}\right)&\mathcal{D}(\beta,M_{7})=(1,2)&\quad\ \ (0,3,0)\\ M_{8}=\left(\begin{smallmatrix}1&1\\ 1&0\end{smallmatrix}\right)&\mathcal{D}(\beta,M_{8})=(1,2)&\quad\ \ (0,3,0)\\ M_{9}=\left(\begin{smallmatrix}1&2\\ 1&0\end{smallmatrix}\right)&\mathcal{D}(\beta,M_{9})=(1,2)&\quad\ \ (0,3,0)\\ M_{10}=\left(\begin{smallmatrix}0&1\\ 1&3\end{smallmatrix}\right)&\mathcal{D}(\beta,M_{10})=(1,2)&\quad\ \ (1,4,2),(2,7,1)\\ M_{11}=\left(\begin{smallmatrix}1&0\\ 1&3\end{smallmatrix}\right)&\mathcal{D}(\beta,M_{11})=(1,4)&\quad\ \ (1,4,0)\\ M_{12}=\left(\begin{smallmatrix}1&1\\ 1&3\end{smallmatrix}\right)&\mathcal{D}(\beta,M_{12})=(1,\frac{5}{2})&\quad\ \ (1,4,1)\\ M_{13}=\left(\begin{smallmatrix}1&0\\ 0&1\end{smallmatrix}\right)&\mathcal{D}(\beta,M_{13})=(1,3)\cup(\frac{7}{2},4)&\quad\ \ (1,4,0),(2,2,3)\\ M_{14}=\left(\begin{smallmatrix}1&0\\ 2&1\end{smallmatrix}\right)&\mathcal{D}(\beta,M_{14})=(\frac{3}{2},4)&\quad\ \ (0,2,3),(1,4,0)\\ M_{15}=\left(\begin{smallmatrix}1&0\\ 1&1\end{smallmatrix}\right)&\mathcal{D}(\beta,M_{15})=(1,2)\cup(\frac{5}{2},4)&\quad\ \ (1,2,3),(1,4,0)\\ M_{16}=\left(\begin{smallmatrix}1&2\\ 0&1\end{smallmatrix}\right)&\mathcal{D}(\beta,M_{16})=(1,2)&\quad\ \ (1,3,1)\\ M_{17}=\left(\begin{smallmatrix}1&1\\ 0&1\end{smallmatrix}\right)&\mathcal{D}(\beta,M_{17})=(1,2)\cup(4,\frac{9}{2})&\quad\ \ (1,2,2),(2,6,1)\\ M_{18}=\left(\begin{smallmatrix}0&1\\ 1&2\end{smallmatrix}\right)&\mathcal{D}(\beta,M_{18})=\emptyset&\quad\ \ (0,3,2),(1,6,0),(2,5,2)\\ M_{19}=\left(\begin{smallmatrix}1&1\\ 1&2\end{smallmatrix}\right)&\mathcal{D}(\beta,M_{19})=(2,3)&\quad\ \ (0,1,2),(1,6,0),(2,3,2)\\ M_{20}=\left(\begin{smallmatrix}1&2\\ 1&2\end{smallmatrix}\right)&\mathcal{D}(\beta,M_{20})=(1,2)&\quad\ \ (1,3,2),(1,6,0),(3,5,2)\\ M_{21}=\left(\begin{smallmatrix}1&2\\ 1&1\end{smallmatrix}\right)&\mathcal{D}(\beta,M_{21})=(3,4)&\quad\ \ (0,2,2),(2,5,1)\\ M_{22}=\left(\begin{smallmatrix}0&1\\ 2&1\end{smallmatrix}\right)&\mathcal{D}(\beta,M_{22})=(1,3)&\quad\ \ (1,5,1)\\ M_{23}=\left(\begin{smallmatrix}1&2\\ 2&1\end{smallmatrix}\right)&\mathcal{D}(\beta,M_{23})=(1,3)&\quad\ \ (2,4,2),(3,7,1)\\ M_{24}=\left(\begin{smallmatrix}1&1\\ 1&1\end{smallmatrix}\right)&\mathcal{D}(\beta,M_{24})=(1,4)&\quad\ \ (3,4,2),(3,8,1)\end{array}

