Guo-Niu Han
email: guoniu.han@unistra.fr

THE THUE-MORSE CONTINUED FRACTIONS IN CHARACTERISTIC 2 ARE ALGEBRAIC

Keywords: 2010 Mathematics Subject Classification. 11J70, 11A55, 11J61, 11T55 Power series field, Continued fraction, Transcendence

Let a, b be distinct, non-constant polynomials in F2 [z]. Let ξ a,b be the power series in F2((z -1 )) whose sequence of partial quotients is the Thue-Morse sequence over {a, b}. We establish that ξ a,b is algebraic of degree 4.

Introduction

An infinite sequence (a n ) n≥0 over a finite alphabet A is k-automatic for an integer k ≥ 2 if it can be generated by a finite automaton which reads the representation in base k of a non-negative integer n from right to left and outputs an element a n in A. It is called automatic if it is k-automatic for some integer k ≥ 2. In 1968, Cobham [START_REF] Aggarwal | Transference theorems in the field of formal power series[END_REF] conjectured that, for any integer b ≥ 2, the base-b expansion of an irrational real number never forms an automatic sequence. This was confirmed in 2007 by Adamczewski and Bugeaud [START_REF] Adamczewski | On the complexity of algebraic numbers I. Expansions in integer bases[END_REF] (see [3,30] for alternative proofs), who also established that if the Hensel expansion of an irrational p-adic number ξ is an automatic sequence, then ξ is transcendental. In a similar spirit, Bugeaud [10] proved that the sequence of partial quotients of an algebraic real number of degree at least 3 never forms an automatic sequence.

Analogous questions can be asked for power series over a finite field, but the answers are different. Throughout the paper, we let q denote a power of a prime number p and F q denote the field with q elements. In 1979 Christol [START_REF] Adamczewski | A short proof of the transcendence of the Thue-Morse continued fractions[END_REF]13] established that a power series in F q ((z -1 )) is algebraic over F q (z) if and only if the sequence of its coefficients is q-automatic (or, equivalently, is p-automatic).

In analogy with the continued fraction algorithm for real numbers, there is a well-studied continued fraction algorithm for power series in F q ((z -1 )), the partial quotients being nonconstant polynomials in F q [z]. In both settings, eventually periodic expansions correspond to quadratic elements, but much more is known on the continued fraction expansion of algebraic power series than on that of algebraic real numbers. In 1949 Mahler [START_REF] Allouche | Sur des points fixes de morphismes d'un monoïde libre[END_REF] established that, for p ≥ 3, the root z -1 + z -p + z -p 2 + . . . in F q ((z -1 )) of the polynomial zX p -zX + 1 has unbounded partial quotients, in the sense that their degrees are unbounded. In the opposite direction, there exist power series which are algebraic of degree at least 3 and have bounded partial quotients (recall that analogous statements have not yet been proved, nor disproved, for real numbers): in 1976, Baum and Sweet [START_REF] Baum | Continued fractions of algebraic power series in characteristic 2[END_REF] proved that the continued fraction expansion of the unique solution ξ BS in F 2 ((z -1 )) of the equation zX 3 + X + z = 0 has all its partial quotients of degree at most 2.

At the end of the 80s, Mendès France asked whether the sequence of partial quotients of an algebraic power series in F q ((z -1 )) is q-automatic, as soon as it takes only finitely many different values. A positive answer has been given by Allouche [5] and Allouche,Bétréma,and Shallit [6] for the algebraic power series of degree at least 3 in F p ((z -1 )) (here, p ≥ 3) which have been constructed by Mills and Robbins [27] and whose partial quotients are polynomials of degree one; see also . However, in 1995 Mkaouar [START_REF] Allouche | Perfect linear complexity profile and apwenian sequences[END_REF] (see also Yao [START_REF] Yao | Critères de non-automaticité et leurs applications[END_REF] for an alternative proof) gave a negative answer to the question of Mendès France by establishing that the sequence of partial quotients of the Baum-Sweet power series ξ BS is morphic, but not automatic. Recall that a sequence is called morphic if it is the image under a coding of a fixed point of a substitution. If the substitution can be chosen of constant length k, then the sequence is k-automatic. The equivalence between the two definitions of automatic sequences given here was established by Cobham [15]; see [7].

All this shows that the sequence of partial quotients of an algebraic power series may or may not be automatic. Conversely, very little is known about the Diophantine nature of a power series whose sequence of partial quotients is automatic, but not ultimately periodic. The only contribution to this question is a recent work of Hu and Han [START_REF] Hu | On the algebraicity of Thue-Morse and period-doubling continued fractions[END_REF]. They proved that, for any distinct nonconstant polynomials a and b in F 2 [z] whose sum of degrees is at most 7, the power series whose sequence of partial quotients is the Thue-Morse sequence written over {a, b} is algebraic of degree 4. Recall that the Thue-Morse word t = t 0 t 1 t 2 . . . = abbabaabbaababbabaababbaabbabaab . . . over {a, b} is defined by t 0 = a, t 2k = t k and t 2k+1 = a (resp., b) if t k = b (resp., a), for k ≥ 0. The Thue-Morse sequence is the most famous automatic sequence and is not ultimately periodic.

To establish their result, Hu and Han made use of a Guess 'n' Prove method and implemented a program which takes a pair (a, b) of distinct non-constant polynomials as input and outputs its minimal defining polynomial and a complete proof. The time needed grows with the degrees of a and b. Hu and Han conjectured that their result holds more generally for every pair (a, b) of distinct non-constant polynomials.

In the present paper, built on [START_REF] Hu | On the algebraicity of Thue-Morse and period-doubling continued fractions[END_REF], we confirm this conjecture. Our main results are stated in Section 2 and proved in Sections 4 and 5. We consider higher degree exponents of approximation in Section 3. Additional remarks are gathered in Section 6.

