Uniform Distribution of the Fractional Part of the Average Prime Divisor

Uniform Distribution of the Fractional Part of the Average Prime Divisor William D. Banks Department of Mathematics, University of Missouri Columbia, MO 652 USA bbanks@math.missouri.edu Moubariz Z. Garaev Instituto de Matemáticas, Universidad acional Autónoma de México C.P. 5880, Morelia, Michoacán, México garaev@matmor.unam.mx Florian Luca Instituto de Matemáticas Universidad acional Autónoma de México C.P. 5880, Morelia, Michoacán, México fluca@matmor.unam.mx Igor E. Shparlinski Department of Computing Macquarie University Sydney, SW 209, Australia igor@ics.mq.edu.au

Abstract We estimate exponential sums with the function ρn) defined as the average of the prime divisors of an integer n 2 we also put ρ) = 0). Our bound implies that the fractional parts of the numbers {ρn) : n } are uniformly distributed over the unit interval. We also estimate the discrepancy of the distribution, and we determine the precise order of the counting function of the set of those positive integer n such that ρn) is an integer. AMS Subject Classification: L07, 37, 60 Introduction In this note, we study some distributional properties of the average prime divisor function ρn), defined for each integer n 2 as follows: ρn) = ωn) p, where, as usual, ωn) denotes the number of distinct prime divisors p of n, and the summation runs over all such primes. We also put ρ) = 0. We obtain nontrivial bounds for exponential sums with ρn) and use these estimates to show that the fractional parts of the numbers {ρn) : n } are uniformly distributed over the closed unit interval [0, ]. In particular, this result implies that ρn) Z for almost all n. Below, we determine the precise order of magnitude of the counting function of those integers n for which ρn) Z. Problems with a similar flavor have been treated previously in [8, 7, 22, 23], in which the authors investigate the counting function of those positive integers n for which n/fn) is an integer, where fn) = τn) the divisor function), or more generally fn) = τ k n) the k-th divisor function which can also be defined as the coefficient of n s in the Dirichlet series for ζs) k, where ζs) is the Riemann zeta function), or fn) = l ωn) for a fixed positive integer l). We remark that the distributional properties of ρn) are rather different from those of the average divisor function ϑn), defined for each integer n as follows: p n ϑn) = σn) τn) = τn) 2 d. d n

Indeed, it is known that ϑn) Z for almost all n ; see [2] and [6]. Throughout this paper, for any real number x > 0 and any integer l, we write log l x for the function defined inductively by log x = max{log x, } where log x is the natural logarithm of x), and log l x = max{loglog l x), } for l >. When l =, we omit the subscript in order to simplify the notation; however, we continue to assume that log x for any x > 0. In what follows, we use the Landau symbol O, as well as the Vinogradov symbols, and with their usual meanings, with the understanding that any implied constants are absolute. We recall that the notations A B, B A and A = OB) are equivalent, and that A B is equivalent to A B A. We always use the letters p and q to denote prime numbers, while m and n always denote positive integers. Acknowledgements. During the preparation of this paper, W. B. was supported in part by SF grant DMS-0070628, F. L. was supported in part by grants SEP-COACYT 37259-E and 37260-E, and I. S. was supported in part by ARC grant DP02459. 2 Preliminary Results Here, we collect some known results that are used in this paper, and we obtain some new auxiliary estimates which may be of independent interest. We start by recalling the Mertens formula see Theorem 4.2 in [], or Theorem 427 in [9]), which asserts that the relation p x p = log 2 x + α + O log x holds for all x 2, where α is an absolute constant. We also use a result of Landau that there exist absolute constants β and γ such that the estimate ) log x ϕn) = β log x + γ + O 2) x n x holds for all x in fact, the error term can be sharpened; see [20]). It is also useful to recall that the Euler function ϕk) satisfies the inequality ϕk) 3 ) ) k log 2 k. 3)

