Abstract

1
Introduction
Throughout this paper, we denote the set of nonnegative integers (resp. positive integers) by N (resp. Z + ). We write the integral and fractional parts of a real number x by [ x ] and { x }, respectively. Moreover, [ x ] is the minimal integer not less than x . We use the Vinogradov symbols ≫ and ≪ , as well as the Landau symbols O , o with their regular meanings. Finally, f ～ g means that the ratio f /g tends to 1 In what follows, we investigate the arithmetical properties of the values of power series f ( X ) at algebraic points. For simplicity, we first consider the case where f ( X ) has the form where ( w ( m )) ∞ = 0 is a sequence of nonnegative integers satisfying w ( m ) < w(m + 1) for any sufficiently large m . We call f(X) a gap series if
1 We say that f(X) is a lacunary series if
Note that if f ( X ) is a lacunary series, then there exists a positive real number δ such that w(m) > (1 + δ) m for any sufficiently large m. In the rest of this secction, suppose that α is an algebraic number with 0 < |α| < 1. In paper [7], Bugeaud posed a problem on the transcendence of the values of power series f ( X ) as follows: If ( w ( m )) m =0 increases sufficiently rapidly, then ∑ m=0 α w(m) is transcendental. Corvaja and Zannier [8] showed that if f(X) = ∑ m=0 Xw(m) is a lacunary series, then ∑ m =0 α w ( m ) is transcendental. For instance, let x , y be real numbers with x > 0 and y > 1. Then two numbers are transcendental. Adamczewski [1] improved the result above in the case of α = β−1 , where β is a Pisot or Salem number. Recall that Pisot numbers are algebraic integers greater than 1 whose conjugates except themselves have absolute values less than 1.  Note that any rational integers greater than 1 are Pisot numbers. Salem numbers are algebraic integers greater than 1 such that the conjugates except themselves have moduli less than 1 and that there exists at least one conjugate with modulus 1. Adamczewski [1] showed that if
m=0 β−w(m) is transcendental for any Pisot or Salem number β . thenアニWe now introduce known results on the algebraic independence of certain lacunary series at fixed algebraic points. First we consider the case where f(X) is a gap series. Durand [10] showed that if α is a real algebraic number with 0 < α < 1, then the continuum set
is algebraically independent. Moreover, Shiokawa [17] gave a criterion for the algebraic independence of the values of certain gap series. Using his criterion, we deduce for general algebraic number α with 0 < | α | < 1 that the set (1.1) is algebraically independent. Next, we consider the case where f ( X ) is not a gap series. Using Mahler's method for algebraic independence, Nishioka [15] proved that the set is algebraically independent. Moreover, Tanaka [18] showed that if positive real numbers w 1 , . . . , w m are linearly independent over Q , then the set
is algebraically independent. On the other hand, it is generally difficult to study algebraic independence in the case where f ( X ) is not lacunary. In Section 2 we review known results on the criteria for transcendence of the value ∑ m = 0 β− w(m), where β is a Pisot or Salem number and ( w ( m )) ∞ = 0 is a certain sequence of nonnegative integers with
In Section 3 we give the main results on the algebraic independence of real numbers. Our results are applicable to the algebraic independence of the two values In the same section we also investigate the linear independence of real numbers applicable to ∑ π = 0 β− l m ♂ ] for a real number ρ > 1. The main criterion for lin- ear independence, which is used to prove the main results, is denoted in Section 4. For the proof of the algebraic independence and the linear independence, we need no functional equation because our criterion is flexible. We prove the main results in Section 5. Moreover, we show the main criterion in Section 6.
