Randomness criterion Σ and its applications

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1 Randomness criterion Σ and its applications Sankhya 80-A, Part , Teturo Kamae, Dong Han Kim and Yu-Mei Xue Abstract The Sigma function, which is the sum of the squares of the number of occurrences of every factor, is a criterion of randomness, measuring specially the uniformity of the block distribution. An infinite word whose prefixes attain asymptotically the smallest possible value of it is called Sigma-random. We prove that the Champernowne word is Sigma-random. We also consider less complex words which have values with asymptotically larger order, Sturmian words and almost 0-words. 1 Introduction Let x = x 1 x 2 {0, 1} be an infinite word over the binary alphabet and ξ {0, 1} + be a finite word, where {0, 1} + = k=1 {0, 1}k. Denote by ξ the length of ξ, that is, ξ = k if ξ {0, 1} k. Let x 1 x 2 x n ξ := #{1 i n ξ + 1 ; ξ = x i x i+1 x i+ ξ 1 }. We say that ξ is a factor of x 1 x 2 x n and of x if x 1 x 2 x n ξ 1 and denote ξ x 1 x 2 x n. For an infinite word x = x 1 x 2 {0, 1} and integer n 1,we define the sigma function by the sum of the number of occurrences of every factors, Σ n x := x 1 x 2 x n 2 ξ, ξ {0,1} + The sigma function Σ n was first introduced by Kamae and Xue in [6] to serve as a randomness criterion. We also consider Σ n as a function on finite words of length n. It is known [6] that lim inf Σ n x n for any x {0, 1}, and with respect to the 1 2, 1 2 -Bernoulli measure on {0, 1}, Σ n x lim n 2 =

2 holds almost surely. More precisely, Σ n x 1 x 2 x n attains the minimum in {0, 1} n if n ξ + 1/2 ξ x 1 x 2 x n ξ n ξ + 1/2 ξ holds for any ξ {0, 1} +. We call such x 1 x 2 x n an equi-distributed word. This is equivalent to say that x 1 x 2 x n ξ x 1 x 2 x n η 1 for any ξ, η {0, 1} + with ξ = η. Using the Bruijn graph, we can construct an equi-distributed word of length n for infinitely many n, but we don t know the answer to the following question: Open problem: For any n = 1, 2,, does an equi-distributed word of length n exist? An infinite word x {0, 1} satisfying 1.1 is called a Σ-random word. We call an infinite word x = x 1 x 2... {0, 1} normal if x i/2 i is a normal number, i.e., every factor of length n appear in x with frequency of 2 n. It is known [6] that any Σ-random word is normal, but the converse is not true. In fact, let Σ n k x = x 1 x 2 x n 2 ξ k = 1, 2, ξ {0,1} k and decompose the limit superior of Σ n x/n 2 into 3 parts: where lim sup Σ n x n 2 S B x + S P x + S L x, 1 S B x = lim lim sup K n 2 Σ n k x, k K 1 S P x = lim lim sup K n 2 Σ n k x, 1 S L x = lim lim sup K n 2 K<k n/k k>n/k Σ n k x. 1.2 Then it is shown [6, Theorem 1] that x is Σ-random if and only if S B x = 1, S P x = 0, S L x = 1 2. On the other hand, x is normal if and only if S B x = 1 [6, Corollary 3]. Let x = x 1 x 2 {0, 1} be normal. Then, for any m 1 < m 2 < and l 1 l 2, y := x 1 x 2 x m1 l 1 x 1 x 2 x m2 l2 2

3 is normal. By Lemma 1 in [7], for any N, we can construct y {0, 1} as above such that S P y > N and S L y > N. Thus, a normal word is not necessarily Σ-random. Remark 1. In Definition 2 of the paper [6], the following explanation of the notation AI, I was missing: For any intervals I, I of {1, 2,, n} with I = I, define AI, I = {{i + j, i + j}; j = 0, 1,, I 1}, where i, i are the first elements of I and I, respectively. In this paper, we consider the Champernowne word. For a positive integer l and j = 0, 1,, 2 l 1, we define w l j the binary word of length l representing j, that is w l j := x 1 x 2 x l {0, 1} l, where j = l x i 2 l i. Let γ l be the concatenation of w l j s for j = 0, 1,, 2 l 1 in the order, that is γ l := w l 0w l 1 w l 2 l 1 {0, 1} l2l. For example, γ 1 = 01, γ 2 = , γ 3 = We define the modified Champernowne word β {0, 1} as their concatenations for l = 1, 2,. That is, β = γ 1 γ 2 γ 3 = {0, 1}. The Champernowne word, which will be denoted by β, is a little different from this. It is defined by arranging binary representations of positive integers in the increasing order. That is, with γ l := w l 2 l 1 w l 2 l w l 2 l 1 instead of γ j, define β = γ 1 γ 2 γ 3. Champernowne[4] constructed his celebrated normal number whose decimal expansion is Champernowne word in the alphabet {0, 1,..., 9}. The normality of the modified Champernowne word β is immediate by the normality of Champernowne word β. For some statistical properties of the Champernowne word, consult [2, 9]. The first theorem is to show the Σ-randomness of the modified Champernowne word β: Theorem 1. The modified Champernowne word β is Σ-random. 3

