On the number of elements with maximal order in the multiplicative group modulo n

Transcription

1 ACTA ARITHMETICA LXXXVI On the number of elements with maximal order in the multiplicative group modulo n by Shuguang Li Athens, Ga.. Introduction. A primitive root modulo the prime p is any integer a coprime to p such that its exponent modulo p is p. There are totally φp primitive roots modulo p in [, p], where φ is Euler s function. It is a natural problem to consider the fraction φp /p, which is the proportion of non-zero residues mod p which are primitive roots. Trivially, we have 0 < φp /p /2. For any real numbers x 2 and u, let D π x, u = πx p x φp /p u where πx is the number of primes up to x. In 974, P. D. T. A. Elliott [2] proved that the limit lim D πx, u = D π u x exists for all real numbers u. The function D π u is continuous and is strictly increasing on the interval [0, /2]. I. J. Schoenberg [2] had earlier considered the distribution problem for φn/n. He proved the existence of lim x x, φn/n u for any real number u, and the continuity of the limit as a function of u. The concept of primitive root modulo a prime can be generalized. This was done by R. D. Carmichael []. He defined a primitive λ-root modulo n as any integer coprime to n and having the maximal exponent modulo n. Thus a primitive root for a prime p is a primitive λ-root modulo p. He also found the following properties see [, 7] of the maximal exponent modulo n, denoted by λn: i λp e = φp e for all primes p and e except p = 2 and e > 2 in which case we have λ2 e = φ2 e /2 = 2 e Mathematics Subject Classification: A27, B05, N36, N37., [3]

2 4 S. Li ii λn = lcm p e n{λp e }. iii Let a be an integer coprime to n and let l a n be the exponent of a modulo n. Then l a n λn. Thus, in this notation, an integer a coprime to n is a primitive λ-root modulo n if and only if l a n = λn. In [3] one can find results concerning the size of the function λn. Let Rn be the number of primitive λ-roots modulo n in [, n]. In this paper we investigate the distribution of the values of Rn/φn, in analogy with that of φp /p considered by Elliott. That is, if we define Dx, u = x Rn/φn u for any real numbers x > 0 and u, what can we say about Dx, u? In particular, does lim x Dx, u exist for all u? Note that the fraction Rn/φn represents the proportion of residue classes mod n that are primitive λ-roots to the total number of residue classes coprime to n. We trivially have 0 < Rn/φn. That this ineuality is nearly best possible is contained in the following result. Theorem. We have lim Rn/φn =, lim n where γ is Euler s constant. Rn/φn/ln ln n = e γ, n The function Dx, u is only interesting for 0 < u <. Perhaps surprisingly, we show that there are values of u 0, where lim x Dx, u does not exist. To attack the uestion we wish we could take a more natural approach, say by working with the first moment of Rn/φn or Rn/n. However these functions are not multiplicative as the function φn/n is, and thus the methods used by Elliott and Schoenberg do not appear to work. Thanks to Pomerance s suggestion, we turn our attention to the first moment of lnrn/φn instead. Though this function is also not multiplicative, it can be approximated by a comparatively simple sum over prime factors of λn. Adopting the notation ln k x, suggested by John Selfridge, for the k-fold iteration of the natural logarithm of x, we will prove our principal results as indicated in the next two theorems. Theorem 2. The maximal order of the function lnrn/φn x