We use πx) to denote the number of primes p x, and we use πx; c, d) to denote the number of primes p x in the fixed arithmetic progression p c mod d). By the classical Page bound see Chapter 20 of [5]), and using partial summation see the remark in Chapter 22 of [5]), it follows that for some absolute constant A > 0, the estimate πx; c, d) = x ϕd) log x + O x exp A )) log x holds provided that d log x and gcdc, d) =. We denote by π ν x) the number of positive integers n x with ωn) = ν. A particular case of the version of the classical Hardy and Ramanujan inequality given in [9] and [2] implies that the estimate 4) π ν x) xlog 2 x) ν ν )! log x 5) holds uniformly for 0.5 log 2 x ν 2 log 2 x see also [0] for a more general statement). We also need the following technical result. Lemma. Let x 3 and x /2 y x. Then, x n x+y τn) ϕn) y log x. x Proof. We have x n x+y τn) ϕn) = x n x+y = 2 ϕn) 2 d n d x /2 x n x+y n 0 mod d) ϕn). x n x+y ϕn) d n d x /2 We now set w = x /2 / log x and split the above sum into two pieces according to whether d < w or d w. 4

For d < w, we use the inequality ϕdm) ϕd)ϕm) together with the estimate 2), getting ϕn) ϕd) ϕm) d<w x n x+y n 0 mod d) d<w x/d m x+y)/d = ) )) x + y d log x β log + O ϕd) x x d<w y x ϕd) y log x. x d<w For the larger values of d, we only need 3) to obtain that log ϕn) 2 x w d x /2 x n x+y n 0 mod d) which completes the proof. = log 2 x w d x /2 x n x+y n 0 mod d) d w d x /2 n x/d m x+y)/d log 2 x log + y/x) d w d x /2 y log 2 x x m logx /2 /w) y log2 2 x, x For integers ν and a, we write Uν; a) for the number of solutions to the congruence u + u 2 + a 0 mod ν), u, u 2 ν, gcdu u 2, ν) =. Lemma 2. The following inequality holds: Uν; a) ν 2 ). p p ν Proof. Let ν = p e...pes s be the prime number factorization of ν. By the Chinese Remainder Theorem, we have s Uν; a) = Up es s ; a). i= 5

For a prime power p e, it is easy to see that Up e ; a) = p e Up; a). Indeed, all solutions to u + u 2 + a 0 mod p e ), u, u 2 p e, gcdu u 2, p) = are of the form u v + wp mod p e ), u 2 v 2 wp mod p e ), where w = 0,...,p e, and v, v 2 ) is a solution to v + v 2 + a 0 mod p), v, v 2 p, gcdv v 2, p) =. Clearly, the last congruence has at least p 2 solutions because for each v =,...,p, v a mod p), the value of v 2 is uniquely determined with v 2 0 mod p). Hence, Up e ; a) p e p 2), which finishes the proof. Lemma 3. For some absolute constant κ > 0, 2 ) = κx + Olog 2 x). p ν x p ν Proof. We have p ν 2 ) = τp) ) = p p p ν d ν τd)µd), d where µd) denotes the Möbius function; we recall that µ) =, µd) = 0 if d 2 is not square-free, and µd) = ) ωd) otherwise. Therefore, 2 ) = τd)µd) = τd)µd) p d d ν x p ν ν x = d x = x d x Using the well-known fact that d ν d y τd)µd) d τd)µd) d 2 d x x d + O) ) + O τd) y log y 6 d x ) τd). d ν x ν 0 mod d)

see Theorem 3.3 in [], or see Theorem 320 in [9] for a much more precise statement), and using partial summation, we have the estimates d x τd) d The latter estimate implies d x τd)µd) d 2 = = Olog 2 x) and d= τd)µd) d 2 d>x τd) d 2 = Ox log x). + Ox log x) = κ + Ox log x), where κ is the constant κ = p 2p ) = 0.3226..., 2 and the product is taken over all primes p. The result follows. 3 Exponential Sums with ρn) Let ex) = exp2πix) for all x R. For any integers a, and with, let S a ) be the exponential sum given by S a ) = eaρn)). Theorem. For every integer a 0, the following inequality holds: n= S a ) a log 2. Proof. Let Pn) denote the largest prime divisor of n 2, and put P) =. As usual, we say that an integer n is y-smooth if and only if Pn) y, and we put ψx, y) = #{ n x : n is y-smooth}. We now choose Q = /u, where u = 2 log 3 log 4, 7