4 Corollary 1. The Champernowne word β is Σ-random. The r-adic versions of Theorem 1 and Corollary 1 hold as well for any r 2. In this case, Σ-random word is defined to be x {0, 1,, r 1} such that Σ n x lim n 2 = r + 1 2r 1 since S B x = 1/r 1 for almost all x with respect to the uniform Bernoulli mesure on {0, 1,, r 1}. Yuval Peres and Benjamin Weiss [12] defined a notion of randomness Poisson random, say for x = x 1 x 2 {0, 1} so as to satisfy the Poisson law of small number. That is, for each m = 0, 1, 2, lim k 1 { } 2 k # ξ {0, 1} k ; x 1 x 2 x n ξ = m = e 1 m!, where we denote n = 2 k + k 1. Suggested by this notion, we define x {0, 1} to be P-random if lim k 1 2 k x 1 x 2 x 2 k +k 1 2 ξ = ξ {0,1} k It should be noted that the Poisson randomness implies neither the Σ- randomness nor the P-randomness since the Poisson randomness is free of replacing a subword of length 2 2k/3 of x nk x nk +1 x nk+1 1 for each k = 1, 2,, where n k = 2 k + k, while it may change the property to be Σ-random or P-random. In fact, by replacing a random subword by a constant word, it increases Σ-value by orders n 2 of the Σ-random word or n of the P-random word, so that the randomness properties are lost. Yuval Peres and Benjamin Weiss [12] proved the following theorem for the case of Poisson random instead of P-random. The proof of our theorem is based on the same idea. Theorem 2. 1 Almost all x {0, 1} are P-random with respect to the 1 2, 1 2 -Bernoulli measure. 2 Any P-random word is normal. 3 Neither the modified Champernowne word β nor the Champernowne word β is P-random. By Theorems 1 and 2, the Σ-randomness does not imply the P-randomness. We don t know whether the P-randomness implies the Σ-randomness or not. For the other extreme, the maximum value of Σ n x is n 2 = n3 3 + on3, 4

5 which is attained by the constant words. It is known by Kamae and Kim [7] that the infinite word x 1 x 2 is eventually periodic if and only if lim Σx 1x 2 x n /n 3 exists and is positive. We are also interested in the intermediate cases, that is, Σ n x increases faster than n 2, but slower than n 3. Especially, we consider the Sturmian words and almost 0-words. Recall that x = x 1 x 2 {0, 1} is a Sturmian word if it has exactly k + 1 factors of length k k = 1, 2,. It is well known [10] that x is a Sturmian word if and only if there exist an irrational number θ 0, 1 called slope and a real number ρ [0, 1 called intercept such that x i = i + 1θ + ρ iθ + ρ for all i = 1, 2,... or 1.4 x i = i + 1θ + ρ iθ + ρ for all i = 1, 2,.... Note that they differs at most two letters. For an irrational slope θ, let a i,2, and p i /q i,2, be the partial quotients and principal convergents of θ, respectively. Then, with p 1 = 1, p 0 = 0, q 1 = 0, q 0 = 1, p i and q i satisfy that p i = a i p i 1 + p i 2, q i = a i q i 1 + q i 2 i = 1, 2,, where a i,2, are the partial quotients of θ. For A, B which depend on some variables, say n, m,, which may satisfy some relations when n, m,, we denote A A B if 0 < lim inf n,m, B A A B if lim n,m, B = 1. A lim sup n,m, B <, For a real number t, we denote t = min n Z t n and t + = max{t, 0}. Theorem 3. Let x = x 1 x 2 {0, 1} be a Sturmian word with the slope θ and a i, p i /q i be the partial quotients and principal convergents of θ. For n = 1, 2,, define m by q m n < q m+1. Then, we have 1 lim inf Σ n x n 2 log n 1, 2 n 2 m 9 a i < Σ n x < 4n m+1 a i

6 The condition that m a i = Om appeared in [11, Corollary 1.65] as a low discrepancy condition of the sequence coming from the irrational rotation. Here, we get the same condition to characterize the minimum increasing order of Σ n. Corollary 2. With the same setting as Theorem 3, Σ n x n 2 log n holds if and only if 1/m m a i is bounded in m, in particular, if a i,2, is bounded. Moreover, if a i,2, is bounded, then Σ n x K<k n/k Σn k x holds with any constant K > 0. In Section 4, we discuss the order of Σ n for x 1 x 2 {0, 1} with n x n = On 1/α for α > 1. We denote by {k i } the increasing sequence given by x ki = 1. In particular, we prove in Corollary 3 that if k i = i α +O1 as i with α 3/2, then Σx 1 x 2 x n n 3 1/α. 2 Champernowne word Define ϕ : {0, 1} + {0, 1, 2, } so that for any x 1 x 2 x l {0, 1} l, ϕx 1 x 2 x l = l x i2 l i. For ξ = x 1 x 2... x l and 1 i j l we write ξ [i,j] = x i x i+1... x j. Lemma 1. For any ξ {0, 1} l γ 1 γ 2 γ k ξ 1. with l 15k 2 k = 1, 2,, we have Proof. Since l 15, our lemma holds trivially for k = 1, 2. We assume k 3. Consider any occurrence of ξ in γ 1 γ 2 γ k to see which of γ j it or a part of it is located in. Since 2 γ j < γ j+1 j = 1, 2,, k 1, there are j 1 and η ξ with η l/3 5k 2 such that at this occurrence, η is located totally in γ j 1. Since j 1 k, η contains at least 5k 2 successive w j 1 i s, and hence, 4k + 1 successive w j 1 i s. Suppose that there exists another occurrence of ξ in γ 1 γ 2 γ k after this. Then, η appears again afterward among at most 2 of consecutive γ j s. Hence one of them, say γ j 2 j 1 j 2 k contains at least 2k successive w j 1 i s, say w j 1 i 1 w j 1 i w j 1 i 1 + 2k 1. This implies that there exists j 2 j 1, i 2 and b {0, 1,, j 2 1} with i 1 < i 2 if j 1 = j 2 such that w j 1 i 1 w j 1 i w j 1 i 1 + 2k 1 b w j 2 i 2 w j 2 i 2 + 1w j 2 i w j 2 2 j where we write A b B if A = c b+1 c l with B = c 1 c 2 c b+1 c l. 6

