HAJIME KANEKO
Abstract. Let α be an integer with α ≥ 2. In this paper we will give new criteria for algebraic independence of numbers ξ = ∑ n =1 α − w (n), where ( w ( n )) π =1 is a strictly increasing sequence of nonnegative integers. Applying our criteria, we deduce algebraic independence of such ξ also for suitable sequences ( w ( n )) ==1 with lim n →∞ w ( n +1)/w ( n ) = 1, which was impossible by early methods. For instance, let l be a positive real number and put
where [ x ] is the integral part of a real number x and fi(n) = exp ((log n) 1+l ) . We prove, using Theorem 2.1, that the uncountable set {ηι | l ≥ 1} is algebraically independent. Moreover, if h and l are distinct positive real numbers, then η h and η l are algebraically independent.
1. I NTRODUCTION
Let α ≥ 2 be an integer. A normal number in base α is a positive number whose base-α digits show a uniform distribution. Namely, all finite words with letters from the alphabet {0, 1, . . . , α – 1} occur with the proper frequency. Borel [3] showed that almost all positive numbers are normal in each integral base. However, it is generally difficult to check whether a given number is normal or not. Borel [4] conjectured that all algebraic irrational numbers are normal in every integral base. This conjecture is still open. There exists no algebraic number that has been proven to be normal. Moreover, no counterexample is known. If Borel's conjecture is true, then nonzero digits in base- α expansions of algebraic irrational numbers appear with average frequency tending to (α − 1)/α. Consequently, for any irrational ξ, if Borel's conjecture is true and if nonzero digits of ξ in base-α occur with average frequency tending to 0, then ξ is transcendental. We now introduce known results about transcendency and algebraic in- dependence of positive numbers whose densities of nonzero digits are low. 1 In this paper, let N be the set of nonnegative integers and Z >1 the set of positive integers. Moreover, we denote the integral part of a real number ξ by [ξ]. Let us use the Vinogradov symbols ≫ and ≪ , as well as the Landau symbols O and o with their regular meanings. Recall that f ≪ 9, 9 ≫ f and f = O (g) are all equivalent and mean that | f | ≤ c | g | holds with some positive constant c. Moreover, f = o(g) (resp. f ~ g) implies that the ratio f/g tends to zero (resp. 1). All implied constants may depend on the given data. We consider the number
where α ≥ 2 is an integer and ( w ( n )) ∞ 1 is a strictly increasing sequence of nonnegative integers. Liouville [9, 10] first showed the existence of tran- scendental numbers in 1844. He obtained the transcendency of the num- ber ∑ n=1 α− n ! by proving what is nowadays called Liouville's inequality. Schmidt [16] generalized this inequality and showed the numbers 71, 72, . . defined by
are algebraically independent. Durand [7] verified for each real algebraic number z with 0 < z < 1 that the uncountable set is algebraically independent. Shiokawa [17] established algebraic indepen- dence of the values of gap series at algebraic points containing those ap- peared in [7] and [16]. However, we can not apply Liouville's method in the case of
Let k ≥ 2 be an integer. Mahler [11] verified that the number ∑ n= 0 α − k n is transcendental. More generally, he proved for each algebraic number z with 0 < | z | < 1 that Φ k (z) = ∑ n =o zk n is transcendental by using the functional equation
Using the Schmidt Subspace Theorem, Corvaja and Zannier [5] generalized Mahler's results above as follows: Assume that ( w ( n )) π=1 is lacunary, that is, satisfies
Then, for every algebraic z with 0 < | z | < 1, the number ∑ n =1 z w( n ) is tran- scendental. Mahler's method is also applicable to algebraic independence theory. Using (1.2), Nishioka [12] showed for each algebraic number z with 0 < |z| < 1 that the values Φ 2 (z), Φ 3 (z), . . . are algebraically independent. For detailed information concerning Mahler's method for transcendence and algebraic independence, see [13]. Now we return to the base-α expansions of algebraic numbers. For posi- tive numbers ξ and R, let λ(α,ξ, R) be the number of nonzero digits among the first (1 + [ R ]) digits of the base- α expansion of ξ . Namely, λ(α,ξ, R) = Card{n ∈ N | n ≤ [R], [ξαn] − α[ξαn−1]̸= 0}, where Card denotes the cardinality. Assume that α = 2. Bailey, Borwein, Crandall, and Pomerance [1] showed for any algebraic irrational ξ that there exists a positive computable constant C (ξ) depending only on ξ satisfying