Results

Let a and b be distinct non-constant polynomials in F q [z] and let ξ q,a,b denote the power series in F q ((z -1 )) whose sequence of partial quotients is the Thue-Morse sequence t over the alphabet {a, b}. Since t is not ultimately periodic, ξ q,a,b is transcendental or algebraic of degree at least 3 over F q (z). Furthermore, as t begins with infinitely palindromes, an argument given in the proof of Theorem 3.2 shows that ξ q,a,b and its square are very well simultaneously approximable by rational fractions with the same denominator and that, consequently, ξ q,a,b cannot be algebraic of degree 3. This proves that ξ q,a,b is transcendental or algebraic of degree at least 4 over F q (z).

Our main result asserts that, in the case q = 2, any Thue-Morse continued fraction ξ 2,a,b = ξ a,b is algebraic of degree 4 over F 2 (z). be the power series in F 2 ((z -1 )) whose sequence of partial quotients is the Thue-Morse sequence t over the alphabet {a, b}. Then, ξ a,b is algebraic of degree 4 over F 2 (z). More precisely, setting

A 0 = b(a + b)(a 2 b 2 + a 2 + b 2 ) + a 2 b 4 , A 1 = ab(a + b)(a 2 b 2 + a 2 + b 2 ), A 2 = a 2 b 2 (a 2 b 2 + a 2 + b 2 ), A 3 = A 1 , A 4 = a(a + b)(a 2 b 2 + a 2 + b 2 ) + a 2 b 4 , we have A 4 ξ 4 a,b + A 3 ξ 3 a,b + A 2 ξ 2 a,b + A 1 ξ a,b + A 0 = 0.
Remark. In Theorem 2.1 the ground field F 2 can be replacd by any field K of characteristic 2, and a and b by any non-constant, distinct polynomials in K[z].

We define an absolute value | • | on the field F q ((z -1 )) by setting |0| = 0 and, for any non-zero power series ξ = ξ(z) = +∞ h=-m a h z -h with a -m ̸ = 0, by setting |ξ| = q m . We write ||ξ|| for the norm of the fractional part of ξ, that is, of the part of the series which comprises only the negative powers of z.

For a power series ξ in F q ((z -1 )), let

C(ξ) = lim inf Q∈Fq[z]\{0} |Q| • ∥Qξ∥ denote its Lagrange constant. Clearly, C(ξ) is an element of L q = {q -k : k ≥ 1} ∪ {0},
and it is positive if and only if the degrees of the partial quotients of ξ are bounded.

If [a 0 ; a 1 , a 2 , . . .] denotes the continued fraction expansion of ξ and p n /q n = [a 0 ; a 1 , a 2 , . . . , a n ] for n ≥ 1, then

|q n ξ -p n | = |q n+1 | -1 = |a n+1 q n | -1 .
Consequently, we have

C(ξ) = 2 -lim sup n→+∞ deg an .
Exactly as in [START_REF] Allouche | Sur des points fixes de morphismes d'un monoïde libre[END_REF], we can check that the root z -1 + z -4 + z -4 2 + . . . in F 2 ((z -1 )) of zX 4 + zX + 1 is algebraic of degree 4 and has unbounded partial quotients, thus its Lagrange constant is 0. Since the degrees of a and b can be arbitrarily chosen in Theorem 2.1, we derive at once the following corollary.

Corollary 2.2. There exist algebraic power series of degree 4 in F 2 ((z -1 )) with an arbitrarily prescribed Lagrange constant in L 2 .

A power series ξ in F q ((z -1 )) is called badly approximable when the degrees of its partial quotients are uniformly bounded. We define the spectrum of a badly approximable power series ξ in F q ((z -1 )) as the set of positive integers k such that ξ has infinitely many partial quotients of degree k.

Problem 2.3. Let N be a finite set of positive integers. Does there exist a badly approximable power series ξ in F q ((z -1 )) which is algebraic of degree at least 3 and whose spectrum is equal to N ? Theorem 2.1 answers positively Problem 2.3 for q = 2 and any set N of cardinality 2.

For a power q of a prime number p the class H(q) of hyperquadratic power series in F q ((z -1 )) has attracted special attention; see e.g. [21,31]. It is composed of the irrational power series ξ in F q ((z -1 )) for which there exist polynomials A, B, C, D in F q [z] such that AD -BC ̸ = 0 and an integer s with

ξ = Aξ p s + B Cξ p s + D .
Once a power series is known to be algebraic, a natural question is to determine whether it belongs to the restricted class of hyperquadratic power series.

In another direction, since the pioneering works of Kolchin [START_REF] Kolchin | Rational approximation to solutions of algebraic differential equations[END_REF] and Osgood [29], formal derivation has been used to study rational approximation to power series. By differentiating with respect to z the minimal defining polynomial satisfied by an algebraic power series ξ in F q ((z -1 )) of degree d, we see that ξ ′ , where the ′ indicates the derivation on F q ((z -1 )) with respect to z extending the derivation on F q [z], can be expressed as a polynomial in ξ of degree at most d -1.

We say that a power series ξ in F q ((z -1 )) satisfies a Riccati differential equation if there are A, B and C in F q (z) such that

ξ ′ = Aξ 2 + Bξ + C.
Clearly, any cubic power series satisfies a Riccati equation. This is also the case for any hyperquadratic power series, but the converse does not hold; see e.g. [22, Section 4].

Proposition 2.4. Let a, b be non-constant, distinct polynomials in F 2 [z]. Then, the Thue-Morse continued fraction ξ a,b in F 2 ((z -1 )) is differential-quadratic: it satisfies the Riccati equation [(ab(a + b))x] ′ = (ab) ′ (1 + x 2 ).
Furthermore, ξ a,b is not hyperquadratic.

Remark. The Riccati equation in Proposition 2.4 is equivalent to [(ab(a+b))x] ′ = [ab(1 + x 2 )] ′ . Its solutions are the power series ξ such that ab(a + b)ξ + ab(1 + ξ 2 ) is a square, that is, such that 1 + (a + b)ξ + ξ 2 is a linear combinaison of terms a i b j , with i and j odd.