7 For each i = i 1, i 1 + 1,, i 1 + 2k 1, define τi {0, 1,, j 2 1} so that w j 1 i τi w j 2 rw j 2 r + 1 for some r in 2.1. Then we have τi+1 τi+j 1 mod j 2. Let i 0 be the minimum h such that τh = min{τi; i = i 1, i 1 + 1,, i 1 + 2k 1} and τ 0 = τi 0. Then we have τ 0 + j 1 < j 2 if j 1 < j 2. Moreover, i 0 i 1 j 2 1 by the minimality of i 0. w j 1 i w j 1 i w j 1 0 i 0 + j 2 τ 0 τ 0 w j 2 i w j 2 i wj 2 0 i 0 + j 1 Figure 1: Overlapping between w j 1 and w j 2 Consider the case where j 1 < j 2. Then, we have τ 0 + j 1 < j 2 and i 0 + j 2 i 1 + j j 2 = i 1 + 2j 2 1 i 1 + 2k 1. There exists i 0 with 0 i 0 < i 0 + j 1 2 j 2 1 such that Note that w j 1 i 0 τ0 w j 2 i 0, w j 1 i 0 + j 2 τ0 w j 2 i 0 + j 1. On the other hand, we have ϕw j 1 i 0 + j 2 = ϕw j 1 i 0 + j 2. ϕ w j 2 i 0 + j 1 [τ0 +1,τ 0 +j 1 ] ϕ w j 2 i j1 0 [τ0 +1,τ 0 +j 1 ] + 2 since j 1 1 and j 2 τ 0 + j 1 1, while ϕ w j 2 i 0 [τ0 +1,τ 0 +j 1 ] + j1 + 1 w j 1 i 0 + j 2 = w j 2 i 0 + j 1 [τ0 +1,τ 0 +j 1 ], w j 1 i 0 = w j 2 i 0 [τ0 +1,τ 0 +j 1 ]. Hence, we have j 2 = ϕ w j 1 i 0 + j 2 ϕw j 1 i 0 = ϕ w j 2 i 0 + j 1 [τ0 +1,τ 0 +j 1 ] ϕ w j 2 i 0 [τ0 +1,τ 0 +j 1 ] j1, 7

8 which is a contradiction since j 1 < j 2. Consider the case j 1 = j 2. If b = 0 in 2.1, then we have a contradiction that w j 1 i 1 = w j 2 i 2 for i 1 < i 2. Assume that b 1. Denote j = j 1 = j 2. By 2.1, we have w j i 1 = w j i 2 [b+1,j] w j i [1,b], w j i = w j i [b+1,j] w j i [1,b]. Note that ϕ w j i 2 [b+1,j] + 1 ϕ w j i [b+1,j] ϕ w j i 1 [1,j b] ϕ w j i [1,j b] mod 2 j b, 2.2 and w j i 2 [b+1,j] = w j i 1 [1,j b], w j i [b+1,j] = w j i [1,j b]. Hence, ϕ w j i 2 [b+1,j] ϕ w j i [b+1,j]. Since 2.2 implies that either ϕ w j i 2 [b+1,j] + 1 = ϕ w j i [b+1,j] or it follows that ϕ w j i 2 [b+1,j] + 1 = 2 j b, ϕ w j i [b+1,j] = 0, ϕ w j i 2 [b+1,j] + 1 = ϕ w j i [b+1,j]. 2.3 By exchanging the roles of i 1 and i 2, we have Since 2.3 implies we have ϕ w j i 1 [j b+1,j] + 1 = ϕ w j i [j b+1,j]. ϕ w j i 1 [1,j b] + 1 = ϕ w j i [1,j b], ϕ w j i b + 1 = ϕ w j i 1 + 1, which contradicts with ϕ w j i = ϕ w j i Thus, we have γ 1 γ 2 γ k ξ 1. Lemma 2. Let β = β 1 β 2 {0, 1} be the modified Champernowne word. Then, for any ξ {0, 1} l l = 1, 2, and n = 1, 2,, we have if l 15log 2 n 2. β 1 β 2 β n ξ 1 8

9 Proof. Lemma 2 holds trivially for n = 1, 2., 10. Assume that n > 10. Note that γ 1 γ 2 γ k = k 12 k Define k = 1, 2, by k 22 k + 2 < n k 12 k Assume that l 15log 2 n 2. Since n > 10, k 3 holds. Hence, n > k 22 k + 2 > 2 k and k < log 2 n holds. Since l 15log 2 n 2 > 15k 2, by Lemma 1, we have which completes the proof. β 1 β 2 β n ξ γ 1 γ 2 γ k ξ 1, Let x 1 x 2 x n {0, 1} n n = 1, 2, and ξ {0, 1} l l = 1, 2,. For integers a, b with 0 b < a, define x 1 x 2 x n a,b ξ := #{1 j n l + 1; ξ j x 1 x 2 x n with j b mod a}. Lemma 3. For any k = 1, 2,, ξ {0, 1} l l = 1, 2, and b {0, 1,, k 1}, we have γ k k,b ξ 2 k l +, and hence, γ k ξ k2 k l +. Proof. Let b = 1, 2,, k 1 and ξ {0, 1} k be given with ξ = w k h. If ξ b w k jw k j for some j with 0 j 2 k 2, then we have j + 1 = 2 k b ϕ w k h [k b+1,k] + ϕ w k h +1 [1,k b], where for ζ {0, 1} + with ζ = l, ζ +1 {0, 1} l is defined so that ϕζ +1 ϕζ + 1 mod 2 l. b w k j w k j + 1 ξ [1,k b] ξ [k b+1,l] η ξ +1 [1,k b] Figure 2: η = w k j + 1 [b+l k,b] {0, 1} k l 9

