A $q$-analog of the Markoff injectivity conjecture holds

Abstract

The elements of Markoff triples are given by coefficients in certain matrix products defined by Christoffel words, and the Markoff injectivity conjecture, a long-standing open problem, is then equivalent to injectivity on Christoffel words. A $q$-analog of these matrix products has been proposed recently by Leclere and Morier-Genoud, and we prove that injectivity on Christoffel words holds for this $q$-analog. The proof is based on the evaluation at $q = \exp(2\pi i/6)$. Other roots of unity provide some information on the original problem, which corresponds to the case $q=1$. We also extend the problem to arbitrary words and provide a large family of pairs of words where injectivity does not hold.

Full text

AQ-ANALOG OF THE MARKOFF
INJECTIVITY CONJECTURE HOLDS
S´
EBASTIEN LABB´
E, M´
ELODIE LAPOINTE, AND WOLFGANG STEINER
Abstract. The elements of Markoff triples are given by coefficients in cer-
tain matrix products defined by Christoffel words, and the Markoff injectivity
conjecture, a long-standing open problem, is then equivalent to injectivity on
Christoffel words. A q-analog of these matrix products has been proposed
recently by Leclere and Morier-Genoud, and we prove that injectivity on
Christoffel words holds for this q-analog. The proof is based on the evalu-
ation at q= exp(2πi/6). Other roots of unity provide some information on
the original problem, which corresponds to the case q= 1. We also extend the
problem to arbitrary words and provide a large family of pairs of words where
injectivity does not hold.
1. Introduction
Christoffel words are words over the alphabet {0,1}defined recursively as follows:
0,1and 01 are Christoffel words and if u, v, uv ∈ {0,1}∗are Christoffel words then
uuv and uvv are Christoffel words [BLRS09]. The shortest Christoffel words are:
0,1,01,001,011,0001,00101,01011,0111,00001,0001001,00100101,0010101,· · ·
Note that these are usually named lower Christoffel words.
A Markoff triple is a positive solution of the Diophantine equation x2+y2+z2=
3xyz [Mar80, Mar79]. Markoff triples can be defined recursively as follows: (1,1,1),
(1,2,1) and (1,5,2) are Markoff triples and if (x, y, z) is a Markoff triple with y≥x
and y≥z, then (x, 3xy −z, y) and (y, 3yz −x, z) are Markoff triples. A list of small
Markoff numbers (elements of a Markoff triple) is
1,2,5,13,29,34,89,169,194,233,433,610,985,1325,1597,2897,4181,· · ·
referenced as sequence A002559 in OEIS [OEI22].
It is known that each Markoff number can be expressed in terms of a Christoffel
word. More precisely, let µbe the monoid homomorphism {0,1}∗→GL2(Z)
defined by
µ(0) = 2 1
1 1and µ(1) = 5 2
2 1.
Each Markoff number is equal to µ(w)12 for some Christoffel word w[Reu09], where
M12 denotes the element above the diagonal in a matrix M=M11 M12
M21 M22 ∈GL2(Z).
2020 Mathematics Subject Classification. Primary 11J06; Secondary 68R15 and 05A30.
Key words and phrases. Markoff number; Christoffel word; q-analog.
This work was supported by the Agence Nationale de la Recherche through the project Codys
(ANR-18-CE40-0007). The second author acknowledges the support of the Natural Sciences and
Engineering Research Council of Canada (NSERC), [funding reference number BP–545242–2020]
and the support of the Fonds de Recherche du Qu´ebec en Science et Technologies.
1
arXiv:2212.09852v1 [math.CO] 19 Dec 2022