for all sufficiently large N . With a suitable positive C (α,ξ) in place of C (ξ) we can prove (1.4) for any integral base α ≥ 2 in the same way. Theorem 1.1. Let α be an integer greater than 1 and ξ > 0 an algebraic irrational number. Then there exist effectively computable positive constants C(α,ξ) and C ′ (α,ξ) depending only on α and ξ such that, for any integer N with N ≥ C ′ (α,ξ),
The idea of the proof of Theorem 1.1 was inspired by the paper of Knight [8]. Let A D X D + A D −1 X D −1 + . . . + A 0 ∈ Z [ X ] be the minimal polynomial of ξ, where A D > 0. In the rest of this section, C 1 (α,ξ) and C 2 (α,ξ) denote effectively computable positive constants depending only on α and ξ. We have (1.6)
We explain the notion of nonzero islands introduced by Knight for another proof of the transcendency of ξο = ∑ n =0 α − 2 n . Let D ′ , A ′ 0 , A ′ 1 , . . . , A ′ D ′ be integers with D ′ ≥ 1 and A ′ D ′ ≥ 1. We show that
For any k with 1 ≤ k ≤ D ′ we have
where τ( m, k ) denotes the number of ways that m can be written as a sum of k powers of 2. Let b be a sufficiently large integer. Put N = (2 P ′ − 1)2 b . Let m be an integer with
Then Lemma 1 in [8] implies that
Hence, considering the carries of the base-α expansion of D ′ ! A ′ D ′ α − N , we deduce the following: there exists an integer m with N ≤ m ≤ N+O(1) such that the m -th digit of the base- α expansion of ω is not zero. In particular, ω̸= 0. Knight used the term nonzero islands to refer nonzero digits which occur from the carries of the base-α expansion of D ′! A ′ D ′ α − N . In [1], the Thue-Siegel-Roth theorem [15] was used in order to find nonzero islands. However, the Thue-Siegel-Roth theorem is ineffective. So, in this paper we use Liouville's inequality instead of the Thue-Siegel-Roth theorem. As a consequence, we obtain the effective lower bounds C ′ (α,ξ) in Theorem 1.1. We give a sketch of the proof of Theorem 1.1 without technical details. For simplicity, assume that 1 ≤ ξ < 2 and write the α -ary expansion of ξ by
Let N be a positive integer. In the same way as the proof of Theorem 7.1 in [1], we can show that there exists an interval I = [ U 1 , U 2 ) ⊂ [0 , N ) satisfying the following four conditions:
(1.9)
for any m with U 1 < m < U 2 . (4) (1.10)
where |I| = U 2 – U 1 is the length of I . Using (1.8) and (1.9), we get (1.11)
Then we have
Let
Then p and q are integers. Thus,
which contradicts Liouville's inequality in the case of N ≥ C 2 (α,ξ) because we have (1.10).
and (1.13)
In what follows, we observe the α -ary expansion of the left-hand side of (1.6), using (1.7). Note that (1.7) is not generally the α-ary expansion of ξ k because α−m ρ ( k , m ) causes carry, O (log ρ ( k , m )) to the higher digits. Recall that A D > 0. Combining (1.10), (1.11), (1.12), and (1.13), we conclude that positive digits left in the α -ary expansion of (1.6), which is a contradic- tion. To explain the details of remaining positive digits, Bailey, Borwein, Crandall, and Pomerance [1] introduced BBP tails. Note that the concept of BBP tails is defined in the paper [2] in order to give rapid algorithms for the computation of the n -th digits of certain transcendental numbers. In the case of α = 2, Rivoal [14] improved the constant C(ξ) for certain classes of algebraic irrational ξ . For example, let ε be an arbitrary positive number and ξ ′ = 0.558 . . . the unique positive zero of the polynomial 8 X ³ — 2X 2 + 4X − 3. Theorem 7.1 in [1] implies for any sufficiently large N that
On the other hand, using Corollary 2 in [14], we obtain λ(2,ξ′, Ν) ≥ (1 − ε) N ¹/3 for all sufficiently large N . Let us consider applications of Theorem 1.1. For each real number k with k > 1, put