2
Transcendental results related to the numbers
of nonzero digits
In this section we review criteria for the transcendence of the value ∑ n = =0 tn β−n where ( t n ) ∞ = 0 is a bounded sequence of nonnegative integers and β is a Pisot or Salem number. First we consider the case where β = b is an integer greater than 1. We denote the base- b expansion of a real number η by where s ( b ) ( η ) = [ η ] and s ( b ) ( η ) ∈ {0,1,..., b − 1} for any positive integer n . We may assume that s( b ) ( η ) ≤ b − 2 for infinitely many n 's. For any positive integer N, put
where Card denotes the cardinality. Borel [5] conjectured for each integral base b ≥ 2 that any algebraic irrational number is normal in base- b , which is still an open problem. For any real number ρ > 1, put If Borel's conjecture is true, then γ(ρ; b − 1 ) is transcendental because γ(ρ; b − 1 ) is a non-normal irrational number in base-b. However, the transcendence of such values is not known except the case of ρ = 2. If ρ = 2, then Duverney, Nishioka, Nishioka, Shiokawa [11] and Bertrand [4] independently proved for any algebraic number α with 0 < |α| < 1 that γ(2; α) is transcendental. Bailey, Borwein, Crandall, and Pomerance [3] gave a criterion for the tran- scendence of real numbers, using lower bounds for the numbers of nonzero digits in the binary expansions of algebraic irrational numbers. Let η be an algebraic irrational number with degree D . Bailey, Borwein, Crandall, and Pomerance [3] showed that there exist positive constants C 1 ( η ) and C 2 ( η ), depending only on η , satisfying
for any integer N with N ≥ C 2 ( η ). Note that C 1 ( η ) is effectively computable but C 2 ( η ) is not. For any integral base b ≥ 2, Adamczewski, Faverjon [2] and Bugeaud [6] gave effective versions of lower bounds for λ b (η; N) as follows: There exist effectively computable positive constants C 3 ( b, η ) and C 4 ( b, η ), depending only on b and η , satisfying
for any integer N with N ≥ C 4 (b, η). Using (2.1), we obtain for any real number ρ > 1 that γ(ρ; b− 1 ) is not an algebraic number of degree less than ρ . In fact, γ(ρ; b−1) is an irrational number satisfying
as N tends to infinity. Thus, (2.1) does not hold if D < ρ . By (2.1), we also deduce a criterion for the transcendence of real numbers as follows: Let η be a positive irrational number. Suppose for any real positive real number ε that
Then η is a transcendental number. Note that the criterion above was essen- tially obtained by Bailey, Borwein, Crandall, and Pomerance [3]. Note that if Em=0 Xw(m) is lacunary, then η = ∑m=0 b-w(m) satisfies (2.2) by
We give another example of transcendental numbers. For any real numbers y > 0 and R ≥ 1, we put ϕ(y; R) := exp ((log R) 1+y) = R(log R) 1 Moreover, we set Note that ξ(y; X) is not lacunary by
4 We get that η := ξ( y ; b − 1 ) is transcendental for any integer b ≥ 2 because η satisfies (2.2). In what follows, we consider the case where β is a general Pisot or Salem number. We introduce results in [14] related to the β -expansion of algebraic numbers. For any formal power series f(X) = ∑ n = 0 t n X n , we put
Moreover, for any nonempty set A of nonnegative integers, we set λ( A; N) := Card( A ∩ [0, N )). We denote the degree of a field extension L/K by [ L : K ]. THEOREM 2.1 ([14]). Let A be a positive integer and let f(X) = ∑ n=0 t n X n be a power series with integral coefficients. Assume that 0 ≤ t n ≤ A for any nonnegative integer n and that there exist infinitely many n’s satisfying t n̸= 0. Let β be a Pisot or Salem number. Suppose that η = f(β− 1 ) is an algebraic number with [Q(β,η) : Q(β)] = D.  Then there exist effectively computable positive constants C 5 (A, β, η) and C 6 (A, β, η), depending only on A, β and η satisfying
for any integer N with N ≥ C 6 (A,β,η). In the rest of this section, let β be a Pisot or Salem number. Using Theorem 2.1, we obtain for any real number ρ > 1 that
by
as N tends to infinity. Note that Theorem 2.1 is applicable to the study of the nonzero digits in the β -expansions of algebraic numbers. We recall the definition of β -expansion defined by Rényi [16] in 1957. Let T β : [0,1) → [0,1) be the β -transformation defined by T β ( x ) = { βx } for x ∈ [0,1). Then the β -expansion of a real number η ∈ [0, 1) is denoted as
for any positive integer N. Applying Theorem 2.1 with B = [β], we deduce that if η ∈ [0,1) is an algebraic number with [Q(β,η) : Q(β)] = D, then
5 for any sufficiently large integer N . Using Theorem 2.1, we also deduce a criterion for the transcendence of real numbers as follows: Let f ( X ) be a power series whose coefficients are bounded nonnegative integers. Suppose that f(X) is not a polynomial and that
