A Note on the Distribution of Numbers with a Minimum Fixed Prime Factor with Exponent Greater than 1

Transcription

1 International Mathematical Forum, Vol. 7, 2012, no. 13, A Note on the Distribution of Numbers with a Minimum Fixed Prime Factor with Exponent Greater than 1 Rafael Jakimczuk División Matemática, Universidad Nacional de Luján Buenos Aires, Argentina jakimczu@mail.unlu.edu.ar In memory of my sister Fedra Marina Jakimczuk ) Abstract In this note we study the distribution of numbers with a minimum fixed prime factor with exponent greater than 1 in their prime factorization. Mathematics Subject Classification: 11A99, 11B99 Keywords: Numbers with a minimum prime factor fixed with exponent greater than 1 1 Preliminary Results The following theorem is sometimes called either the principle of inclusionexclusion or the principle of cross-classification. We now enunciate the principle. Theorem 1.1 Let S be a set of N distinct elements, and let S 1,...,S r be arbitrary subsets of S containing N 1,...,N r elements, respectively. For 1 i < j <... < l r, let S ij...l be the intersection of S i,s j,...,s l and let N ij...l be the number of elements of S ij...l. Then the number K of elements of S not in any of S 1,...,S r is K = N N i N ij N ijk 1) r N 12...r. 1) 1 i r 1 i<j r 1 i<j<k r In particular if S = r S i then K =0and consequently N = N i N ij N ijk 1) r1 N 12...r. 2) 1 i r 1 i<j r 1 i<j<k r

2 610 R. Jakimczuk Proof. See, for example, either [1, page 233] or [2, page 84]. We also need the following well-known limit [1, Chapter XVII, Theorem 280] n lim 1 1 ) = 6 n π, 3) 2 where p i is the i-th prime number. In this note as usual). denotes the integer-part function. Note that 2 Main Results 0 x x < 1. In this section p n denotes the n-th prime number. Then p 1 =2,p 2 =3,p 3 = 5,p 4 =7,p 5 =11,... A positive integer n is quadratfrei if it is either a product of different primes or 1. For example, n = 2 and n = are quadratfrei. Let Q 1 be the set of quadratfrei numbers, it is well-known [1, Chapter XVIII, Theorem 333] that this set has positive density 6. That is, if Q π 2 1 x) is the number of quadratfrei numbers not exceeding x we have Q 1 x) lim = 6 x x π. 2 Let Q 2 be the set of not quadratfrei numbers. That is, the set of numbers such that in their prime factorization there exists a prime with exponent greater than 1. The density of this set will be 1 6. That is, if Q π 2 2 x) is the number of not quadratfrei numbers not exceeding x we have Q 2 x) lim =1 6 x x π. 2 Let us consider the set β ph of all positive integers such that in their prime factorization p h is the minimum prime with exponent greater than 1. Note that if i j then the sets β pi and β pj are disjoint. On the other hand β ph = Q 2. That is, the sets β ph h =1, 2,...) are a partition of Q 2. The density of Q 2 is see above)1 6. We shall prove that the set β π 2 ph h =1, 2,...) has positive density D ph and that the sum of the infinite positive densities is 1 6. That π 2 is, D ph =1 6. Consequently the sum of the densities of the infinite π 2 sets β ph equals the density of the union of these infinite sets. Let β ph x) be the number of numbers in the set β ph that do not exceed x and let δ ph x) be the set of numbers in the set β ph that do not exceed x. Then the number of elements in the set δ ph x) isβ ph x). We have the following theorem.