3 Elements with maximal order 5 is less than or eual to c ln 5 x for some constant c > 0, and greater than or eual to c ln 6 x for another constant c > 0. On the other hand, the minimal order of the function is less than or eual to a constant. As a conseuence of Theorem 2 we have Theorem 3. There is a positive constant c and an unbounded set of numbers x such that for each u in 0,, we have Dx, u c/ ln u. On the other hand, there are positive constants δ, b and an unbounded set of numbers x with Dx, ln 5 x b δ. Thus for some positive constant u 0, lim x Dx, u does not exist for all u with 0 < u < u 0. It follows immediately from Theorem 3 that the seuence of distribution functions Dn, u does not converge weakly. Suppose that it does. Let Du be the distribution function to which Dn, u converges weakly [4]. Then Du is discontinuous in 0, u 0 by Theorem 3, which contradicts the fact that the set of discontinuities of Du is countable a well-known property of monotone functions. In a forthcoming paper [8] we will prove, by a more complicated argument, that the constant δ in Theorem 3 can be taken as anything less than. This result will be shown to be relevant to the study of the integers n for which a fixed integer a is a primitive λ-root, in analogy with Artin s conjecture for primes. The author would like to take this opportunity to express his heartfelt thanks to Carl Pomerance for patient advice concerning problems in this paper and remarkable comments. Without Pomerance s help this project would not have survived. The author is indebted to Andrew Granville and Vsevolod Lev for their suggestions. The author would also like to thank the referee for a comment regarding Theorem The closed form for Rn and a few properties of Rn. Throughout this paper we always use p and to represent primes and k, m, n to represent natural numbers. We give an explicit formula for Rn in the next theorem, to which a different approach can be found in [9]. Then we will study some deeper features of the function. Theorem 2.. Let n := #{prime p : p e n and v λp e } for a prime with v λn, except the case 2 3 n and 2 λn, when 2 n := + #{prime p : p n}. Then Rn = φn. n λn P r o o f. For any integer m let Nm be the number of elements of order m in Z/nZ. We claim that Nm is a multiplicative function.

4 6 S. Li Suppose k, l =. Clearly the product of two elements, of orders k and l respectively, has order kl. Conversely, every element of order kl can be written uniuely as a product of an element of order k and an element of order l. Thus Nkl = NkNl. As a conseuence we have Nλn = v λn Nv. To compute N v it is convenient to factor Z/nZ into a direct sum of cyclic groups. Let C m denote a cyclic group of order m. Let n = p e... pe r r 2 e where p,..., p r are distinct odd primes and e,..., e r are positive integers. By the Chinese remainder theorem we have Z/nZ = Cλp e... C λp e r r Z/2 e Z. If e = 0 the last summand drops off. Otherwise Z/2 e Z = { Cλ2 e if e = or 2, C λ2 e C 2 if e 3. We have factored Z/nZ into a direct sum of r + r cyclic groups, where r = 0, or 2. Let n i be the order of the ith summand, so that Z/nZ = Cn C n2... C nr+r. The number of solutions to x m = in C k is m, k, so the number of solutions to x v = in Z/nZ is given by r+r i= v, n i. Similarly x v = has r+r i= v, n i solutions in Z/nZ. Thus N v = r+r i= r+r v, n i v, n i. Note that the second product is n times the first product, since { v, n i = v, n i if v n i, v, n i if v n i. v λn λn i= Thus, N v = n r+r i= v, n i and so Nλn = N v = n But for the double product we have r+r v λn i= v, n i = r+r which completes the proof. i= v λn v, n i = r+r r+r v λn i= λn, n i = i= v, n i. r+r i= n i = φn, We note that n 0 and euality holds if and only if λn is not divisible by. Let us define the function rn := Rn/φn. Then rn =