for any positive real number ε . Then f ( β− 1 ) is transcendental. Note that the criterion above was already obtained in [13] and that the criterion is applicable even if the representation ∑ n =0 t n β− n does not coincide with the β -expansion of f ( β− 1 ). In the same way as the case where β = b ≥ 2 is an integer, we obtain for any positive real number y that ξ(y; β− 1 ) is transcendental. In the end of this section we introduce a corollary of Theorem 2.1, which we need to prove our criteria for linear independence. COROLLARY 2.2. Let A be a positive integer and f(X) a nonpolynomial power series whose coefficients are bounded nonnegative integers. Assume that there exists a positive real number δ satisfying
for infinitely many integer R ≥ 0. Then, for any Pisot or Salem number β, we have
3
Main results
3.1
Results on algebraic independence
We use the same notation as Section 2. THEOREM 3.1. Let β be a Pisot or Salem number. Then the continuum set
is algebraically independent. Note that if β = b is an integer greater than 1, then the algebraic indepen- dence of (3.1) was proved in [12]. Moreover, Theorem 1.4 in [12] implies that if y 1 and y 2 are distinct positive real numbers, then the two values ξ( y 1 , b − 1 ) and ξ( y 2 , b − 1 ) are algebraically independent. However, the algebraic independence of the set {ξ(y;b-¹) | y ∈ R, y > 0} is unknown. Next, we generalize Theorem 1.4 in [12] as follows: THEOREM 3.2. Let y1 and y2 be distinct positive real numbers. Then the two values ξ(y1; β−1) and ξ(y2; β−1) are algebraically independent for any Pisot or Salem number β We now give further results on the algebraic independence of two values. THEOREM 3.3. For any Pisot or Salem number β the two values are algebraically independent. Many results on the algebraic independence treat the values of power series with same types, for instances, the values of Fredholm series ∑ m =0 X k ™ for k = 2,3, . . . (see also Section 1). On the other hand, our results are applicable to the values of two power series of different types. We now introduce a typical example as follows: THEOREM 3.4. Let y and x be real numbers with y > 0,x > 1. Then, the two values ξ(y; β− 1) and ∑ ∞ = 0 β− l x m ] are algebraically independent for any Pisot or Salem number β .
3.2
Results on linear independence
Let F be the set of nonpolynomial power series g( X ) with bounded nonnegative integral coefficients satisfying the following two assumptions: 1. For an arbitrary positive real number ε , we have λ( S (g); R ) = o (R c ) as R tends to infinity. 2. There exists a positive constant C such that [R, CR] ∩ S(g) ≠ 0 for any sufficiently large R. In this subsection, we study arithmetical properties of the values of power series with different types. In particular, we study the linear independence of the values f( β− 1 ) k 1 g( β− 1 ) k 2 ( k 1 , k 2 ∈ N ), where β is a Pisot or Salem number, f(X) = γ(ρ; X) for some ρ > 1, and g(X) ∈ F . In order to state our results, we give a lemma on the zeros of certain polynomials. For any positive integer k, put G k (X) := (1 − X) k + (k − 1) X − 1. LEMMA 3.5. Suppose that k ≥ 3. Then the following holds: 1) There exists a unique zero σκ of Gk(X) on the interval (0,1). 2) Let x be a real number with 0 < x < 1. Then G k ( x ) < 0 (resp. G k ( x ) > 0) if and only if x < σκ (resp. x > σκ). 3) (σκ) ξ=3 is strictly decreasing. Proof. Observe that G k (X) = −k (1 − X)k−1 + k − 1 is monotone increasing on the interval (0,1) and that G k ( X ) has a unique zero σ k on (0,1). Thus, G k ( X ) is monotonically decreasing on (0, σ k ) and monotonically increasing on ( σ k , 1). Hence, the first and second statements of the lemma follow from G k (0) = 0 and G k (1) = k − 2 > 0. Next, we assume that k ≥ 4. Using
we get
Hence, we obtain σ k < σ k –1 by the second statement of the lemma.
THEOREM 3.6. Let A be a positive integer and ρ a real number. Suppose that
Then, for any g(X) ∈ F and any Pisot or Salem number β, the set {γ(ρ;β−1)k1 g(β−1)k2 | k1,k2 ∈ N,k1 ≤ A} is linearly independent over Q(β). We give numerical examples of σ − 1 (n ≥ 4) as follows: σ − 1 = 5.278 . . . ,  σ − 1 = 8.942 . . . ,  σ − 1 = 13.60 . . . COROLLARY 3.7. Let A, ρ be as in Theorem 3.6. 1) For any real number y > 1 and any Pisot or Salem number β , the set
is linearly independent over Q(β). 2) For any positive real number y and any Pisot or Salem number β, the set {γ(ρ;β−1)k1ξ(y;β−1)k2 | k1,k2 ∈ N,k1 ≤ A} is linearly independent over Q(β). Using the asymptotic behavior of the sequence (σ m ) π = 3 , we deduce the following: COROLLARY 3.8. Let ε be an arbitrary positive real number. Then there exists an effectively computable positive constant A 0 (ε), depending only on ε satisfying the following: Let A be an integer with A ≥ A 0 (ε) and ρ a real number with ρ > (ε + 1/2) A ² . Then, for any g(X) ∈ F and any Pisot or Salem number β, the set
is linearly independent over Q(β).