10 Hence, 2.4 holds at most for one j if b = 1, 2,, k 1 is given. Moreover, ξ 0 w k j holds exactly for one j with 0 j 2 k 1, say j = h. This implies that γ k k,b ξ 1 if ξ k for any b = 0, 1,, k 1. Let l = ξ < k. If b + l k, then it is clear that γ k k,b ξ = 2 k l. Let b + l > k. If 2.4 holds for some j with 0 j 2 k 2, then we have j + 1 = 2 2k b l ϕ ξ [k b+1,l] + 2 k b ϕ w k j + 1 [b+l k,b] + ϕ ξ +1 [1,k b]. Hence, there are exactly 2 k l many of j satisfying 2.4 if b = 0, 1,, k 1 is given, since in the above w k j +1 [b+l k+1,b] can be any element in {0, 1} k l. Thus, we have γ k k,b ξ 2 k l +, and γ k ξ k2 k l + for any ξ {0, 1} l l = 1, 2,, which completes the proof. Lemma 4. For any k = 1, 2,, ξ {0, 1} l l = 1, 2, and η {0, 1} +, we have γ k η ξ η ξ k2 k l Proof. Let b {1, 2,, k}. Note that γ k is a multiple of k. Thus, γ k η k,b ξ η k,b ξ is the number of occurrences of ξ in γ k η starting at, say j, with j b mod k and 1 j k2 k. The number of such j with j k2 k 1 is at most 2 k l + as shown in the proof of Lemma 3. Since there can be at most one more occurrence at j = k2 k 1 + b, we have γ k η k,b ξ η k,b ξ 2 k l Thus, γ k η ξ η ξ k2 k l Lemma 5. Let β = β 1 β 2 {0, 1} be the modified Champernowne word. Then, for any ξ {0, 1} l l = 1, 2, and n = 1, 2,, we have β 1 β 2 β n ξ 4n 2 l + log 2 n 2. Proof. Lemma 5 is verified directly for n = 1, 2,, 10. Assume that n 11. Let k 22 k + 2 < n k 12 k for some k 3. Then, β 1 β 2 β n is a prefix of γ 1 γ 2 γ k. By lemmas 3 and 4, we have k 1 β 1 β 2 β n ξ γ 1 γ 2 γ k ξ = γ j γ j+1 γ k ξ γ j+1 γ k ξ + γ k ξ j=1 k 1 j2 j l k2 k l + j=1 k k l k 12 k l+1 + k2 + l 2 k 2 2 3l k > l 2k 12 k l + k 2 4n 2 l + log 2 n 2. 10

11 Proof of Theorem 1: To prove that the modified Champernowne word β = β 1 β 2 {0, 1} is Σ-random, by [6] it is sufficient to prove that lim sup Σ n β n We have an upper bound of 3 terms which are a little different from 1.2: lim sup Σ n β n 2 S B + S P + S L, where 1 S B = lim lim sup K n 2 Σ n k β, k K S P 1 = lim lim sup K n 2 Σ n k β, K<k 15log 2 n 2 S L = lim sup 1 n 2 k>15log 2 n 2 Σ n k β. Since β is known [5] to be normal, we have S B = 1 by [6]. By Lemma 2, if k 15log 2 n 2, we have Σ n k β = Hence, we have ξ {0,1} k β 1 β 2 β n 2 ξ = ξ {0,1} k β 1 β 2 β n ξ = n k + 1. Σ n k β k>15log 2 n 2 n n k + 1 = k=1 nn Thus, S L 1/2. By Lemma 5, we have S P 1 = lim lim sup K n 2 Σ n k β K<k 15log 2 n 2 1 4n 2 lim lim sup K n 2 min{2 k, n} 2 2 k + log 2 n 4 K<k 15log 2 n nlog lim + lim sup 2 n 6 K 2K n 2 = 0 Thus, we have lim sup Σ n β n = 3 2, 11

12 which completes the proof. Proof of Corollary 1: Lemmas 1, 3 and 4 hold for the Champernowne word β = γ 1 γ 2 γ 3. As for Lemmas 2 and 5, since γ i = γ i /2 for i = 1, 2,, we have the same results for β only by replacing log 2 n by log 2 n + 1. Thus, we establish Corollary 1. Proof of Theorem 2: 1 Let X 1 X 2 be independent random variables with P X i = 0 = P X i = 1 = 1/2 i = 1, 2,. For k = 1, 2,, let n = 2 k + k 1. Denoting Z k = X 1 X 2 X n 2 ξ, ξ {0,1} k we prove that E[Z k ] = 2 2 k 1+o1 and V[Z k ] = Ok2 k, so that 1/2 k Z k 2 holds almost surely by the law of large number. We follow the proofs in Section 2 of [6] note Remark 1. It holds that E[Z k ] = E = E = 2 k i,j=1 = 2 k + ξ {0,1} k 2 k 1 Xi X i+k 1 =ξ ξ {0,1} k i,j=1 2 2 k 1 Xi X i+k 1 =ξ1 Xj X j+k 1 =ξ P [X i X i+k 1 = X j X j+k 1 ] 2 k i,j=1 i j P [X i X i+k 1 = X j X j+k 1 ] = 2 k + 2 k 2 k 12 k = 2n1 + o1, where P[X i X i+k 1 = X j X j+k 1 ] = 2 k holds if i j [6]. Let I be the set of intervals in {1, 2,, n + k 1} of length k. For I = {i, i + 1,, i + k 1} I, let X[I] = X i X i+1 X i+k 1. Then, by the same argument of the proof of Lemma 9 of [6] we have V[Z k ] = I,I,J,J I I,I,J,J are coupled + I,I,J,J I I,I,J,J are connected Cov 1 X[I]=X[I ], 1 X[J]=X[J ] Cov 1 X[I]=X[I ], 1 X[J]=X[J ] n 2 kp[x[i] = X[I ]] + On = n 2 k2 k + On = Ok2 k. 12

13 Here, we say I, I, J, J I with I = I and J = J are coupled if I J = I J 1 and I J I J = or I J = I J 1 and I J I J =. The intervals I, I, J, J I are connected if I 1 I 2, I 2 I 3 and I 3 I 4 for some rearrangement I 1, I 2, I 3, I 4 of I, I, J, J. 2 For k = 1, 2,, let n = 2 k + k 1. Let x = x 1 x 2 {0, 1} be P-random. Let Yk x be the random variable on the probability space {0, 1} k, P k with the normalized counting measure P k on {0, 1} k such that Y x k ξ = x 1x 2 x 2 k +k 1 ξ for any ξ {0, 1} k. Then, since E[Yk x2 ] 2 as k, the random variables Yk x for k = 1, 2, are uniformly integrable. Let { Norε, h, k := η {0, 1} k ; η ξ k h } 2 h < ε for any ξ {0, 1}h be the set of ε, k-normal words. Then for any ε, δ > 0 and h, there exists k 0 such that for any k k 0, #Norε, h, k > 1 δ2 k. Take any S {0, 1} k. Then for any m 0, we have 1 2 k #{1 i 2k ; x i x i+1 x i+k 1 S} = 1 2 k x 1 x 2 x n ξ = mp k S {Yk x = m} ξ S m 0 P k S + m=m 0 +1 m=1 mp k Y x k = m. Since the random variables Yk x k = 1, 2, are uniformly integrable, m=m 0 +1 mp kyk x = m 0 as m 0 uniformly in k. Take any ε > 0 and h. Choose m 0 so that mp k Yk x = m < ε k = 1, 2,. 3 m=m 0 +1 Choose δ > 0 so that m 0 δ < ε/3. Let k max{k 0, 3h/ε}, where k 0 is determined as above corresponding to h, ε, δ. Then, P k Norε, h, k > 1 δ holds. Hence, 1/2 k #{1 i 2 k ; x i x i+1 x i+k 1 / Norε, h, k} m 0 δ + m=m 0 +1 mp k Y x k 2ε = m < 3. 13