The strategy of the proof of Theorem 2.1 is the following. After a careful study of the minimal defining polynomials P a,b (X) of ξ a,b found by Hu and Han [START_REF] Hu | On the algebraicity of Thue-Morse and period-doubling continued fractions[END_REF] in the case where the sum of the degrees of a and b is at most 7, we have guessed the coefficients, expressed in terms of a and b, of the quartic polynomial P a,b (X) which vanishes at ξ a,b , for any distinct, non-constant polynomials a and b. It then only remained for us to check that P a,b (ξ a,b ) = 0. This step is, however, much more difficult than it may seem to be. Denoting by p ℓ /q ℓ (we drop the letters a, b) the ℓ-th convergent of ξ a,b for ℓ ≥ 1, it is sufficient to check that P a,b (p ℓ /q ℓ ) tends to 0 as ℓ tends to infinity along a subsequence of the integers. We focus on the indices ℓ which are powers of 4. We heavily use the properties of symmetry of the Thue-Morse sequence and we proceed by induction to show that |P a,b (p

4 k /q 4 k )| is, for k ≥ 1, less than |q 4 k | -2
times some constant independent of k.

Higher degree exponents of approximation

Beside the rational approximation to a power series ξ in F q ((z -1 )), we often consider the simultaneous rational approximation of ξ, ξ 2 , . . . , ξ n by rational fractions with the same denominator, as well as small values of the linear form b 0 + b 1 ξ + . . . + b n ξ n with coefficients in F q [z]. This leads us to introduce the exponents of approximation w n and λ n , defined below. For a survey of recent results on these exponents evaluated at real numbers, the reader is directed to [11].

The height H(P ) of a polynomial

P (X) = b n (z)X n + . . . + b 1 (z)X + b 0 (z) over F q [z
] is the maximum of the absolute values of its coefficients, that is, of

|b 0 |, |b 1 |, . . . , |b n |. Recall that the 'fractional part' ∥ • ∥ is defined by +∞ h=-m a h z -h = +∞ h=1 a h z -h ,
for every power series ξ = +∞ h=-m a h z -h in F q ((z -1 )). Definition 3.1. Let ξ be in F q ((z -1 )). Let n ≥ 1 be an integer. We let w n (ξ) denote the supremum of the real numbers w for which

0 < |P (ξ)| < H(P ) -w
has infinitely many solutions in polynomials P (X) over F q [z] of degree at most n. We let λ n (ξ) denote the supremum of the real numbers λ for which

0 < max{∥Q(z)ξ∥, . . . , ∥Q(z)ξ n ∥} < q -λ deg(Q)
has infinitely many solutions in polynomials Q(z) in F q [z]. For positive real numbers w, λ, set

B n (ξ, w) = lim inf H(P )→+∞ H(P ) w • |P (ξ)| and B ′ n (ξ, λ) = lim inf |Q|→+∞ |Q| λ • max{∥Q(z)ξ∥, . . . , ∥Q(z)ξ n ∥}.
Let ξ be an algebraic power series in F q ((z -1 )) of degree d ≥ 2. Let n ≥ 1 be an integer. We briefly show how a Liouville-type argument allows us to bound w n (ξ) from above. Denote by ξ 1 = ξ, ξ 2 , . . . , ξ d the Galois conjugates of ξ. Let P (X) be a non-zero polynomial in F q [z](X). Then, the product P (ξ 1 ) . . . P (ξ d ) is a nonzero element of F q (z), whose absolute value is bounded from below by c 1 (ξ), which (as c 2 (ξ) and c 3 (ξ) below) is positive and depends only on ξ. Since |P (ξ j )|, j = 2, . . . , d, are bounded from above by c 2 (ξ) times H(P ), we get that

|P (ξ)| > c 3 (ξ)H(P ) -d+1
and we derive that

(3.1) w n (ξ) ≤ d -1, n ≥ 1.
In the particular case of ξ a,b , this gives w 3 (ξ a,b ) ≤ 3, by Theorem 2.1. Actually, this inequality is an equality.

Theorem 3.2. The Thue-Morse power series ξ a,b in F 2 ((z -1 )) satisfies Proof. Since the continued fraction expansion of ξ a,b begins with arbitrarily large palindromes, ξ a,b and its square are simultaneously well approximable by rational fractions with the same denominator. This argument appeared in [START_REF]Mathématiques, 7, rue René Descartes, 67084 Strasbourg[END_REF] and we recall it below for the sake of completeness. For k ≥ 0, let p k /q k denote the k-th convergent to ξ a,b . Recall that the Thue-Morse word t = t 0 t 1 t 2 . . . over {a, b} is the fixed point starting with a of the uniform morphism τ defined by τ (a) = ab and τ (b) = ba. Since the words τ 2 (a) = abba and τ 2 (b) = baab, and the prefix of length 4 of t = abba . . . are palindromes, every prefix of t of length a power of 4 is a palindrome. Put

λ 2 (ξ a,b ) = 1, w 2 (ξ a,b ) = 3, w n (ξ a,b ) = 3, λ n (ξ a,b ) = 1 3 , n ≥ 3.
M a = a 1 1 0 , M b = b 1 1 0 .
Then, for k ≥ 1, we have

M t 0 M t 1 . . . M t k-1 = q k q k-1 p k p k-1 .
Take k = 4 ℓ for some positive integer ℓ. The matrix M t 0 M t 1 . . . M t k-1 is symmetric since M a and M b are symmetric and t 0 t 1 . . . t k-1 is a palindrome. Consequently, p k = q k-1 . It then follows from t k-1 = a, t k = b and the theory of

continued fractions in F 2 ((z -1 )) that ξ a,b - p k q k = 1 |b||q k | 2 , ξ a,b - p k-1 q k-1 = 1 |a||q k-1 | 2 = 1 |q k-1 | • |q k | , thus ξ 2 a,b - p k-1 q k = ξ a,b - p k q k ξ a,b + p k-1 q k-1 + ξ a,b q k-1 q k = 1 |q k | 2 , since |ξ a,b | = 1/|a|. We conclude that max{∥q k ξ a,b ∥, ∥q k ξ 2 a,b ∥} = 1 |q k | .
This gives λ 2 (ξ a,b ) ≥ 1 and there is equality since λ 1 (ξ a,b ) = 1. Furthermore, the quantity B ′ 2 (ξ a,b , 1) is finite, since ξ a,b has bounded partial quotients. By the power series field analogue of a classical transference inequality established by Aggarwal [4], we immediately obtain that B 2 (ξ a,b , 3) is finite, thus

w 2 (ξ a,b ) ≥ 3.
This lower bound holds as well for w 2 (ξ q,a,b ). Combined with (3.1), this shows that ξ q,a,b is transcendental or algebraic of degree at least 4, thereby establishing the assertion stated at the end of the first paragraph of Section 2.