4
Main criterion for linear independence
We denote the Let k be a nonnegative integer and f(X) ∈ Z[[X]]\Z[X]. Minkowski sum of S ( f ) by
Moreover, for any ( k 1 , . . . , k r ) ∈ N r and f 1 ( X ), . . . , f r ( X ) ∈ Z [[X]]\Z [X], we set
REMARK 1. Suppose that 0 ∈ S ( f i ) for i = 1 , . . , r . Then, for any ( k 1 , . . , k r ) ∈ Nr and (k'1,...,k',) E Nr with k, > k', for any i = 1,...,r, we have Let A be a nonempty set of nonnegative integers and R a real number with R > min A. Then we put θ(R; A) := max{n ∈ A | n < R}. THEOREM 4.1. Let A,r be integers with A ≥ 1 and r ≥ 2. Let fi(X) = En=0 ti(n)Xn(i = 1,...,r) be nonpolynomial power series with integral coeffi- cients satisfying 0 ≤ t i (n) ≤ C 7 for any i = 1,...,r and n ≥ 0, where C 7 is a positive constant. We assume that f 1 (X), . . . , f r (X) satisfy the following three assumptions: 1. There exists a positive real number δ satisfying
as R tends to infinity. Moreover, for any i = 2, . . . , r and any real number ε, we have λ( S ( f i ); R ) = o ( λ( S ( f i − 1 ); R ) as R tends to infinity. 2. There exist positive constants C 8 , C 9 such that [R,C 8 R] ∩ S (f,) ≠ 0 for any real number R with R ≥ C 9 3. Let k 1 , . . . , k r be nonnegative integers. Suppose that
for any sufficiently large R, depending only on k 1 , . Then, for any Pisot or Salem number β, the set
is linearly independent over Q(β). REMARK 2. Theorem 4.1 is a generalization of Theorem 2.1 in [12] be- cause Theorem 4.1 is applicable to more general linear independence results. The third assumption of Theorem 2.1 in [12] is essentially as follows: Let k 1 , . . . , k r − 1 , k r be nonnegative integers. Suppose that there exists a positive integer κ = κ(k 1 , . . . , k r − 1 ), depending only on k 1 , . . . , k r − 1 , satisfying
for any sufficiently large R , depending only on k 1 , . . . ,k r − 1 ,k r . In Theorem 4.1 we only consider the case of κ = 1 + k r − 1 for simplicity, which is sufficient to prove our main results.
Proof of main results
5
In this section we prove results in Section 3, using Theorem 4.1.
Proof of results on algebraic independence
5.1
Proof of Theorem 3.1. Let y 1 , y 2 , . . , y r be real numbers with 1 ≤ y 1 < y 2 < < yr . We show that fi(X) := ξ(yi; X) (i = 1, . . . ,r) fulfill the assumptions of Theorem 4.1 for any positive integer A. Recall that we proved Theorem 1.3 in [12], showing for any integer b ≥ 2 that f 1 ( b − 1 ), . . . , f r ( b − 1 ) satisfy the assump- tions of Theorem 2.1 in [12]. In particular, we verified the third assumption with κ = 1 + k r − 1 . In the same way, we can check that f 1 ( X ), . . . , f ( X ) fulfill the assumptions of Theorem 4.1. ● Proof of Theorem 3.2. We can show Theorem 3.2 in the same way as the proof of Theorem 1.4 in [12] by checking the assumptions of Theorem 4.1. ● Proof of Theorem 3.3. Put a(R) := exp ((log R)(log log R)) = R log log R for R ≥ 3 and
Then f 1 ( X ) and f 2 ( X ) satisfy the second assumption of Theorem 4.1. We denote the inverse function of α( R ) by b( R ) . For simplicity, let ϕ( R ) := ϕ( 1; R ). Note that the inverse function of ϕ(R) is
Then we have (log b( R ))(log log b( R )) = log R = (log ψ( R ))².