5 Elements with maximal order 7 prime / n. The word prime will be suppressed as we always use to denote a prime. We now prove Theorem in the introduction. Proof of Theorem. It is trivial that rn. Let us prove that lim n rn =. Let x be any large number and ε be a small positive constant. Let B = {primes p x : gcdp, P x ε = and p 3 mod 4}, where P z = 2<p z p for any z. Then there is a positive constant δ such that for all sufficiently large x we have x #B δ ln x 2. This result can be obtained by applying Theorem 7.4 of [5] to sieve the set A = {p : p x and p 3 mod 4} with the set P = {primes p > 2}, taking κ =, α = /2 and z = x ε. Here we can choose any ε < /4. If p B and is a prime factor of p other than 2, then > x ε. But p x. Thus p has at most /ε odd prime factors, counting multiplicity. Choose [ln x] such primes p i B, i =,..., [ln x]. Let n x = [ln x] i= p i. Then by definition of rn, rn x = λn x n x 2 [ln x] x ε [ln x]/ε, which has limit as x goes to infinity. Thus we proved lim n rn =. To see the lower bound, first note that rn, λn Nn where Nn is chosen to be the least number such that the product of the primes up to Nn is greater than or eual to λn. From prime number theory, Nn = + o ln λn + o ln n. Thus, from Mertens theorem, rn e γ + o/ln 2 n. We claim this ineuality is sharp. For any given large x let m = ln x. Then from prime number theory again, we have ln m ln x. Thus x /2 m x 2 if x is sufficiently large. By Linnik s theorem, there is a prime p such that p mod m and p m c for some absolute constant c. With this choice of p we have x /2 p x 2c. Thus by Mertens theorem, rp = p ln x = + o e γ = + o e γ ln 2 p. Send x to infinity to see that there are infinitely many such primes. So we proved our claim and our theorem.

6 8 S. Li Our goal is to study the distribution of values of the function rn. As we noted before, we can write rn = = exp n + O, : n= where the O is less than or eual to zero. It is convenient to introduce the function fn :=. : n= Thus, rn e fn crn for an absolute constant c. We see that the distribution of values of rn is dominated by its counterpart of fn. It is important to understand the behavior of the function fn. Our strategy is to study the first moment of fn, the sum fn. We note that fn = : n= = x n= where x because n = implies that λn n. We close this section by showing that the contribution to this sum from the terms with > is negligible. Theorem 2.2. As x, λn > x = O. ln 3 x Let us mention the following lemma before proving Theorem 2.2. Lemma 2.3 see [0, ]. For any integer k 2 and any x 2, p = ln 2 x ln k φk + O φk p x p mod k where the implied constant is uniform. P r o o f of Theorem 2.2. Notice that for a prime λn, either 2 n or p for some prime p n. Note that > x 2 n > 3, x 2 ln 3 x.

7 Thus 2 We have > >y λn > = Elements with maximal order 9 < x p x p mod 2 = y ln y + o and Using these facts and Lemma 2.3 we have x p x p mod p n > = x > >y p x p mod p n ln 2 p ln2 x + O x + O ln 2. 2 x = + o. y ln Our theorem follows by substituting the above estimate in 2. x = O. ln 3 x 3. The first moment of fn. When the inner sum of has the following bounds. Theorem 3.. There exist positive constants c and c 2 so that, for all large numbers x and any prime, we have c 2 x c x {ln 3 x/ln }, ln 3 x/ln n= where {y} denotes the fractional part of the real number y and y the minimal distance from y to the integers, that is, y = min n Z { y n }. Before proving Theorem 3. let us look at its conseuences. Theorem 3.2. For the positive constant c in Theorem 3. we have fn 2c x + ln 3 x/ln for all large x. P r o o f. The theorem follows immediately by combining Theorems 2.2 and 3., and the observation that 2 + ln 3 x/ln in. Although we can bound the first moment of fn from below in a similar way, we are not able to use these estimates to get the correct order of magnitude. We would need this to show there are many values of n for which fn is large. We get around this problem by introducing a smaller

8 20 S. Li function fn for which we are able to find the correct order of magnitude for its average order. Let if n > e fn e, := ln 4 n, n= 0 otherwise. Note that every term in the sum for fn is also in the sum for fn, so fn fn. For the new function we have Theorem 3.3. For the positive constant c 2 in Theorem 3., fn 2 c 2x 2 {ln 3 x/ln } for all large x. ln 4 x P r o o f. The proof follows almost immediately from Theorem 3.. One just needs to check that the contribution from pairs n, with ln 4 n < ln 4 x is negligible. Let us turn to the proof of Theorem 3.. We need some preparations. Lemma 3.4. We have = k n= p x /2 k p where the error is non-negative. n= m x/p m,p k x/p= x ln + O P r o o f. We will work out the above formula for, n= in such a way that detailed explanations for the formulae marked by numbers to their left are supplied in the subseuent discussion: = k = k = k = k k λn n= p x k p p x k p p x /2 k p r m x/p r m,p k x/p r = m x/p m,p k x/p= m x/p m,p k x/p=, x + O 2 x + O 2 x ln + O.

