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R. D A ℓ ( S d A A sgvnby given is D ( T OF ETS = ) ( n A s ( = ) = ) A 1.TeDrcltdniyi hnex- then is density Dirichlet The [1]. = ) lim s ↓ 1 n n lim ∑ → = ∞ ( I lim s s 1 NTEGERS ↓ ∞ α − N ∑ A B ( 1 log a n A ) ( i ( s ≤ , s s n ∑ B n ) ) ∈ n n A ) A , a and 1 , i n ⊂ 1 . s , B α etost.The sets. two be sa bcsaof abscissa an is § mined fined rich- ting this ase (8) (7) (6) (5) rm nd he 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 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 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,...}. We can split A into two disjoint sets as shown below: dens(A) > 0. Then dens(A ⊕ m)= dens(A), (49) A = {a1,a2,...,aN−1,aN ,aN+1,...} (35) = {a1,a2,...,aN−1}∪{aN,aN+1,...} (36) where m is a positive integer. = AN ∪ AN, (37) 1 2 Proof: The proof follows by finite induction. We have already N N proven that dens(A ⊕ 1)= dens(A). Therefore, we have that where A1 is a finite set and A2 is a ‘tail’ set starting at the element aN. By the Heavy Tail property, we have that ∀N dens(A)= N N dens(A ⊕ (m + 1)) = dens((A ⊕ m) ⊕ 1) (50) dens(A2 ). Forthe same reason, ∀N, dens(A⊕1)= dens(A2 ⊕1). Note that, for s > 1, if kn ≥ n + 1, then the following inequal- = dens(A ⊕ m)= dens(A). (51) ities hold:  ∑ 1 ∑ 1 ∑ 1 n∈AN (kn)s n∈AN (n+1)s n∈AN ns One possible application for a density on a set of natural num- 2 ≤ 2 ≤ 2 . (38) ζ(s) ζ(s) ζ(s) bers A is to interpret it as the chance of choosing a natural num- ber in A when all natural numbers are equally likely to be chosen. In other words, for the aboveinequality to be valid, we must have Interestingly, as we show next the above measure of uncertainty does not obey all the axioms of Kolmogorov since it is not σ- n + 1 N k ≥ , ∀n ∈ A2 . (39) additive. Additionally, we show that it is impossible to define a n finite σ-additive translation invariant measure on (N,2N). This For instance, let k = (aN + 1)/aN. Consequently, we establish result emphasizes an important point that there are reasonable that measures of uncertainty that do not satisfy the formal standard definition of a probability measure. ∑ 1 n AN a +1 1 1 ∈ 2 ( N n)s ∑ N s ∑ N s aN n∈A (n+1) n∈A n σ ≤ 2 ≤ 2 . (40) Theorem 1 There is no -additive measure defined on the mea- ζ ζ ζ N (s) (s) (s) surable space (N,2 ) such that:

3 1. 0 < µ(N) < ∞; and Corollary 7 (Geometric Progressions) Let G = {ar,ar2,ar3,...}, where a and r > 1 are positive integers. µ 2. is translation invariant. Then dens(G)= 0. (55) Proof: Suppose that µ is translation invariant. Then every single- ω ω ton set must have the same measure. Let 1 < 2 be any natural Proof: Notice that, for r > 1, numbers. Since {ω1}⊕ (ω2 − ω1)= {ω2}, we have ∞ 1 1 ∞ 1 1 ∞ 1 n µ ω µ ω ω ω µ ω ∑ = ∑ < ∑ < ∞ (56) ({ 2})= ({ 1}⊕ ( 2 − 1)) = ({ 1}), (arn)s as rns as r n=1 n=1 n=1   where the last equality follows from translation invariance. Let Thus, it also follows from the preceding discussion that µ µ N ∞ µ µ c = ({1}). If c = 0, then ( ) 6= 0 = ∑i 1 ({i}). Thus is = ∞ dens(G)= 0.  not σ-additive. If c > 0, then µ(N) < ∞ = ∑ µ({i}), which i=1 Let M = {p,2p,3p,...} be an arithmetic progression, where also implies that µ is not σ-additive.  p p ∈ N. For example, M2 = {2,4,6,...}, M5 = {5,10,15,...}, etc. Regarding the cardinality of these sets, we have that kM k = ℵ . Proposition 8 (Criterion for Zero Density) Let A = p 0 ∞ 1 {a1,a2,...,an,an+1,...}. If lims↓1 ∑ s converges, then n=1 an Proposition 9 (Arithmetic Progressions) For a fixed integer p, dens(A)= 0. the density of the set Mp is given by