Let d be a natural number with 2 ≤ d < k . (1.5) implies that ν k is not an algebraic number of degree at most d . Moreover, using Theorem 1.1, we deduce criteria for transcendence. Corollary 1.2. Let α be an integer greater than 1 and ξ a positive irrational number. Assume for an arbitrary positive number ε that (1.14) λ(α,ξ, N ) = σ( N ε ). Then ξ is transcendental. For instance, the numbers are transcendental by Corollary 1.2 because these numbers fulfill (1.14). Moreover, for positive numbers l and x with x ≥ 1, let
Then the number
is transcendental by Corollary 1.2 because η satisfies (1.14). In fact, it is easily seen that, for any ε > 0,
as R tends to infinity. Note that η does not satisfy inequality (1.3). Thus, we cannot prove, using the result of Corvaja and Zannier, that the number ηι is transcendental. In fact, for a real number x with x > 1, we have
to infinity, we obtain
The main purpose of this paper is to deduce algebraic independence of certain classes of numbers ξ which satisfy (1.14). We will introduce criteria for algebraic independence in Theorem 2.1. We prove this theorem in Section 4. Our method is quite flexible because we do not use functional equation. As a consequence of Theorem 2.1 we deduce algebraic independence of the values η for real numbers l with l ≥ 1. Theorem 1.3. The uncountable set { η | l ≥ 1} is algebraically indepen- dent. We can not prove, using Theorem 2.1, that the uncountable set { ηι | l > 0} is algebraically independent. However, we deduce that any two elements of this set are algebraically independent. Theorem 1.4. Let h and l be distinct positive real numbers. Then η, and ηι are algebraically independent. We verify Theorems 1.3 and 1.4 in Section 3. 2. Criteria for algebraic independence Let α ≥ 2 be an integer and ξ a positive number. For each integer m, put
which is the α-ary expansion of ξ. Set S(ξ) = { m ∈ N | t(ξ, m)̸= 0}. Recall for R > 0 that λ(α, ξ, R) = Card{ n ∈ S(ξ) | n ≤ R}. Note that if 1 ≤ ξ < α , then 0 ∈ S (ξ). For each positive integer α , let αS(ξ) = { n 1 + . . . + n a | n 1 , . . . , n a ∈ S(ξ)} . For convenience, let OS(ξ) = {0}. Moreover, for any positive numbers ξι, . . . , ξ r and nonnegative integers a 1 , . . . , a r , let
where R is a real number with R > min{n E A}. Assume that 1 ≤ ξ 1 , . . . ,ξ r < α . Let (α 1 , . . ,α τ ), (α ′ 1 , . . ,α ′ ) ∈ N r , where a i ≥ a i for every i with 1 ≤ i ≤ r . Then we have
because S(ξ 1 ), . . . , S(ξ r ) ∋ 0. We now state criteria for algebraic indepen- dence. Theorem 2.1. Let ξ 1 , . . , ξ r be positive irrational numbers. Suppose that these numbers satisfy the following three assumptions: (1) For an arbitrary positive ε, we have
and, for h = 2, . . . , r, λ(α, ξһ, R) = σ(λ(α, ξһ–ι, R)ε). as R tends to infinity. (2) There exists a positive constant C 1 such that
for every sufficiently large real number R. (3) Let a 1 , . . . , a r − 1 , a r be any nonnegative integers. If r ≥ 2 , then there exist a positive integer κ = κ(α 1 , . . . ,ατ − 1 ) and a positive constant C 2 (a1, . . . , a r ), where κ depends only on a1, . . . , a r −1 and C 2 (a1, . . . , a r ) only on a1,...,ar, such that
Then ξ 1 , . . . , ξ r  are  algebraically  independent. Remark 2.2. In the case of r = 1, Theorem 2.1 follows from Corollary 1.2. We verify Theorem 2.1 in Section 4. In the rest of this section we give a sketch of the proof of Theorem 2.1 without technical details in the case where r = 2 and κ(α 1 ) = 1 + α 1 for all a 1 ≥ 0, where κ(α 1 ) is defined in the third assumption of Theorem 2.1. For simplicity, suppose that 1 ≤ ξ 1 , ξ 2 < 2. If ξ 1 and ξ 2 are algebraically dependent, then there exists a nonzero polynomial P(X 1 , X 2 ) ∈ Z [X] such that
Let
where Λ is a nonempty finite subset of N 2 and A k a nonzero integer for each k ∈ Λ. We search nonzero islands of the α -ary expansion of the left-hand side of (2.1). For any k = (a,b) ∈ Λ, we get
Observe that ρ(k,m) > 0 if and only if m ∈ aS(ξ₁) + bS(ξ₂). In the proof of Theorem 1.1, we used the relation
