Transcendence of stammering continued fractions. Yann BUGEAUD

Transcription

1 Transcendence of stammering continued fractions Yann BUGEAUD To the memory of Af van der Poorten Abstract. Let θ = [0; a 1, a 2,...] be an agebraic number of degree at east three. Recenty, we have estabished that the sequence of partia quotients (a ) 1 of θ is not too simpe and cannot be generated by a finite automaton. In this expository paper, we point out the main ingredients of the proof and we briefy survey earier works. 1. Introduction It is widey beieved that the continued fraction expansion of an irrationa agebraic number 1 θ = θ + [0; a 1, a 2,..., a,...] = θ + 1 a 1 + a either is eventuay periodic (and we know that this is the case if, and ony if, θ is a quadratic irrationa), or contains arbitrariy arge partia quotients. Here, and in a what foows, x and x denote, respectivey, the integer part and the upper integer part of the rea number x. A preiminary step consists in providing expicit exampes of transcendenta continued fractions. The first resut of this type goes back to Liouvie [16], who constructed transcendenta rea numbers with a very fast growing sequence of partia quotients. His key too is the so-caed Liouvie inequaity which asserts that, if θ is a rea agebraic number of degree d 2, then there exists a positive constant c 1 (θ) such that θ p/q c 1 (θ) q d, for every rationa number p/q with q 1. Subsequenty, various authors used deeper transcendence criteria from Diophantine approximation to construct other casses of transcendenta continued fractions. Of particuar 2000 Mathematics Subject Cassification : 11J70, 11J81, 11J87. Keywords: continued fractions, transcendence. 1

2 interest is the work of Maiet [19] (see aso Section 34 of [22]), who was the first to give expicit exampes of transcendenta continued fractions with bounded partia quotients. A particuar case of Maiet s resut asserts that if (a ) 1 is a non-eventuay periodic sequence of positive integers at most equa to M, and if there is an increasing sequence ( n ) n 1 such that a n = a n +1 =... = a nn = 1, for n 1, then the rea number α = [0; a 1, a 2,...] is transcendenta. His proof is based on a genera form of the Liouvie inequaity which imits the approximation of rea agebraic numbers θ of degree d 3 by quadratic irrationas. More precisey, Maiet showed that there exists a positive constant c 2 (θ) such that θ γ c 2 (θ) H(γ) d, for every rea quadratic number γ. (1.1) Here, and everywhere in the present text, H(P ) denotes the height of the integer poynomia P (X), that is, the maximum of the absoute vaues of its coefficients; furthermore, H(γ) denotes the height of the agebraic number γ, that is, the height of its minima defining poynomia over Z. A rapid (and rough) cacuation shows that the height of the quadratic irrationa rea number α n := [0; a 1,..., a n 1, 1], where the notation 1 means that the partia quotient 1 is repeated infinitey many times, satisfies H(α n ) n i=1 (a i + 2) 2 (M + 2) 2 n. (1.2) This provides us with infinitey many very good approximations to α. Indeed, by construction, for n 1, the first n n partia quotients of α and α n are the same, thus we derive from (1.2) and (4.4) beow that α α n < 2 2 n n 4 H(α n ) n(og 2)/(2 og(m+2)). (1.3) It then foows from (1.1) that α cannot be agebraic of degree 3. As (a ) 1 is infinite and not eventuay periodic, α is transcendenta. A. Baker [9] used in 1962 Roth s theorem for number fieds obtained by LeVeque to strongy improve upon the resuts of Maiet and make them more expicit. He observed that, when infinitey many of the quadratic approximations found by Maiet ie in the same quadratic number fied, one can repace the use of (1.1) by that of LeVeque s Theorem, which asserts that, for any given rea number fied K, any positive rea number ε, and any rea agebraic number θ ying outside K, there exists a positive constant c 3 (θ, ε) such that θ γ c 3 (θ, ε) H(γ) 2 ε, for every rea agebraic number γ in K. (1.4) 2

