On a Density for Sets of Integers R. J. Cintra∗ L. C. Rˆego† H. M. de Oliveira‡ R. M. Campello de Souza§ Abstract A relationship between the Riemann zeta function and a density on integer sets is explored. Several properties of the examined density are derived. Keywords Number theory, probability theory, arithmetization 1 INTRODUCTION of A is given by 1 Several measures for the density of sets have been discussed in ∑a ≤n ℓd(A)= lim i ai . (5) the literature [1–6]. Presumably the most employed tool for eval- n→∞ logn uating the density of sets is the asymptotic density, also referred to as natural density. The asymptotic density is expressed by In [1], Bell and Burris bring a good exposition on the Dirich- let density. The Dirichlet density is defined as the limit of the kA ∩{1,2,3,...,n}k d(A)= lim , (1) ratio between two Dirichlet series. Let A ⊂ B be two sets. The n→∞ n generating series of A is given by provided that such a limit does exist. The symbol k·k denotes ∞ N(A,n) the cardinality, and A is a set of integers. Analogously, the lower A(s)= ∑ , (6) ns and upper asymptotic densities are defined by n=1 kA ∩{1,2,3,...,n}k where N(A,n) is a counting function that returns the number of d(A)= liminf , (2) elements in A of norm n [1]. The Dirichlet density is then ex- n→∞ n kA ∩{1,2,3,...,n}k pressed by d¯(A)= limsup , (3) n→∞ n A(s) D(A)= lim , (7) The asymptotic density is said to exist if and only if both the s↓α B(s) lower and upper asymptotic densities do exist and are equal. where B(s) is the generating series of B and α is an abscissa of Although the asymptotic density does not always exist, convergence[1]. In [6], we also find the Dirichlet density defined the Schnirelmann density [2, 3] is always well-defined. The as Schnirelmann density is defined as 1 D(A)= lim(s − 1) ∑ , (8) arXiv:1502.02601v1 [stat.ME] 9 Feb 2015 s↓1 ns kA ∩{1,2,3,...,n}k n∈A δ(A)= inf . (4) n≥1 n whenever the limit exists. This density admits lower and up- per versions, simply by replacing the above limit by liminf and Interestingly, this density is highly sensitive to the initial ele- limsup, respectively. ments of sequence. For instance, if 1 6∈ A, then δ (A)= 0 [7]. The aim of this work is to investigate the properties of the Another interesting tool is the logarithmic density [4–6]. Let Dirichlet density as defined in Equation 7 in the particular case A = {a ,a ,a ,...} be a set of integers. The logarithmic density 1 2 3 where the set B is the set of natural numbers. This induces a ∗R. J. Cintra is with the Signal Processing Group, Departamento de Es- density based on the Riemann zeta function. tat´ıstica, Universidade Federal de Pernambuco. Email: [email protected] †L. C. Rˆego is with Departamento de Estat´ıstica, Universidade Federal de Pernambuco. Email: [email protected] 2 ADENSITY FOR SETS OF INTEGERS ‡H. M. de Olveira was with the Departamento de Eletrˆonica e Sistemas Uni- versidade Federal de Pernambuco. Now he is with the Signal Processing Group, In this section, we investigate the particular case of the Dirich- Departamento de Estat´ıstica, Universidade Federal de Pernambuco. Email: let function, applied for subsets of the natural numbers. In this [email protected] §R. M. Campello de Souza is with the Departamento de Eletrˆonica e Sistemas case, taking into consideration the usual norm, where the norm Universidade Federal de Pernambuco. Email: [email protected] of any natural number is equal to its absolute value, the counting 1 function N(·,·) becomes Proof: dens(Ac)= dens(N − A)= dens(N) − dens(N ∩ A)= 1 − dens(A). 