3 Numbers with a minimum prime with exponent greater than Theorem 2.1 The following asymptotic formula holds h1 β ph x) = 1 1 )) 1 x O1). 4) Consequently the numbers in the set β ph Proof. We have see 1)) have positive density h1 D ph = 1 1 )) 1. 5) x β ph x) = x x x p 2 h 1 i h1 p 2 h p2 i 1 i<j h1 p 2 h p2 i p 2 j 1 i<j<k h1 p 2 h p2 i p 2 jp 2 k 1) h1 x = x x x p 2 h p2 1 p 2 h1 p 2 h 1 i h1 p 2 h p2 i 1 i<j h1 p 2 h p2 i p 2 j x 1) h1 x O1) 1 i<j<k h1 p 2 h p2 i p 2 jp 2 k p 2 h p2 1 p 2 h1 = x h1 1 1 )) O1). 6) p 2 h The theorem is proved. We now prove that the sum of the infinite positive densities is 1 6 π 2. We give two proof of this theorem. Theorem 2.2 The following formula holds, D ph = h1 1 1 )) 1 = ) ) 1 1 ) )1 1 ) 1 1 ) =1 6 2 π. 7) 2 First proof. Without difficulty can be proved by mathematical induction the following equality n h1 1 1 )) 1 Equations 3) and 8) give 7). The theorem is proved. p 2 h n =1 1 1 ). 8) Second proof. Let χ ph x) be the set of numbers multiples of p 2 h that do not

4 612 R. Jakimczuk exceed x. Let Nx) be the number of elements in n χ ph x). We have see 2)) Nx) = x x x 1 i n 1 i<j n p 2 j 1 i<j<k n p 2 jp 2 k 1) n1 x = x x x p 2 1p 2 2 p 2 n 1 i n 1 i<j n p 2 j 1 i<j<k n p 2 jp 2 k 1) n1 x n O1) = )) x O1). p 2 1p 2 2 p 2 n Consequently Nx) n lim =1 1 1 ). 9) x x On the other hand, we have the equality n n χ ph x) = δ ph x). Note that if i j then the sets δ pi x) and δ pj x) are disjoints. Consequently we have see 4)) Therefore n n h1 Nx) = β ph x) = 1 1 )) 1 x O1). Nx) lim x x n h1 = 1 1 )) 1. 10) Equations 9) and 10) give 8). Finally, equations 3) and 8) give 7). The theorem is proved. We now prove the following theorem. Theorem 2.3 The following asymptotic formula holds D ph 6 π 2 h 2 log 2 h. 11) Proof. We have Prime Number Theorem) p h h log h. Consequently see 5) and 3)) we have lim h D ph 6 π 2 h 2 log 2 h = lim h h1 )) π 2 h 2 log 2 h 1 p 2 h =1.

5 Numbers with a minimum prime with exponent greater than This proves 11). The theorem is proved. To finish, we study a simple partition of the set N of positive integers. We shall prove that each set in this partition has positive density and the sum of the infinite densities is 1. That is, the density of the union N of these sets. Let us consider a fixed prime p. Now consider the set A p k k =0, 1,...)of the numbers such that in their prime factorization appear p k. If k = 0 then A p k is the set of numbers not multiple of p. Note that if k i k j then A p k i and A p k j are disjoint. On the other hand k=0 A p k = N. Therefore the infinite sets A p k k =0, 1,...) are a partition of N. Let A p kx) be the number of numbers in the set A p k not exceeding x. We have the following theorem. Theorem 2.4 The following asymptotic formula holds A p kx) = p 1 x O1) k =0, 1,...) 12) pk1 Consequently the numbers in the set A p k have positive density D p k = p 1 p k1 k =0, 1,...) 13) Proof. The numbers in the set A p k are the numbers in the following p 1 linear forms p k pn r) r =1, 2,...,p 1) n =1, 2, 3,...) Consider the p 1 inequalities p k pn r) x r =1, 2,...,p 1) The solutions to these inequalities are x n =1, 2,..., p r k1 p r =1, 2,...,p 1) Consequently p1 x A p kx) = p r k1 p r=1 = p 1 x O1). pk1 The theorem is proved. We now prove that the sum of the infinite positive densities is 1.

6 614 R. Jakimczuk Theorem 2.5 The following formula holds D p k =1. k=0 Proof. We have see 13)) D p k = k=0 k=0 p 1 p = p 1 k1 p 1p 1p 1 ) 2 = p 1 p p =1. The theorem is proved. ACKNOWLEDGEMENTS. The author is very grateful to Universidad Nacional de Luján. References [1] G. H. Hardy and E. M. Wright, An Introduction to the Theory of Numbers, Fourth Edition, Oxford, [2] W. J. LeVeque, Topics in Number Theory, Addison-Wesley, Received: September, 2011