3 This is ceary reevant for the exampe mentioned above, since a the α n beong to the quadratic fied Q( 5). In particuar, it foows from (1.3) and (1.4) that if (a ) 1 is a non-eventuay periodic sequence of positive integers at most equa to M, and if there is an increasing sequence ( n ) n 1 such that a n = a n +1 =... = a κn = 1, for n 1 and some rea number κ > 4(og(M + 2))/(og 2), then the rea number α = [0; a 1, a 2,...] is transcendenta. Subsequenty, further transcendence resuts have been obtained by appying a coroary to the Schmidt Subspace Theorem which states that, for any positive rea number ε and any rea agebraic number θ of degree at east 3, there exists a positive constant c 4 (θ, ε) such that θ γ c 4 (θ, ε) H(γ) 3 ε, for every rea quadratic number γ; (1.5) see Coroary 3.2 beow. The difference between (1.4) and (1.5) is that one takes into account every rea quadratic number in (1.5), whie the approximants in (1.4) a beong to the same number fied. By means of (1.5), Davison [14], Quefféec [23] and other authors [15, 7] estabished the transcendence of severa famiies of continued fractions with bounded partia quotients. In particuar, the rea number whose sequence of partia quotients is the Thue Morse sequence or any Sturmian or quasi Sturmian sequence is transcendenta [23, 7]. The next step, initiated in [1], has been the use of the Schmidt Subspace Theorem, instead of its coroary (1.5), to get severa combinatoria transcendence criteria for continued fraction expansions [1, 3, 4, 5]. Recenty in [10], we have shown how a sight modification of their proofs aows us to consideraby improve two of these criteria. In the present survey, we focus on the new combinatoria transcendence criterion for stammering continued fractions estabished in [10] and expain the two main ingredients of its proof. We aso point out some of its appications, incuding that to the Cobham Loxton van der Poorten conjecture for automatic continued fraction expansions. 2. Recent resuts Throughout this note, we identify a sequence a = (a ) 1 of positive integers with the infinite word a 1 a 2... a..., as we denoted by a. This shoud not cause any confusion. For n 1, we denote by p(n, a) the number of distinct bocks of n consecutive etters occurring in the word a, that is, p(n, a) := Card{a a +n : 0}. The function n p(n, a) is caed the compexity function of a. A we-known resut of Morse and Hedund [20, 21] asserts that p(n, a) n + 1 for n 1, uness a is utimatey periodic (in which case there exists a constant C such that p(n, a) C for n 1). 3

4 Let α be an irrationa rea number and write α = α + [0; a 1, a 2,...]. Let a denote the infinite word a 1 a 2... A natura way to measure the intrinsic compexity of α is to count the number p(n, α) := p(n, a) of distinct bocks of given ength n in the word a. Let α be a rea agebraic number of degree at east three. A first step towards a proof that α has unbounded partia quotients woud be to get a good ower bound for p(n, α). Theorem 1.1 of [10], reproduced beow, asserts that the compexity function of an agebraic number of degree at east three cannot increase too sowy. Theorem 2.1. Let a = (a ) 1 be a sequence of positive integers which is not utimatey periodic. If the rea number is agebraic, then α := [0; a 1, a 2,..., a,...] im n + p(n, α) n = +. (2.1) Theorem 2.1 improves Theorem 7 from [7] and Theorem 4 from [1], where im p(n, α) n = + n + was proved instead of (2.1). This gives a positive answer to Probem 3 of [1]. An infinite sequence a = (a ) 1 is an automatic sequence if it can be generated by a finite automaton, that is, if there exists an integer k 2 such that a is a finitestate function of the representation of in base k, for every 1. We refer the reader to [8] for a more precise definition and exampes of automatic sequences. Let b 2 be an integer. In 1968, Cobham [12] asked whether a rea number whose b-ary expansion can be generated by a finite automaton is aways either rationa or transcendenta. After severa attempts by Cobham himsef and by Loxton and van der Poorten [17], Loxton and van der Poorten [18] asserted in 1988 that the b-ary expansion of an irrationa agebraic number cannot be generated by a finite automaton. The proof proposed in [18], which rests on a method introduced by Maher, contains unfortunatey a gap. A positive answer to Cobham s question was finay given in [2], by means of the combinatoria transcendence criterion estabished in [6]. Since the compexity function of every automatic sequence a satisfies p(n, a) = O(n) (this was proved by Cobham [13] in 1972), Theorem 2.1 impies straightforwardy the next resut. Theorem 2.2. The continued fraction expansion of an agebraic number of degree at east three cannot be generated by a finite automaton. Before stating our combinatoria transcendence criterion for continued fractions, we introduce some notation. The ength of a word W, that is, the number of etters composing W, is denoted by W. For any positive integer k, we write W k for the word W... W 4