Since We keep the notation of Theorem 2.1 and check that ξ a,b is a root of the equation

A 4 X 4 + . . . + A 1 X + A 0 = 0.
In view of the discussion at the beginning of Section 2, this implies that ξ a,b is algebraic of degree 4. The computation is easier if we replace a and b by their inverses, that is, if we consider the continued fraction

(4.1) ζ = [0; a -1 , b -1 , b -1 , a -1 , b -1 , a -1 , a -1 , b -1 , . . .]. Definition 4.1. Set M 0 (a, b) = a a -1 1 1 0 = 1 a a 0 and M k (a, b) = M k-1 (a, b) • M k-1 (b, a) 2 • M k-1 (a, b), k ≥ 1, N k (a, b) = M k-1 (a, b) • M k-1 (b, a), k ≥ 1.
Observe that

M 1 (a, b) = a 2 b 2 + b 2 + 1 a 2 b + ab 2 + a a 2 b + ab 2 + a a 2 b 2 + a 2
and that, for k ≥ 1, the matrix M k (a, b) is symmetric and of the form

1 + • • • + (ab) 2 2k-1 a + • • • + (a + b) (2 2k-1 +1)/3 (ab) 2 2k-2 a + • • • + (a + b) (2 2k-1 +1)/3 (ab) 2 2k-2 a 2 + • • • + (ab) 2 2k-1 .
An immediate induction shows that its entries are polynomials in a and b, whose degrees in a (resp., in b) are at most equal to 2 2k-1 . It follows from the definition of N k (a, b) that its upper left and lower right entries are symmetrical in a and b, while its upper right entry is obtained from its lower left one by exchanging a and b. We introduce some further notation.

Definition 4.2. Write M k (a, b) = Q k (a, b) P k (a, b) P k (a, b) R k (a, b) , k ≥ 0,
and

N k (a, b) = U k (a, b) V k (a, b) V k (b, a) W k (a, b) , k ≥ 1,
To shorten the notation, we simply write

M k = Q k P k P k R k , N k = U k V k V k W k , k ≥ 1,
and observe that

U k = U k , W k = W k , k ≥ 1. Set τ = 1 + a + b and define n k = 2 2k-1 , k ≥ 1.
First, we establish several relations between the entries of M k and M k .

Proposition 4.3. For k ≥ 1, we have

Q k + R k = τ (2n k +2)/3 , P k + P k = (1 + τ )τ (2n k -4)/3 , Q k + R k = τ (2n k -4)/3 .
Proof. We proceed by induction. Check that

Q 1 + R 1 = a 2 + b 2 + 1 = τ 2 , P 1 + P 1 = a + b = 1 + τ = (1 + τ )τ 0 ,
and

Q 1 + R 1 = 2a 2 b 2 + 2b 2 + 1 = 1 = τ 0 . Set m k = (2n k + 2)/3 = (2 2k + 2)/3, k ≥ 0,
and observe that

m k = 4m k-1 -2, k ≥ 1. Since M k = N k N k , we have Q k = U 2 k + V 2 k , R k = V 2 k + W 2 k , P k = U k V k + V k W k , k ≥ 1.
Let k ≥ 2 be an integer and assume that the proposition holds for the index k -1. Since τ is invariant by exchanging a and b, it follows from the induction hypothesis that

R k-1 + Q k-1 = R k-1 + Q k-1 = τ m k-1 .
Consequently, we get

V k + V k = Q k-1 P k-1 + P k-1 R k-1 + Q k-1 P k-1 + P k-1 R k-1 = P k-1 (Q k-1 + R k-1 ) + P k-1 ( Q k-1 + R k-1 ) = (P k-1 + P k-1 )(Q k-1 + R k-1 ) = (1 + τ )τ m k-1 -2 τ m k-1 .
Likewise, we have

U k + W k = Q k-1 Q k-1 + R k-1 R k-1 = Q k-1 (τ m k-1 + R k-1 ) + R k-1 (Q k-1 + τ m k-1 ) = τ m k-1 (Q k-1 + R k-1 ) = τ 2m k-1 -2 .
This gives

P k + P k = (V k + V k )(U k + W k ) = (1 + τ )τ 4m k-1 -4 = (1 + τ )τ m k -2 ,
as expected. We get

Q k + R k = U 2 k + V 2 k + V 2 k + W 2 k = (U k + W k ) 2 + (V k + V k ) 2 = τ 4m k-1 -4 + (1 + τ 2 )τ 4m k-1 -4 = τ 4m k-1 -2 = τ m k .
Finally, we check that

Q k + R k = U 2 k + W 2 k = (U k + W k ) 2 = τ 4m k-1 -4 = τ m k -2 ,
and the proof is complete. □

In the next lemma, we express P k , Q k , R k in terms of a new auxiliary quantity denoted by Z k . Lemma 4.4. Let k be a positive integer and define

Z k = U k + V k .
Then, we have

Q k = Z 2 k , R k = Z 2 k + τ (2n k +2)/3 , P k = Z 2 k + τ (n k +1)/3 Z k + (ab) n k , and Z k = Z k + (1 + τ )τ (n k -2)/3 . Proof. Since Q k = U 2 k + V 2 k , we get immediately that Q k = Z 2 k .
The expression of R k in terms of Z k follows then from Proposition 4.3.