λ(S(f 1 ); R) ～ b(R) = o(R 2 )
Moreover, using
we get, for any ε,
Hence, we showed that the first assumption of Theorem 4.1 is satisfied for any positive integer A. In what follows, we verify that the third assumption of Theorem 4.1 is fulfilled. It suffices to show the following: For any positive integer k and any real number ε with 0 < ε < 1, we have
for any sufficiently large R , depending only on k and ε , where the implied constant in the symbol ≪ does not depend on R, but on k and ε . In fact, let k 1 and k 2 be any nonnegative integers. Using (5.3) with k = 1 + k 1 and ε = 1/2, we get
as R tends to infinity, where for the last inequality we use (5.1) and (5.2). We first consider the case of k = 1. Note that
Using the mean value theorem for log a ( x ), we see for m ≥ 3 that a(m) ≤ a(m + 1) ≪ a(m). Using the mean value theorem for a ( x ), we get that there exists a real number ρ with 0 < ρ < 1 satisfying
Thus, we obtain
which implies (5.3) in the case of k = 1. Next we assume that (5.3) holds for a fixed positive integer k and an arbitrary positive real number ε . In what follows, we verify (5.3) for k +1 with fixed ε < 1. Put R 0 := R − θ(R;kS(f1)). It suffices to consider the case of
In fact, suppose that (5.4) does not hold. Since 0 ∈ S (f 1 ) by the definition of f 1 (X), we have θ(R;kS(f1)) ∈ kS(f1) ⊂ (k + 1)S(f1) by Remark 1. Thus, we get
which implies (5.3). In what follows, we assume that (5.4) is satisfied. In particular, applying (5.1) to (5.4), we see
for any sufficiently large R. Moreover, the inductive hypothesis implies that
Observing that we see
Thus, we get
where for the last inequality we apply (5.3) with k = 1 because R 0 is sufficiently large by (5.5). By (5.5) and (5.6), we obtain
Using the fact that (log a(x))/(log x) is ultimately increasing with x = b(R) > x' = b(R1-c/4), we get
Consequently,
which implies (5.3). Proof of Theorem 3.4. Let
Then f 1 ( X ) and f 2 ( X ) satisfy the second assumption of Theorem 4.1. We denote the inverse function of ϕ(2y; R) by
Let Then we have
as R tends to infinity. Thus, f 1 ( X ) and f 2 ( X ) satisfy the first and third as- sumptions of Theorem 4.1 because we checked in the proof of Theorem 3.2 that □ f 1 (X) and f 2 (X) fulfill the same assumptions.
5.2
Proof of results on linear independence
Proof of Theorem 3.6. We show that the assumptions of Theorem 4.1 are sat- isfied, where A is defined as in Theorem 3.6, r = 2, f 1 (X) := γ(ρ;X), and f 2 ( X ) := g ( X ). The second assumption follows from the second assumption of F. In order to check the first assumption, it suffices to show that
by (2.3) and the first assumption of F . We may assume that A ≥ 4 by (3.2). Using
we get by A ≥ 4 that
Hence, we obtain
which implies (5.9) by (3.2) and the second statement of Lemma 3.5. In what follows, we check the third assumption of Theorem 4.1. The following lemma was inspired by the results of Daniel [9]. LEMMA 5.1. Let k be a positive integer. Then
for any R ≥ 1, where the implied constant in the symbol O does not depend on R, but on k. Proof. First we consider the case of k = 1. Using the mean value theorem, we see that
for any positive integer m . For any sufficiently large R , take a positive integer m with [m ρ ] < R ≤ [(m + 1) ρ ] Then we get
by (5.11). Next, we assume that (5.10) holds for a positive integer k. Let R 0 := R − θ(R;kS(f 1 )) ∈ Z + . The inductive hypothesis implies that
In the same way as the proof of (5.3), putting η := θ(R;kS(f1)) + θ(R0;S(f1)), we see
Consequently, using (5.10) with k = 1 and R = R 0 , we obtain
by (5.12).
