Prime-like sequences leading to the construction of normal numbers

Prime-like sequences leading to the construction of normal numbers Jean-Marie De Koninck 1 and Imre Kátai 2 Abstract Given an integer q 2, a q-normal number is an irrational number η such that any reassigned sequence of k digits occurs in the q-ary exansion of η at the exected frequency, namely 1/q k. Given an integer q 3, we consider the sequence of rimes reduced modulo q and examine various ossibilities of constructing normal numbers using this sequence. We create a sequence of indeendent random variables that mimics the sequence of rimes and then show that for almost all outcomes this allows to obtain a normal number. AMS Subject Classification numbers: 11K16, 11A41, 11N13 Key words: normal numbers, rimes, arithmetic rogression Édition du 11 setembre 2012 1 Introduction Given an integer q 2, a q-normal number, or simly a normal number, is an irrational number whose q-ary exansion is such that any reassigned sequence, of length k 1, of base q digits from this exansion, occurs at the exected frequency, namely 1/q k. In earlier aers [3], [4], [5], [6], we used the comlexity of the factorization of integers to create large families of normal numbers. In this aer, given an integer q 3, we consider the sequence of rimes reduced modulo q and examine various ossibilities of constructing normal numbers using this sequence. Let A q := {0, 1,..., q 1}. Given an integer t 1, an exression of the form i 1 i 2... i t, where each i j A q, is called a finite word of length t. The symbol Λ will denote the emty word. We let A t q stand for the set of all words of length t and A q stand for the set of all the words regardless of their length. An infinite sequence of digits a 1 a 2..., where each a i A q, is called an infinite word. An infinite sequence of base q digits a 1 a 2... will be said to be a normal sequence if any reassigned sequence of k digits occurs at the exected frequency of 1/q k. 1 Research suorted in art by a grant from NSERC. 2 Research suorted by ELTE IK. 1

Given a fixed integer q 3, let { Λ if n, q) 1, f q n) = l if n l mod q), l, q) = 1. Further, letting ϕ stand for the Euler function, set B ϕq) = {l 1,..., l ϕq) } be the set of reduced residues modulo q. Let stand for the set of all rimes, writing 1 < 2 < for the sequence of consecutive rimes, and consider the infinite word ξ q = f q 1 )f q 2 )f q 3 )... We first state the following conjecture. Conjecture 1. The word ξ q is a normal sequence over B ϕq) in the sense that given any integer k 1 and any word β = r 1... r k Bϕq) k, then, setting ξ q N) = f q 1 )f q 2 )... f q N ) for each N N and we have M N ξ q β) := #{γ 1, γ 2 ) ξ N) q = γ 1 βγ 2 }, M N ξ q β) lim N N = 1 ϕq) k. Now, with the above notation, consider the following weaker conjecture. Conjecture 2. For every finite word β, there exists a ositive integer N such that M N ξ q β) > 0. Remark 1. Observe that, in 2000, Shiu [10] rovided some hoe in the direction of a roof of this last conjecture by roving that given any ositive integer k, there exists a string of congruent rimes of length k, that is a set of consecutive rimes n+1 < n+2 < < n+k where i stands for the i-th rime) such that n+1 n+2 n+k a mod q), for some ositive integer n, for any given modulus q and ositive integer a relatively rime with q. 2

Let ε n be a real function which tends monotonically to 0 as n but in such a way that log log n)ε n as n. Letting n) stand for the smallest rime factor of n, consider the set 1.1) N εn) := {n N : n) > n εn } = {n 1, n 2,...}. We then have the following conjecture. Conjecture 3. Let n 1 < n 2 < be the sequence defined in 1.1). Then the infinite word ξ q := f q n 1 )f q n 2 )... is a normal sequence over the set {l mod q : l, q) = 1}. Although the roblem of generating normal numbers using the sequence of rimes does seem inaccessible, we will nevertheless manage to create large families of normal numbers, in the direction of Conjectures 1, 2 and 3, but this time using rime-like sequences. 2 Main results Theorem 1. Let n 1 < n 2 < be the sequence defined in 1.1). Then the infinite word η q := res q n 1 )res q n 2 )..., where res q n) = l if n l mod q), contains every finite word whose digits belong to B ϕq) infinitely often. Remark 2. It is now convenient to recall a famous conjecture concerning the distribution of rimes. Let F 1,..., F g be distinct irreducible olynomials in Z[x] with ositive leading coefficients) and assume that the roduct F := F 1 F g has no fixed rime divisor. Then the famous Hyothesis H of Schinzel and Sierinski [9] states that there exist infinitely many integers n such that each F i n) i = 1,..., g) is a rime number. The following quantitative form of Hyothesis H was later given by Bateman and Horn [1],[2]): Bateman-Horn Hyothesis) If QF 1,..., F g ; x) stands for the number of ositive integers n x such that each F i n) i = 1,..., g) is a rime number, then QF 1,..., F g ; x) = 1 + o1)) CF 1,..., F g ) h 1 h g where h i = deg F i and CF 1,..., F g ) = x log g x 1 1 ) g 1 ρ) ) ), 3 x ),