2 S. LABB´
E, M. LAPOINTE, AND W. STEINER
For example, the Markoff number 194 is associated with the Christoffel word
00101 as it is the entry at position (1,2) in the matrix
µ(00101) = 2 1
1 12 1
1 15 2
2 12 1
1 15 2
2 1=463 194
284 119.
Whether the map w7→ µ(w)12 provides a bijection between Christoffel words and
Markoff numbers is a question (stated differently in [Fro13]) that has remained open
for more than 100 years [Aig13]. The conjecture can be expressed in terms of the
injectivity of the map w7→ µ(w)12 [Reu19, §3.3].
Conjecture 1.1 (Markoff Injectivity Conjecture).The map w7→ µ(w)12 is injec-
tive on the set of Christoffel words.
In [LL22], a q-analog of the Markoff Injectivity Conjecture was considered based
on the q-analog of µ(0) and µ(1) proposed in [LMG21], which in terms of
Lq=q0
q1and Rq=q1
0 1,
can be written as
µq(0) = RqLq=q+q21
q1,
µq(1) = RqRqLqLq=q+ 2q2+q3+q41 + q
q+q21.
It extends to a morphism of monoids µq:{0,1}∗→GL2(Z[q±1]). The q-analog
satisfies that µ1(w) = µ(w) for every w∈ {0,1}∗. Thus if wis a Christoffel word,
then the entry above the diagonal µq(w)12 is a polynomial of indeterminate qwith
nonnegative integer coefficients such that it is a Markoff number when evaluated
at q= 1. For example,
µq(00101)12 = 1+4q+10q2+18q3+27q4+33q5+33q6+29q7+21q8+12q9+5q10 +q11
which, when evaluated at q= 1, is equal to
µ1(00101)12 = 1 + 4 + 10 + 18 + 27 + 33 + 33 + 29 + 21 + 12 + 5 + 1 = 194.
In this work, we prove the q-analog of the Markoff Injectivity Conjecture.
Theorem 1.2. The map w7→ µq(w)12 is injective on the set of Christoffel words.
Theorem 1.2 is proved in Section 2. In Section 3, we give examples where the
map w7→ µq(w)12 is not injective when considered on the language {0,1}∗.
2. Proof of Theorem 1.2
The main idea of this section is to evaluate the polynomial µq(w)12 at primitive
root of unity ζk= exp(2πi/k), in particular when k= 6.
First, we observe that when w∈ {0,1}∗, the matrix µζ6(w) can be expressed in
terms of ζ6, the length |w|of wand the number |w|1of occurrences of 1in w.
Lemma 2.1. For every w∈ {0,1}∗, we have
(1) µζ6(w) = ζ|w|+|w|1
6 |w| −|w|−|w|1
−|w|1−|w|ζ6+|w|1|w|
|w|+|w|1−|w|1+1 0
0 1.

AQ-ANALOG OF THE MARKOFF INJECTIVITY CONJECTURE HOLDS 3
ζ6
1
|w|+|w|1≡2
mod 6
|w|+|w|1≡3
mod 6
|w|+|w|1≡4
mod 6
|w|+|w|1≡5
mod 6
|w|+|w|1≡0
mod 6
|w|+|w|1≡1
mod 6
µq(011)12
Figure 1. A partition of the complex plane C\ {0}into six dis-
joint cones spanned by the vectors ζk
6and ζk+1
6,k∈ {0,1,2,3,4,5}.
For every w∈ {0,1}∗\ {ε},µq(w)12 lies in the cone corresponding
to |w|+|w|1mod 6. For instance, µq(011) = ζ5
6(3−5ζ6)=3ζ5
6−5
and |w|+|w|1≡5 mod 6.
Proof. The proof is done by recurrence on the length of w. We have µζ6(ε) = ( 1 0
0 1 ).
Thus, the formula works for w=ε. If (1) holds for w, then we have
µζ6(w0) = ζ|w|+|w|1
6|w|1+1+|w|ζ6|w| − (|w|+|w|1)ζ6
|w|+|w|1− |w|1ζ61− |w|1− |w|ζ6ζ61+ζ61−ζ6
1 1−ζ6
=ζ|w|+|w|1+1
6|w|1+ 1 + (|w|+1) ζ6|w|+ 1 −(|w|+|w|1+1) ζ6
|w|+|w|1+ 1 − |w|1ζ61− |w|1−(|w|+1) ζ6,
µζ6(w1) = ζ|w|+|w|1
6|w|1+1+|w|ζ6|w| − (|w|+|w|1)ζ6
|w|+|w|1− |w|1ζ61− |w|1− |w|ζ6ζ2
62+ζ61−2ζ6
2−ζ6−ζ6
=ζ|w|+|w|1+2
6|w|1+ 2 + (|w|+1) ζ6|w|+ 1 −(|w|+|w|1+2) ζ6
|w|+|w|1+ 2 −(|w|1+1) ζ6−|w|1−(|w|+1) ζ6,
hence (1) holds for w0and w1.
In particular, Equation (1) implies that the entry above the diagonal is
(2) µζ6(w)12 =ζ|w|+|w|1
6|w| − (|w|+|w|1)ζ6∈C.
The next result shows that when w∈ {0,1}∗\ {ε}, the number µζ6(w)12 lies in one
of the six cones of angle π
3that partition the complex plane according to the value
of |w|+|w|1, see Figure 1.
Lemma 2.2. For every w∈ {0,1}∗\ {ε}, we have
µζ6(w)12 ∈ρ·eiθ |ρ > 0,(|w|+|w|1+ 4)π
3< θ ≤(|w|+|w|1+ 5) π
3.
Moreover, w=εif and only if µζ6(w)12 = 0.
Proof. Let w∈ {0,1}∗\ {ε}. Since |w|+|w|1≥ |w|>0, then observe that
|w| − (|w|+|w|1)ζ6∈ρ·eiθ |ρ > 0,4π
3< θ ≤5π
3.