in order to find nonzero islands. Indeed, (1.8) implies that ( k − 1) S ( ξ ) ⊂ kS(ξ) (see also (1.13)). On the other hand, let (a 1 , a 2 ), (a ′ 1 , a ′ 2 ) ∈ Λ. Then, in general, neither α 1 S(ξ 1 ) + α 2 S(ξ 2 ) ⊂ α ′ 1 S(ξ 1 ) + α ′ 2 S(ξ 2 ) nor α ′ 1 S(ξ 1 ) + a',S(ξ2) ⊂ α 1 S (ξ1) + α 2 S (ξ2) holds. This is the most different point between the proofs of Theorems 1.1 and 2.1. Let ➤ be the lexicographical order in N². Namely, (α 1 ,α 2 ) ≥ (α ′ ,α ′ 2 ), if α 1 > α ′ 1 , or if α 1 = α ′ 1 and α 2 > α ′ 2 . Let g = (91,92) ∈ A be the greatest element of A with respect to ¬. Without loss of generality, we may assume that Ag > 0. For any (a 1 ,a 2 ) ∈ Λ, if α 1 = g 1 , then a 2 ≤ g 2 . Thus, we have a 1 S(ξ 1 ) + α 2 S(ξ 2 ) ⊂ g 1 S(ξ 1 ) + g 2 S(ξ 2 ) . If a 1 < g 1 , then the relation above does not hold generally. However, by the third assumption of Theorem 2.1, the set a 1 S(ξ 1 ) + a 2 S(ξ 2 ) is approximated by (1 + a 1 ) S (ξ 1 ) because κ (a 1 ) = a 1 + 1. Moreover, we have (1 + a 1 ) S (ξ 1 ) ⊂ 91S(ξ1). Based on the observation above, we give nonzero islands in Section 4.4. Let N be a sufficiently large integer. We construct an interval J = [ T 1 , T 2 ) ⊂ [0, N ) satisfying the following three conditions: (1) T 1 E p 1 S (ξ 1 ) + p 2 S (ξ 2 ) for some (p 1 ,p 2 ) E A with p 1 < g 1 . (2) If T 2 < N, then T 2 ∈ q 1 S(ξ 1 ) + q 2 S(ξ 2 ) for some (q 1 , q 2 ) ∈ Λ with 91 < 91. (3) Let m be any integer with T 1 < m < T 2 and let ( a 1 , a 2 ) ∈ Λ with α 1 < 9 1 . Then m̸= α 1 S(ξ 1 ) + α 2 S(ξ 2 ). Since p 1 S(ξ 1 ) + p 2 S(ξ 2 ) and q 1 S(ξ 1 ) + q 2 S(ξ 2 ) are approximated by g 1 S(ξ 1 ), we get a subinterval I = [ R 1 , R 2 ) of J satisfying the following three condi- tions:
(3) Let m be any integer with R 1 < m < R 2 and let k = (a 1 ,a 2 ) ∈ Λ with g > k. Then (2.3) m & a1S($1) + a2S($2). Denote the length of I by | I | = R 2 − R 1 . Using the second assumption of Theorem 2.1, we deduce that there is an m 0 ∈ N satisfying m 0 ∈ S (ξ 2 ) and
In particular, (2.5)
Now we observe the α -ary expansion of the left-hand side of (2.1), using (2.2). Recall that Ag > 0 and that α-mρ(k,m) causes carry, O (log(k,m)) to the higher digits. Hence, combining (2.3), (2.4), and (2.5), we conclude that positive digits left in the α -ary expansion of (2.1), which is a contradiction. To explain the details of remaining positive digits, we introduce BBP tails Y R in the last of Section 4.2.
3. Proof of main results
Proof of Theorem 1.3. Let { ηι , ηι , . . , ηι } be any finite subset of { ηι | l ≥ 1}. Without loss of generality, we may assume that l 1 < l 2 < · · · < l r . Let
for i = 1,...,r. Then S(ξ i ) Э 0 for i = 1,...,r. We check that ξ 1,...,ξ r satisfy the assumptions of Theorem 2.1. Let l be a positive number. In Section 1 we proved that, for any sufficiently large x ,
Therefore, we verified the second assumption with C 1 = 1/2. For any posi- tive numbers l and x with x ≥ 1, put
Note that g l is the inverse function of f l . Namely, for any x ≥ 1, we have fi(gi(x)) = x . Let i and j be integers with 1 ≤ i,j ≤ r . As we mentioned in Section 1, for any ε > 0, (3.3)
Thus the first assumption is fulfilled by (3.3) and (3.4). Finally we check the third assumption. We introduce the results of Daniel [6]. Let ( μ n ) ∞ 1 be the strictly increasing sequence of those positive integers that can be represented as the sum of three cubes of positive integers. Then Daniel showed that
In the same way as the proof of the result above, we get the following:
where
Proof. We prove Lemma 3.1 by induction on the value a 1 + · · · + a r . Assume that a 1 + · · · + a r = 1. Then there exists an integer h with 1 ≤ h ≤ r and ah = 1. We have
there exists ρ = ρ( x ) ∈ (0, 1) such that
for sufficiently large R. If R is sufficiently large, then there exists m ∈ N such that
we obtain
which implies (3.5) in the case of a 1 + . . . + a r = 1. Next, assume that a 1 + ··· + a r ≥ 2. Let
Put
Then we deduce, using the case of α 1 + · · · + α r = 1, that there exists a positive constant C satisfying
Note that
because
Let
Then since
we get
Taking the logarithm of G(R), we deduce that the function G(R) is mono-
and hence
with
Therefore we proved Theorem 1.3.