with ρ) denoting the number of solutions of F 1 n) F g n) 0 mod ). Theorem 2. Let β be an arbitrary word belonging to B k ϕq) and let ξ q be defined as in Conjecture 3. If the Bateman-Horn Hyothesis holds, then Let λ m = M N ξ q β) as N. { 0 if m = 1, 2,..., 10, 1/ log m if m 11. Let ξ m be a sequence of indeendent random variables defined by P ξ m = 1) = λ m and P ξ m = 0) = 1 λ m. Let Ω be the set of all ossible events ω in this robability sace. Let ω be a articular outcome, say m 1, m 2,..., that is one for which ξ mj = 1 for j = 1, 2,... and ξ l = 0 if l {m 1, m 2,...}. Now, for a fixed integer q 3, set res q m) = l if m l mod q), with l A q. Then, let η q ω) be the real number whose q-ary exansion is given by We then have the following result. η q ω) = 0.res q m 1 )res q m 2 )... Theorem 3. The number η q ω) is a q-normal number for almost all outcomes ω. 3 Preliminary results For here on, the letter c will be used to denote a ositive constant, but not necessarily the same at each occurrence. Lemma 1. Let q 2, k 1 and M 1 be fixed integers. Given any nonnegative integer n < q M, write its q-ary exansion as n = M 1 j=0 ε j n)q j, ε j n) A q and, given any word α = b 1... b k A k q, set E α n) := #{j {0, 1,..., M k} : ε j n)... ε j+k 1 n) = α}. Then, there exists a constant c = ck, q) such that E α n) M ) 2 c q M M. q k 0 n<q M 4

Proof. Let Then, Σ 1 := Similarly, Σ 2 := = = { 1 if c1,..., c fc 1,..., c k ) = k ) = b 1,..., b k ), 0 otherwise. 0 n<q M E α n) = 0 n<q M E α n) 2 0 n<q M M k 1 j 1 =0 0 n<q M j 1 j 2 k + = Σ 2,1 + Σ 2,2, 0 n<q M M k 1 j 2 =0 0 n<q M j 1 j 2 >k M k 1 j=0 fε j n),..., ε j+k 1 n)) = q M k M k). fε j1 n),..., ε j1 +k 1n)) fε j2 n),..., ε j2 +k 1n)) fε j1 n),..., ε j1 +k 1n)) fε j2 n),..., ε j2 +k 1n)) say. On the one hand, it is clear that fε j1 n),..., ε j1 +k 1n)) fε j2 n),..., ε j2 +k 1n)) 3.1) 0 Σ 2,1 2k + 1)q M k M k) cq M M. On the other hand, to estimate Σ 2,2, first observe that for fixed j 1, j 2 with j 1 j 2 > k, we have to sum 1 over those n [0, q M 1[ for which ε j1 n)... ε j1 +k 1n) = α = ε j2 n)... ε j2 +k 1n). But this occurs exactly for q M 2k many n s. Thus, 3.2) Σ 2,2 = q M 2k 1 = q M 2k M 2 + Oq M M). j 1 j 2 >k 0 j 1,j 2 M k 1 In light of 3.1) and 3.2), it follows that E α n) Mq ) 2 = Σ k 2 2 Mq Σ k 1 + M 2 0 n<q M qm q2k = q M 2k M 2 + Oq M M) 2 M 2 = Oq M M), thus comleting the roof of the lemma. 5 q 2k qm + M 2 qm q2k