4 S. LABB´
E, M. LAPOINTE, AND W. STEINER
Since ζ6=eiπ
3, from Equation (2), we have
µζ6(w)12 =ζ|w|+|w|1
6|w| − (|w|+|w|1)ζ6
∈eiπ
3(|w|+|w|1)ρ·eiθ |ρ > 0,4π
3< θ ≤5π
3
=ρ·eiθ |ρ > 0,(|w|+|w|1+4)π
3< θ ≤(|w|+|w|1+5)π
3.
We have µζ6(w)12 = 0 if w=εand, from above, µζ6(w)12 6= 0 if w∈ {0,1}∗\{ε}.
Thus, if µζ6(w)12 = 0, then w=ε.
The next result shows that we can recover the number of 0’s and 1’s occurring
in a word w∈ {0,1}∗from the the polynomial µq(w)12 evaluated at q=ζ6.
Proposition 2.3. Let w, w0∈ {0,1}∗. If µζ6(w)12 =µζ6(w0)12, then |w|0=|w0|0
and |w|1=|w0|1.
Proof. If µζ6(w)12 =µζ6(w0)12 = 0, then from Lemma 2.2, we have w=ε=w0,
thus |w|0=0=|w0|0and |w|1=0=|w0|1. Now, assume that µζ6(w)12 =
µζ6(w0)12 6= 0. From Lemma 2.2, we have
µζ6(w)12 ∈ρ·eiθ |ρ > 0,(|w|+|w|1+ 4)π
3< θ ≤(|w0|+|w0|1+ 5) π
3,
µζ6(w0)12 ∈ρ·eiθ |ρ > 0,(|w0|+|w0|1+ 4)π
3< θ ≤(|w0|+|w0|1+ 5) π
3,
which are two disjoint cones in the complex plane when |w|+|w|16≡ |w0|+|w0|1
mod 6. Since µζ6(w)12 =µζ6(w0)12 , the two cones must intersect and be equal.
Therefore, we have |w|+|w|1≡ |w0|+|w0|1mod 6. From Lemma 2.1, we have
|w0| − (|w0|+|w0|1)ζ6=ζ−|w0|−|w0|1
6µζ6(w0)12
=ζ−|w0|−|w0|1
6µζ6(w)12
=ζ−|w0|−|w0|1
6ζ|w|+|w|1
6|w| − (|w|+|w|1)ζ6
=|w| − (|w|+|w|1)ζ6.
This implies that |w0|=|w|and |w0|+|w0|1=|w|+|w|1. Then |w|1=|w0|1and
|w|0=|w|−|w|1=|w0|−|w0|1=|w0|0.
We may now prove the main result.
Proof of Theorem 1.2. We want to show the injectivity of the map w7→ µq(w)12
over the set of Christoffel words. Let w, w0∈ {0,1}∗be two Christoffel words
such that µq(w)12 =µq(w0)12. In particular, we have µζ6(w)12 =µζ6(w0)12. From
Proposition 2.3, |w|0=|w0|0and |w|1=|w0|1. Thus w=w0because there exists
a unique Christoffel word with a fixed number of 0’s and 1’s. Therefore the map
w7→ µζ6(w)12 is injective over the set of Christoffel words, and so is the map
w7→ µq(w)12.
Remark 2.4. Note that the monoid generated by µζk(0) and µζk(1) is a finite
group when k∈ {2,3,4,5}, and we observe the following relations between the
residue class of µ1(w)12 (mod k) and µζk(w)12 for w∈ {0,1}∗(we leave the proof
to the reader):
•µ1(w)12 (mod 2) ≡(0 if and only if µ−1(w)12 = 0,
1 if and only if µ−1(w)12 ∈ {−1,1},