5 (k times repeated concatenation of the word W ). More generay, for any positive rea number x, we denote by W x the word W x W, where W is the prefix of W of ength (x x ) W. Let a = (a ) 1 be a sequence of positive integers. We say that a satisfies Condition ( ) if a is not utimatey periodic and if there exist w > 1 and two sequences of finite words (U n ) n 1, (V n ) n 1 such that: (i) for every n 1, the word U n V w n is a prefix of the word a; (ii) the sequence ( U n / V n ) n 1 is bounded; (iii) the sequence ( V n ) n 1 is increasing. Equivaenty, the word a satisfies Condition ( ) if there exists a positive rea number ε such that, for arbitrariy arge integers N, the prefix a 1 a 2... a N of a contains two disjoint occurrences of a word of ength εn. The key too for the proofs of Theorems 2.1 and 2.2 is the foowing combinatoria transcendence criterion. Theorem 2.3. Let a = (a ) 1 be a sequence of positive integers. Let (p /q ) 1 denote the sequence of convergents to the rea number α := [0; a 1, a 2,..., a,...]. Assume that the sequence (q 1/ ) 1 is bounded. If a satisfies Condition ( ), then α is transcendenta. Theorem 2.3 was estabished in [10]. Its proof uses the Schmidt Subspace Theorem; see Theorem 3.1 beow. Consequenty, the proofs of Theorems 2.1 and 2.2 rest utimatey on the Schmidt Subspace Theorem. This is aso the case for the simiar resuts on expansions of irrationa agebraic numbers to an integer base; see [2, 6]. A simpe combinatoria study (see e.g. [10]) shows that if (2.1) does not hod for a rea number α := [0; a 1, a 2,..., a,...], then the sequence (a ) 1 either is utimatey periodic, or satisfies Condition ( ) above. In the atter case, Theorem 2.3 impies that α is transcendenta. This shows that Theorem 2.1 is a consequence of Theorem The Schmidt Subspace Theorem The proof of Theorem 2.3 rests on the Schmidt Subspace Theorem. Theorem 3.1 (W. M. Schmidt). Let m 2 be an integer. Let L 1,..., L m be ineary independent inear forms in x = (x 1,..., x m ) with agebraic coefficients. Let ε be a positive rea number. Then, the set of soutions x = (x 1,..., x m ) in Z m to the inequaity L 1 (x) L m (x) (max{ x 1,..., x m }) ε ies in finitey many proper subspaces of Q m. Proof. See e.g. [25, 26]. 5

6 Roth s theorem (that is, (1.4) with K = Q) is equivaent to the case m = 2 of Theorem 3.1. We point out an immediate consequence of the case m = 3 of Theorem 3.1, which extends Roth s theorem to approximation by quadratic numbers. Coroary 3.2. Let θ be a rea agebraic number of degree at east 3. Let ε be a positive rea number. Then, there are ony finitey many integer poynomias P (X) of degree at most 2 such that P (θ) < H(P ) 2 ε. Consequenty, there exists a positive constant c(θ, ε) such that θ γ > c(θ, ε)h(γ) 3 ε, for any agebraic number γ of degree at most 2. Proof. By Theorem 3.1 appied with the three inear forms X 2 θ 2 + X 1 θ + X 0, X 1, X 2, the set of integer tripes (x 0, x 1, x 2 ) satisfying x 2 θ 2 + x 1 θ + x 0 x 1 x 2 (max{ x 0, x 1, x 2 }) ε (3.1) ies in finitey many proper subspaces of Q 3. If x 1 x 2 = 0, then, by Roth s theorem (that is, (1.4) for K = Q), there are ony finitey many integers y 0, y 2, z 0, z 1 such that y 2 z 1 0 and y 2 y 2 θ 2 + y 0 < (max{ y 0, y 2 }) ε, z 1 z 1 θ + z 0 < (max{ z 0, z 1 }) ε. Consequenty, we can assume that x 1 and x 2 are both non-zero. Let a 0 X 0 + a 1 X 1 + a 2 X 2 = 0 denote a proper subspace of Q 3, with a 0, a 1, a 2 in Z and a 0 0. If (3.1) and a 0 x 0 + a 1 x 1 + a 2 x 2 = 0 hod for an integer tripe (x 0, x 1, x 2 ) with x 1 x 2 0, then x 2 θ 2 + x 1 θ + x 0 = x 2 (θ 2 a 2 /a 0 ) + x 1 (θ a 1 /a 0 ). By Roth s theorem, there are ony finitey many integer pairs (x 1, x 2 ) such that x 1 x 2 0 and x 2 (θ 2 a 2 /a 0 ) + x 1 (θ a 1 /a 0 ) x 1 x 2 (max{ x 1, x 2 }) ε Consequenty, the tripe (x 0, x 1, x 2 ) is ying in a finite set, which depends on a 0, a 1, a 2. This proves the first statement of the coroary. The second statement foows immediatey since there is an absoute constant c such that, for any integer poynomia P (X), we have P (θ) c H(P ) θ γ, where γ is the root of P (X) which is the cosest to θ. 4. Auxiiary resuts on continued fractions 6