9 Elements with maximal order 2 To see why the euality of 3 is true, let us first notice that k λn implies either k+ n or k p for some prime p n. Since the 4-fold sum in 3 counts the exact number of the positive integers n x with n = and a prime factor p for which k p where k λn, the difference between this 4-fold sum in 3 and the sum just before it is bounded by x 2. 2 n The difference between the 4-fold sum in 3 and the sum in 4 is x p r x p 2 x 2k = k r 2 k k p x k p x /2 <p x k p p x k p x 2. By Lemma 2.3, the difference between the sum in 4 and the sum in 5 is bounded by x p = x ln2 x /2 k ln k + O k x ln. k k It is easy to notice that the errors in 3, 4 and 5 are all non-negative. Therefore we proved Lemma 3.4. For the sake of convenience we introduce a few notations of sieve methods. Let A be the set of positive integers up to x. Let P k be the set of primes congruent to modulo k. Let P kz := p for any z x. Then SA, P k, y := p z p mod k n A n, P k y= and W z := p z p P k. p Lemma 3.5. i With the notations introduced above we have SA, P k, x xw x uniformly for all, k and x. ii Let z = expln x/. As x we have SA, P k, z = x + O W z ln x uniformly for all and k.

10 22 S. Li P r o o f. See Theorems 2.2 and 7.2 in [5]. Lemma 3.6. There are absolute positive constants c and x 0 so that provided that k and x x 0. SA, P k, x cx, P r o o f. Let z = expln x/. Write SA, P k, x = SA, P k, z E where E is the number of positive integers n x such that gcdn, P kz = and gcdn, P kx/p kz >. Then by the condition k and Lemma 2.3, E x p z<p x p P k p n z<p x p P k ln2 x ln 2 z k ln x = x k / + O ln3 x k = O where the implied constant is independent of and k. By Lemma 3.5ii we have uniformly as x, where W z = p SA, P k, z = xw z + o p z p P k = exp = exp p z ln 2 z k + O p P k k ln k, p + O k, by Lemma 2.3 again. But ln 2 z = ln 3 x and k. Thus W z c for some constant c > 0 independent of and k. Therefore, if x is sufficiently large, SA, P k, x c x + o ox cx for some constant c with 0 < c < c. This ends the proof. Proof of Theorem 3.. First we will show that c x ln 3 x/ln n= for some positive constant c. First we claim that

11 6 n= x k Elements with maximal order 23 k exp k + O x ln. To see this let us look at the innermost sum in Lemma 3.4. Noting that p x /2, by Lemma 3.5 and Lemma 2.3, we have m x/p m,p k x/p= x p prime l x/p l mod k = xp l exp l x/p l mod k l + O 2k = x p exp ln 2x/p k ln k + O k xp exp k uniformly in and p. Now put this result into the sum in Lemma 3.4 and use Lemma 2.3 again to find that n= x k ln2 x /2 k ln + O exp k k k + O x ln Since k k ln /k = Oln /, we have 6. Next we divide the sum on the right side of 6 into two sums, according to whether k > or k. Let M be the minimal integer so that M >. Then M = ln 3 x/ln {ln 3 x/ln } +, which euals [ln 3 x/ln ] +. Then k > k exp k / k > k = M / 2 M = 2. {ln 3 x/ln } Let L be the maximal integer so that L. Then [ ] ln3 x L = = ln { } 3 x ln ln ln3 x. ln Noting that k, we have,.