1, if n ∈ A, N(A,n)= (9) Proposition 2 (Monotonicity) The discussed density is a mono- 0, otherwise. ( tone function, i.e., if A ⊂ B, then dens(A) ≤ dens(B). Definition 1 The density dens of a subset A ⊂ N is given by Proof: We have that ∞ ∑ N(A,n) 1 , n=1 ns ∑ 1 dens(A) lim ∞ (10) n∈B ns s↓1 ∑ N(N,n) 1 dens(B)= lim (19) n=1 ns s↓1 ζ(s) 1 ∑n A s 1 1 ∈ n ∑n A s ∑n B A s = lim ∞ (11) ∈ n ∈ − n s↓1 ∑ 1 = lim + lim (20) n=1 ns s↓1 ζ(s) s↓1 ζ(s) ∑ 1 ∑ 1 n∈A ns n∈A ns = lim ζ , s > 1, (12) ≥ lim (21) s↓1 (s) s↓1 ζ(s) if the limit exists. The quantity ζ(·) denotes the Riemann zeta = dens(A). (22) function [8]. Proposition 1 The following assertions hold true: Proposition 3 (Finite Sets) Every finite subset of N has density 1. dens(N)= 1 zero. 2. dens(A) ≥ 0 (nonnegativity) Proof: Let A be a finite set. For s > 1, it yields 1 3. if dens(A) and dens(B) exist and A∩B = ∅, then dens(A∪ 1 1 ∞ ∑ s ≤ ∑ = constant < . (23) B) exists and is equal to n∈A n n∈A n dens(A)+ dens(B) (additivity). Thus, 1 ∑ s constant Proof: Statements 1 and 2 cab be trivially checked. The additiv- dens(A)= lim n∈A n ≤ lim = 0. (24) ity property can be derived as follows: s↓1 ζ(s) s↓1 ζ(s) 1 ∑ s dens(A ∪ B)= lim n∈A∪B n (13) As a consequence, sets A ⊂ N of nonzero density must be infinite. s↓1 ζ(s) 1 1 ∑ s ∑ s Corollary 4 The density of a singleton is zero. = lim n∈A n + lim n∈B n (14) s↓1 ζ(s) s↓1 ζ(s) Proposition 4 (Union) Let A and B be two sets of integers. Then = dens(A)+ dens(B). (15) the density of A ∪ B is given by dens(A ∪ B)= dens(A)+ dens(B) − dens(A ∩ B). (25) Corollary 1 The density of the null set is zero. Proof: Observe that A ∪ B = A ∪ (B − A), and A ∩ (B − A)= ∅. Then, it follows directly from the properties of the density mea- ∅ N ∅ N Proof: In fact, dens( )= 0, since 1 = dens( )= dens( ∪ )= sure that dens(∅)+ dens(N)= dens(∅)+ 1. dens(A ∪ B)= dens(A ∪ (B − A)) (26) Corollary 2 dens(B − A)= dens(B) − dens(A ∩ B). = dens(A)+ dens(B − A) (27) Proof: The proof is straightforward: = dens(A)+ dens(B) − dens(A ∩ B). (28) dens(B)= dens((A ∩ B) ∪ (Ac ∩ B)) (16) = dens(A ∩ B)+ dens(Ac ∩ B) (17) Proposition 5 (Heavy Tail) Let A = = dens(A ∩ B)+ dens(B − A), (18) {a1,a2,...,aN−1,aN,aN+1,...}. IfA1 = {a1,a2,...,aN−1} and A = {a ,a ,...}, then since A ∩ B and Ac ∩ B are disjoint. 2 N N+1 dens(A)= dens(A ). (29) Corollary 3 For every A, dens(Ac)= 1 − dens(A), where Ac is 2 the complement of A. Proof: Observe that A = A1 ∪ A2 and A1 and A2 are disjoint. 1For ease of exposition, in the following results, we assume that the densities Therefore, dens(A)= dens(A1)+ dens(A2). Since A1 is a finite of the relevant sets always do exist. set, dens(A1)= 0. 2 Now consider the following operation m ⊗ A , {ma | a ∈ Taking the limit as s ↓ 1, the above upper bound becomes A,m ∈ N}. This can be interpreted as a dilation operation on 1 ∑ N s the elements of A. n∈A2 n N lim = dens(A2 )= dens(A). (41) In [4, 5, 9], Erd¨os et al. examined the density of the set of s↓1 ζ(s) multiples m ⊗ A, showing the existence of a logarithmic density equal to its lower asymptotic density. Herein we investigate fur- Now examining the lower bound and using the dilation property, ther this matter, evaluating the density of sets of multiples. we have that 1 ∑ N a +1 Proposition 6 (Dilation) Let A be a set, such as dens(A) > 0. n∈A2 ( N n)s aN Then lim ζ (42) 1 s↓1 (s) dens(m ⊗ A)= dens(A). (30) aN N m = dens(A2 ) (43) aN + 1 Proof: This result follows directly from the definition of the dis- a = N dens(A). (44) cussed density: aN + 1 ∑ 1 ε n∈A (mn)s Since ∀ > 0, ∃N such that dens(m ⊗ A)= lim (31) s↓1 ζ(s) a N dens(A) > dens(A) − ε, (45) 1 1 s ∑ s aN + 1 = lim m n∈A n (32) s↓1 ζ(s) and as 1 1 = dens(A). (33) ∑ N s m n∈A2 (n+1) N lim ζ = dens(A2 ⊕ 1)= dens(A ⊕ 1), (46) s↓1 (s) Let A ⊕ m , {a + m | a ∈ A,m ∈ N}. This process is called a it follows that ∀ε > 0 we have that translation of A by m units [10, p.49]. Our aim is to show that the discussed density is translation invariant, i.e., dens(A ⊕ m)= dens(A) − ε ≤ dens(A ⊕ 1) ≤ dens(A). (47) dens(A), m > 0. Before that we need the following lemma. Therefore, letting ε → 0, it follows that Lemma 1 (Unitary Translation) Let A be a set, such as dens(A) > 0. Then dens(A ⊕ 1)= dens(A). (48) dens(A ⊕ 1)= dens(A). (34) Proposition 7 (Translation Invariance) Let A be a set, such as Proof: Let A = {a1,a2,...,aN−1,aN ,aN+1,...}.