Proof of Theorem 1.4. Without loss of generality, we may assume that h < l. Let ξ 1 = η h − [η h ] + 1 and ξ 2 = η ι − [η l] + 1. Note that 1 < ξ 1 ,ξ 2 < 2 and that S (ξ 1 ), S (ξ 2 ) ∋ 0. In the same way as in the proof of Theorem 1.3, we can verify that ξ 1 and ξ 2 satisfy the first and second assumptions of Theorem 2.1 with C 1 = 1/2. In what follows, we prove that the third assumption is satisfied. Let gι( x ) be defined by (3.2). Lemma 3.2. Let b be a positive integer. Then, for any positive number ε, we have
for R ≥ 2. Proof. We show (3.9) by induction on b. Assume that b = 1. In the same way as in the proof of Lemma 3.1, we deduce that there exists a positive constant C satisfying
which implies (3.9) because, for any positive ε,
as R tends to infinity. Suppose that b ≥ 2. Without loss of generality, we may assume that R ′ ≥ 2 and that ε < 1. In particular, we have
By the induction hypothesis, R' = θ(R';(b = 1)S(ξ₁)) ≪ R'gʰ(R')−b+1+ε/₃ =: H(R'). We obtain, taking the logarithm of H ( R ), that the function H ( R ) is mono- tone increasing for sufficiently large R. Hence,
0
Let
Then For all sufficiently large R , we obtain
and hence
Combining (3.11) and the inequality above, we proved (3.9). □ Let (a 1 , a 2 ) ∈ N 2 . Then, applying Lemma 3.2 with b = a 1 +1 and ε = 1/2, we get
as R tends to infinity. Therefore we checked the third assumption of Theo- ■ rem 2.1 with κ = κ(α 1 ) = α 1 + 1 and hence verified Theorem 1.4.
4. Proof of Theorem 2.1
4.1. Base- α expansions of powers of real numbers. We prove The- orem 2.1 by induction on r . Using Corollary 1.2, we deduce the case of r = 1. In what follows, suppose that r ≥ 2. Without loss of generality we may assume that 1 ≤ ξ 1 , . . ,ξ r < 2. In fact, ξ 1 , . . . ,ξ r are algebraically independent if and only if ξ ′ , . . . ,ξ ′ are algebraically independent, where
For simplicity, let
For convenience, put N 0 = {0}. Let ξ > 0,b ∈ N,x = (x 1 , . . . ,x b ) ∈ N b , and s = (s1, . . . , sb) ∈ Zb. Put
have
where
Note that, for each m ∈ N, ρ(k, m) > 0 if and only if m ∈ ∑ i =1 α i S (ξ i ). It is easily seen that
We now check the following: Lemma 4.1. Let k = (a1, . . . ,ar) ∈ Nr\{(0, . . . ,0)} and N ∈ N. (1)
(2) Card{m ∈ N | m ≤ N, ρ(k, m) > 0} ≤ (α − 1) | k | λ(N) k. Proof. Put, for i = 1,...,r,
Then
lows from the first one because ρ (k, m) ∈ N for each m ∈ N. ■ 4.2. Auxiliary functions. We define the lexicographical order ➤ on Nr as follows. For any k = ( α 1 , . . . , α r ), k ′ = ( α ′ 1 , . . . , α ′ ) with k̸= k ′ , there exists a positive l such that the first (l − 1) symbols in k = (a 1 , . . . ,a r ) and k' = ( α' 1 , . . . ,α' , ) coincide, but their lth symbols are different. Then k = (a 1 , . . . , a r ) > k ′ = (a ′ 1 , . . , a ′ ) if and only if a l > a ′ . Each nonzero polynomial Q(X) ∈ Z [ X 1 , . . . , X r ] is uniquely written as
where Λ(Q) is a finite subset of N r determined by Q , B k a nonzero integer and X = ( X 1 , . . , X r ). Recall for k = ( α 1 , . . . , α r ) that X k = X 1 1 . . . X 1 r . Let g(Q) = (g1(Q), . . . , gr(Q)) be the greatest element of Λ(Q) with respect to >. Moreover, put
where k = (α 1 , . . . ,α r ). We define the number e(Q) as follows. If Λ 3 (Q) is empty, then put e(Q) = 0. Otherwise, let
Now assume that ξ 1 , . . . , ξ r are algebraically dependent. Then there exists a nonzero polynomial P(X) ∈ Z [ X 1 , . . . , X r ] such that
By the induction hypothesis ξ 2 , . . . , ξ r are algebraically independent. Thus, the degree of P(X) in X 1 is positive. Namely, g 1 (P) ≥ 1. Without loss of generality, we may assume that (4.3)
In what follows, let