7 Cassica references on the theory of continued fractions incude [22, 26]. Let α := [0; a 1, a 2,...] be a rea irrationa number. Set p 1 = q 0 = 1 and q 1 = p 0 = 0. For 1, set p /q = [0; a 1, a 2,..., a ] and note that and q = a q 1 + q 2. The theory of continued fraction impies that q α p < q 1 +1, for 1, (4.1) q +h q ( 2) h 1, for h, 1. (4.2) It foows from (4.1) that, if two rea irrationa numbers α and α have the same first partia quotients for some integer 1, then α α 2q 2, (4.3) where q denotes the denominator of the -th convergent to α, and by (4.2). In this and the next sections, we use the notation α α 2 2, (4.4) [0; a 1,..., a r, a r+1,..., a r+s ] := [0; U, V ], where U = a 1... a r and V = a r+1... a r+s, to indicate that the bock of partia quotients a r+1,..., a r+s is repeated infinitey many times. We aso denote by ζ the Gaois conjugate of a quadratic rea number ζ. We reproduce beow Lemma 6.1 from [11]. Lemma 4.1. Let α be a quadratic rea number with utimatey periodic continued fraction expansion α = [0; a 1,..., a r, a r+1,..., a r+s ], with r 3 and s 1, and denote by α its Gaois conjugate. Let (p /q ) 1 denote the sequence of convergents to α. There exists an absoute constant κ such that, if a r a r+s, then we have α α κ a 2 r max{a r 2, a r 1 } qr 2. Lemma 4.1 is an easy consequence of the theorem of Gaois (see [22], page 83) which states that the Gaois conjugate of is the quadratic number [a r+1 ; a r+2,..., a r+s, a r+1 ] [0; a r+s,..., a r+2, a r+1 ]. Athough we do not use it in the computation (5.9) beow, it can be considered as a key observation for the proof of Theorem

8 5. Transcendence criterion for stammering continued fractions In this section, we expain the main ingredients of the proof of Theorem 2.3. Let a = (a ) 1 be a sequence of positive integers. Let w and w be non-negative rea numbers with w > 1. We say that a satisfies Condition ( ) w,w if a is not utimatey periodic and if there exist two sequences of finite words (U n ) n 1, (V n ) n 1 such that: (i) for every n 1, the word U n V w n is a prefix of the word a; (ii) the sequence ( U n / V n ) n 1 is bounded from above by w ; (iii) the sequence ( V n ) n 1 is increasing. Theorem 5.1. Let a = (a ) 1 be a sequence of positive integers. Let (p /q ) 1 denote the sequence of convergents to the rea number α := [0; a 1, a 2,..., a,...]. Assume that the sequence (q 1/ ) 1 is bounded. If there exist non-negative rea numbers w and w with w > 1 such that a satisfies Condition ( ) w,w, then α is transcendenta. Theorem 5.1, estabished in [10], improves Theorem 2 from [1] and Theorem 3.1 from [5], where the assumption w > ((2og M/ og m) 1)w + 1 (5.1) was required, with M = im sup + q 1/ and m = im inf + q 1/. Furthermore, it contains Theorem 3.2 from [3]. The reader is directed to [10] for a compete proof of Theorem 5.1. We content ourseves to expain how Theorem 3.1 and Coroary 3.2 can be appied to prove the transcendence of famiies of stammering continued fractions. We compare the various resuts obtained under the assumption that the sequence (q 1/ ) 1 converges, which makes the comparisons easier. Assume that the rea numbers w and w are fixed, as we as the sequences (U n ) n 1 and (V n ) n 1 occurring in the definition of Condition ( ) w,w. Set r n = U n and s n = V n, for n 1. We assume that the rea number α := [0; a 1, a 2,...] is agebraic of degree at east three. Throughout this section, the numerica constants impied in are absoute. We observe that α admits infinitey many good quadratic approximants obtained by truncating its continued fraction expansion and competing by periodicity. With the above notation, for n 1, the rea number α is cose to the quadratic number α n = [0; U n, V n ]. Namey, since the first r n + ws n partia quotients of α and of α n are the same, we deduce from (4.3) that α α n 2q 2 r n + ws n. (5.2) 8