The matrix M k is the product, in a suitable order, of n k copies of M 0 (a, b) and n k copies of M 0 (b, a), whose determinants are respectively equal to a 2 and b 2 . The computation of the determinant of M k then gives the relation

Q k R k + P 2 k = (ab) 2n k . Consequently, P 2 k = Q k (Q k + τ (2n k +2)/3 ) + (ab) 2n k , thus P k = Z 2 k + τ (n k +1)/3 Z k + (ab) n k . It follows from Proposition 4.3 that Q k + Q k = τ (2n k +2)/3 + τ (2n k -4)/3 = (1 + τ 2 )τ (2n k -4)/3 ,
which implies the last equality of the lemma. □

We are now in position to express Z k+1 in terms of n k and Z k .

Proposition 4.5. For k ≥ 1, we have

Z k+1 = τ n k Z k + τ (2n k -1)/3 (ab) n k + (ab) 2n k .
Proof. We proceed by induction. Observe that

Z 1 = U 1 + V 1 = ab + b + 1 and Z 2 = a 4 b 4 + a 3 b + a 2 b 2 + a 2 + ab + ab 3 + b 2 + 1 + a 3 b 2 + a 2 b + a 2 b 3 + b + b 3 . Since τ 2 Z 1 + τ (ab) 2 + (ab) 4 = (1 + a 2 + b 2 )(ab + b + 1) + (1 + a + b)a 2 b 2 + a 4 b 4 ,
we see that the proposition holds for k = 1. Assuming that it holds for a positive integer k, we get

Z k+1 = P k P k + Q k Q k + Q k P k + P k R k = P k ( Q k + R k ) + (P k + Q k )( Q k + P k ) = Z 2 k + τ (n k +1)/3 Z k + (ab) n k τ (2n k +2)/3 + τ (n k +1)/3 Z k + (ab) n k τ (n k +1)/3 Z k + (ab) n k = τ (2n k +2)/3 Z 2 k + τ n k +1 Z k + (ab) n k τ (2n k +2)/3 + τ (n k +1)/3 Z k τ (n k +1)/3 Z k + (1 + τ )τ (2n k -1)/3 + (ab) n k + (ab) n k τ (n k +1)/3 Z k + (1 + τ )τ (2n k -1)/3 + (ab) n k = τ n k +1 Z k + (ab) n k τ (2n k +2)/3 + τ n k Z k (1 + τ ) + (ab) n k (1 + τ )τ (2n k -1)/3 + (ab) n k = τ n k Z k + τ (2n k -1)/3 (ab) n k + (ab) 2n k .
This completes the proof. □

We deduce from Proposition 4.5 an equation of degree 4 satisfied by Z k .

Proposition 4.6. For k ≥ 1, we have

Z 4 k + τ (2n k -1)/3 Z 2 k + (a + b)τ n k -1 Z k + (b(a + b)τ 2 + a 2 )τ (4n k -8)/3 + (a + b)(ab) n k τ (2n k -4)/3 + (ab) 2n k = 0. Proof. For k ≥ 1, set ∆ k = Z 4 k + τ (2n k -1)/3 Z 2 k + (a + b)τ n k -1 Z k + (b(a + b)τ 2 + a 2 )τ (4n k -8)/3 . Note that ∆ 1 = (1 + b 4 + a 4 b 4 ) + (1 + a + b)(1 + b 2 + a 2 b 2 + (a + b)(1 + b + ab)) + b(a + b)(1 + a 2 + b 2 ) + a 2 = a 4 b 4 + a 2 b 3 + a 3 b 2 = (a + b)(ab) n 1 τ (2n 1 -4)/3 + (ab) 2n 1 .
Thus, Proposition 4.6 holds for k = 1. Let k be a positive integer. Since n k+1 = 4n k , it follows from Proposition 4.5 that

∆ k+1 = Z 4 k+1 + τ (8n k -1)/3 Z 2 k+1 + (a + b)τ 4n k -1 Z k+1 + (b(a + b)τ 2 + a 2 )τ (16n k -8)/3 = τ n k Z k + τ (2n k -1)/3 (ab) n k + (ab) 2n k 4 + τ (8n k -1)/3 τ n k Z k + τ (2n k -1)/3 (ab) n k + (ab) 2n k 2 + (a + b)τ 4n k -1 τ n k Z k + τ (2n k -1)/3 (ab) n k + (ab) 2n k + (b(a + b)τ 2 + a 2 )τ (16n k -8)/3 = τ 4n k Z 4 k + τ (8n k -4)/3 (ab) 4n k + (ab) 8n k + τ (8n k -1)/3 τ 2n k Z 2 k + τ (4n k -2)/3 (ab) 2n k + (ab) 4n k + (a + b)τ 4n k -1 τ n k Z k + τ (2n k -1)/3 (ab) n k + (ab) 2n k + (b(a + b)τ 2 + a 2 )τ (16n k -8)/3 = τ 4n k Z 4 k + τ (8n k -1)/3 (τ 2n k Z 2 k ) + (a + b)τ 4n k -1 (τ n k Z k ) + (b(a + b)τ 2 + a 2 )τ (16n k -8)/3 + τ (8n k -4)/3 (ab) 4n k + τ (8n k -1)/3 (τ (4n k -2)/3 (ab) 2n k + (ab) 4n k ) + (a + b)τ 4n k -1 (τ (2n k -1)/3 (ab) n k + (ab) 2n k ) + (ab) 8n k = τ 4n k Z 4 k + τ (2n k -1)/3 Z 2 k + (a + b)τ n k -1 Z k + (b(a + b)τ 2 + a 2 )τ (4n k -8)/3 + τ (8n k -4)/3 (1 + τ )(ab) 4n k + τ (8n k -1)/3 τ (4n k -2)/3 (ab) 2n k + (a + b)τ 4n k -1 (τ (2n k -1)/3 (ab) n k + (ab) 2n k ) + (ab) 8n k = τ 4n k ∆ k + τ (8n k -4)/3 (a + b)(ab) 4n k + τ 4n k (ab) 2n k + (a + b)τ 4n k -1 (τ (2n k -1)/3 (ab) n k ) + (ab) 8n k . Assuming that ∆ k = (a + b)(ab) n k τ (2n k -4)/3 + (ab) 2n k , we deduce that ∆ k+1 = τ 4n k (a + b)(ab) n k τ (2n k -4)/3 + τ 4n k (ab) 2n k + τ (8n k -4)/3 (a + b)(ab) 4n k + τ 4n k (ab) 2n k + (a + b)τ 4n k -1 (τ (2n k -1)/3 (ab) n k ) + (ab) 8n k = τ (8n k -4)/3 (a + b)(ab) 4n k + (ab) 8n k .
Since n k+1 = 4n k , this shows that Proposition 4.6 holds for every positive integer k. □ Set