14 Thus, x 1 x 2 x n Norε, h, n for n = 2 k + k 1 and x 1 x 2 x n Nor2ε, h, n for any sufficiently large n not necessarily related to some k. Since this holds for arbitrary ε > 0 and any sufficiently large n corresponding to ε > 0, we have x 1 x 2 x n h+1 ξ lim = 2 h n h + 1 for any ξ {0, 1} h. Since h is arbitrary, this implies that x is normal. 3 Let β = β 1 β 2 {0, 1} be the modified Champernowne word, that is, β = γ 1 γ 2. For k, let n = 2 k + k 1 and l be the integer such that l 12 l n < l2 l Then, β 1 β 2 β n contains γ l and l k holds. Let L = k l. Then, 0 < L log 2 k holds. Take any ξ {0, 1} k of the form ξ = η0ζη with η {0, 1} L. Then, we have γ l ξ k 2L 1. This is because for any decomposition ζ = ζ ζ with ζ, ζ {0, 1}, we have ξ ζ w l jw l j +1, where j = ϕζ η0ζ and j + 1 = ϕζ η0ζ +1 0ζ +1 is the word in {0, 1} ζ +1 such that ϕ0ζ +1 = ϕ0ζ + 1. The number of ξ {0, 1} k of the above form is 2 k L 1 since we can determine ζη arbitrary. These ξ satisfy β 1 β 2 β n ξ k 2L 1. Hence we have Σ n k β 1β 2 β n k 2L k L 1 = k 1+o1 n. Thus, lim k 1/nΣ n k β 1β 2 β n = 2 does not hold, and β is not P- random. The proof is same for the Champerwowne word β. 3 Sturmian words Proof of 1 of Theorem 3: Let x = x 1 x 2 {0, 1} be a Sturmian word. Since the number of factors of x of length k is k + 1, by Jensen s inequality Σ n k x = n k x 1 x 2 x n 2 n k + 12 ξ k + 1 = k + 1 k + 1 ξ {0,1} k holds for k = 1, 2,. Thus, n Σ n n k x = k + 1 k=1 n + 2 n log 2 Thus, we have n n n k + 1 k + 1 k=1 3 2 n2 5 2 n n2 log n 4 + log 2n 2. lim inf Σ n x n 2 log n 1. 14

15 Let x = x 1 x 2 be a Sturmian word with slope θ and intercept ρ see 1.4. Assume that x i = i + 1θ + ρ iθ + ρ. Then, x i = 0 or 1 according to iθ + ρ [0, 1 θ or [1 θ, 0, where P = {[0, 1 θ, [1 θ, 0} is the partition of T = R/Z. For each positive integer k, the partition of T divided by k + 1 points 0, θ, 2θ,..., k 1θ, kθ is the least common refinement of P, P θ,, P kθ, and hence each element of the partition determine a subword factor of x of length k. The three distance theorem[14, 13] states that that the partition has intervals of three length: For each q i < k, the k + 1 intervals in T = R/Z divided by the points { jθ; 0 j k} consists of k q i + 1 intervals of length q i θ, q i l 1 intervals of length q i 1 θ c 1 q i θ and l + 1 intervals of length q i 1 θ c q i θ. where c = k q i 1 q i and l = k cq i q i 1. Each interval determine a subword of x of length k. See [1] for the detail. As for the number of visit to the subwords corresponding to these intervals, we have the following estimates. Lemma 6. Let d be an integer with 0 d < a j+1. Let J be an interval in T with J = q j 1 θ d q j θ. The number of visits t say to J of an orbit of the rotation by θ among the times 1, 2,, n satisfies t n J dq j + q j 1 J + 1. Proof. Let T be the rotation by θ on T. Consider the induced map T J of T on J. Then T J is also an irrational rotation. The return time on J is q j on a subinterval of length q j 1 θ d + 1 q j θ and d + 1q j + q j 1 on the remaining subinterval of length q j θ. Note that T J with respect to the partition corresponding to the return time produces a balanced word, and hence, the total time until k-th return deviates at most dq j + q j 1 for any k = 1, 2,. Let u J. Then, n with T n u J such that just t number of T u, T 2 u,, T n u are in J are in between t/ J +dq j +q j 1 and t/ J dq j q j 1, since t/ J is the average and dq j + q j 1 is the deviation bound. Therefore, in the general case that {u, T n u} J does not necessarily hold, if t number of u, T u, T 2 u,, T n u are in J, then there exist 0 n 1 < n 2 n with T n 1 J and T n 2 J such that t 1 number of T n1+1 u, T n1+2 u,, T n 2 u are in J. Therefore, n n 2 n 1 t 1/ J dq j q j 1. Moreover, there exist n 1 0 < n n 2 with T n 1 J and T n 2 J such that t + 1 number of T n 1 +1 u, T n 1 +2 u,, T n 2 u are in J. Hence, n n 2 n 1 t + 1/ J + dq j + q j 1. Therefore for a given n, the number of t = 0, 1,, n such that T t u J is estimated as n dq j q j 1 J 1 t n + dq j + q j 1 J