A unique strongly connected component of the graph Γβ\Gamma_{\beta} is depicted in Figure 4. Using our computer program we can see that for the Sturmian sequence 𝐮\mathbf{u} associated to θ=[0,3,1,1,1,2¯]\theta=[0,3,\overline{1,1,1,2}], the balanced sequence 𝐯=colour(𝐮,𝐲,𝐲)\mathbf{v}=\mathrm{colour}(\mathbf{u},\mathbf{y},\mathbf{y}^{\prime}) with Per𝐲=3,Per𝐲=4\mathrm{Per}\ \mathbf{y}=3,\ \mathrm{Per}\ \mathbf{y}^{\prime}=4 has E(𝐯)=43E^{*}(\mathbf{v})=\frac{4}{3}.

If we search for 𝐯=colour(𝐮,𝐲,𝐲)\mathbf{v}=\mathrm{colour}(\mathbf{u},\mathbf{y},\mathbf{y}^{\prime}) such that E(𝐯)<43E^{*}(\mathbf{v})<\frac{4}{3}, i.e., if we choose β<13\beta_{-}<\tfrac{1}{3}, then by Theorem 31 we have to exclude also the solution m=0m=0, k==3k=\ell=3, which reduces 𝒟(β,M10)\mathcal{D}(\beta_{-},M_{10}) to \emptyset, therefore no strongly connected component remains in the graph. To summarize, there is no 𝐯=colour(𝐮,𝐲,𝐲)\mathbf{v}=\mathrm{colour}(\mathbf{u},\mathbf{y},\mathbf{y}^{\prime}) with E(𝐯)<43E^{*}(\mathbf{v})<\frac{4}{3} for Per𝐲=3,Per𝐲=4\mathrm{Per}\,\mathbf{y}=3,\ \mathrm{Per}\,\mathbf{y}^{\prime}=4.

(1211)\left(\begin{array}[]{cc}1&2\\ 1&1\\ \end{array}\right)(1101)\left(\begin{array}[]{cc}1&1\\ 0&1\\ \end{array}\right)(1210)\left(\begin{array}[]{cc}1&2\\ 1&0\\ \end{array}\right)(0110)\left(\begin{array}[]{cc}0&1\\ 1&0\\ \end{array}\right)(1011)\left(\begin{array}[]{cc}1&0\\ 1&1\\ \end{array}\right)(1201)\left(\begin{array}[]{cc}1&2\\ 0&1\\ \end{array}\right)(1001)\left(\begin{array}[]{cc}1&0\\ 0&1\\ \end{array}\right)(0113)\left(\begin{array}[]{cc}0&1\\ 1&3\\ \end{array}\right)(1110)\left(\begin{array}[]{cc}1&1\\ 1&0\\ \end{array}\right)31231141311
Figure 4. The strongly connected component of the graph of admissible tails for β=13\beta=\frac{1}{3} and P=3,P=4P=3,\ P^{\prime}=4.

Let us apply first the forward reduction on the graph from Figure 4. Consider the vertex M17=(1101)M_{17}=\left(\begin{smallmatrix}1&1\\ 0&1\end{smallmatrix}\right) and the outgoing edge labeled by 4. Each prolongation to an infinite path has the next edge label equal to 1. Therefore the corresponding δ=[4,1,](92,5)\delta=[4,1,\ldots]\in(\frac{9}{2},5), i.e., δ𝒟(β,M17)=(1,2)(4,92)\delta\not\in\mathcal{D}(\beta,M_{17})=(1,2)\cup(4,\frac{9}{2}). Consequently, the edge labeled by 4 may be erased.

Next, we apply the backward reduction. Consider the vertex M21=(1211)M_{21}=\left(\begin{smallmatrix}1&2\\ 1&1\end{smallmatrix}\right). The sequence of edge labels of each ingoing path ends in 3,1,13,1,1. For m=1,k=6,=1m=1,k=6,\ell=1 satisfying M21(10m1)(k)=(00)mod(34)M_{21}\left(\begin{smallmatrix}1&0\\ m&1\end{smallmatrix}\right)\left(\begin{smallmatrix}\ell\\ k\end{smallmatrix}\right)=\left(\begin{smallmatrix}0\\ 0\end{smallmatrix}\right)\mod\left(\begin{smallmatrix}3\\ 4\end{smallmatrix}\right) we have [0,1,1,3,]J=[59,47][0,1,1,3,\ldots]\in J=[\frac{5}{9},\frac{4}{7}] and min{fm,k,(x):xJ}=2+5/97+5/9=2368>β=13\min\{f_{m,k,\ell}(x):x\in J\}=\frac{2+5/9}{7+5/9}=\frac{23}{68}>\beta=\frac{1}{3}. Therefore, by Remark 34 the triplet m=1,k=6,=1m=1,k=6,\ell=1 reduces 𝒟(β,M21)\mathcal{D}(\beta,M_{21}) to 𝒟(β,M21)=(3,4)(1,72)=(3,72)\mathcal{D}^{\prime}(\beta,M_{21})=(3,4)\cap(1,\frac{7}{2})=(3,\frac{7}{2}).

