Study of some equivalence classes of primes

Notes on Number Theory and Discrete Mathematics Print ISSN 3-532, Online ISSN 2367-8275 Vol 23, 27, No 2, 2 29 Study of some equivalence classes of primes Sadani Idir Department of Mathematics University of Mouloud Mammeri 5 Tizi-Ouzou, Algeria e-mail: sadaniidir@yahoofr Received: 3 April 26 Accepted: 3 March 27 Abstract: In this paper, we introduce an equivalence relation on P for studying some specific classes of prime numbers This relation and the famous prime number theorem allows us to estimate the number of prime numbers of each equivalence class, the number of the different equivalence classes and to show some other results Keywords: Equivalence relation, Prime number theorem, Recurrence relation AMS Classification: A4 Introduction Let P be the set of prime numbers, and for all x R, let πx be the prime-counting function The prime number theorem which was proved independently by de la Vallée Poussin [], and Hadamard [2] in 896, states that: πx x, as x + ln x We can give an equivalent statement for this theorem as, for example, let p n denote the n-th prime number Then π n = p n n ln n as n + 2 We define the following functions: f n x = n times n times {}}{{}}{ f f fx and f n y = f f f y, 2

with f x = x and o is the composition operator The aim of this paper is to construct an equivalence relation on the set of primes, using the restriction of the prime-counting function π to P, for studying some classes of primes The following results illustrate asymptotic distribution of a number of functions that we have proposed 2 Main results We start with the following obvious lemma: Lemma 2 Let π be a restriction of π to P Then, π : P N is a bijection and its inverse is π : N P Remark Throughout this paper, we simply use the notation π to designate the restriction of π to the set P instead of using π The proof of the following theorem is obvious Theorem 2 We define the relation on the set of prime numbers P defined by: if p and q are two prime numbers, p q if and only if there exists n Z such that p = π n q Then, is an equivalence relation The elements of the equivalence class ṗ are defined by: ṗ = {, π 2 p, πp, p, π p, π 2 p, } 2 The smallest element p of ṗ is the prime number which verify πp is not prime and it is called origin of the class ṗ Example 3 = {2, 3, 5,, 3, 27, 79, 538, 527, } with origin the number 2 since π2 = is not a prime 7 = {7, 7, 59, 277, 787, } with origin the number 7 since π7 = 4 is not a prime Notation We denote by P x the set of all origins p x defined by: P x = {2, 7, 3, 9, 23, 29, 37, 43, 47, } And we denote simply by P the set of all origins p Theorem 22 Let A be a finite set of increasing sequence of consecutive primes defined by: A = {p i, i n and p = 5} Then, there exists at least one class ṗ A, p A, such that ṗ = {p} 22

Before giving the proof of this result, we give an illustrative example Example 2 Let A be a set which defined by A = {5, 7,, 3, 7} The class of the integer 5 is 5 = {5, } and its cardinality is greater than We notice that 7 / 5, therefore the integer 7 constitutes the origin of a new class which is 7 = {7, 7} and its cardinality is greater than It only remains to see that the prime number 3 does not belong to the classes 2 and 7 Thus the prime number 3 constitutes the origin of a new class 3 and clearly its cardinality is Proof of theorem 22 Assuming that each class p i, i n, containing at least two elements, and supposing that πp n is prime According to the prime number theorem, the interval [πp n, p n ] contains πp n π 2 p n = k prime numbers This leads to the two following cases: If k =, ie, there exists only one prime number in [πp n, p n ], namely q Therefore, we obtain π q / A and πq is not a prime number since πq and πp n are consecutive Then q contains only one element in A which is the prime number q If k >, ie, there exist at least two prime numbers in [πp n, p n ], namely q,, q k We suppose that q t, t {,, k}, is a prime number Then q t+ is not prime and since π q t+ / A, then, the unique element of q t+ is q t+ Finally, in both cases, there exist at least one class ṗ, p A, such that p is a unique element of this class in A, ie, ṗ = {p} The proof now is completed We have the following definition Definition Let A be the set defined as in the Theorem 22 and p is a prime number belonging to A We say that ṗ is an outside class of A, if πp is not prime in A and π p / A ie, ṗ = in A In the case where ṗ 2, we say that ṗ is an inside class of A Lemma 22 Let x n be a sequence defined by the recursive relation: x n+ = px n = xn ln x n, x e 278288 Proof Then px = e ln px i = e ln p i+ x, 3 We set ln ln x = ln 2 x and we have x = px = x ln x = ln x = ln x ln 2 x x 2 = px = x ln x = ln x 2 = ln x ln 2 x x n+ = px n = x n = ln x n+ = ln x n ln 2 x n ln x n 23

And combining all these, we obtain which is equivalent to Then, passing to the limit, we obtain lim n + Consequently, ln x n+ = ln px n = ln x x x = lim x n+ n px n = x x x n+ = n ln x i lim n + and since lim x n = e n + The formula 4 is obviously equivalent to px = e ln px i, which is the desired result n n ln 2 x i, n ln x i = lim ln x ln px i n e = ln x i = ln x ln px i, 4 Lemma 23 Let x > We define the number of prime numbers belonging to the class p less than or equal to x as follows π iter x, p = = p p x p p x p=π l p, l N Then, π iter x, p is approximately equal to the number of iterations n of the sequence x n+ = px n with x = x and the value of n verify x n p and x n+ < p, as x + Proof From the prime number theorem, we have πx px = x lnx, x πn x p n x, x Next, we can choose an integer n such that π n x = p with πp not a prime number, it follows that p p n x, x And, π iter x, p = = p p x p p x p=π l p p p x p p l p = nx,p l= = nx, p as x, such that nx, p is the number of iterations n which depends obviously on x and p 24