where n is a nonnegative integer and the right-hand side of the equality above is defined in the third assumption of Theorem 2.1. Let m and n be integers with 0 ≤ m ≤ n. Then, for any positive number R , we have
because
So, if necessary, by increasing κ ( n ), we may assume that κ ( n ) ≥ 1 for any n ∈ N and that the sequence ( κ ( n )) ∞ = 0 is monotone increasing. Lemma 4.2. There is a nonzero polynomial F(Xr-1,Xr) E Z[Xr-1,Xr] such that
Proof. We define the nonzero polynomial σ( X r − 1 , X r ) ∈ Z [ X r − 1 , X r ] as fol- lows. If r = 2, then put
If r ≥ 3, then P( X ) is uniquely written as
where Γ is a finite subset of Nr−² and ϕ k ( Xr−1, Xr ) ∈ Z [ Xr−1, Xr ] a nonzero polynomial. Note that 1 := (g1(P), . . . , gr-2(P)) ∈ Γ. Now put
Let
where σ i (X r ) ∈ Z [X r ] with σ b (X r ) ≠ 0. We show for any integer n with n ≥ b that there is a nonzero polynomial ψ( n )( X r −1, X r ) ∈ Z [ X r −1, X r ] satisfying the following: σ( X r −1, X r )ψ( n )( X r −1, X r ) is written as
where ψ,{n}(Xr) ∈ Z[Xr] for i = 0,1,...,b with ψ{n}(Xr)̸= 0. In the case of b = 0, it is clear that ψ( η )( X r − 1, X r ) = X r − 1 satisfies (4.6). Suppose that b ≥ 1. We check (4.6) by induction on n. If n = b, then putting ψ(b)(Xr−1, Xr) = 1, we get (4.6). Assume that n ≥ b+1. Then the induction hypothesis implies that
fulfills (4.6). ψ(n)(Xr-1,Xr) ≠ 0 because σ b(Xr)ψ(n-1)(Xr-1,Xr) ≠ 0. Let w = max{0, b — 1}. In what follows, we verify that F(Xr_1,Xr) := ψ(κ(w))(Xr_1,Xr) satisfies the statement of Lemma 4.2. Using (4.5) and (4.6), we deduce that the first ( r − 2) symbols of g( P ) and g( FP ) coincide in the case of r ≥ 3. Moreover, we obtain (4.7)
and (4.8)
In fact, if Λ 3 ( FP ) is not empty, then by (4.4), (4.5), and (4.6), we get e(FP) ≤ b − 1. Hence, combining (4.7) and (4.8), we conclude that 9r-1(FP) ≥ κ(e(FP)) because the sequence (κ( n )) π =0 is monotone increasing.
For simplicity, put
Recall that, for i = 1, 2, . . . , r − 2,
SO
for each n ∈ N. Let
where Ak is a nonzero integer. Then
Note that, for each k ∈ Λ, we have |k| ≥ 1 because X, divides P( X ). Without loss of generality, we may assume that Ak 0 ≥ 1. Lemma 4.3. Λ 1 and Λ 2 are not empty. Proof. First suppose that Λ 2 is empty. Then, for each k = (α 1 , . . . ,α τ ), we have a 1 = g 1 , . . . , α r − 1 = g r − 1 . Thus, (4.9) implies that ξ r is an algebraic number, which contradicts to the induction hypothesis. Next, assume that Λ 1 is empty. Then we get
By (4.10), the greatest element of Φ( F ( X r − 1 , X r ) P ( X )) with respect to the lexicographical order on Nr−1 is ( g1, . . . , gr−1). So, Φ(F(Xr−1, Xr)P(X)) is not zero. Namely, X r − 1 does not divide F ( X r − 1 , X r ) P ( X ), which contra- □ dicts (4.3). Let
Denote the greatest element of Λ 1 and Λ 2 with respect to ≥ by k 1 and k 2 , respectively. Let Then, for each k ∈ Λ 2 , we have k ← e. In fact, if Λ 3 is empty, then there exists a positive l with l ≤ r − 2 such that the first (l − 1) symbols in e and k2 coincide, but the l th symbol of e is greater than that of k2. Otherwise, k2 is written as
with a < D , and so k 2 ← e. By the first assumption of Theorem 2.1, for any k ∈ Λ 2 , (4.11)
Lemma 4.2 implies (4.12) 9r-1 ≥ k(e(FP)). Let Ξ be the set of nonnegative integers N such that, for every integer n with 0 ≤ n ≤ N, (4.13)
Note that Ξ is an infinite set. In fact, by the first assumption of Theorem 2.1, we have
If necessary, by increasing C 2 (e), we may assume that λ r (n) ≥ 5 for every n ∈ N with n ≥ C 2 (e), where C 2 (e) is defined in the third assumption of Theorem 2.1. For simplicity, let
Lemma 4.4. Let M and E be any positive real numbers with
and
Then
Proof. Using
we get
Note that M + E ≥ C 2 (e). Thus, using (4.12) and the third assumption of Theorem 2.1 with
we deduce that
which implies Lemma 4.4.