Finally, we apply once more the forward reduction. Consider again the vertex M21M_{21}, now with 𝒟(β,M21)=(3,72)\mathcal{D}^{\prime}(\beta,M_{21})=(3,\frac{7}{2}), and its unique outgoing edge labeled by 3. Each prolongation to an infinite path has the next edge label equal to 1. Therefore the corresponding δ=[3,1,](72,4)\delta=[3,1,\ldots]\in(\frac{7}{2},4), i.e., δ𝒟(β,M21)=(3,72)\delta\not\in\mathcal{D}^{\prime}(\beta,M_{21})=(3,\frac{7}{2}). Thus, the edge labeled by 3 may be erased. Consequently, the graph is reduced so that it contains a unique cycle labeled by 1,1,1,2¯\overline{1,1,1,2}, which proves that it is the only (1+13)(1+\frac{1}{3})-admissible suffix of θ\theta.

14. Asymptotic repetitive threshold for senary to denary balanced sequences

In order to find RTB(d)RTB^{*}(d), we first detect possible periods P=Per𝐲,P=Per𝐲P=\mathrm{Per}\,\mathbf{y},P^{\prime}=\mathrm{Per}\,\mathbf{y}^{\prime} which give a dd-ary balanced sequence. For this task we use Remark 18. Then some pairs are eliminated due to Proposition 21 and Lemma 23.

The starting parameter β\beta for the construction of graphs of (1+β)(1+\beta)-admissible tails can be chosen to be the value of the minimal critical exponent since by definition E(𝐯)E(𝐯)E(\mathbf{v})\geq E^{*}(\mathbf{v}) for every sequence 𝐯\mathbf{v}.

By Proposition 14 we have E(𝐯)1+1PPE^{*}(\mathbf{v})\geq 1+\frac{1}{P\,P^{\prime}} for every balanced sequence 𝐯=colour(𝐮,𝐲,𝐲)\mathbf{v}=\mathrm{colour}(\mathbf{u},\mathbf{y},\mathbf{y}^{\prime}). Therefore when searching for E(𝐯)1+βE^{*}(\mathbf{v})\leq 1+\beta we have to consider only the pairs (P,P)(P,P^{\prime}) such that 1βPP\frac{1}{\beta}\leq{P\,P^{\prime}}.

As soon as we find 𝐯\mathbf{v} with E(𝐯)<1+βE^{*}(\mathbf{v})<1+\beta, we lower β\beta to β=E(𝐯)1\beta^{\prime}=E^{*}(\mathbf{v})-1 and construct graphs of (1+β)(1+\beta^{\prime})-admissible tails for the remaining pairs (P,P)(P,P^{\prime}).

Using the above described procedure, we have found the minimal asymptotic critical exponent of dd-ary balanced sequences up to d=10d=10. They are listed in Table 1.