AQ-ANALOG OF THE MARKOFF INJECTIVITY CONJECTURE HOLDS 5
•µ1(w)12 (mod 3) ≡




0 if and only if µζ3(w)12 = 0,
1 if and only if µζ3(w)12 ∈ {1, ζ3, ζ2
3},
2 if and only if µζ3(w)12 ∈ {−1,−ζ3,−ζ2
3},
•µ1(w)12 (mod 4) ≡




0 if and only if µi(w)12 = 0,
1 or 3 if and only if µi(w)12 ∈ {±1,±i},
2 if and only if µi(w)12 ∈ {1±i, −1±i}.
The value µ1(w)12 mod 5 can also be deduced from µζ5(w)12, which takes 31 dis-
tinct values in the complex plane, see Figure 2. For k≥6, the monoid generated by
µζk(0) and µζk(1) is infinite, and we have not found relations between the residue
class of µ1(w)12 (mod k) and µζk(w)12.
1.5 1.0 0.5 0.5 1.0 1.5
1.5
1.0
0.5
0.5
1.0
1.5
A
0
A
1
A
2
A
3
A
4
Figure 2. For w∈ {0,1}∗,µζ5(w)12 takes 31 different values. The
set Ak={µζ5(w)12 |µ1(w)12 ≡kmod 5, w ∈ {0,1}∗}consists
of the vertices of a regular pentagon when k∈ {1,2,3,4}, of the
vertices of a regular decagon and the origin when k= 0.
3. w7→ µq(w)12 is not injective on {0,1}∗
In this section, we provide a list of pairs of words over the alphabet {0,1}for
which w7→ µq(w)12 is not injective. For example, 00011 and 01001 have the same
image as we have
µq(00011)12 = 1+4q+10q2+19q3+27q4+33q5+34q6+29q7+21q8+12q9+5q10+q11
=µq(01001)12.

6 S. LABB´
E, M. LAPOINTE, AND W. STEINER
The section contains two results: Theorem 3.1 and Theorem 3.2. All pairs of
words we know of are of form of Equation (4) or Equation (7). So we believe they
completely describe the pairs of words x, y ∈ {0,1}∗such that µq(x)12 =µq(y)12.
3.1. First result. To state the results, we need the two involutions w7→ ewand
w7→ won {0,1}∗which are defined by ew=wk· · · w1and w=wk· · · w1if w=
w1· · · wk, with 0=1,1=0, i.e., ewis the mirror image of wand wis obtained
from ewby exchanging 0and 1. Also, more generally, we consider images of the
homomorphism
Mq:{0,1}∗→GL2(Z[q±1]),07→ Lq,17→ Rq
which will be used to prove identities for µqsince µq(0) = Mq(10) and µq(1) =
Mq(1100).
Theorem 3.1. For all w∈ {0,1}∗,k, m, n ≥0, we have
Mq0k1w10m12 =Mq0k1w10n12,(3)
µq(0w1)12 =µq(0ew1)12 .(4)
Proof. We have Mq(0kw0m)12 =qkMq(w)12 for all w∈ {0,1}∗,k, m ≥0, because
(1,0) Lq= (q, 0) and Lqt
(1,0) = t
(1,0). Hence, it suffices to prove (3) for k=m=
n= 0. Since
QqLqQ−1
q=t
Rq,and QqRqQ−1
q=t
Lq,with Qq=q0
0 1,
we have, for w=w1· · · w`∈ {0,1}∗,
(5) QqMq(w)Q−1
q=t
Mq(w1)· · · t
Mq(w`) = t
Mq(w`· · · w1) = t
Mq(w)
and thus
Mq(1w1)12 =RqQ−1
q
t
Mq(w)QqRq12 = (1,1) t
Mq(w)t
(q, 1) = (q, 1) Mq(w)t
(1,1)
=Mq(1w1)12,
using that 1 ×1 matrices are invariant under transposition. This proves (3).
Let σ:{0,1}∗→ {0,1}∗be the homomorphism given by σ(0) = 10 and σ(1) =
1100. Then we have µq(w) = Mq(σ(w)) and σ(w) = σ(ew) for all w∈ {0,1}∗, thus
µq(0w1)12 =Mq10σ(w)110012 =Mq10σ(w)110012 =Mq10σ(ew)110012
=µq(0ew1)12 ,
where we have used (3) and 0w1=0w1for the second equation. 
Recall that if 0w1∈ {0,1}∗is a Christoffel word, then wis a palindrome.
Therefore Theorem 3.1 is compatible with Theorem 1.2.
3.2. Second result. We obtain more identities by images of the homomorphisms
(with w∈ {0,1}∗)
ϕw:{0,1,2,3}∗→ {0,1}∗,07→ w0110w0110,27→ w0110w1001,
17→ w1001w1001,37→ w1001w0110,
ψw:{0,1,2,3}∗→ {0,1}∗,07→ w01 ew01,27→ w01 ew10,
17→ w10 ew10,37→ w10 ew01.
We extend the involution w7→ wto {0,1,2,3}∗by setting 2=3and 3=2.