16 By Lemma 6, we deduce the following Lemma. Lemma 7. Let q i < k and k n. Then, the factors ξ of x 1 x 2 x n with ξ = k satisfy one of the following conditions: k q i + 1 many of them satisfy x 1 x 2 x n ξ n k + 1 q i θ q i q i θ + 1, q i l 1 many of them satisfy x 1 x 2 x n ξ n k + 1 q i 1 θ c 1 q i θ l + 1 many of them satisfy c 1q i + q i 1 q i 1 θ c 1 q i θ + 1, x 1 x 2 x n ξ n k + 1 q i 1 θ c q i θ where c = k q i 1 q i and l = k cq i q i 1. For n = 1, 2,, define m = 1, 2, by Lemma 8. We have cq i + q i 1 q i 1 θ c q i θ + 1, q m n < q m Σ n x 1 x 2 x n < 4n2 3 m+1 a i Proof. Let q i < k and k n. By Lemma 7, we have Σ n k x k q i + 1A q i l 1A l + 1A 2 3, with A 1 = n k q i q i θ + 1, A 2 = n k c 1q i + q i 1 q i 1 θ c 1 q i θ + 1, A 3 = n k cq i + q i 1 q i 1 θ c q i θ + 1, where c = k q i 1 /q i. Since 1 q i θ = q i θ < q i /q i+2, we have 1 1 q i < q i θ < q i+2 16

17 Thus, we have q i 1 θ c 1 q i θ < 1 k qi q i q i = 1 1 k k + q i q i+2 qi q i and < 1 q i 1 k q i q i+2 Assume that n. Using these inequalities, we have A 1 < n k + q i n + 1, 1 q i q i+2 max{a 2, A 3 } n k cq i + q i 1 q i 1 θ c 1 q i θ < n k n 1 k + 7n. q i q i Combining them, we have n 2 2n Σ n k x < k q i q i Therefore, for i = 0, 1,, m 1, we have q i+1 Σ n k x < q2 i+1 n n2 2 3q i k=q i +1 4n2 + 65n 2 < 4n2 a i q i 3 q i 1 k + 7n n n2 q i + 65n 2 < 4n2 a i Now assume that n <. This is the case when i = m and k > q m. Then, we have n n n 2 Σ n k x < k q m q m+1 k=q m+1 k=q m+1 + n2 2 n k=q m+1 q m 2n q m n n2 q m+1 q m+1 3q m 1 k + 7n 2 q m+1 q m n n2 q m q m+1 4n2 q m n 2 < 4n2 a m n 2 3q m 3 < 4n2 a m

18 Thus, adding the above terms, we get Σ n x = Σ n 1 x + < n 2 + 4n2 3 n k=2 m i=0 which completes the proof. m 1 Σ n k x = Σn 1 x + a i n2 3 i=0 k=q i +1 Σ n k x + m+1 a i + 50, Lemma 9. Under the same setting as Theorem 3, we have Σ n x > a i. n2 m n k=q m+1 Σ n k x Proof. Let q i k < and ξ be a factor of x of length k. Let k n. By Lemma 7, we have with Σ n k x q i l 1B l + 1B 3 2 +, B 2 = n + k 1 c 1q i q i 1 q i 1 θ c 1 q i θ 1, B 3 = n k + 1 cq i q i 1 q i 1 θ c q i θ 1, where c = k q i 1 q i. Then, we have min{b 2, B 3 } n k + 1 cq i q i 1 q i 1 θ c q i θ 1 > n 2k 1 1 k 1 > nqi 1 k 3k, q i q i where we used 3.2 as follows: q i 1 θ c q i θ > 1 1 q i 1 k q i 1 1 = 1 1 k. q i q i q i Since q i+2 2q i holds for i = 0, 1, 2,, either holds. Assume that Then, since 2q i or q i+2 2 2q i, n 16q i and n n n + 3 q i n 1 2 n 3q i n 1 2 n 3 16 n > n 10, 18

19 we have 1 k=q i Σ n k x > 1 k=q i q i = 1 1 q i k=q i n q i 1 k 3k 2 q i + n n + 3 where u = n /n + 3. Hence we have 1 Σ n k x > 1 3q i n + 3 k=q i a i+1 3 We conclude that 1 n + 3 n n + 3 q i q i k 1 u + q i q i n n + 3 q i 3 3 a i+1 n 12n 10 Σ n k x > a i+1 + a i k=q i n 2 n n = a i n2. 2 k dk, for any i = 0, 1, 2, if both n 16 and n q i+2 are satisfied, and hence, if n 16q i+2 is satisfied. This is because, if 2q i is not satisfied, then a i+1 = 1 and q i+2 2 is satisfied. Hence, q i+2 1 k=q i Σ n k x q i+2 1 Σ n k x a i n2 1 + a i n2 = a i+1 + a i k= The proof is same for the other case that q i+2 2 is not satisfied. Hence, we have Σ n x > n a i qi+2 n. i=0 Since q i+2 2q i holds for any i = 0, 1, 2, and q m n, we have 16q i+2 n for any i m 10. Thus, we have which completes the proof. Σ n x > n2 m i=0 a i+1, Proof of 2 of Theorem 3: Clear from Lemmas 8 and 9. n 2. 19