Table 1. The asymptotic repetitive threshold RTB(d)RTB^{*}(d) for alphabets of size d10d\leq 10 and the parameters of a dd-ary balanced sequence 𝐯=colour(𝐮,𝐲,𝐲)\mathbf{v}=\mathrm{colour}(\mathbf{u},\mathbf{y},\mathbf{y}^{\prime}) for which the threshold RTB(d)RTB^{*}(d) is reached: θ\theta is the slope of a Sturmian sequence 𝐮\mathbf{u}, P=Per𝐲P=\mathrm{Per}\,\mathbf{y} and P=Per𝐲P^{\prime}=\mathrm{Per}\,\mathbf{y}^{\prime}.
dd RTB(d)RTB^{*}(d) θ\theta PP PP^{\prime}
2 2+5+123.6180342+\frac{\sqrt{5}+1}{2}\doteq 3.618034 [0,1¯][0,\overline{1}] 1 1
3 2+122.7071072+\frac{1}{\sqrt{2}}\doteq 2.707107 [0,1,2¯][0,1,\overline{2}] 1 2
4 1+5+141.8090171+\frac{\sqrt{5}+1}{4}\doteq 1.809017 [0,1¯][0,\overline{1}] 2 2
5 32=1.5\frac{3}{2}=1.5 [0,1,2¯][0,1,\overline{2}] 2 4
6 75+365801.239835\frac{75+3\sqrt{65}}{80}\doteq 1.239835 [0,4,1,2,1,1,1¯][0,4,\overline{1,2,1,1,1}] 1 16
7 49+577641.140950\frac{49+\sqrt{577}}{64}\doteq 1.140950 [0,5,1,1,1,1,5,2¯][0,5,1,\overline{1,1,1,5,2}] 1 32
8 1+35161.0477461+\frac{3-\sqrt{5}}{16}\doteq 1.047746 [0,1¯][0,\overline{1}] 8 8
9 2120161.032992\frac{21-\sqrt{20}}{16}\doteq 1.032992 [0,1,4¯][0,1,\overline{4}] 8 16
10 3642173041.0146027\frac{364-21\sqrt{7}}{304}\doteq 1.0146027 [0,6,1,1,1,1,2,1,2,1,1,1¯][0,6,\overline{1,1,1,1,2,1,2,1,1,1}] 4 64

Let us comment properties of the graphs Γβ\Gamma_{\beta} with the optimal value β=RTB(d)1\beta=RTB^{*}(d)-1.

2d52\leq d\leq 5 :

It was sufficient to use the graphs of admissible tails without reduction, as explained in Section 12.

6d96\leq d\leq 9 :

The reduction was necessary. After reduction of the graph Γβ\Gamma_{\beta} constructed for the pair (P,P)(P,P^{\prime}) from Table 1, there is always a unique strongly connected component in the form of a cycle, i.e., the period of the continued fraction was determined uniquely by the graph. For all other pairs (P,P)(P,P^{\prime}) of admissible periods the graph has no strongly connected component. Let us point out that for 6 letters the forward reduction suffices, while for more letters both the forward and the backward reduction is needed. For 6 letters, the unique (1+β)(1+\beta)-admissible tail corresponds to the bold cycle in Figure 3.

d=10d=10 :

For (P,P)(4,64)(P,P^{\prime})\neq(4,64) the graph has no strongly connected component. For P=4,P=64P=4,P^{\prime}=64, a unique strongly connected component of Γβ\Gamma_{\beta} after reduction is depicted in Figure 5. Thus, there are two oriented cycles sharing two vertices. Hence a “human intervention” is needed to pick up the suitable continued fraction. Using our computer program we obtain: E(𝐯^)=RTB(10)E^{*}(\hat{\mathbf{v}})=RTB^{*}(10) for the balanced sequence 𝐯^=colour(𝐮^,𝐲,𝐲)\hat{\mathbf{v}}=\mathrm{colour}(\hat{\mathbf{u}},\mathbf{y},\mathbf{y}^{\prime}), where θ^=[0,6,1,1,1,1,2,1,2,1,1,1¯]\hat{\theta}=[0,6,\overline{1,1,1,1,2,1,2,1,1,1}].

(0025)\left(\begin{array}[]{cc}0&0\\ 2&5\\ \end{array}\right)(00111)\left(\begin{array}[]{cc}0&0\\ 1&11\\ \end{array}\right)(0014)\left(\begin{array}[]{cc}0&0\\ 1&4\\ \end{array}\right)(0041)\left(\begin{array}[]{cc}0&0\\ 4&1\\ \end{array}\right)(0016)\left(\begin{array}[]{cc}0&0\\ 1&6\\ \end{array}\right)(00113)\left(\begin{array}[]{cc}0&0\\ 1&13\\ \end{array}\right)(0043)\left(\begin{array}[]{cc}0&0\\ 4&3\\ \end{array}\right)(00112)\left(\begin{array}[]{cc}0&0\\ 1&12\\ \end{array}\right)112111121
Figure 5. The strongly connected component of the graph of admissible tails for β=60217304\beta=\frac{60-21\sqrt{7}}{304} and P=4,P=64P=4,\ P^{\prime}=64.