AQ-ANALOG OF THE MARKOFF INJECTIVITY CONJECTURE HOLDS 7
Theorem 3.2. For all w∈ {0,1}∗,v∈ {0,1,2,3}∗,k, m, n ≥0, we have
Mq0k1ϕw(v)w10m12 =Mq0k1ϕw(v)w10n12.(6)
µq0ψw(v)w112 =µq0ψw(v)w112.(7)
For the proof of the theorem, we decompose ϕw=ηw◦τwith
ηw:{0,1,2,3}∗→ {0,1}∗,07→ w0110,27→ w0110,
17→ w1001,37→ w1001,
τ:{0,1,2,3}∗→ {0,1,2,3}∗,07→ 02,27→ 12,
17→ 13,37→ 03,
and we use the homomorphism
η0
w:{0,1,2,3}∗→ {0,1}∗,07→ 0110w, 27→ 0110w,
17→ 1001w, 37→ 1001w,
satisfying ϕw(v)w=ηw(τ(v))w=wη0
w(τ(v)). We have to show that the difference
∆w(v) = Mq1ηw(v)w112 −Mq1wη0
w(v)112
is zero for all v∈τ({0,1,2,3}∗) = {02,03,12,13}∗= ({0,1}{2,3})∗.
Lemma 3.3. Let a∈ {2,3},v∈({0,1}{2,3})∗,w∈ {0,1}∗. If ∆w(v)=0, then
∆w(u0uav)=∆w(u1uav)
for all u∈({0,1}{2,3})∗and
∆w(u2uav)=∆w(u3uav)
for all u∈({0,1}{2,3})∗{0,1}.
Proof. Assume first that |u|is even. Then
∆w(u0¯uav)−∆w(u1¯uav)
=Mq1ηw(u0¯uav)w112 −Mq1ηw(u1¯uav)w112
+Mq1wη0
w(¯v¯au0¯u)112 −Mq1wη0
w(¯v¯au1¯u)112
=Mq1ηw(u)wMq(0110)−Mq(1001)Mqηw(¯uav)w112
+Mq1wη0
w(¯v¯au)Mq(0110)−Mq(1001)Mqwη0
w(¯u)112
= (q3+ 1) Mq1ηw(u)wSQqMqηw(¯uav)w112
+ (q3+ 1) Mq1wη0
w(¯v¯au)SQqMqwη0
w(¯u)112
= (q3+ 1) q|ηw(u)w|RqSQqMqw−1ηw(av)w112
+ (q3+ 1) q|ηw(u)w|Mq1wη0
w(¯v¯a)w−1SQqRq12
= (q3+ 1) q|ηw(u)w|daMq1ηw(v)w112 −Mq1wη0
w(¯v)112 
= (q3+ 1) q|ηw(u)w|da∆w(v),
with d2=−qand d3=q4. Here, we use for the third equation that
Mq(0110) = Mq(1001)+(q3+1) SQq,where S=0−1
1 0 , Qq=q0
0 1.
For the fourth equation, we use that, by (5),
Mq(z)S QqMq(z) = Mq(z)St
Mq(z)Qq= det(Mq(z)) SQq=q|z|SQq