22 Using (4.1) and (4.9), we get, for each R ∈ N,
SO
Let N ∈ N. In what follows, we estimate the number y( N ) of R ∈ N satisfying R ≤ N and Y R > 0, namely,
4.3. Bounds for y ( N ). First, we consider upper bounds for y ( N ). Lemma 4.5.
as N tends to infinity. Proof. For k ∈ Λ and R ∈ N, let
In particular,
By the first assumption of Theorem 2.1, we have (4.16)
Let K = [ D log α N ], where log α N = (log N )/(log α ) and [ x ] is the smallest integer greater than or equal to a real number x . Then by (4.15), (4.16) and the first statement of Lemma 4.1, we get
Since Y (k, R) ≥ 0,
Using YR E Z, we obtain
Next, we estimate lower bounds for y( N ). Lemma 4.6. Let N ∈ N be sufficiently large and I = [U 1 , U 2 ] an interval with I ⊂ [0, N ). Suppose that ρ ( k, x ) = 0 for any integer x ∈ ( U 1 , U 2 ) and k ∈ Λ\{k 0}. Moreover, assume that there exists U ∈ N satisfying
and
Then Yn > 0 for any n ∈ [U 1 , U ). Proof. We prove Lemma 4.6 by induction on n . First we consider the case of n = U − 1. Using (4.14), A k o ≥ 1, and the assumptions on I and U , we obtain
for all sufficiently large N . Next, suppose that Y n > 0 for some n ∈ N with 1 + U 1 ≤ n ≤ U − 1. Then by Ak ρ (k 0 , n ) ≥ 0 we get
4.4. Completion of the proof of Theorem 2.1. We construct intervals I = [ U 1 , U 2 ) satisfying the assumptions of Lemma 4.6. Using (4.11) and the second statement of Lemma 4.1, we deduce the following: Let N E E be sufficiently large. Then the number of nonnegative integers T with T ≤ N such that there exists a k ∈ Λ 2 with ρ(k, T ) > 0 is at most
Set T 1+ r = N and
For any interval I ⊂ R, let |I| denote its length. Then we have (4.18)
Moreover, put
Lemma 4.7. Let N ∈ Ξ be sufficiently large. (1)
Proof. By (4.17) and (4.18)
which implies the first statement of Lemma 4.7. We now check the sec- ond statement. Take positive integers N 0 < N 1 satisfying N i > C 2 (e) and ρ(k 2 , N i ) > 0 for i = 0, 1. If N > N 1 , then there exists j 0 = j 0 ( N ) with
by the definition of T 1 , T 2 , . . . , T 1 + τ . Let J ( j ) ∈ J 1 \J 2 . Then j ≤ j 0 . Hence, for any N ∈ Ξ with N ≥ 6 N 1 ,
By Lemma 4.1 the number of nonnegative integers R with R ≤ N such that there exists a k ∈ Λ 1 with ρ(k, R) > 0 is at most
where C 3 is a positive constant. Say these R ′ s are 0 = R 1 < R 2 < · · · < R µ , where (4.19)
Let R1+μ = N and
Then
Put
(1)
Proof. We check the first statement. For any J = [ T j , T 1+ j ) ∈ J 2 , we have C 2 (e) ≤ T j < T 1+j ≤ N . If N ∈ Ξ is sufficiently large, then by (4.13)
So, using (4.12) and the third assumption of Theorem 2.1 with
we obtain
we get ρ(k 1 , θ(T 1+ j )) > 0. Thus, by the definition of R 1 , R 2 , . . . , R 1+μ 2 (4.20) θ(T1+ j ) = Ri for some i ∈ N. Consequently, we put
we get
Consequently, using Lemma 4.7, we conclude that
□ In what follows, we show that each interval I E I 2 satisfies the assump- tions of Lemma 4.6. The first assumption of Theorem 2.1 implies that, for any k E Λ, (4.24)
By the second assumption of Theorem 2.1, there exists a positive constant C 4 such that, for any real number R with R ≥ C 4 ,
Moreover, by (4.24) there is a positive constant C 5 such that, for each nat- ural number N with N ≥ C 5 ,
SO
Moreover, by (4.27) and (4.28)
Hence, for any integer x with x ∈ ( R i , R i +1 ) and k ∈ Λ\{k 0 }, we have ρ(k, x) = 0 because Λ\{k 0} = Λ 1 ∪ Λ 2 . Thus, by Lemma 4.6, Y n > 0 for any n ∈ N with
Hence, using Lemma 4.8, we conclude that
which contradicts the statement of Lemma 4.5. Therefore we proved Theo- rem 2.1.