20 Definition 1. For x 1 x 2 x n {0, 1} n, define Λx 1 x 2 x n = max{ η 2 l : η l x 1 x 2 x n } Lemma 10. [7] For any x 1 x 2 x n {0, 1} n, it holds that Σx 1 x 2 x n Λx 1x 2 x n. 48 Let x = x 1 x 2 {0, 1} be a Sturmian word with the slope θ and a i, and p i /q i, be the partial quotients and principal convergents of θ. Put p 0 = 0, q 0 = 1. For each n = 1, 2,, let m be satisfying q m n < q m+1. There is a sequence of subwords w 1 = 0, w 0 = 1, w 1 = w 0 a1 1 w 1 and w i+1 = w i a i+1 w i 1 i = 1, 2,. Note that w i = q i. For any i = 1, 2,, it holds that x is a one sided infinite word contained in a biinfinite word which is obtained by concatenating w i and w i 1 in a way that w i -blocks are either w a i+1 i or w a i+1+1 i which are separated by isolated w i 1, that is, something like w i 1 w a i+1 i w i 1 w a i+1+1 i w i 1 w a i+1 i w i 1 w a i+1 i w i 1 w a i+1+1 i w i 1. Lemma If m+1 a i log n, then log q m+1 = Olog q m holds. 2 Assume that Σ n x n 2 log n. Then, we have a m+1 = Olog n. It also holds that a m+1 i = Olog n for any fixed i = 0, 1, 2,. Proof. 1 Assume that m+1 a i log n. Then we have logq m+1 1 log q m, since both of n = q m+1 1 and n = q m correspond to the same m. Hence, log q m+1 = Olog q m. 2 Assume that Σ n x n 2 log n. If n = q m+1 1, then we have a m+1 q m n < a m+1 + 1q m. We may assume that a m+1 is sufficiently large, since otherwise, we have nothing to prove. Since x 1 x 2 x n contains wm l with l = a m+1 /2, by Lemma 11, we have Σ n x q2 ml 3 48 q2 ma m n2 a m Since Σ n x n 2 log n, we have a m+1 = Olog n for this special n = q m+1 1, say n 1. The smallest possible n corresponding to m is q m. Let n 2 = q m. Then, since n 2 = q m q m+1 a m = n a m and a m+1 = Olog n 1, we have log n 2 logn loga m log n 1 log log n 1 + C with some constant C. Hence, log n 1 = Olog n 2 and a m+1 = Olog n 2. Thus, a m+1 = Olog n for general n with q m n < q m+1. 20

21 The last part is clear since a m+1 i = Olog q m+1 i and q m+1 i q m+1. Proof of Corollary 2: Consider the following statements: m a i = Om, 3.3 Σ n x n 2 log n, 3.4 m+1 a i log n. 3.5 m m a i = O loga i To prove Corollary 2, we ll show that 3.3 implies 3.4, 3.4 implies 3.5, 3.5 implies 3.6, and 3.6 implies : Assume 3.3. By Theorem 3 1, it is sufficient to prove that Σ n x = On 2 log n. Let m 1 2l m. Then we have n q 2l l l a 2i 1 a 2i + 1 a 2i 1 + 1a 2i + 1 Hence by 3.3 and Theorem 3 2, we have lim sup Σ n x n 2 lim sup log n C lim sup 4/3n 2 m+1 a i + 50 n 2 1/2 log l a 2i 1 + 1a 2i + 1 m+1 a i m 1 loga C lim sup i + 1 with some positive constant C. Thus, Σ n x = On 2 log n m+1 a i m 1 log 2 < : Assume 3.4. Then, by Theorem 3 2, we have > lim sup n 2 m 9 a i 24000Σ n C lim sup x n 2 m 9 a i n 2 log n = C lim sup m 9 a i log n with some constant C > 0. Hence, m 9 a i = Olog n. Moreover, by Lemma 12, m+1 i=m 8 a i = Olog n. Thus, we have m+1 a i = Olog n. 21

22 On the other hand, by Theorem 3 2, we have > lim sup C lim sup Σ n x 4/3n 2 m n 2 log n n 2 m+1 a i i=0 a i = C lim sup log n m+1 a i with some constant C > 0. Thus, log n = O m+1 a i, and together with m+1 a i = Olog n, we have m+1 a i log n : Assume 3.5. Then since m+1 log n < log q m+1 < log a i + 1 = we get : m+1 a i = O m+1 loga i + 1 m+1. loga i + 1. Suppose that 3.3 does not hold, so that we have lim sup m 1 m m a m =. By applying Jensen s inequality to fx := logx + 1, we have 1 m log a i m loga i + 1. m m It follows from this that lim sup m m a i m loga lim sup i + 1 m 1 log 1 m m m a i m a i + 1 =, since x/ logx + 1 as x. Thus, 3.6 does not hold. Assume that a i,2, is bounded. Recall that in the proof of Lemma 9, we have q i+1 Σ n k x < 4n2 a i i = 0, 1,, m. 3 k=q i +1 Take an arbitrary K > 0. Since q i+2 2q i holds for any i = 0, 1, 2,, the number of i = 0, 1,, m + 1 such that q i < K or > n/k is bounded by a constant independent of n. Therefore, k K Σn k x + k>n/k Σn k x is bounded from above by a bounded number of sum of the 22

23 terms k=q i +1 Σn k x or n we have Σ n k x + k K k=q m+1 Σn k n k>n/k x. Hence, by the above inequalities, Σ n k x = On2. By 1 of Theorem 3, this implies that K<k n/k Σn k x Σn x, which completes the proof. 4 Almost 0-words Through out this section, we fix x = x 1 x 2 {0, 1} with x i =. For n = 1, 2,, denote N = Nx, n = x 1 x 2 x n 1 as a function of n. Let {k 1 < k 2 < } = {i 1; x i = 1}. Let Fn x : [0, [0, 1] be the probability distribution function of number of 0 s between two consecutive 1 s in x 1 x 2... x n, i.e., Fn x t := #{0 i N 1 ; k i+1 k i 1 t} + 1 n kn t, N + 1 where we put k 0 = 0. Denoting δf x n t = F x n t F x n t 0, we have Theorem 4. Assume that n N α, and sup t = max{max{k i+1 k i ; 0 i N 1}, n k N } N α 1 δfn xt>0 supn + 1δFn x t = max{#{i; k i+1 k i = s} + 1 n kn =s; s > 0} N β t with α, β such that α β Then, we have Σ n x n 3 1/α. Corollary 3. Assume that k i = i α + O1 for some α 3/2 as i. Then, we have Σ n x n 3 1/α. Define Ξ n 0 x = ξ {0,1} + ξ 1 =0 x 1 x 2 x n 2 ξ, Ξn 1 x = ξ {0,1} + ξ 1 1 x 1 x 2 x n 2 ξ. Lemma 12. We have 2 Ξ n 0 x = t s N + 1dFn t x. s=1 0 23