8 S. LABB´
E, M. LAPOINTE, AND W. STEINER
for all z∈ {0,1}∗, in particular for z=ηw(u)w(with z=ηw(u)w) and for z=
wη0
w(u) (with z=wη0
w(u)). For the fifth equation, we use that
(1,0) RqSQqMq(w−1ηw(2)) = (1,0) RqSQqMq(0110) = −(q2, q) = −q(1,0) Rq,
(1,0) RqSQqMq(w−1ηw(3)) = (1,0) RqSQqMq(1001)=(q5, q4) = q4(1,0) Mq(1),
Mq(η0
w(3)w−1SQqRqt
(0,1) = Mq(1001)SQqRqt
(0,1) = t
(q, q) = q Mq(1)t
(0,1),
Mq(η0
w(2)w−1SQqRqt
(0,1) = Mq(0110)SQqRqt
(0,1) = −t
(q4, q4) = −q4Rqt
(0,1).
Therefore, ∆w(v) = 0 implies that ∆w(u0¯uav)=∆w(u1¯uav).
The proof of ∆w(u2¯uav)=∆w(u3¯uav) for odd |u|runs along the same lines. 
Lemma 3.4. For all v∈({0,1}{2,3})∗,w∈ {0,1}∗, we have ∆w(v) = 0.
Proof. We proceed by induction on the length of v. The statement is trivially true
for |v|= 0. Supose that it is true up to length k−1 and consider it for length k.
We claim that the value of ∆w(v1· · · v2k) does not depend on the choice of
v1· · · vj, for any j≤k. The claim is true for j= 1, by Lemma 3.3 with u=εand
the induction hypothesis. If the claim is true up to j−1, then it gives together with
Lemma 3.3, for any u1· · · uj∈({0,1}{2,3})∗∪({0,1}{2,3})∗{0,1}, that
∆w(u1· · · ujvj+1 · · · v2k)=∆w(vj+1 · · · v2j−1ujvj+1 · · · v2k)
= ∆w(vj+1 · · · v2j−1vjvj+1 · · · v2k)=∆w(v1· · · v2k).
This proves the claim.
Since ηw(uu)w=wη0
w(uu) for all u∈({0,1}{2,3})∗∪({0,1}{2,3})∗{0,1}, we
have ∆w(vk+1 · · · v2kvk+1 · · · v2k) = 0, thus ∆w(v1· · · v2k) = 0 for all v1· · · v2k∈
({0,1}{2,3})∗.
Proof of Theorem 3.2. As for (3), it suffices to prove (6) for k=m=n= 0. Since
ϕw(v) = ηw(τ(v)) and ϕw(v)w=wη0
w(τ(v)) for all w∈ {0,1}∗,v∈ {0,1,2,3}∗,
Lemma 3.4 implies that (6) holds.
Let σbe as in the proof of Theorem 3.1. Then
µq0ψw(v)w112 =Mq(1η0σ(w)1(v)0σ(w)1100) = Mq(1η0σ(w)1(v)0σ(w)1100)
=µq0ψw(v)w112,
using that 0σ(ψw(v))1=η0σ(w)1(v), and using (6) for the second equation. 
The equation Mq(x)12 =Mq(y)12 has many solutions x, y ∈ {0,1}∗which are
not of the form of Equation (3) or (6), for example
Mq(110000011)12 = 1+2q+3q2+4q3+4q4+4q5+3q6+2q7+q8=Mq(100010001)12,
but we believe that Equations (4) and (7) are complete.
Question 3.5. Do there exist x, y ∈ {0,1}∗satisfying µq(x)12 =µq(y)12 which are
not given by Equation (4) or (7)?
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(S. Labb´e) Univ. Bordeaux, CNRS, Bordeaux INP, LaBRI, UMR 5800, F-33400, Tal-
ence, France
Email address:sebastien.labbe@labri.fr
URL:http://www.slabbe.org/
(M. Lapointe) Universit´
e de Moncton, D´
epartement de math´
ematiques et de statis-
tique, 18 avenue Antonine-Maillet, Moncton NB E1A 3E9, Canada
Email address:melodie.lapointe@umoncton.ca
URL:http://lapointemelodie.github.io/
(W. Steiner) Universit´
e Paris Cit´
e, CNRS, IRIF, F–75006 Paris, France
Email address:steiner@irif.fr
URL:https://www.irif.fr/~steiner