and we denote by E the set of Q-smooth positive integers n. According to Corollary.3 of [] see also [3]), we have the bound #E = ψ, Q) u u+ou) log 2. ext, we denote by E 2 the set of the positive integers n not in E such that Pn) 2 n. Clearly, ow let #E 2 p>q p 2 /Q log 2. k = δ) log 2 and K = + δ) log 2, where δ > 0 is a sufficiently small absolute constant to be chosen later, and let E 3 denote the set of positive integers n such that either ωn) < k or ωn) > K. By the Turán-Kubilius inequality see [4, 24]), it follows that #E 3 log 2. Let E 4 denote the set of positive integers n / log 2. Of course, #E 4 log 2. ow let n be a positive integer not in 4 i=e i. This integer n has a unique representation of the form n = mp, where m is such that m < /Q and Pm)m > / log 2, and p = Pn) is a prime number in the half-open interval p L m, where { } L m = max Q, Pm), and L m = L m, /m]. m log 2 Let E 5 be the set of those n such that L m = Q. In this case, Q log 2 < m < Q. 8

When m is fixed, p /m can take at most π/m) values. Thus, the number of elements n E 5 is ) #E 5 π m /Q log 2 )<m</q /Q log 2 )<m</q log Q u log /Q log 2 )<m</q log m log/m) m ) log Q Q log 2 u log 3 log log2 3 log log 4 )) log 2. Finally, let E 6 be the set of those positive integers n which are not in 5 i=e i and such that L m = Pm). In this case, Pm) m log 2, so we see immediately that p = Pn) Pm) log 2. Thus, E 6 is contained in the set of all those positive integers n which are divisible by two primes q < p, such that p q log 2 and p > Q. In particular, q Q/ log 2 > Q /2. Fix q and p. The number of such n is O/pq). Hence, using ), we derive that for each q the total number T q ) of such n with some prime p in the interval q, q log 2 ] can be estimated from above as T q ) q q<p q log 2 = q log + log 3 log q p = q log 2q log 2 ) log 2 q) + O ) ) + O log 3 q log q q log q. q log q Summing the above inequality over all q > Q /2, we get that #E 6 T q ) log 3 q log q log 3 log Q q>q /2 q>q /2 u log 3 log log2 3 log log 4 log 2. ) 9

Thus for each i =,...,6, we have the estimate #E i log 2. 6) We now let be the set of positive integers n which do not belong to any of the sets E i for i =,...,6. We see from 6) that S a x) = ) eaρn)) + O. 7) log n 2 Each n admits a unique representation of the form n = pm, where p > Pm). Moreover, in this case, p L m = /m log 2 ), /m]. Let M be the set of permissible values for m. Let ν be the subset of n with ωn) = ν. ote that each such n is of the form n = pm, where m M has ωm) = ν and p L m. We write M ν for the set of those m M with ωm) = ν. ote that # = m M πl m ), and for k ν K, we have # ν = m M ν πl m ), 8) where we have used πl m ) to denote the number of primes in the interval L m. With the above notations, we can write n eaρn)) = = K eaρmp)) p L m K eaρm)ν )/ν) eap/ν). m M ν p L m m M ν Therefore, by the Page bound 4), for any integers b and d with gcdb, d) =, and with d log x, we have x ebp/d) = ebc/d) + O xd exp A )) log x. ϕd) log x p x c d gcdc,d)= 0