Proof: It follows directly from the definition of dens(A).  1 dens(M )= . (57) p p Definition 2 (Sparse Set) A zero density set is said to be a sparse set. Proof: Note that Mp = p ⊗ N. Thus, according to Proposition 6, we have that As matter of fact, any criterion that ensure the convergence of ∞ 1 1 1 lims 1 ∑ s can be taken into consideration. In the next result, N N ↓ n=1 an dens(Mp)= dens(p ⊗ )= dens( )= . (58) we utilize the Ratio Test for convergence of series. p p  Corollary 5 Let A = {a1,a2,...,an,an+1,...}. If an Consequently, we have, for instance, dens(M1)= dens(N)= 1, limn→∞ < 1, then dens(A)= 0. an+1 dens(M2)= 1/2, dens(M5)= 1/5 etc. Proof: Observe that for s > 1 One can derive a physical interpretation for the density of Mp. Consider discrete-time signals Mp[n] characterized by a se- ∞ 1 1 1 quence of discrete-time impulses associated to the sets M . The ∑ < ∑ = ∑ . (52) p s signals M [n] are built according to a binary function that returns n∈A n n∈A n n=1 an p one if n is a multiple of p, and zero otherwise. For example, we According to the Ratio Test [11, p.68], the series on the right- can have M3[n] and M5[n], associated to M3 and M5, as depicted hand side of the above inequality converges whenever in Figure 1. In terms of signal analysis, dens(Mp) has some cor- respondence to the average value of Mp[n]. 1 a a lim n+1 lim n 1 (53) ∞ 1 = ∞ < . n→ n→ an+1 3 DENSITY OF PARTICULAR SETS an ∑ 1 In this section, some special sets have their density examined. Thus, by series dominance, it follows that n∈A ns also con- verges. And finally, this implies that the density of A must be We focus our attention in three notablesets: (i) set of prime num- zero.  bers, (ii) Fibonacci sequence, and (iii) set of square-free integers.

Lemma 2 (Powers of the Set Elements) Let A = {a1,a2,...} 3.1 SET OF PRIME NUMBERS m , m m and m > 1 be an integer. The density of A {a1 ,a2 ,...} is zero. Let P = {2,3,5,7,11,13,17,19,23,...} be the set of prime num- bers. Proof: Notice that, for s > 1 and m > 1, Proposition 10 The prime numbers set is sparse. ∞ 1 ∞ 1 ∞ 1 ∑ < ∑ < ∑ < ∞ (54) ζ ′ , (am)s am nm Proof: Utilizing the definition of the prime zeta function, (s) n=1 n n=1 n n=1 ζ ′ ∑ p−s, we have that dens P lim (s) . Addition- i : pi is prime i ( )= s↓1 ζ (s) Thus, it follows from the preceding discussion that dens(A)= 0. ally observe that ζ ′(s) < ∞ for s > 1. Taking in account that  ∞ ζ ′ ζ (ks) Corollary 6 The density of the set of perfect squares is zero. ln (s)= ∑ , (59) k=1 k

4 1.2 1.2

1 1

0.8 0.8 [n] 3 0.6 0.6 p[n] M

0.4 0.4

0.2 0.2

0 0 0 3 6 9 12 15 18 21 0 5 10 15 20 25 30 n n (a) Figure 2: Discrete-time signal P[n] corresponding to the set of

1.2 prime numbers.

1 Since the density is zero, it indicates that the associated 0.8 discrete-time signal P[n], as shown in Figure 2, has null average [n] 5 0.6 value. M