a 2 = (1 + a + b) 2 , a 0 = a(a + b)a 2 + a 2 , a 1 = a 3 = (a + b)a 2 , a 4 = b(a + b)a 2 + a 2 .
To prove that [START_REF] Adamczewski | On the complexity of algebraic numbers I. Expansions in integer bases[END_REF], it is sufficient to show that the absolute value of

a 4 ζ 4 + a 3 ζ 3 + a 2 ζ 2 + a 1 ζ + a 0 = 0, with ζ as in (4.
δ k := a 4 P 4 k + a 3 P 3 k Q k + a 2 P 2 k Q 2 k + a 1 P k Q 3 k + a 0 Q 4 k
tends to 0 as k tends to infinity. The next proposition is more precise.

Proposition 4.7. For k ≥ 1, there exists a polynomial

T k (X, Y ) with coefficients in F 2 such that δ k = (ab) 2n k T k (a, b).
Proof. We express δ k in terms of Z k . A lengthy computation based on Lemma 4.4 gives:

δ k = a 4 P 4 k + a 3 P 3 k Q k + a 2 P 2 k Q 2 k + a 1 P k Q 3 k + a 0 Q 4 k = a 4 (P 4 k + Q 4 k ) + (a + b) 2 τ 2 Q 4 k + (a + b)τ 2 P k Q k (P 2 k + Q 2 k ) + τ 2 P 2 k Q 2 k = a 4 (τ (n k +1)/3 Z k + (ab) n k ) 4 + (a + b) 2 τ 2 Q 4 k + (a + b)τ 2 P k Q k (τ (n k +1)/3 Z k + (ab) n k ) 2 + τ 2 P 2 k Q 2 k = a 4 (τ (n k +1)/3 Z k + (ab) n k ) 4 + (a + b) 2 τ 2 Z 8 k + (a + b)τ 2 P k Z 2 k (τ (n k +1)/3 Z k + (ab) n k ) 2 + τ 2 P 2 k Z 4 k = a 4 (τ (n k +1)/3 Z k + (ab) n k ) 4 + (a + b) 2 τ 2 Z 8 k + (a + b)τ 2 (Z 2 k + τ (n k +1)/3 Z k + (ab) n k )Z 2 k (τ (n k +1)/3 Z k + (ab) n k ) 2 + τ 2 (Z 2 k + τ (n k +1)/3 Z k + (ab) n k ) 2 Z 4 k = a 4 τ (4n k +4)/3 Z 4 k + a 4 (ab) 4n k + (a + b) 2 τ 2 Z 8 k + (a + b)τ 2 Z 2 k (Z 2 k + τ (n k +1)/3 Z k + (ab) n k )(τ (2n k +2)/3 Z 2 k + (ab) 2n k ) + τ 2 (Z 4 k + τ (2n k +2)/3 Z 2 k + (ab) 2n k )Z 4 k = a 4 τ (4n k +4)/3 Z 4 k + a 4 (ab) 4n k + (a + b) 2 τ 2 Z 8 k + (a + b)τ 2 Z 2 k τ (2n k +2)/3 Z 4 k + τ 3n k +1 Z 3 k + τ (2n k +2)/3 Z 2 k (ab) n k + (a + b)τ 2 Z 2 k (ab) 2n k Z 2 k + (ab) 2n k τ (n k +1)/3 Z k + (ab) 3n k + τ 2 (Z 8 k + τ (2n k +2)/3 Z 6 k + (ab) 2n k Z 4 k ) = a 4 τ (4n k +4)/3 Z 4 k + a 4 (ab) 4n k + (a + b) 2 τ 2 Z 8 k + (a + b)τ 2 τ (2n k +2)/3 Z 6 k + (a + b)τ 2 τ 3n k +1 Z 5 k + (a + b)τ 2 τ (2n k +2)/3 Z 4 k (ab) n k + (a + b)τ 2 (ab) 2n k Z 4 k + (a + b)τ 2 (ab) 2n k τ (n k +1)/3 Z 3 k + (a + b)τ 2 Z 2 k (ab) 3n k + τ 2 Z 8 k + τ 2 τ (2n k +2)/3 Z 6 k + τ 2 (ab) 2n k Z 4 k = a 4 τ (4n k +4)/3 + τ 2 (ab) 2n k + (a + b)τ 2 τ (2n k +2)/3 (ab) n k + (a + b)τ 2 (ab) 2n k Z 4 k + a 4 (ab) 4n k + [(a + b) 2 τ 2 + τ 2 ]Z 8 k + [(a + b)τ 2 τ (2n k +2)/3 + τ 2 τ (2n k +2)/3 ]Z 6 k + (a + b)τ 2 τ 3n k +1 Z 5 k + (a + b)τ 2 (ab) 2n k τ (n k +1)/3 Z 3 k + (a + b)τ 2 Z 2 k (ab) 3n k = (b(a + b)τ 2 + a 2 )τ (4n k +4)/3 + (a + b)τ (2n k +8)/3 (ab) n k + τ 3 (ab) 2n k Z 4 k + (b(a + b)τ 2 + a 2 )(ab) 4n k + τ 4 Z 8 k + τ (2n k +8)/3 Z 6 k + (a + b)τ 3n k +3 Z 5 k + (a + b)(ab) 2n k τ (n k +7)/3 Z 3 k + (a + b)τ 2 Z 2 k (ab) 3n k = τ 4 Z 8 k + τ (2n k +11)/3 Z 6 k + (a + b)τ n k +3 Z 5 k + [(b(a + b)τ 2 + a 2 )τ (4n k -5)/3 + (a + b)τ (2n k -1)/3 (ab) n k + (ab) 2n k ]τ 3 Z 4 k + (a + b)(ab) 2n k τ (n k +7)/3 Z 3 k + (a + b)τ 2 (ab) 3n k Z 2 k + (b(a + b)τ 2 + a 2 )(ab) 4n k .
This can be rewritten as