Theorem 23 Let x > We define the function ηx, p as follows: Then, we have ηx, p = ln ln p p p x, ηx, p ln x + oln x Proof In view of the proof of Lemma 23, we have, p = π n x p n x implies that, ln x which is equivalent to n ln ln x + ln ln πx i = ln ln x + n ln πx i ln x n ln ln πx i = Finally, for all fixed p and x, we have ln p / ln x, then ln ln p ln x + oln x ie, ηx, p ln x + oln x Theorem 24 We have, 2 Proof p p x, p p π iter x, p π iter x π iter x, p ηx, p ln ln x, x n ln px i x p = x p, ln ln p ln x ln p p p x, ln x ln x ln ln x + o, x ln ln x For the proof of the first formula, we have, on the one hand ηx, p = ln ln p ln ln x p p x, p p = π iterx, p ln ln x p p x, Then we obtain π iter x, p ηx, p ln ln x On the other hand, for all x > e, x δ > p with < δ <, we have ηx, p ln ln x δ x δ <p x, = ln δ + ln ln x x δ <p x, = ln δ + ln ln xπ iter x, p π iter x δ, p ln δ + ln ln xπ iter x, p ln x δ 25

Then π iter x, p ln x δ + ηx, p ln δ + ln ln x Now, according to Lemma 23, ln x δ = oπ iter x, p, and then for x sufficiently large depending on δ, ln x δ δπ iter x, p, and thus π iter x, p ηx, p δln δ + ln ln x Now, for all ɛ >, we can choose δ more near to, for this δ, so that /δ = + ɛ, and for x sufficiently large, we have 2 According to Theorem 23, we have π iter x, p ηx, p ln ln x ηx, p ln ln x π iter x, p < + ɛ ηx, p ln ln x ln x ln x ln ln x + o π iter x, x ln ln x Definition 2 Let x be a positive real number We denote by π c x the number of different classes ṗ such that 2 p x Precisely, π c x :=, p P p x 2 We denote by θ x the function defined by θ x = p x ln p Example 3 In the interval [2, ], the value 2 represents the origin of the class 2 but the values 3, 5, do not, since they belong to the same class 2 The value 7 represent the origin of the class 7 Thus in this case, we have π c x = 2 We have the following result: Theorem 25 We have, lim π cx = + x 2 Let p [2, x] Then π c x = πx ππx = πx π 2 x 5 Proof Suppose that the number of different classes is finite as x We know that 26 p i = 6

Next, let p k i p k, where k is finite by hypothesis We obtain = p i p k k< i Therefore, we have we deduce p k i which contradicts formula 6 <, and since the second sum has a finite number of terms, p k k< i <, 2 To find the value of π c x means that we estimate the number of origins p Clearly, the prime number p is an origin that means πp is not a prime number Then, let p be between 2 and x, and let p,x P be the greatest prime number in [2, x], therefore πp,x is not a prime number and for all integer l, π l πp,x + > p,x So, we only have to search the numbers which are not primes and less than πp,x Thus, we have The number of the even numbers less than or equal to p,x equal to πp,x 2 the number of the odd numbers less than or equal to p,x equal to πp,x ππp 2,x such that ππp,x is the number of the prime numbers less than or equal to p,x Next, we add the two quantities, we obtain, since p,x x, the quantity π c x which is equal to π c x = πx ππx We easily obtain the following consequence Corollary 2 Let p be a prime and suppose that p = π n p be an origin Then Proof We have n 2 p = πp π c π i p π c x = πx ππx π c πx = ππx ππ 2 x π c π n 2 x = ππ n x ππ n x Now, we add these equations together, we obtain the desired result For all initial value y e, we define the following sequence: y n+ = y n ln y n 7 This sequence is stationary for y = e and increasing divergent to infinite for y > e and as n It is clear that, inductively, y n y ln y ln y ln y = y ln y n, then we have the following consequence: 27

Lemma 24 We have, π c x ln 2 x + p x ln p = ln x where p represents the origin of the classes ṗ ln 2 x + θ x ln x, Proof Since for all p = p n, n >, we have p > p ln n p, then ln x ln p ln ln p ln ln p = ln ln p = ln ln p ln ln p p x p ln n p x,n p,n ln x ln p p x ln ln p p x According to the following formula we have ln x ln p = ln xπ iter x p x p x fp n x ln ln p p x fn x ln n 2 ln x, therefore, the inequality is obtained directly by substitution Proposition 2 We have, 2 x ln ln x θ x x ln x ln ln x ln x θ x x as x ln p ft dt 8 ln t ln x ln ln x 9 Proof We have πx ππx ln 2 x + p x ln p ln x p x ln p πx ππx ln x ln x Moreover, since And recalling that we obtain ln p p x p x x ln x ln p ln x = π c x ln x p x p x πx x ln x, x ln x = x ln xln x ln ln x ln x The second inequality is obtained in the same way, x ln p ln x ln x x ln xln x ln ln x 2 It is enough to tend x to + in the inequality 9 = x ln ln x ln x ln ln x ln x ln x ln ln x 28

3 Future work Our future work is to generalize this equivalence relation by introducing a new function with φ a bijection defined in N to N φ π : P N, 2 Study of the following hypothesis: let p P Suppose that the 2-tuple p, p + h are primes infinitely often for all h 2N Then, there exists at least an integer h among these integers h verify the following equation: πp + h πp = c infinitely often, such that πp + h and πp are primes and c < h fixed References [] De la Vallée Poussin, C J 896 Recherche analytique sur la théorie des nombres, Ann Soc Sci Bruxelle, 2, 83 256 [2] Hadamard, J 896 Sur la distribution des zéros de la fonction ζs et ses conséquences arithmétiques, Bull Soc Math France, 24, 99 22 29