The sum over c is the classical Ramanujan sum and since gcdb, d) = ) is equal to µd), where, as before, µd) denotes the Möbius function see, for example, Theorem 272 of [9]). Consequently, writing ν a = ν/ gcda, ν), and noting that n ν a ν 2 log 2 < log m M ν ) Q < log log 2 m log 2 holds for all m M and all sufficiently large, we derive that K eaρn)) ϕν a ) πl m) + ) ν a m exp 0.5A log Q ) ). We have K m M ν ν a m exp 0.5A log Q ) log 2 exp 0.5A log Q ) m< log log 2 exp 0.5A log Q ) m log 2. In particular, by 8), n eaρn)) K We now substitute the inequality ) ν ϕν a ) = ϕ gcdν, a) ϕν a ) # ν + log 2. 9) ϕν) gcdν, a) ϕν) a in 9), and use the trivial inequality # ν π ν ) in combination with 5), to get that n eaρn)) a log K log 2 ) ν ϕν)ν )! + log 2. 0) It is easy to show that the Stirling formula implies that for any positive real number x and any positive integer ν, the following inequality holds: x ν ν! x /2 exp x )) 2 λ2 + Oλ 3 ),

where λ = ν/x. In particular, we see that if δ is sufficiently small, then log 2 ) ν log ν )! log2 exp ν log ) 2 2. ) 3 log 2 For an integer l, let I l be the interval defined by I l = Using ), we derive that K log 2 ) ν ϕν)ν )! [ log 2 + l log 2, log 2 + l + ) log 2 log log2 δ log 2 l δ log 2 + ]. exp l 2 /3) ν I l ϕν). Furthermore, by 2), we have ϕν) = β log log2 + l + ) ) ) log 2 ν I l log 2 + l log3 + O log 2 log 2 2) log2. Combining the preceding two estimates and substituting into 0), the proof is completed. 4 Uniformity of Distribution of ρn) For a sequence of points in the half-open interval [0, ), the discrepancy of the sequence is defined as = sup Au) u, 0 u< where Au) is the number of points of the sequence in the interval [0, u). Let denote the discrepancy of the sequence consisting of the fractional parts of the numbers ρn) for n =,...,. In this section, we give an upper bound for. 2

Theorem 2. For, the following bound holds: log2 3 log 2. Proof. According to the well-known Erdős-Turán relation between the discrepancy and the appropriate exponential sums see [6, 5]), we find that for any H : H + a S a). a H We now choose u, Q, k and K as in the proof of Theorem. Using the bound 9), we derive a H a S a) a H a K ϕν a ) # ν + log H log 2. Changing the order of summation and then collecting together the values of a with the same value of gcda, ν) = d for each divisor d of ν, and making use of the inequality ϕν a ) ϕν)/ gcdν, a), we deduce that a H a K ϕν a ) # ν K K log H ϕν) # ν ϕν) # ν K d d ν a H gcda,ν)=d d d ν τν) ϕν) # ν. b H/d a db We now proceed as in the proof of Theorem, and using Lemma instead of 2), we obtain K τν) ϕν) # ν log 3 log 2. Therefore, a H a S a) log 3 log H log 2. 3

Putting everything together, gives the estimate H + log 3 log H log 2. Choosing H = log 2, we obtain the stated result. 5 Integrality of ρn) Let R) denote the number of positive integers n with ρn) Z. Theorem 3. The following bound holds R) Proof. For an integer m, we denote am) = q m log 2. q. The proof of the upper bound follows easily from the proof of Theorem. Indeed, with the notations of Theorem, we may assume that n, since the number of n not in is at most O/ log 2 ). Write n = mp and fix m. Then, in order for ρn) to be an integer it is necessary that the prime p L m satisfies p am) mod ν), where ν = ωm)+. Since L m is large enough with respect to ν = Olog 2 ), we can apply the Page bound 4) as in the proof of Theorem, and we see that the number S m of such choices for p is bounded by S m ϕν) πl m) + O m exp 0.5A ) ) log Q in fact, there are no such primes when gcdam), ν) > ). Summing up the above bound up over all the possible choices of m M, and using 8), we find that the estimate #R) m M S m K ϕν) # ν + log 2 4