Put log R x := (log x )/(log R ). Using Lemma 5.1 with k = 1 + k 1 , we get
as R tends to infinity. Moreover, using (2.3) and the first assumption of F , we see
Thus, we obtain
as R tends to infinity. For the proof of (4.2), it suffices to show that
If k 1 = 0 or k 1 = 1, then (5.13) is clear by G 1 (X) = − X and G 2 (X) = − X (1 − X). If k 1 = 2, then we have G 3 (X) = − X(1 − 3X + X 2 ) and σ 3 = (3 − √ 5 )/ 2. By (5.9) and (4.1), we get which implies (5.13) by the second statement of Lemma 3.5. Finally, suppose that k 1 ≥ 3. Using (3.2), (4.1), and the third statement of Lemma 3.5, we obtain
which means (5.13). Therefore, we proved Theorem 3.6. Proof of Corollary 3.7. The first statement of Corollary 3.7 follows from Theo- rem 3.6 by The second statement of the corollary is similarly verified by ξ(y; X) ∈ F. In fact, the first assumption of F follows from the fact that, for any real number M , Moreover, using the mean value theorem for log ϕ( y ; R ), we can show that
Proof of Corollary 3.8. By Theorem 3.6 and the second statement of Lemma 3.5, it suffices to show that (ε + 1/2) A ² > σ − 1 , namely, for any sufficiently large A , depending only on ε > 0. We now fix an arbitrary positive real number ε . In the proof of Corollary 3.8, the implied constant in the symbol O does not depend on A , but on ε . Observe that
口
Proof of Theorem 4.1
6
Put
assumption is easily checked. Moreover, the first and third assumptions are also seen by

17
and, for h = 1, . . . , r,
as R tends to infinity. For the proof of Theorem 4.1, it suffices to show that { Fi(β− 1 )k1 F2(β− 1 )k2 . . . F,(β− 1 )kr | k1,k2,...,k, ∈ N,k1 ≤ A} is linearly independent over Q ( β ). In particular, rewriting f i ( X ) by f i ( X ) for i = 1, . . . , r , we may assume that fi(0)̸= 0 for any i = 1, . . . , r . For simplicity, put, for i = 1,...,r, ξi := fi(β−1), Si := S(fi), λi(R) := λ(S(fi); R). Using Corollary 2.2 and the first assumption of Theorem 4.1, we see that
and that ξ 2 , . . . , ξ r are transcendental. We introduce notation for the proof of Theorem 4.1. For any nonempty subset A of N and any positive integer k, let Ak denote the n-fold Cartesian product. For convenience, set A 0 := {0}. Let k ∈ N and p = (p1,...,pk) ∈ Nk . We put
and, for i = 1, . . . , r
We calculate ξ k in the same way as the proof of Theorem 2.1 in [12].  The method was inspired by the proof of Theorem 7.1 in [3]. Let k ∈ N r\{(0, . . ,0)}. Then we have
where
Note that ρ(k; m) is positive if and only if
We see that
We give an analogue of Lemma 4.1 in [12]. LEMMA 6.1. Let k ∈ Nr\{(0,.. .,0)} and let N ∈ Z +. Then we have
and
Proof. We see that (6.4) follows from (6.3) because ρ(k; m) ∈ N for any m. Put S(i; N) := Si ∩ [0, N) for i = 1, . . . , r . Then we get
which implies (6.3).
Assume that the set {ξ k | k = (k 1 , . . ,k r ) ∈ N r ,k 1 ≤ A} is linearly inde- pendent over Q( β ). Then there exists P( X 1 , . . , X r ) ∈ Z[ β ][ X 1 , . . , X r ]\Z[ β ] such that the degree of P(X 1 , . . . , X r ) in X 1 is at most A and that
Let D be the total degree of P ( X 1 , . . . , X r ). Without loss of generality, we may assume that Xr(-1 + Xr) divides P(Xr,...,Xr) and that if r ≥ 3, then Xr-1 divides P(X 1 , . . , X , ).  Put
where Λ is a nonempty finite subset of Nr and Ak ∈ Z [β]\{0} for any k ∈ Λ . For any k = ( k 1 , . . , k r ) ∈ Λ , we have k r ≥ 1 because X r divides P( X 1 , . . , X r ). Moreover, if r ≥ 3, then

19
because X r − 1 divides P(X 1 , . . , X r ). The lexicographic order ➤ on N r is defined as follows: Let k = ( k 1 , . . . , k r ) and k' = (k'1,...,k',) be distinct elements of Nr. Put l := min{i | 1 ≤ i ≤ r, k i̸= k ′ i } . Then k > k ′ if and only if k l > k ′ . The first assumption of Theorem 4.1 implies that if k > k', then
as N tends to infinity. Let g = ( g 1 , . . . , g r ) be the greatest element of Λ with respect to > . Without loss of generality, we may assume that
We see that
In fact, (6.10) follows from (6.7) if r ≥ 3. Suppose that r = 2. Then g 1 is the degree of P(X 1 , X 2 ) in X 1 . Thus, g 1 is positive because ξ 2 is transcendental. Putting A1 := {k = (k1, . . . ,k7-1,k7) | k1 = 91, . . . ,k7-1 = 9r-1,k7 < 9r} and we see Λ = {g} ∪ Λ 1 ∪ Λ 2 .  Using the fact that ξ r is transcendental and that −1 + X r divides P ( X 1 , . . . , X r ), we obtain the following lemma, applying the same method as the proof of Lemma 4.3 in [12] with F(Xr-1,Xr) = 1: LEMMA 6.2. A1 and A2 are not empty. Set