Lemma 2. Given a fixed ositive integer R, consider the word κ = c 1... c R A R q. Fix another word α = b 1... b k A k q, with k R. Let K 1 stand for the number of solutions γ 1, γ 2 ) of κ = γ 1 αγ 2, that is the number of those j s for which c j+1... c j+k = α. Then, given fixed indices i 1,..., i H, let K 2 be the number of solutions of c j+1... c j+k = α for which {j + 1,..., j + k} {i 1,..., i H } = holds. Then, Proof. The roof is obvious. 0 K 1 K 2 2kH. Lemma 3 Borel-Cantelli Lemma). Let {E n } n N be an infinite sequence of events in some robability sace. Assuming that the sum of the robabilities of the E n s is finite, that is, P E n ) < +, then the robability that n=1 infinitely many of them occur is 0. Proof. For a roof of this result, see the book of Janos Galambos [8]. Lemma 4. Let F 1,..., F g be distinct irreducible olynomials in Z[x] with ositive leading coefficients) and set F := F 1 F g. Let ρ) stand for the number of solutions of F n) 0 mod ) and assume that ρ) < for all rimes. Write n) for the smallest rime factor of the integer n 2 and assume that u and x are real numbers satisfying u 1 and x 1/u 2. Then, #{n x : F i n) = q i for i = 1,..., k} = x 1 ρ) ) <x { 1/u 1 + O F ex ulog u log log 3u log k 2))) + O F ex } log x)). Proof. This is Theorem 2.6 in the book of Halbertsam and Richert [7]. 4 Proof of Theorem 1 Theorem 1 is essentially a consequence of Lemma 4. Indeed, letting a 1 <... < a k be ositive integers corime to q and considering the roduct of linear olynomials we have that 4.1) F n) := qn + a 1 ) qn + a k ), #{n [x, 2x] : F n)) > 2qx + a k ) εx } = 1 + o1))x 6 <x εx 1 ρ) ).

If n is counted in 4.1), we certainly have that qn+a j ) > qn+a j ) ε qn+a j for j = 1,..., k. On the other hand, the desired numbers qn + a j, j = 1,..., k, are consecutive integers with no small rime factors for all but a negligible number. Indeed, if they were not consecutive, then there would be an integer b a 1, a k ) such that qn + b) > x εx. Set G b n) := qn + b. Then we would have 4.2) #{n [x, 2x] : F n)g b n)) > x εx } = 1 + o1))x 1 ρ ) b), <x εx <x εx where ρ b ) stands for the number of solutions of F n)g b n) 0 mod ). Since ρ) = k recall that the F i s are linear) and ρ b ) = k + 1 if q, > a k, it follows that we have the following two oosite inequalities: 1 ρ) ) Ca 1,..., a k ) ε x log x) k, <x εx 1 ρ ) b) Ca 1,..., a k ) ε x log x) k 1. Now, for the choice of b, we clearly have a k a 1 + 1 k ossible values. We have thus roved that for every large number x, there is at least one n [x, 2x] for which the numbers qn+a 1,..., qn+a k are consecutive integers without small rime factors, that is for which qn + a j ) > qn + a j ) ε qn+a j, thus comleting the roof of Theorem 1. 5 Proof of Theorem 2 The roof of Theorem 2 is almost similar to the one of Theorem 1. Indeed assume that the Bateman-Horn Hyothesis holds see Remark 2 above). Then, let a 1 be a ositive integer such that a 1 b 1 mod q) and a 1 0 mod D), where D = π, where π are rimes. Similarly, let a 2 be a π k π q ositive integer such that a 2 b 2 mod q) and a 2 0 mod D), with a 2 > a 1. Continuing in this manner, that is if a 1,..., a l 1 have been chosen, we let a l b l mod q) with D a l and a l > a l 1. Then, alying the Bateman-Horn Hyothesis, we get that if 0 < a 1 < < a k are k integers satisfying a j, q) = 1 for j = 1,..., k, then for each ositive integer n, setting F n) = qn + a 1 ) qn + a k ), letting ρm) = #{ν mod m) : F ν) 0 mod m)}, 7