In Figure 5, θ^\hat{\theta} corresponds to an infinite path going alternatively through the right hand cycle, then the left hand cycle, and again the right hand cycle, then the left hand cycle, and so on. The following two arguments show that if θ\theta does not have a suffix corresponding to such alternation of the right and left hand cycle, then E(𝐯)>1+βE^{*}(\mathbf{v})>1+\beta. First we observe, that every path in the graph from Figure 5 uses infinitely many times the vertices named (0016)\left(\begin{smallmatrix}0&0\\ 1&6\end{smallmatrix}\right) and (0025)\left(\begin{smallmatrix}0&0\\ 2&5\end{smallmatrix}\right).

  • •:

    Let us explain that any θ\theta corresponding to an infinite path going infinitely many times twice consecutively through the left hand cycle gives rise to a colouring 𝐯\mathbf{v} with E(𝐯)>1+βE^{*}(\mathbf{v})>1+\beta.

    Since P=4,P=64P=4,P^{\prime}=64, we have H=4H=4 and Y=1Y=1, Y=16Y^{\prime}=16. Assume θ=[0,a1,a2,a3,]\theta=[0,a_{1},a_{2},a_{3},\ldots] and there exist infinitely many NN such that ANA_{N} belongs to the class with (Y,Y)(Y,Y^{\prime})-name (0016)\left(\begin{smallmatrix}0&0\\ 1&6\end{smallmatrix}\right), where we arrive to ANA_{N} using the left hand cycle and we leave ANA_{N} using again the left hand cycle. In this case δN=[aN+1,aN+2,]=[1,2,1,1,1,1,](2921,1813)\delta_{N}=[a_{N+1},a_{N+2},\ldots]=[1,2,1,1,1,1,\ldots]\in(\frac{29}{21},\frac{18}{13}) and QN1QN=[0,aN,,a2,a1]=[0,1,1,1,2,1,](1219,711)\frac{Q_{N-1}}{Q_{N}}=[0,a_{N},\ldots,a_{2},a_{1}]=[0,1,1,1,2,1,\ldots]\in(\frac{12}{19},\frac{7}{11}). Then (0016)(1419)=(00)mod(116)\left(\begin{smallmatrix}0&0\\ 1&6\end{smallmatrix}\right)\left(\begin{smallmatrix}14\\ 19\end{smallmatrix}\right)=\left(\begin{smallmatrix}0\\ 0\end{smallmatrix}\right)\mod\left(\begin{smallmatrix}1\\ 16\end{smallmatrix}\right), therefore k=419=76,=414=56k=4\cdot 19=76,\ell=4\cdot 14=56 belongs to 𝒮1(N,0){\mathcal{S}}_{1}(N,0). It belongs to 𝒮2(N,0){\mathcal{S}}_{2}(N,0) as the inequality |56δN76|<δN+1|56\delta_{N}-76|<\delta_{N}+1 is satisfied as well. Using (3) we obtain

    ΦN1+QN1QN76+56QN1QN1+121976+561219>1+β.\Phi_{N}\geq\frac{1+\frac{Q_{N-1}}{Q_{N}}}{76+56\frac{Q_{N-1}}{Q_{N}}}\geq\frac{1+\frac{12}{19}}{76+56\frac{12}{19}}>1+\beta.

    By Proposition 20 E(𝐯)=1+lim supΦN>1+βE^{*}(\mathbf{v})=1+\limsup\Phi_{N}>1+\beta.

  • •:

    Let us show that also any θ\theta corresponding to an infinite path going infinitely many times twice consecutively through the right hand cycle gives rise to a colouring 𝐯\mathbf{v} with E(𝐯)>1+βE^{*}(\mathbf{v})>1+\beta.