9 Furthermore, α n is root of the quadratic poynomia (see e.g. [22]) P n (X) := (q rn 1q rn +s n q rn q rn +s n 1)X 2 (q rn 1p rn +s n q rn p rn +s n 1 + p rn 1q rn +s n p rn q rn +s n 1)X + (p rn 1p rn +s n p rn p rn +s n 1), and we deduce that H(α n ) H(P n ) 2q rn q rn +s n. Consequenty, α α n H(α n ) 2(og q rn+ wsn )/(og q r n q rn+sn ). Assuming that (q 1/ ) 1 converges, our assumption that α is agebraic contradicts the ast assertion of Coroary 3.2 when im sup n + r n + ws n 2r n + s n > 3 2, that is, when w > 2w + 3/2. (5.3) This is the approach foowed in [23, 7]. It appies for instance when (a ) 1 is the Thue Morse sequence t := on {1, 2} defined as the fixed point beginning by 1 of the morphism τ defined by τ(1) = 12 and τ(2) = 21. Indeed, for each n 1, the prefix of ength 5 2 n of t is equa to its prefix of ength 3 2 n raised to the power 5/3. Thus, we can take w = 5/3 and w = 0. The fact that the sequence (q 1/ ) 1 converges in this case has been estabished in [24]. The main new ingredient in [1] is the use of Theorem 3.1 with m = 4, instead of Coroary 3.2, which is deduced from Theorem 3.1 with m = 3. Let us now expain to which system of four inear forms we appy Theorem 3.1. By (4.1), we have and, ikewise, (q rn 1q rn +s n q rn q rn +s n 1)α (q rn 1p rn +s n q rn p rn +s n 1) q rn 1 q rn +s n α p rn +s n + q rn q rn +s n 1α p rn +s n 1 (5.4) 2 q rn qr 1 n +s n (q rn 1q rn +s n q rn q rn +s n 1)α (p rn 1q rn +s n p rn q rn +s n 1) q rn +s n q rn 1α p rn 1 + q rn +s n 1 q rn α p rn 2 q 1 r n q rn +s n. (5.5) Furthermore, we have P n (α) H(P n ) α α n q rn q rn +s n q 2 r n + ws n. (5.6) 9

10 We consider the four ineary independent inear forms L 1 (X 1, X 2, X 3, X 4 ) = α 2 X 1 α(x 2 + X 3 ) + X 4, L 2 (X 1, X 2, X 3, X 4 ) = αx 1 X 2, L 3 (X 1, X 2, X 3, X 4 ) = αx 1 X 3, L 4 (X 1, X 2, X 3, X 4 ) = X 1. Instead of treating the coefficient of X in P n (X) as a singe variabe, we cut it into two variabes. Evauating these inear forms on the quadrupe z n := (q rn 1q rn +s n q rn q rn +s n 1, q rn 1p rn +s n q rn p rn +s n 1, p rn 1q rn +s n p rn q rn +s n 1, p rn 1p rn +s n p rn p rn +s n 1), it foows from (5.4), (5.5) and (5.6) that L j (z n ) qr 2 n qr 2 n +s n q 2 r n + ws n. (5.7) 1 j 4 Again on the assumption that (q 1/ ) 1 converges, we are abe to appy Theorem 3.1 (and, with some additiona work, deduce that α is transcendenta) ony when that is, when im sup n + r n + ws n 2r n + s n > 1, w > w + 1. (5.8) This is precisey the inequaity (5.1) with m = M. The novety in [10] is the observation that the estimate (5.6) can be consideraby improved when r n is arge. Namey, using (5.2), (5.4), and (5.5), we get P n (α) = P n (α) P n (α n ) = (q rn 1q rn +s n q rn q rn +s n 1)(α α n )(α + α n ) (q rn 1p rn +s n q rn p rn +s n 1 + p rn 1q rn +s n p rn q rn +s n 1)(α α n ) = α α n (q rn 1q rn +s n q rn q rn +s n 1)α (q rn 1p rn +s n q rn p rn +s n 1) + (q rn 1q rn +s n q rn q rn +s n 1)α (p rn 1q rn +s n p rn q rn +s n 1) + (q rn 1q rn +s n q rn q rn +s n 1)(α n α) α α n (q rn qr 1 n +s n + qr 1 n q rn +s n + q rn q rn +s n α α n ) α α n qr 1 n q rn +s n qr 1 n q rn +s n q 2 r n + ws n. (5.9) Compared to the estimate (5.6), which was used in [1], we gain a factor q 2 r n. As we wi see beow, this aows us eventuay to repace the assumption (5.8) by (5.11) beow. The improvement can be expained by Lemma 4.1. Indeed, since P n (α) H(P n ) α α n α α n, 10