A CKNOWLEDGEMENTS
I am deeply grateful to Prof. Masayoshi Hata for many suggestions and for improving the language of this paper. I would like to thank Prof. Yann Bugeaud for careful reading of the manuscript and for giving useful ad- vice. I would like to express my gratitude to Prof. Masanori Katsurada and Prof. Takaaki Tanaka for careful reading and correcting the manuscript. I would like to thank the referee for the constructive and fruitful comments and suggestions on the manuscript. This work is supported by the JSPS fellowship.
R EFERENCES
[1] D. H. Bailey, J. M. Borwein, R. E. Crandall and C. Pomerance, On the binary expansions of algebraic numbers, J. Théor. Nombres Bordeaux 16 (2004), 487–518. [2] D. H. Bailey, P. B. Borwein, S. Plouffe, On the rapid computation of various polylogarithmic constants, Mathematics of Computation 66 (1997), 903-913. Borel, Les probabilités dénombrables et leurs applications [3] É. arithmétiques, Rend. circ. Mat. Palermo 27 (1909), 247–271. [4] É. Borel, Sur les chiffres décimaux de √2 et divers problèmes de prob- abilités en chaîne, C. R. Acad. Sci. Paris 230 (1950), 591–593. [5] P. Corvaja and U. Zannier, Some new applications of the subspace theorem, Compositio Math. 131 (2002), 319–340. [6] S. Daniel, On gaps between numbers that are sums of three cubes, Mathematika 44 (1997), 1–13. [7] A. Durand, Indépendance algébrique de nombres complexes et critère de transcendance, Compositio Math. 35 (1977), 259–267. [8] M. J. Knight, An 'ocean of zeros' proof that a certain non-Liouville number is transcendental, American Mathematical Monthly 98 (1991), 947-949. [9] J. Liouville, Remarques relatives 1° à des classes très-étendues de quan- tités dont la valeur n'est ni rationnelle ni même réducible à des irra- tionnelles algébri-ques; 2° à un passage du livre des Principes où New- ton calcule l'action exercée par une sphère sur un point extérieur, C. R. Acad. Sci. Paris 18 (1844), 883–885. [10] J. Liouville, Nouvelle démonstration d'un théorème sur les irra- tionnelles algébriques, C. R. Acad. Sci. Paris 18 (1844), 910–911. [11] K. Mahler, Arithmetische Eigenschaften der Lösungen einer Klasse von Funktionalgleichungen, Math. Ann. 101  (1929), 342–366. [12] K. Nishioka, Algebraic independence by Mahler's method and S -unit equations, Compositio Math. 92 (1994), 87–110. [13] K. Nishioka, Mahler Functions and Transcendence, Lecture Notes in Math. 1631, Springer, 1996. [14] T. Rivoal, On the bits counting function of real numbers, J. Aust. Math. Soc. 85 (2008), 95–111. [15] K. Roth, Rational approximations to algebraic numbers, Mathematika 2 (1955), 1-20. Corrigendum, pp. 168. [16] W. M. Schmidt, Simultaneous approximation and algebraic indepen- dence of numbers, Bull. Amer. Math. Soc. 68 (1962), 475–478. [17] I. Shiokawa, Algebraic independence of certain gap series, Arch. Math. 38 (1982), 438–442. DEPARTMENT OF MATHEMATICS, COLLEGE OF SCIENCE AND TECHNOLOGY, NIHON U NIVERSITY, 1-8-14 K ANDA -S URUGADAI, C HIYODA - KU , T OKYO 101-8308, J APAN E-mail address: kanekoha@math.cst.nihon-u.ac.jp