0.4 3.2 FIBONACCI SEQUENCE 0.2 The Fibonacci sequence constitutes another interesting subject 0 of investigation. Fibonacci numbers are constructed according 0 5 10 15 20 25 n to the following recursive equation u[n]= u[n − 1]+ u[n − 2], (b) n > 1, for u[0]= u[1]= 1, where u[n] denotes the nth Fibonacci number [12, p.160]. This procedure results in the Fibonacci set F 1 2 3 5 8 13 21 34 55 . Figure 1: Discrete-time signals M3[n] and M5[n] associated to the = { , , , , , , , , ,...} sets M3 and M5, respectively. Proposition 11 The Fibonacci set is sparse. ε for > 0 we have that Proof: It is known that the sum of the reciprocals of the Fi- ∞ ζ ′ ε bonacci numbers converges to a constant (reciprocal Fibonacci ζ ε (k(1 + )) ln (1 + )= ∑ (60) constant), whose value is approximately 3.35988566... [13]. k=1 k ∞ Consequently, for s > 1, we have that ζ ′(k(1 + ε)) = ζ ′(1 + ε)+ ∑ . (61) k ∑∞ 1 k=2 n=2 u[n]s dens(F)= lim (67) Now consider the following inequalities: s↓1 ζ(s) ∞ 1 ζ ′ ε ∑ ζ ′ ε ζ ε ∑∞ (k(1+ )) n=1 u[n] 3.35988566... (1 + ) ln (1 + ) − k=2 k < lim = lim = 0. (68) 0 ≤ = (62) s↓1 ζ(s) s↓1 ζ(s) ζ(1 + ε) ζ(1 + ε) ζ ε ∞ ζ ′(k(1+ε))  ln (1 + ) k = ζ ε − ∑ ζ ε . (63) (1 + ) k=2 (1 + ) lnζ(1 + ε) 3.3 SET OF SQUARE-FREE INTEGERS ≤ ζ ε . (64) (1 + ) The set of square-free numbers can be defined as S , {n ∈ N µ µ As ε → 0, we have that ζ(1 + ε) → ∞, therefore | | (n)| = 1}, where (·) is the M¨obius function, which is given by lnζ(1 + ε) lim = 0, (65) ε→0 ζ(1 + ε) 1, if n = 1, 0, if p2|n, for some p, because limx→∞ ln(x)/x = 0. Therefore, we can apply the µ  (n)= r (69) Squeeze Theorem once more, and find that (−1) , if n is the product of distinct  ζ ′(1 + ε) prime numbers. lim = 0. (66)  ε→0 ζ 1 ε  ( + ) Thus the first elements of S are given by  1,2,3,5,6,7,10,11,13,14,15,...

5 Proposition 12 The density of the set of square-free integers is 1 ζ (2) . 0.5

∞ µ ζ ) ∑ | (n)| (s) T 2 Proof: It is known [14] that n=1 ns = ζ (2s) , thus it follows 0.4 easily that: ), d'(M

T 2 0.3 1 ∑ s dens(S)= lim n∈S n (70) 0.2 s↓1 ζ(s) dens'(M ∞ |µ(n)| 0.1 ∑ s lim n=1 n (71) = 0 s↓1 ζ(s) 0 1 2 3 10 10 10 10 ζ (s) T ζ (2s) 1 = lim ζ = lim ζ (72) s↓1 (s) s↓1 (2s) Figure 3: Computational result for the approximate density 1 6 for the truncated prime set MT , according to (i) the function = = . (73) 2 ζ(2) π2 dens′(PT ) (bold curve), and (ii) the function d′(PT ) (thin line). Observe the convergence to 1/2. 

Definition 4 (Approximate Density) The approximate density 4 COMPUTATIONAL RESULTS for a truncated set AT is given by

Definition 3 Let A be a set of positive integers. The truncated 1 ∑ T set AT is defined by dens′(AT )= n∈A n . (80) ∑T 1 n=1 n AT = A ∩{1,2,3,...,T − 1,T}, (74) Again, whenever the limit order of Equation 77 can be inter- where T is a positive integer. changed, we have that dens(A)= lim dens′(AT ). (81) In other words, AT contains the elements of A which are no T →∞ greater than T . The quantity dens′(·) furnishes a computationally feasible way The concept of truncated sets can furnish computational ap- to investigate the behavior of dens(·). proximations for the numerical value of some densities. Con- In the following, we obtain computational approximations for sider, for example, the quantity the asymptotic density and the discussed density. The approx- kAT k imate densities are numerically evaluated as T increases in the d′(AT ) , . (75) range from 1 to 1000. Now we investigate (i) arithmetic pro- kNT k gressions, (ii) the set of prime numbers, and (iii) the Fibonacci This expression d′(·) can be interpreted in a frequentist way as sequence. the ratio between favorable cases and possible cases. Clearly, the asymptotic density can be expressed in terms of d′(·): 4.1 ARITHMETIC PROGRESSIONS Considering an arithmetic progressions M , we have that: d(A)= lim d′(AT ). (76) q T →∞ ′ T 1 d(Mq)= lim d (Mq )= , (82) In a similar fashion, we can consider a version of the discussed T →∞ q density for truncated sets as defined below. First, observe that and discussed density dens(·) can be expressed as the following dou- ble limit ′ T dens(Mq)= lim dens (Mq ). (83) T →∞ 1 ∑ T s dens(A)= lim lim n∈A n . (77) Both quantities limits converge to the same quantity 1/q. Intu- s↓1 T →∞ ∑T 1 n=1 ns itively, we have that the chance of “selecting” a multiple of q among all integers is 1/q. This is exactly the density of M . Fig- Restricted to the class of sets for which the above limits can have q ure 3 shows the result of computational calculations of d′(·) and their order interchanged, we find that ′ T dens (·) for M2 . ∑ 1 n∈AT ns dens(A)= lim lim (78) 4.2 PRIME NUMBERS SETS T →∞ s↓1 ∑T 1 n=1 ns T 1 Consider the truncated set of prime numbers P = {p : ∑ T = lim n∈A n . (79) p is a prime, p < T). One may easily compute both d′(PT ) and T →∞ ∑T 1 n=1 n ! dens′(PT ) for any finite T .