so that ρm) = 0 if m, q) > 1 and ρ) < for each rime, and further setting Π x :=, qx+a k we have that, as x, letting µ stand for the Moebius function, 1 = µδ) = µδ) 1 n x δ Π x n x F n),πx)=1 5.1) δ F n),π x) = 1 + o1))x δ Π x µδ)ρδ) δ x = 1 + o1))c log k x, n x F n) 0 mod δ) = 1 + o1))x qx+a k 1 ρ) ) where c is a ositive constant which deends only on a 1,..., a k. Now, we can show that almost all rime solutions π 1 < < π k reresent a chain of consecutive rimes. To see this, assume the contrary, that is that the rimes π 1 < < π k are not consecutive, meaning that there exists a rime π satisfying π 1 < π < π k and π {π 2,..., π k 1 }. Assume that π l < π < π l+1 for some l {1,..., k 1}. We then have π 2 = π 1 + a 2 a 1, π 3 = π 1 + a 3 a 1,. =. π l = π 1 + a l a 1,. =. π k = π 1 + a k a 1, π = π 1 + d, where a l a 1 < d < a l+1 a 1. We can now find an uer bound for the number of such k + 1 tules. Indeed, by using the Brun-Selberg sieve, one can obtain that the number x of such solutions u to x is no larger than c, which in light of 5.1) log k+1 x roves our claim, thus comleting the roof of Theorem 2. 6 Proof of Theorem 3 Let N 3. Choose a ositive integer R which is such that S := q N + qr < q N+1. Then, set Y N,S = ξ n and θ N,S = ξ n ξ n+1 + + ξ n+q 1 ). q N n<s 8 q N n<s

Then EY N,S ) = q N n<s 1 S log n = log S qn log q N ) ) q N + O, N 2 while EY N,S EY N,S )) 2 c q N n<s From the Tchebyshev inequality, we then get that ) q 6.1) P Y N,S EY N,S ) > N+1 log q N+1 Similarly ) q 6.2) P θ N N,S > c 1 N 2 < 1 N 2. λ n c qn N. < c N 2. Now, given T integers n 1 < < n T located in the interval [q N, S 1], consider the set B n1,...,n T = {ω : ξ nj = 1 for j = 1,..., T and ξ m = 0 if m {n 1,..., n T }, m [q N, S 1]}. Further set σ n = λ n 1 λ n = 1 logn/e) and W N,S = q N n<s 1 λ n ). From the above definitions of B n1,...,n T, σ n and W N,S, it follows that 6.3) P B n1,...,n T ) = σ n1 σ nt W N,S. Let us now introduce the intervals so that J a = [q N + aq, q N + a + 1)q 1] a = 0, 1,..., R 1), [q N, S] = q N q N 1 1 Given n 1,..., n T, let J M1,..., J MH be those intervals which contain at least two elements say, J Mj contains k j 2 elements) from the set {n 1,..., n T }. a=0 J a. 9

Then it follows from 6.2) that P B n1,...,n T ) < c 1 H N. 2 j=1 k j cq N /N 2 Let us now consider those n 1,..., n T for which H j=1 k j < cq N /N 2. Consider those elements amongst n 1,..., n T which have k j 2 fixed elements in J Mj j = 1,..., H) and exactly one element in the intervals J a1,..., J au. Define the quantities L and U by H k j = L and U = T L. j=1 Here 0 a 1 < < a U R 1 are such that {a 1,..., a U } {M 1,..., M H } =. We shall denote that articular set of n 1,..., n T by Da 1,..., a UN ), that is a set that contains exactly q U disjoint sets. Since, for j = 1,..., q 1, we have c 0 σ n σ n+j n log 2 n c 1 q N N, 2 it follows from 6.3) that P B n1,...,n T ) = U σ q N +a j q j=1 H l=1 )) T 1 σ q N +M l q W N,S 1 + O. N 2 q N Since T/N 2 q N 0 as N, it follows that )) T 1 1 + O = 1 + o1) N 2 q N as N. All this means that 6.4) P B n1,...,n T ) = 1 + o1))p B n 1,...,n T ) if n 1,..., n T and n 1,..., n T belong to the same set Da 1,..., a U ). For a given outcome ω, we now consider 6.5) η N,S) q ω) := res q n 1 )... res q n T ) and we let F β η q N,S) ω)) be the number of occurrences of the word β as a subword in η q N,S) ω). Now, setting Z N,S := { ω : Y N,S EY N,S ) > N 10 q N+1 log q N+1 }