12 24 S. Li k k n= exp k / Put these results back in 6. We have x + O ln 3 x/ln = = k 2! k k / 2 2 L < 2L 2. {ln 3 x/ln } x ln x ln 3 x/ln, since ln 3 x/ln /2. Thus we proved one half of Theorem 3.. Secondly we will prove that n= c 2 x {ln 3 x/ln } for some positive constant c 2 independent of. By Lemma 3.4, noting that the error in it is non-negative, we have n= k with k > k with k > = cx k with k > p x /2 k p p x /2 k p m x/p m,p k x/p= c x p ln2 x /2 k ln + O k ln2 x M ln cx + O M M, where M is the smallest integer so that M >. We obtain 7 and 8 by using Lemmas 3.6 and 2.3, respectively. So what is left to be explained is 9. Notice that the sum in 8 can be written as so that 9 follows. k M k + O k M k ln k, k

13 Elements with maximal order 25 Since M = ln 3 x/ln {ln 3 x/ln } + and, we have M ln M ln 3 x/ln ln ln 3 x/ln {ln 3 x/ln }+ = ln 3 x {ln 3 x/ln }. Therefore, choosing c 2 = c/2, we have ln2 x M ln cx + O n= M M = cx {ln 3 x/ln } + o c 2 x {ln 3 x/ln }, for x sufficiently large. We have proved Theorem 3.. and 4. Two crucial series. As one can see from Theorems 3.2 and 3.3, it is necessary to understand the series + ln T/ln 2 {ln T/ln } T ln 2 T in the course of estimating the first moments of fn and fn. This section is dedicated to the study of some features of the two series. Theorem 4.. For T e ee, we have ln + ln T/ln 3 T. T First let us mention the following well known fact in prime number theory. Lemma 4.2. For all x 2, p = ln ln x + c 3 + Oexp c 4 ln x /2, p x where c 3 and c 4 > 0 are constants. P r o o f of Theorem 4.. Write the series as the sum of s and s 2 as follows: + + ln T/ln = s + ln T/ln + s 2, Q Q< T where Q will be determined later. We will use the trivial estimate s = Oln 2 Q. For s 2, let us write s 2 = + + ln T/ln + ln T/ln Q< T ln T/ln >ln A/ln = s 2 + s 2 2, Q< T ln T/ln ln A/ln

14 26 S. Li where A > e will be chosen so that ln A/ln Q < /2 but the exact value for A will be determined later. Clearly s 2 = O ln 2 T A while s 2 2 =, + k ln T/ln Q< T k ln T/ln ln A/ln where there is at most one integer k for each prime as our A satisfies ln A/ln Q < /2. Thus, by Lemma 4.2, s 2 2 = k ln T +ln A/ln Q k ln T +ln A/ln Q T/A /k AT /k /k T ln 2 AT /k ln 2 +O A exp c 4 k ln T. A Since ln A/ln Q < /2 and Q T, we have ln A/ln T < /2, so that /k T ln 2 AT /k ln T + ln A ln 2 = ln A ln T ln A = ln + 2 ln A ln T ln A < ln + 4 ln A < 4 ln A ln T ln T and Hence s 2 2 k ln T ln T ln A ln Q ln Q A ln T + ln A 3. ln T + ln A 4 ln A ln Q ln T + O exp c 4 ln Q. /2 3 Now choose Q so that ln Q = 3/c 2 4ln 2 T 2, and choose A = ln 2 T 2. If T is sufficiently large, we have Q T and ln A/ln Q < /2. We thus have s 2 ln T + ln A 4 ln A 2 ln Q ln T + O ln3 T = O ln T ln 2 T 2. At the same time, s ln2 T 2 = O A Then T = O ln 2 T and so s 2 = O. ln 2 T + ln T/ln = s + s 2 = Oln 2 Q = Oln 3 T. This concludes the proof of Theorem 4..