δ k = τ 4 ∆ k Z 4 k + [(a + b)τ (2n k -1)/3 (ab) n k + (ab) 2n k ]τ 3 Z 4 k + (a + b)(ab) 2n k τ (n k +7)/3 Z 3 k + (a + b)τ 2 (ab) 3n k Z 2 k + (b(a + b)τ 2 + a 2 )(ab) 4n k ,
where

∆ k = Z 4 k + τ (2n k -1)/3 Z 2 k + (a + b)τ n k -1 Z k + (b(a + b)τ 2 + a 2
)τ (4n k -8)/3 has been already introduced in the proof of Proposition 4.6.

It follows from Proposition 4.6 that

τ 4 ∆ k + (a + b)τ (2n k -1)/3 (ab) n k τ 3 = τ 4 (ab) 2n k .
Consequently, we get

δ k = τ 4 (ab) 2n k Z 4 k + (ab) 2n k τ 3 Z 4 k + (a + b)(ab) 2n k τ (n k +1)/3+2 Z 3 k + (a + b)τ 2 (ab) 3n k Z 2 k + (b(a + b)τ 2 + a 2 )(ab) 4n k .
This shows that δ k can be written as (ab) 2n k times a polynomial in a and b, which depends on k. □

To conclude the proof of Theorem 2.1, we explain the relationship between P k , Q k and the convergents p ℓ /q ℓ = p ℓ (a, b)/q ℓ (a, b) of the Thue-Morse continued fraction ξ a,b = [0; a, b, b, a, b, a, a, b, . . .].

We have

q 4 k (a, b) = (ab) n k Q k (a -1 , b -1 ), p 4 k (a, b) = (ab) n k P k (a -1 , b -1 ), k ≥ 1,
and we check that

A j = (ab) 4 a j (a -1 , b -1 ), 0 ≤ j ≤ 4. For k ≥ 1, put ε 4 k =A 4 p 4 k q 4 k 4 + A 3 p 4 k q 4 k 3 + A 2 p 4 k q 4 k 2 + A 1 p 4 k q 4 k + A 0 = (ab) -2n k +4 T k (a -1 , b -1 ) (ab) -4n k q 4 4 k = (ab) 2n k +4 T k (a -1 , b -1 ) q 4 4 k . Set d = deg a + deg b. Since |q 4 k | = 2 dn k , |T k (a -1 , b -1 )| ≤ 1, |ab| 2n k = 2 2dn k , we get |ε 4 k | ≤ 2 -2dn k = |q 4 k | -2 .
Recalling that |ξ a,b -p

4 k /q 4 k | < |q 4 k | -2 , we derive that |A 4 ξ 4 a,b + A 3 ξ 3 a,b + A 2 ξ 2 a,b + A 1 ξ a,b + A 0 | ≤ max{|A 1 |, . . . , |A 4 |} • 2 -2dn k . Since n k is arbitrarily large, this gives that A 4 ξ 4 a,b + A 3 ξ 3 a,b + A 2 ξ 2 a,b + A 1 ξ a,b + A 0 = 0,
and concludes the proof of the theorem.

Proof of Proposition 2.4 Since

A 0 + A 4 = (a + b) 2 (ab + a + b) 2 is a square, its derivative is 0 and we get that (A 0 ) ′ = (A 4 ) ′ . By differentiating the minimal defining polynomial A 4 X 4 + . . .

+ A 1 X + A 0 of ξ a,b , we obtain 4 j=0 (A j ) ′ X j + A 1 X ′ (1 + X 2 ) = 0, hence (A 0 ) ′ (1 + X 4 ) + (A 1 ) ′ (X + X 3 ) = A 1 X ′ (1 + X 2 ) and (A 0 ) ′ (1 + X 2 ) + (A 1 ) ′ X = A 1 X ′ , that is (A 0 ) ′ (1 + X 2 ) = (A 1 X) ′ . Since (A 0 ) ′ = (ab) ′ (a + b + ab) 2 , A 1 = ab(a + b)(a + b + ab) 2 , the last equation becomes [(ab(a + b))X] ′ = (ab) ′ (1 + X 2 ).
This establishes that ξ a,b is differential-quadratic and satisfies a simple Riccati equation.

Assume now that there are an integer s ≥ 2 and polynomials A, B, C and D in

. , c 2 s -3 in F 2 [z] such that (A 4 X 4 +. . .+A 1 X +A 0 ) (c 2 s -3 X 2 s -3 +. . .+c 1 X +c 0 ) = AX 2 s +1 +BX 2 s +CX +D.
We thus get 2 s -2 linear forms in c 0 , . . . , c 2 s -3 which vanish. The associated matrix is a pentadiagonal (if s ≥ 3, the case s = 2 being immediate) Toeplitz matrix of the form

M s =          A 2 A 1 A 0 0 0 0 • • • 0 0 0 A 3 A 2 A 1 A 0 0 0 • • • 0 0 0 A 4 A 3 A 2 A 1 A 0 0 • • • 0 0 0 0 A 4 A 3 A 2 A 1 A 0 • • • 0 0 0 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 0 0 0 0 0 0 • • • A 4 A 3 A 2          .
To compute its determinant, we simply expand it and observe that A 2 is the only coefficient of the minimal defining polynomial of ξ a,b which involves the term a 4 b 4 . All the other terms involved are of the form a i b j , with 0 ≤ i, j ≤ 4 and i + j ≤ 7. Consequently, the determinant of M s is equal to (a 4 b 4 ) 2 s -2 plus a linear combination of terms a i b j , with 0 ≤ i, j ≤ 4(2 s -2) and (i, j) ̸ = (4(2 s -2), 4(2 s -2)). In particular, it does not vanish and the system of equations has only the trivial solution c 0 = . . . = c 2 s -3 = 0. This shows that ξ a,b cannot be hyperquadratic.