Recall that the degree g1 of P(X1,...,Xr) in X1 is at most A. Thus, we can apply the third assumption of Theorem 4.1 with k = ( k 1 , . . , k r ) = e. In fact, we see
Hence, there exits a positive constant C 12 satisfying the following: For any integer R with R ≥ C 12 , we have
and
In what follows, we set for simplicity. Using (6.11) and (6.12), we obtain the following lemma in the same way as the proof of Lemma 4.4 in [12]:
20
LEMMA 6.3.  Let M, E be real numbers with
Then
Using k 1 ≤ A and the first assumption of Theorem 4.1, we get
Thus, the set
as N tends to infinity. For the proof of (6.13), it suffices to check
by (6.8). If gi > ki for some i < r — 2, then (6.14) holds. Suppose that gi = ki for any i ≤ r − 2. Then we get − 1 + gr− 1 ≥ kr− 1 and 1 + D > kr by k ∈ Λ 2 , which implies (6.14). Combining (6.5), (6.6), and (6.1), we get
For an arbitrary nonnegative integer R , multiplying β R to the both-hand sides of the equality above, we obtain
Putting
we see that Y R is an algebraic integer because β is a Pisot or Salem number. LEMMA 6.4. There exist positive integers C 13 and C 14 satisfying the follow- ing: For any integer R with R ≥ C14, we have
Proof. Let d be the degree of β and let σ 1 , σ 2 , . . . , σ d be the conjugate embed- dings of Q(β) into C such that σι(γ) = γ for any γ ∈ Q(β). Set C 15 := max{|σ i (A k )| | i = 1, . . . ,d,k ∈ A} Let 2 ≤ i ≤ d. Using (6.15) and (6.2), and | β i | ≤ 1, we get
|σ(Yʀ)| ≤ RD+2, Hence, if Y R̸= 0, then we obtain
In the case of β = 2 and r = 1, Bailey, Borwein, Crandall, and Pomerance estimated the numbers y N of positive Y R with R < N in order to give lower bounds for the nonzero digits in binary expansions (Theorem 7.1 in [3]). More- over, if β = b > 1 is a rational integer and r ≥ 2, then y N is applied to prove a criterion for algebraic independence (Theorem 2.1 in [12]). Now, we put, for N ∈ Z +,
In the case where β is a Pisot or Salem number and r = 1, then y N is estimated to give lower bounds for the numbers of nonzero digits in β -expansions (Theorem 2.2 in [14]). In what follows, we calculate upper and lower bounds for y N , which gives contradiction. First, we estimate upper bounds for y N in Lemma 6.5. Next, we give lower bounds for y N in Lemma 6.10, estimating upper bounds for R – θ( R; Ω) in Lemma 6.9, where
In what follows, we assume that N is a sufficiently large integer satisfying

LEMMA 6.5. We have
as N tends to infinity. Proof. Put
where log β N = (log N )/(log β ). Then we see
and
where
for k ∈ Λ. (k 1 , k 2 , . . . , k , ) ∈ Λ that
as N tends to infinity. Observe that
Using (6.3), we get
Thus, the first assumption of Theorem 4.1 with ε = δ/(2D) implies that
Using (6.2), we see
Note for any m ∈ N that
by (6.17). Hence, we obtain
In what follows, we estimate lower bounds for y N in the case where N ∈ Ξ is sufficiently large. Recall that Λ 2 is not empty by Lemma 6.2 and that 0 ∈ S i for i = 1, . . . , r . In particular, for any k ∈ Λ, we have ρ(k; 0) > 0. Put
If N is sufficiently large, then (6.4) and (6.13) imply that For convenience, put T 1+ - := N . Set
where J(j) is an interval of R defined by J(j) = [Tj,T1+j) for 1 ≤ j ≤ τ. In what follows, we denote the length of a bounded interval I of R by |I|. Then we have Let

24
LEMMA 6.6. If N ∈ Ξ is sufficiently large, then we have Recall that A 1 is not empty by Lemma 6.2. Let k 1 be the maximal element of Λ 1 with respect to >. Set
and R 1+μ := N . Then (6.4) implies that
where C 16 is a positive constant. Let
where I(i) is an interval of R defined by I(i) = [ Ri, Ri+1 ) for 1 < i < μ. Set
for i = 1, . . . , μ . Observe that and that
Set
In the same way as the proof of Lemma 4.8 in [12], we obtain the following, using (6.12) and Lemmas 6.3 and 6.6. LEMMA 6.7. For any sufficiently large N ∈ Ξ, we have
In what follows, we assume that N ∈ Ξ satisfies
Let 1 ≤ i ≤ μ with I(i) ∈ I 2 and let R ∈ ( R i , R i +1 ). We now show that
for any k ∈ Λ 1 ∪ Λ 2 = Λ\{g}. In fact, if k ∈ Λ 1 , then (6.25) follows from the definition of R 1 , . . . , R μ +1. Suppose that k ∈ Λ 2 . By the definition of I 2 , we have I(i) ⊂ J(j) for some j with 1 ≤ j ≤ τ, and so R ∈ (Tj,Tj+j). Thus, we get (6.25). Applying the first assumption of Theorem 4.1 with ε = δ/(2D), we see by 91 ≤ A that
as N ∈ Ξ tends to infinity. Thus, we obtain for any sufficiently large N ∈ Ξ that
We can apply the second assumption of Theorem 4.1 with
by (6.26) and (6.24). Thus, we get that there exists V ( N, i ) E S, with Put M = M(N,i) := Ri + V(N,i).