6 0.8 1

0.8 0.6 ) T ), li(T)/T

T 0.6 ), d'(F

0.4 T ), d'(P

T 0.4

0.2 dens'(F

dens'(P 0.2

0 0 0 1 2 3 0 1 2 3 10 10 10 10 10 10 10 10 T T

Figure 4: Computational result for the approximate density Figure 5: Density of the truncated Fibonacci set: dens′(FT ) for the truncated prime set PT , according to (i) the function (bold line) and d′(FT ) (thin line). dens′(PT ) (bold curve), (ii) the function d′(PT ) (thin line), and (iii) the approximation based on the Prime Number Theorem REFERENCES (dotted line). [1] J. P. Bell and S. N. Burris, “Dirichlet density extends global asymptotic density in multiplicative systems,” Preprint published ′ T π(T ) An alternative path for the estimation of d (P )= T , where at htt://www.math.uwaterloo.ca/∼snburris/htdocs/MYWORKS/, the function π(x) represents the number of primes that do not 2006. exceed x, is to utilize the approximation for π(T) given by the [2] R. L. Duncan, “Some continuity properties of the Schnirelmann Prime Number Theorem [12,15, p.336]: density,” Pacific Journal of Mathematics, vol. 26, no. 1, pp. 57– 58, 1968. π(T) ≈ li(T ), (84) [3] ——, “Some continuity properties of the Schnirelmann density II,” Pacific Journal of Mathematics, vol. 32, no. 1, pp. 65–67, where li(·) denotes the logarithmic integral [16]. The curves dis- 1970. played in Figure 4 correspond to the calculation of dens′(PT ), [4] P. Erd¨os, “On the density of some of integers,” Bulletin d′(PT ) and li(T )/T . As expected, all curves decay to zero. of the American Mathematical Society, vol. 54, no. 8, pp. 685–692, Aug. 1948. [5] H. Davenport and P. Erd¨os, “On sequences of positive integers,” 4.3 FIBONACCI SEQUENCE Journal of the Indian Mathematical Society, vol. 15, pp. 19–24, 1951. T Now consider the truncated Fibonacci set F = [6] R. Ahlswede and L. H. Khachatrian, “Number theoretic correla- {1,2,3,5,8,13,21,...,T }. The result of calculating tion inequalities for Dirichlet densities,” Journal of Number The- ory, vol. 63, pp. 34–46, 1997. kFT k d′(FT )= (85) [7] I. Niven and H. S. Zickerman, An Introduction to the Theory of kNT k Numbers. John Wiley & Sons, Inc., 1960. [8] I. S. Gradshteyn and I. M. Ryzhik, Table of Integrals, Series, and and Products, 4th ed. New York: Academic Press, 1965. 1 ∑ T [9] P. Erd¨os, R. R. Hall, and G. Tenebaum, “On the densities of sets dens′(FT )= n∈F n , (86) ∑T 1 of multiples,” Journal f¨ur die reine und angewandte Mathematik, n=1 n vol. 454, pp. 119–141, 1994. are shown in Figure 5. Both curvestend to zero as T grows. This [10] W. Rudin, Real and Complex Analysis. McGraw-Hill, 1970. fact is expected since we have already verified that dens(F)= 0. [11] S. Abbott, Understanding Analysis. Springer, 2001. ′ The small convergencerate of dens (·) is typical of problems that [12] D. M. Burton, Elementary . McGraw-Hill, 1998. deal with the harmonic series. [13] R. Andr´e-Jeannin, “Irrationalit´ede la somme des inverses de cer- taines suites r´ecurrentes,” in C. R. Acad. Sci. Paris S´er. I Math., vol. 308, 1989, pp. 539–541. 5 CONCLUSION [14] M. R. Schroeder, Number Theory in Science and Communication, 3rd ed., ser. Springer Series in Information Sciences, T. S. Huang, A density for infinite sets of integers was investigated. It was T. Kohonen, and M. R. Schroeder, Eds. Springer, 1997, vol. 7. shown that the suggested density shares several properties of usual probability. A computational simulation of the proposed [15] H. Riesel, Prime Numbers and Computer Methods for Factoriza- tion. Birkh¨auser, 1985. approximate density for finite sets was addressed. An open topic is the characterization of sets for which the limiting value of the [16] M. Abramowitz and I. Segun, Handbook of Mathematical Func- tions. New York: Dover, 1968. approximate density equals their density.

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