15 Elements with maximal order 27 Theorem 4. gives us an upper bound for the series mentioned at the beginning of the section. Next let us investigate the normal value of the series. We note that T = + ln T/ln = T ln T/ln <ln 2 /ln + T ln T/ln ln 2 /ln T ln T/ln <ln 2 /ln T ln T/ln <ln 2 /ln + ln T/ln + + ln T/ln T ln T/ln ln 2 /ln + O ln where the last euality follows from the fact that /p ln p <, p running over primes. This suggests considering the average value of the following function. Let us define gt := e t t/ln <ln 2 /ln for all t 0. Thus the above argument shows that 0 gt + O. + t/ln e t Lemma 4.3. There is a positive constant c 5 so that, for all y > 0, P r o o f. By definition, [y] gk = k= k y [y] gk c 5 y. k= e k k/ln <ln 2 /ln e y k y k/ln <ln 2 /ln Since ln 2 /ln < /2, if k/ln < ln 2 /ln then there is a uniue integer l with l ln 2 ln < k ln < l + ln 2 ln..

16 28 S. Li For the fixed integer l there are at most 2 ln 2 + integers k so that k ln = k ln l < ln 2 ln. Therefore [y] k= gk y/ln 2 ln 2 + y e y l=0 e y as the last sum is bounded. Thus we proved Lemma 4.3. ln 2 ln y, Theorem 4.4. There is a positive number c 6 so that, for all y, [y] k= c 6y. + k/ln e k P r o o f. This follows from 0 and Lemma 4.3. Definition. Let S be a set of natural numbers. If lim x /x#{n x : n S} exists, we call it the density of S. Otherwise we call the corresponding upper or lower limit the upper or lower density of S, respectively. As a corollary of Theorem 4.4 we have Theorem 4.5. Let c 6 be the same constant as in Theorem 4.4, and let b > c 6 be any number. Then the set S b of k N with b + k/ln e k has positive upper density. P r o o f. Suppose that S b has density zero. Then N \ S b has density one. On the other hand, for any y > 0, b. + k/ln k y e k k y k S b Then by Theorem 4.4, we have c 6 b y. k y k S b Send y to infinity, we have c 6 b, a contradiction. Therefore S b has positive upper density. We are done. In the rest of the section we will consider the second series mentioned at the beginning of this section. Our attention will focus on how big the series could be when T is large. The next lemma is an elementary result.

17 Elements with maximal order 29 Lemma 4.6. If 5 and ln 2T/ln < ln 2/ln, then 2 {ln T/ln } 4. For a given T let Q be the largest prime less than or eual to ln 2 T such that all primes with 5 Q satisfy ln 2T ln ln 2 ln. We want to know how large Q can be as a function of T. By the following result on simultaneous Diophantine approximation we can say something about this uestion. Lemma 4.7. Let ξ,..., ξ l be any l real numbers. Then, for any integer N >, there exists a positive integer m N l such that mξ i < /N for all i =,..., l. P r o o f. See the proof of Theorem 200 in [6]. Let Q be any large prime. Consider the l irrationals /ln 5, /ln 7,..., /ln Q where l = πq 2. Let N = ln Q/ln 2. Then by Lemma 4.7 there exists an integer m, m N l, so that every prime with 5 Q satisfies m/ln < /N ln 2/ln Q. Since m/ln 5 < ln 2/ln Q and ln 5 is irrational it follows that m = mq as Q. By the definition of N we have N = ln Q/ln 2 + O when Q is sufficiently large. But l = πq 2, so for Q sufficiently large by the prime number theorem we have Q > l ln N ln m. Now choose T such that 2T = e m. Clearly m > ln T. Thus Q > ln 2 T. With this choice of T all primes with 5 ln 2 T satisfy ln 2T/ln < ln 2/ln. Thus by Lemma 4.6, ln 2 T 2 {ln T/ln } 5 ln 2 T ln 2T/ln <ln 2/ln 5 ln 2 T 4 = ln 4 T 4 Therefore we have proved the following theorem. 2 {ln T/ln } + O. Theorem 4.8. There is an unbounded set of numbers T for which 2 {ln T/ln } 5 ln 4 T. ln 2 T 5. Proofs of Theorems 2 and 3. Recall the function rn = Rn/φn from Section 2. We can restate Theorem 2 in the following form. These estimates are also bounds for the first moment of fn and that of fn.