holds, and this leads to the desired upper bound using arguments towards the end of the proof of Theorem. We now turn to the proof of the lower bound. Suppose that is sufficiently large. Let us fix ν J where J is the interval [ 0.5 log 2 /4 ), 2 log 2 /4 ) ]. Choose a positive integer m /4 with ωm) = ν 2. ow take a solution u, u 2 ) to the congruence u + u 2 + am) 0 mod ν) u, u 2 ν, gcdu u 2, ν) =. 3) We now select a prime p in the interval /4 < p < /3 such that p u mod ν) and a prime p 2 in the interval /3 < p 2 < /mp such that p u mod ν). We remark that /mp > /4 /3 = 5/2 > /3 ; in particular, p 2 > p. It is clear from the above construction that n = p p 2 m is in R), and that different choices of m, p, p 2 lead to different values for n R). It remains to count how many numbers n can be obtained by this process. Assume that m, u, u 2, p are fixed. Because ν = ωm) + 2 = Olog 2 ), the Page bound 4) applies. Therefore, there are π/mp ) π /3 ) ϕν) ) + O mp exp 0.5A log ) mp ν log possible values for p 2 here we have used the trivial estimate ϕν) ν). Therefore, if m, u, u 2 are fixed the above construction gives R m,u,u 2 ) integers n R), where R m,u,u 2 ) mν log /4 <p /3 p u mod ν) p. Using the Page bound 4) again and partial summation, we derive that R m,u,u 2 ) log2 /3 ) log mνϕν) log 2 /4 ) ) mν log log 2. Here we have used the estimate ϕν) log 2. Recalling Lemma 2, we see 5

that the contribution to R) from each fixed value of m is u,u 2 R m,u,u 2 ) Uν; am)) mν log log 2 m log log 2 p ν 2 ), p where the sum is taken over all solutions u, u 2 ) to 3). Thus, we finally obtain that R) 2 ) log log 2 p m. ν J p ν m< /4 ωm)=ν 2 We now apply 5) which, combined with the Stirling formula, yields #{m /4 : ωm) = ν 2} /4 log 2 ) ν 2 ν 2)! log /4 log log2. Hence, by partial summation and for an appropriate constant c > 0, we have m< /4 ωm)=ν 2 Therefore, m R) = log /4 c log /log 2 ) /2 ) m< /4 m log 2 ) 3/2 + c log log2 ) ) + O log 0.5log 2 /4 ) ν 2log 2 /4 ) p ν Applying Lemma 3, we complete the proof. 6 Remarks log log2. 2 ). p Clearly, using the bound of Theorem directly in the proof of Theorem 2 leads to a weaker bound about the square root of the present bound). On 6

the other hand, in the proof of Theorem itself we can probably work much more carefully with ν a and obtain a better dependence on a. It is easy to see that the results of Theorems, 2 and 3 also hold for the function ρn) = p ord p n, Ωn) where Ωn) is the number of prime divisors p of n, each one counted with multiplicity ord p n one recalls that the Turán-Kubilius inequality [4, 24] applies to Ωn) as well). In particular, n= p n ea ρn)) a log 2. Similar but weaker results can be obtained for ρ s n) = ωn) p n p s and ρf; n) = ωn) fp), where s is an arbitrary positive integer, and f is an arbitrary polynomial with integer coefficients and positive degree. Indeed, such results can be achieved by following closely the present proofs of Theorem and Theorem 2, replacing bounds on Ramanujan sums by bounds on sums of the form ebc s /d) and ebfc)/d), c d gcdc,d)= c d gcdc,d)= which can be easily derived from the bounds on Gauss sums leading to the bound d /2+o) for former sums) and bounds on sums with general polynomials leading to the bound d /deg f+o) for the latter sums). See [4, 2, 3] for more details. The same arguments can also be used to study exponential sums and the distribution of fractional parts of the functions fn)/ωn), fn)/ωn) and fn)/τn), where f is a polynomial with integer coefficients. We remark that it has been shown in [6] that almost all of these values do not lie in Z. The methods used in the proof of Theorem 3 can be modified to the study of the counting function of those integers n for which mρn) Z, where m is a fixed integer. p n 7

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