    Assume there exist infinitely many NN such that ANA_{N} belongs to the class with (Y,Y)(Y,Y^{\prime})-name (0025)\left(\begin{smallmatrix}0&0\\ 2&5\end{smallmatrix}\right), where we leave ANA_{N} using twice consecutively the right hand cycle. Hence δN=[aN+1,aN+2,]=[1,1,1,2,1,1,1,1,2,1,](305193,177112)\delta_{N}=[a_{N+1},a_{N+2},\ldots]=[1,1,1,2,1,1,1,1,2,1,\ldots]\in(\frac{305}{193},\frac{177}{112}) and QN1QN=[0,aN,,a2,a1]=[0,1,a,]\frac{Q_{N-1}}{Q_{N}}=[0,a_{N},\ldots,a_{2},a_{1}]=[0,1,a,\ldots], where a=1a=1 or a=2a=2, i.e., QN1QN(12,34)\frac{Q_{N-1}}{Q_{N}}\in(\frac{1}{2},\frac{3}{4}). The solution k=72,=44k=72,\ell=44 belongs to 𝒮(N,0)=𝒮1(N,0)𝒮2(N,0){\mathcal{S}}(N,0)={\mathcal{S}}_{1}(N,0)\cap{\mathcal{S}}_{2}(N,0). Using (3) we obtain

    ΦN1+QN1QN72+44QN1QN1+1272+4412>1+β.\Phi_{N}\geq\frac{1+\frac{Q_{N-1}}{Q_{N}}}{72+44\frac{Q_{N-1}}{Q_{N}}}\geq\frac{1+\frac{1}{2}}{72+44\frac{1}{2}}>1+\beta.

    By Proposition 20 E(𝐯)=1+lim supΦN>1+βE^{*}(\mathbf{v})=1+\limsup\Phi_{N}>1+\beta.

To sum up, only θ\theta having the continued fraction with the same period as θ^=[0,6,1,1,1,1,2,1,2,1,1,1¯]\hat{\theta}=[0,6,\overline{1,1,1,1,2,1,2,1,1,1}] can give the minimal value of E(𝐯)E^{*}(\mathbf{v}).

15. Comments and open questions

Let us first make some comments on complexity of computation of the asymptotic repetitive threshold. We were not able to determine RTB(d)RTB^{*}(d) for d11d\geq 11 with our computer because of time complexity of the computation. The bigger dd, the larger the number of pairs (P,P)(P,P^{\prime}) that have to be treated. The number of such pairs grows exponentially. Moreover, even the number of vertices in the graph grows exponentially with dd. For example, for every dd we have to consider P=1P=1 and P=2d2P^{\prime}=2^{d-2}. The number of vertices in the corresponding graph Γβ\Gamma_{\beta} = the number of classes of equivalence \equiv is equal to 32d33\cdot 2^{d-3}.

The closer the value 1+β1+\beta to RTB(d)RTB^{*}(d) the more time consuming determination of 𝒟(β,A)\mathcal{D}(\beta,A). The reason is the fact that more triplets m,k,m,k,\ell satisfy Properties 𝔓1\mathfrak{P}1 and 𝔓3\mathfrak{P}3. Hence, it would be useful to have a good upper bound on RTB(d)RTB^{*}(d) so that we do not have to repeat the computation for more values β\beta.

Time complexity is not the only obstacle to revealing RTB(d)RTB^{*}(d) for d11d\geq 11. We are afraid that for larger dd our method will not give a unique period of the continued fraction corresponding to the minimal E(𝐯)E^{*}(\mathbf{v}) and that “human intervention” will be needed similarly as in the case d=10d=10.

Let us conclude with some other open problems connected to the topic.

  • We conjecture that RTB(d)<1+qdRTB^{*}(d)<1+q^{d} for some positive q<1q<1. But from the obtained values RTB(d)RTB^{*}(d), we are not able to derive a conjecture on the precise value RTB(d)RTB^{*}(d). We observe at least that if (P,P)(P,P^{\prime}) is an optimal pair of periods for dd even, then (P,2P)(P,2P^{\prime}) is optimal for d+1d+1. It works for d=2,4,6,8d=2,4,6,8.

  • We believe that RTB(d)RTB^{*}(d) always belongs to a quadratic field, but we have no proof of this fact.

  • RTB(d)RTB^{*}(d) is defined to be inf{E(𝐮):𝐮 d-ary balanced}\inf\{E^{*}(\mathbf{u}):\mathbf{u}\text{ \ $d$-ary balanced}\}. We can see for d10d\leq 10 that RTB(d)RTB^{*}(d) is minimum of the set and it is not its accumulation point. The question is whether RTB(d)RTB^{*}(d) may be an accumulation point. And if not, what is the second, third, etc. smallest element of the set.

References