18 30 S. Li Theorem 5.. There exist positive constants c 7, c 8 and c 9 so that i for all sufficiently large x we have ln rn c 7 x ln 5 x; ii there is an unbounded set of numbers x for which ln rn c 8 x ln 6 x; iii there is an unbounded set of numbers x for which ln rn c 9 x. P r o o f. By the definition of fn in Section 2 we have fn = ln rn+ O, where the O is non-positive. Since 0 < rn, we have ln rn + Ox, fn = where the Ox is non-positive. Noticing that fn fn, we can get Theorem 5. by applying Theorems 3.2, 3.3, 4., 4.5 and 4.8. Corollary 5.2. Let c 9 be the constant in Theorem 5.. There is an unbounded set of numbers x such that for all u with 0 < u <. P r o o f. By definition, rn u Dx, u c 9 / ln u ln rn x ln u Dx, u for all u with 0 < u <. Thus, the corollary follows from Theorem 5.iii. Corollary 5.3. There is a positive constant c 0 with the property that, for any positive constant b < c 0, the set S b = {n N : rn ln 5 n b } has positive upper density. Consider the set S b = {n N : fn b ln 6 n}. Since ln rn = fn + O with the O being non-positive and fn fn, we have S b S b. It is enough to show that there is a constant c 0 such that, for all b with 0 < b < c 0, S b has a positive upper density.

19 Elements with maximal order 3 From the definition of fn we trivially have fn 2 ln 6 n when n is large enough. Thus we have, for x sufficiently large, fn = fn + fn b ln 6 x + 2 ln 6 x. n S b n S b n S b n S b On the other hand, we choose c 0 = c 2 /0 where c 2 is the constant in Theorem 3.. Then, by Theorems 3.3 and 4.8, there exists an unbounded set of real numbers x for which fn c 0 x ln 6 x. Let b be any constant with 0 < b < c 0. Then combining the above we have for such numbers x, c 0 b + 2 b + 2. x x x n S b n S b n S b Thus the upper density of S b is at least c 0 b/2, which is positive. Thus we have proved the corollary. Corollary 5.4. There exist positive constants δ, b and an unbounded set of numbers x with Dx, ln 5 x b δ. P r o o f. This follows immediately from the definition of Dx, u and from Corollary 5.3. Theorem 5.5. There exists a positive number u 0 such that, for each u 0, u 0, the function Dx, u does not have a limit as x. Thus the function rn does not have distribution function. P r o o f. This is a corollary of Corollaries 5.2 and 5.4. Note that Theorem 3 in the introduction follows from Corollaries 5.2, 5.4 and Theorem 5.5. References [] R. D. Carmichael, The Theory of Numbers, Wiley, New York, 94. [2] P. D. T. A. Elliott, On the limiting distribution of fp+ for non-negative additive functions, Acta Math , [3] P. Erdős, C. Pomerance and E. Schmutz, Carmichael s lambda function, Acta Arith , [4] J. Galambos, Advanced Probability Theory, 2nd ed., Dekker, New York, 995. [5] H. Halberstam and H. E. Richert, Sieve Methods, Academic Press, New York, 974.

20 32 S. Li [6] G. H. Hardy and B. M. Wright, An Introduction to the Theory of Numbers, Oxford Univ. Press, London, 960. [7] W. J. LeVeue, Topics in Number Theory, Vol. I, Addison-Wesley, Reading, Mass., 956. [8] S. L i, Artin s conjecture on average for composite moduli, preprint. [9] G. Martin, The least prime primitive root and the shifted sieve, Acta Arith , [0] K. K. Norton, On the number of restricted prime factors of an integer I, Illinois J. Math , [] C. Pomerance, On the distribution of amicable numbers, J. Reine Angew. Math. 293/ , [2] I. J. Schoenberg, On asymptotic distributions of arithmetical functions, Trans. Amer. Math. Soc , Department of Mathematics University of Georgia Athens, Georgia U.S.A. sli@math.uga.edu Received on and in revised form on