By the definition of R i , there exists k r Using Remark 1, we see
Thus, we get
by V ( N , i ) ∈ S r . LEMMA 6.8. Let N E E be sufficiently large and let 1 ≤ i ≤ μ with I(i) E I 2 . Then YR > 0 for any R with R i ≤ R < M . Proof. We prove Lemma 6.8 by induction on R. First we show that YM-1 > 0. We see
By (6.28)
We now estimate upper bounds for | S 4 |. Let m be an integer with −1 + [2D log β N]. 1 < m < (6.31) Using (6.27) and (6.26), we get
for sufficiently large N ∈ Ξ and R i +1 > m + M − 1 > R i . Thus, applying (6.25) with R = m + M − 1 for any m with (6.31), we obtain by (6.2) that
Therefore, (6.17) implies that
as N tends to infinity. In particular, if N ∈ Ξ is sufficiently large, then
Combining (6.29), (6.30), and (6.32), we deduce that if N E Ξ is sufficiently large, then YM-1 > 0. Next, we assume that Y R > 0 for some R with R i < R < M . Using (6.25), we see
By the inductive hypothesis
Therefore, we proved Lemma 6.8.
Recall that Ω is defined in (6.16). LEMMA 6.9. Let N ∈ Ξ be sufficiently large and let 1 ≤ i ≤ μ with I(i) ∈ I 2 . Let R be an integer with R i + 4C 13 log 3 N ≤ R ≤ M. Then we have
Proof. Put R 1 := θ( R ; Ω). In the same way as the proof of (6.33), we see for any integer n with R i < n < R i+1 that
First, we consider the case of YR ≥ 1. Then (6.35) implies that
and that R − R 1 = 1, which implies (6.34). In what follows, we may assume that 0 < YR < 1 by Lemma 6.8. Let S := [C 13 log β N] . Suppose for any integer m with 0 ≤ m ≤ S that
Noting M > R > R − 1 > · · > R − S > R i , we get by (6.35) that 1 > YR = βYR-1 = ··· = β S YR-S = β 1 + S YR-S-1 > 0, where we use Lemma 6.8 for the last inequality by R i < R − S − 1 < M . So we get
Since – S – 1 > 2C13 log 3 N > 1 for any sufficiently large N , we apply Lemma 6.4 as follows:
Thus, we obtain
a contradiction. Hence, there exists an integer m ′ with 0 ≤ m ′ ≤ S satisfying ρ( g ; R − m ′ ) ≥ 1. Applying (6.35) with n = R − m ′ , we get by Y R m' > 0 that
where for the last inequality we use (6.9). Hence, we deduce that
R 1 ≤ m ′ + 1 ≤ 2C 13 log 3 N .

LEMMA 6.10.

Proof. Let N ∈ Ξ be sufficiently large and let 1 ≤ i ≤ μ with I(i) ∈ I 2 . Note that
by (6.26). Combining (6.27), (6.36), and Lemma 6.9, we see that there exists a constant C 17 such that
Therefore, using (6.22) and (6.23), we obtain
Finally, we deduce a contradiction from Lemma 6.5 and 6.10, which proves Theorem 4.1.
Acknowledgements
This work was supported by JSPS KAKENHI Grant Number 15K17505.