and it follows from 6.1) and 6.2) that q V N,S := {ω N : θ N,S > c 1 P Z N,S ) + P V N,S ) < N 2 }, c N 2. Now, assume that ω Z N,S V N,S. Then, recall definition 6.5) and set S β n 1,..., n T ) = #{η N,S) q = γ 1 βγ 2 : γ 1, γ 2 A q}. In the set Dn 1,..., n T ), the elements from J aj can be written as q N + qa j + l j, where l j A q j = 1,..., U). Furthermore, write each integer µ q U 1 as µ = l 1 + l 2 q + + l U q U 1. Then we get E β µ) = #{γ 1, γ 2 ) A q A q : l 1... l U = γ 1 βγ 2 }. Using Lemma 2, we then obtain that S β n 1,..., n T ) E β µ) 2kH + 1). Now Da 1,..., a U ) is characterized by choosing all ossible values l 1... l U A U q. Hence, letting δ N = 1/N, we can aly Lemma 1 and obtain that the number of those l 1... l U for which E βµ) U > Uδ q k N is less than cqn. Hence, from 6.4), we obtain that 6.6) Uδ 2 N n 1,...,n T ) Da 1,...,a U ) E β µ) U q k >Uδ N P B n1,...,n T ) < c 1 EY N,S )δ 2 N n 1,...,n T ) Da 1,...,a U ) P B n1,...,n T ). Now, summing the inequality 6.6) over all ossible values of n 1,..., n T for which ω Z N,S V N,S, the new right hand side of 6.6) is then no c 1 larger than EY N,S ) N 2. Collecting the above inequalities, we obtain that if { K N,S := ω : ω Z N,S, E βη q N,S) ) T 2T }, q k N 11

then Hence, if we let P K N,S ) < c N 2. S j = q N + q N j log N j = 1,..., m N ), where m N = q 2 q) log N, we then have mn ) ) 6.7) P KN,Sj Z N,Sj V N,Sj < c log N. N 2 j=1 Let Q be the set of those ω which belong to infinitely many of the sets K N,Sj Z N,Sj V N,Sj. Now, summing 6.7) on N = 1,...,, we obtain a finite sum. We may therefore aly the Borel-Cantelli Lemma Lemma 3) and conclude that P Q) = 0. Now let M 1 { < M 2 < be the sequence of integers which are} the members of the set q N a + log N qn 1 : a = 0,..., m N, N = 3, 4,..., and let ω Q. Then, regarding the sequence we have that ξ R = res q m 1 )... res q m R ), 6.8) Since 1 M j+1 M j F β ξ Mj ω)) M j 1 q k j ). 1 as j, it follows from 6.8) that the relation F β ξ n ω)) n 1 q k n ) also holds. Since this assertion is true for every finite word β, the roof of the theorem is comlete. References [1] P.T. Bateman and R.A. Horn, A heuristic asymtotic formula concerning the distribution of rime numbers, Math. Com. 16 1962), 363-367. [2] P.T. Bateman and R.A. Horn, Primes reresented by irreducible olynomials in one variable, Proc. Symos. Pure Math. 8 1965), 119-135. 12

[3] J.M. De Koninck and I. Kátai, Construction of normal numbers by classified rime divisors of integers, Functiones et Aroximatio 45.2 2011), 231 253. [4] J.M. De Koninck and I. Kátai, Construction of normal numbers by classified rime divisors of integers II, Functiones et Aroximatio to aear). [5] J.M. De Koninck and I. Kátai, Normal numbers created from rimes and olynomials, Uniform Distribution Theory 7 2012), no.2, 1 20. [6] J.M. De Koninck and I. Kátai, Some new methods for constructing normal numbers, Annales des Sciences Mathématiques du Québec to aear). [7] H. Halberstam and H.E. Richert, Sieve Methods, Academic Press, New York, 1974. [8] J. Galambos, Introductory Probability Theory, Marcel Dekker, New York, 1984. [9] A. Schinzel and W. Sierinski, Sur certaines hyothèses concernant les nombres remiers, Acta Arith. 4 1958), 185-208; Corrigendum: ibid. 5 1959), 259. [10] D. Shiu, Strings of congruent rimes, J. London Math. Soc. 2) 61 2000), no.2, 359-363. Jean-Marie De Koninck Dé. de mathématiques et de statistique Université Laval Québec Québec G1V 0A6 Canada jmdk@mat.ulaval.ca Imre Kátai Comuter Algebra Deartment Eötvös Loránd University 1117 Budaest Pázmány Péter Sétány I/C Hungary katai@comalg.inf.elte.hu JMDK, le 11 setembre 2012; fichier: normal-rimes-2012-fea.tex 13