11 where α n denotes the Gaois conjugate of α n, we get an improvement on (5.6) when α n is cose to α, that is, when α n is cose to α n. And Lemma 4.1 precisey asserts that this situation hods when r n is arge. Using (5.9) we can sighty improve (5.3) by appying the first statement of Coroary 3.2 instead of the second one. Indeed, we can concude that α is transcendenta when there exist ε > 0 and arbitrariy arge integers n such that q 1 r n q rn +s n q 2 r n + ws n q2 r n q 2 r n +s n < (q rn q rn +s n ) ε. If (q 1/ ) 1 converges, this shows that the assumption w > w + 3/2, (5.10) is enough to deduce that α is transcendenta. By combining the use of Theorem 3.1 with m = 4 and (5.9), we are abe to improve (5.8) in the same way as (5.10) improves (5.3). Namey, we have L j (z n ) qr 2 n +s n q 2 r n + ws n 1 j 4 2 (w 1)s n if n is sufficienty arge, where we have set δ = (q rn q rn +s n ) δ(w 1)s n/(2r n +s n ), og im sup + q 1/ Thus, with ε = δ(w 1)/(2w + 2), which is positive when we see that 1 j 4. w > 1, (5.11) L j (z n ) (q rn q rn +s n ) ε hods for any sufficienty arge integer n. We can then appy Theorem 3.1 to prove that α is transcendenta. The detais are given in [10]. 6. Two open questions We concude this survey by two open questions. Let (a ) 1 be the sequence defined by a = 2 if is a perfect square, and a = 1 otherwise. Is the rea number [0; a 1, a 2,...] = [0; 2, 1, 1, 2, 1, 1,...] 11

12 transcendenta? Theorem 2.3 cannot be appied in this case since the sequence of squares grows too sowy. Theorem 2.2 asserts that automatic continued fractions are transcendenta or quadratic. Conjecturay, the same hods for morphic continued fractions (see [8] for a precise definition). Since there exist morphic words a = (a ) 1 whose compexity function n p(n, a) grows as fast as a constant times n 2, Theorem 2.3 is not strong enough to give a positive answer to this conjecture. References [1] B. Adamczewski and Y. Bugeaud, On the compexity of agebraic numbers, II. Continued fractions, Acta Math. 195 (2005), [2] B. Adamczewski and Y. Bugeaud, On the compexity of agebraic numbers, I. Expansions in integer bases, Ann. of Math. 165 (2007), [3] B. Adamczewski and Y. Bugeaud, On the Maiet Baker continued fractions, J. reine angew. Math. 606 (2007), [4] B. Adamczewski and Y. Bugeaud, Paindromic continued fractions, Ann. Inst. Fourier (Grenobe) 57 (2007), [5] B. Adamczewski, Y. Bugeaud, and L. Davison, Continued fractions and transcendenta numbers, Ann. Inst. Fourier (Grenobe) 56 (2006), [6] B. Adamczewski, Y. Bugeaud et F. Luca, Sur a compexité des nombres agébriques, C. R. Acad. Sci. Paris 339 (2004), [7] J.-P. Aouche, J. L. Davison, M. Quefféec, and L. Q. Zamboni, Transcendence of Sturmian or morphic continued fractions, J. Number Theory 91 (2001), [8] J.-P. Aouche and J. Shait, Automatic Sequences: Theory, Appications, Generaizations, Cambridge University Press, Cambridge, [9] A. Baker, Continued fractions of transcendenta numbers, Mathematika 9 (1962), 1 8. [10] Y. Bugeaud, Automatic continued fractions are transcendenta or quadratic. Preprint. [11] Y. Bugeaud, Continued fractions with ow compexity: Transcendence measures and quadratic approximation, Compos. Math. 148 (2012), [12] A. Cobham, On the Hartmanis-Stearns probem for a cass of tag machines. In: Conference Record of 1968 Ninth Annua Symposium on Switching and Automata Theory, Schenectady, New York (1968), [13] A. Cobham, Uniform tag sequences, Math. Systems Theory 6 (1972),

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