24 Proof. Consider ξ = 0 s for s = 1, 2,. Then, 0 t ξ = t s for t = 0, 1, 2,. Therefore, x 1 x 2 x n ξ = Hence, we have ξ {0,1} + ξ 1 =0 N 1 i=0 0 k i+1 k i 1 ξ + 0 n k N ξ = 0 t s+1 + N +1dF x n t. x 1 x 2 x n 2 ξ = 2 t s N + 1dFn t x. s=1 0 Lemma 13. Assume that n N α and Assume also that sup N + 1δFn x t = ON β t [0, sup t N α 1 for some α 1. δfn xt>0 for some 0 β 1. Then, we have Ξ n 1 x = On2+β/α. Proof. Assume the conditions in our lemma. Since x 1 x 2 x n ξ N for any ξ {0, 1} + with ξ 1 = 1, we have ξ {0,1} + ξ 1 =1 x 1 x 2 x n 2 ξ N 2 #{ξ {0, 1} + ; ξ 1 = 1, ξ x 1 x 2 x n } N 2 sup δf x n t>0 t N 2 N α 1 2 = N 2α n 2. Let ξ {0, 1} + satisfy ξ 1 2. Then, ξ can be written uniquely as ξ = 0 a 10 b 1. Since each occurrence of such ξ in x 1 x 2 x n corresponds to different occurrence of 10 b 1 in x 1 x 2 x n. Hence, Therefore, ξ {0,1} + ξ 1 2 x 1 x 2 x n ξ N + 1δFn x b sup N + 1δFn x t. t [0, x 1 x 2 x n 2 ξ sup t [0, N + 1δF x n t ξ {0,1} + x 1 x 2 x n ξ sup N + 1δFn x t n 2 = ON β n 2 = On 2+β/α. t [0, 24

25 Thus, we have Ξ n 1 x = ξ {0,1} + ξ 1 =1 x 1 x 2 x n 2 ξ + ξ {0,1} + ξ 1 2 x 1 x 2 x n 2 ξ = On2+β/α. Proof of Theorem 4: Assume that x {0, 1} satisfies the conditions in Theorem 4. By Lemma 13, we have 2 Ξ n 0 x = t s N + 1dFn t x. s=1 Under the conditions N + 1dF x n t = N + 1, 0 tn + 1dF x n t + N = n for the probability distribution F x n t on [0, t 0 ], where t 0 := sup δf x n t>0 t N α 1, we ll show that an upper and a lower estimates of Ξ n 0 x coincide in the asymptotical order. That is, 2 Ξ n 0 x = t s N + 1dFn x t = s=1 0 t 0 t0 s=1 t 0 s=1 0 2 t 0 s N + 1dFn x t t 0 s N t 3 0N 2 N 3α 1 n 3 1/α. On the other hand, Ξ n 0 x = t s N + 1dFn x t = s=1 0 tn + 1dFn x t s= s 1N + 1dF x n t N α 1 n N s 1N N α sn 2 s=1 s=0 N α 1 N 2 N α 1 s 2 N 2 N 3α 1 = N 3α 1 n 3 1/α. s=

26 Thus, we have Ξ n 0 x n3 1/α. We already proved that Σ n 1 x = On2+β/α. Since α β + 1, we have 2 + β/α 2 + α 1/α = 3 1/α. Thus, Σ n x = Ξ n 0 x + Σ n 1 x n 3 1/α. Proof of Corollary 3: Let x {0, 1} satisfy the condition in Corollary 3. Then, it satisfies the conditions in Theorem 4 with the same α and β = 2 α +. Since α 3/2, we have α β + 1. Hence by Theorem 5, Σ n x n 3 1/α. Acknowledgement: The authors would like to thank Osaka City University for inviting two of the authors and giving opportunity of the collaboration. This work is partially supported by NSFC grant No and No and NRF-2015R1A2A2A We thank also the referees for their valuable suggestions. In fact, the conditions in Lemma 1 and succeeding Lemmas are improved by one of the suggestions. Also, some of nontrivial errors are corrected by the suggestions. References [1] P. Alessandri and V. Berthé, Three distance theorems and combinatorics on words, Enseign. Math , pp [2] A. Belshaw and P. Borwein, Champernowne Number, Strong Normality, and the X Chromosome, Computational and Analytical Mathematics, pp , Springer Proc. Math. Stat., 50. Springer, New York [3] V. Berthé, S. Ferenczi, C. Mauduit and A. Siegel ed., Substitutions in dynamics, arithmetics and combinatorics, Lecture Notes in Mathematics 1794, Springer-Verlag, [4] D.G. Champernowne, The construction of decimals normal in the scale of ten, J. London Math. Soc pp [5] P. Erdős and A. Rényi, On a new law of large numbers, J. Analyse Math , pp [6] T. Kamae and Y-M. Xue, An easy criterion for randomness, Sankhya A , pp

27 [7] T. Kamae and D.H. Kim, A characterization of eventual periodicity, Theoret. Comput. Sci , pp. 1 8 [8] D.H. Kim, Return time complexity of Sturmian sequences, Theoret. Comput. Sci , pp [9] C. Mauduit and A. Sárközy, On finite pseudorandom binary sequences: II. The Champernowne, Rudin-Shapiro, and Thue-Morse sequences, a further construction, J. Number Theory pp [10] M. Morse and G.A. Hedlund, Symbolic dynamics II: Sturmian sequences, Amer. J. Math , pp [11] M. Drmota and R. Tichy. Sequences, discrepancies and applications. Lecture Notes in Mathematics, Springer-Verlag, Berlin, [12] Yuval Peres and Benjamin Weiss, Private communication [13] N.B. Slater, Gaps and steps for the sequence nq mod 1, Proc. Cambridge Phil. Soc pp [14] V.T. Sós, On the distribution mod 1 of the sequence nα., Ann. Univ. Sci. Budap. Rolando Eötvös, Sect. Math , pp Teturo Kamae Advanced Mathematics Institute, Osaka City University Osaka, Japan kamae@apost.plala.or.jp home page Dong Han Kim Department of Mathematics Education, Dongguk University - Seoul, Seoul 04620, Republic of Korea kim2010@dongguk.edu Yu-Mei Xue School of Mathematics and System Sciences & LMIB, BeiHang University Beijing , PR China yxue@buaa.edu.cn 27