Concluding remarks

We gather in this concluding section several additional results, without their proofs.

6.1. Explicit expressions. By Proposition 4.5, we can derive an explicit formula for Z k , namely

Z k = τ (2 2k-1 -2)/3 1 + b + 2k-2 j=0 τ (2-2 j+1 -χ(j))/3 (ab) 2 j , k ≥ 1,
where χ(j) := j (mod 2). Then, by Lemma 4.4, we obtain the following explicit formula for P k /Q k . Proposition 6.1. For k ≥ 2, we have We stress that Theorem 6.2 implies that ξ a,b is algebraic. (i,j)∈S a i b j = a -1 + a -2 b -1 + a -2 b -3 + • • • 6.3. Jacobi continued fraction for formal power series. In the field of formal power series F((x)) over a field F, the Jacobi continued fraction J(u, v) defined by two sequences u = (u n ) n≥1 and v = (v n ) n≥0 with v n ̸ = 0 for all n ≥ 0 is the infinite continued fraction

P k Q k = a +
J(u, v) = v 0 1 + u 1 x - v 1 x 2 1 + u 2 x - v 2 x 2 1 + u 3 x - v 3 x 2 . . .
. The basic properties of Jacobi continued fractions can be found in [16,32]. The Hankel determinants H n (J(u, v)) of J(u, v) can be calculated by means of the following fundamental relation, first stated by Heilermann in 1846 [START_REF] Heilermann | Über die Verwandlung der Reihen in Kettenbrüche[END_REF]:

H n (J(u, v)) = v n 0 v n-1 1 v n-2 2 • • • v 2 n-2 v n-1 .
Assume that F is the field F 2 . If the Jacobi continued fraction exists, then v j = 1 for j ≥ 0. In this case, the Hankel determinants H n are all equal to 1. By [8,[START_REF] Guo | Criteria for apwenian sequences[END_REF], the sequence (c n ) n≥0 of the coefficients of J(u, v) = n≥0 c n x n is apwenian. This means that the following relations hold: , where (u 1 , u 2 , . . .) = (1, 0, 0, 1, 0, 1, 1, 0, . . .) is the Thue-Morse sequence over {0, 1}. Since ω(z -1 ) z = ξ z+1,z , it follows from Theorem 2.1 that g 4 ω(x) 4 + g 3 ω(x) 3 + g 2 ω(x) 2 + g 1 ω(x) + g 0 = 0, where g 0 = x 5 + x 3 + x 2 + x + 1, g 1 = x 6 + x 5 + x 4 + x 3 + x 2 + x, g 2 = x 6 + 1, g 3 = x 8 + x 7 + x 6 + x 5 + x 4 + x 3 , g 4 = x 10 + x 9 + x 8 + x 7 + x 5 + x 4 .

Consequently, ω(x) is algebraic of degree 4.

Theorem 2 . 1 .

 21 Let a, b be non-constant, distinct polynomials in F 2 [z] and let ξ a,b = [0; a, b, b, a, b, a, a, b, b, a, a, b, a, b, b, a, . . .]

  Moreover, there are positive constants c 4 , c 5 , depending only on a and b, such that |P (ξ a,b )| > c 4 H(P ) -3 , for every nonzero P (X) in F 2 [z](X) of degree ≤ 3, and there are polynomials P (X) in F 2 [z](X) of degree 2 of arbitrarily large height such that |P (ξ a,b )| < c 5 H(P ) -3 .

  ξ a,b is algebraic of degree 4, by Theorem 2.1, the Liouville-type result obtained below Definition 3.1 combined with the lower bound w 2 (ξ a,b ) ≥ 3 implies that w 2 (ξ a,b ) = w 3 (ξ a,b ) = 3 and w n (ξ a,b ) = 3, λ n (ξ a,b ) = 1 3 , n ≥ 3, the value of λ 3 (ξ a,b ) being a consequence of a transference inequality established in [4]. □ 4. Proof of Theorem 2.1

  For a polynomial M (a, b) (or a matrix M (a, b) whose coefficients are polynomials) in the variables a and b, write M (a, b) the image of M (a, b) under the involution which exchanges a and b, that is, M (a, b) = M (b, a).

  F 2 [z] such that ξ a,b satisfies Aξ 2 s +1 a,b + Bξ 2 s a,b + Cξ a,b + D = 0.Then, the polynomial AX 2 s +1 +BX 2 s +CX+D must be a multiple of the minimal defining polynomial A 4 X 4 +. . .+A 1 X +A 0 of ξ a,b and there exist c 0 , c 1 , . .

a 2 bτ ( 2 - 2

 222 + ab 2 + (a + b)α k 1 + b 2 + a 2 b 2 + α k + α 2 2j+1 )/3 (ab) 2 2j .Notice that α k satisfies the relationα 4 k + τ 2 α k + τ (8-2 2k+1 )/3 (ab) 2 2k + a 4 b 4 = 0.Proposition 6.1 implies the following theorem by taking the limit as k goes to infinite. 2 2j+1 )/3 (ab) 2 2j . Then, α is algebraic and it satisfiesα 4 + τ 2 α + a 4 b 4 = 0.Furthermore, the Thue-Morse continued fraction ξ a,b can be expressed as ξ a,b = a + a 2 b + ab 2 + (a + b)α 1 + b 2 + a 2 b 2 + α + α 2 .

Figure 1 . 2 .

 12 Figure 1. Coefficients of ξ a,b

c 0 = 1 1 1 + u 1 x + x 2 1 + u 2 x + x 2 1 + u 3 x + x 2

 11222 and c n ≡ c 2n+1 + c 2n+2 mod 2, n ≥ 0. Consider the Jacobi continued fraction ω(x) defined by the Thue-Morse sequence ω(x) =

Acknowledgement

The authors thank Alain Lasjaunias for useful discussion.