Uniform Distribution of the Fractional Part
of the Average Prime Divisor
William D. Banks
Department of Mathematics, University of Missouri
Columbia, MO 65211 USA
bbanks@math.missouri.edu
Moubariz Z. Garaev
Instituto de Matemáticas,
Universidad Nacional Autónoma de México
C.P. 58180, Morelia, Michoacán, México
garaev@matmor.unam.mx
Florian Luca
Instituto de Matemáticas
Universidad Nacional Autónoma de México
C.P. 58180, Morelia, Michoacán, México
fluca@matmor.unam.mx
Igor E. Shparlinski
Department of Computing
Macquarie University
Sydney, NSW 2109, Australia
igor@ics.mq.edu.au
1
Abstract
We estimate exponential sums with the function ρ(n) defined as
the average of the prime divisors of an integer n ≥ 2 (we also put
ρ(1) = 0). Our bound implies that the fractional parts of the numbers
{ρ(n) : n ≥ 1} 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 ρ(n) is an integer.
AMS Subject Classification: 11L07, 11N37, 11N60
1
Introduction
In this note, we study some distributional properties of the average prime
divisor function ρ(n), defined for each integer n ≥ 2 as follows:
ρ(n) =
1 X
p,
ω(n)
p|n
where, as usual, ω(n) denotes the number of distinct prime divisors p of
n, and the summation runs over all such primes. We also put ρ(1) = 0.
We obtain nontrivial bounds for exponential sums with ρ(n) and use these
estimates to show that the fractional parts of the numbers {ρ(n) : n ≥ 1}
are uniformly distributed over the closed unit interval [0, 1].
In particular, this result implies that ρ(n) 6∈ Z for almost all n ≥ 1. Below, we determine the precise order of magnitude of the counting function of
those integers n ≥ 1 for which ρ(n) ∈ Z. Problems with a similar flavor have
been treated previously in [8, 17, 22, 23], in which the authors investigate the
counting function of those positive integers n for which n/f (n) is an integer,
where f (n) = τ (n) (the divisor function), or more generally f (n) = τk (n)
(the k-th divisor function which can also be defined as the coefficient of n−s
in the Dirichlet series for ζ(s)k , where ζ(s) is the Riemann zeta function),
or f (n) = ℓω(n) (for a fixed positive integer ℓ). We remark that the distributional properties of ρ(n) are rather different from those of the average divisor
function ϑ(n), defined for each integer n ≥ 1 as follows:
ϑ(n) =
σ(n)
1 X
d.
=
τ (n)
τ (n)
d|n
2
Indeed, it is known that ϑ(n) ∈ Z for almost all n ≥ 1; see [2] and [16].
Throughout this paper, for any real number x > 0 and any integer ℓ ≥ 1,
we write logℓ x for the function defined inductively by log1 x = max{log x, 1}
(where log x is the natural logarithm of x), and logℓ x = max{log(logℓ−1 x), 1}
for ℓ > 1. When ℓ = 1, we omit the subscript in order to simplify the
notation; however, we continue to assume that log x ≥ 1 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 = O(B) 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 NSF grant DMS-0070628, F. L. was supported in part
by grants SEP-CONACYT 37259-E and 37260-E, and I. S. was supported in
part by ARC grant DP0211459.
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 4.12 in [1], or
Theorem 427 in [9]), which asserts that the relation
X1
1
(1)
= log2 x + α + O
p
log x
p≤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
X 1
log x
(2)
= β log x + γ + O
ϕ(n)
x
n≤x
holds for all x ≥ 1 (in fact, the error term can be sharpened; see [20]). It is
also useful to recall that the Euler function ϕ(k) satisfies the inequality
ϕ(k) ≫
3
k
.
log2 k
(3)
We use π(x) to denote the number of primes p ≤ x, and we use π(x; 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
p
x
π(x; c, d) =
+ O x exp −A log x
(4)
ϕ(d) log x
√
holds provided that 1 ≤ d ≤ log x and gcd(c, d) = 1.
We denote by πν (x) the number of positive integers n ≤ x with ω(n) = ν.
A particular case of the version of the classical Hardy and Ramanujan
inequality given in [19] and [21] implies that the estimate
πν (x) ≍
x(log2 x)ν−1
(ν − 1)! log x
(5)
holds uniformly for 0.5 log2 x ≤ ν ≤ 2 log2 x (see also [10] for a more general
statement).
We also need the following technical result.
Lemma 1. Let x ≥ 3 and x1/2 ≤ y ≤ x. Then,
X
τ (n)
y log x
≪
.
ϕ(n)
x
x≤n≤x+y
Proof. We have
X
τ (n)
=
ϕ(n)
x≤n≤x+y
X
1 X
1 X
1
1≤2
ϕ(n)
ϕ(n)
x≤n≤x+y
x≤n≤x+y
= 2
X
X
d≤x1/2
d|n
X
x≤n≤x+y
n≡0 (mod d)
d|n
d≤x1/2
1
.
ϕ(n)
We now set w = x1/2 / 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 ϕ(dm) ≥ ϕ(d)ϕ(m) together with the
estimate (2), getting
X
X
X
X 1
1
1
≤
ϕ(n)
ϕ(d)
ϕ(m)
d<w x≤n≤x+y
d<w
n≡0
(mod d)
x/d≤m≤(x+y)/d
X 1 x+y
d log x
β log
+O
=
ϕ(d)
x
x
d<w
yX 1
y log x
≪
≪
.
x d<w ϕ(d)
x
For the larger values of d, we only need (3) to obtain that
X
X
X
X
1
1
≪ log2 x
ϕ(n)
n
1/2
1/2
x≤n≤x+y
x≤n≤x+y
w≤d≤x
n≡0
w≤d≤x
(mod d)
= log2 x
X
w≤d≤x1/2
n≡0
1
d
X
x/d≤m≤(x+y)/d
≪ log2 x log(1 + y/x)
≪
(mod d)
X
w≤d≤x
1
m
1
d
1/2
y log2 x
y log22 x
log(x1/2 /w) ≪
,
x
x
which completes the proof.
For integers ν ≥ 1 and a, we write U(ν; a) for the number of solutions to
the congruence
u1 + u2 + a ≡ 0 (mod ν),
1 ≤ u1 , u2 ≤ ν,
gcd(u1u2 , ν) = 1.
Lemma 2. The following inequality holds:
Y
2
.
1−
U(ν; a) ≥ ν
p
p|ν
Proof. Let ν = pe11 . . . pess be the prime number factorization of ν. By the
Chinese Remainder Theorem, we have
U(ν; a) =
s
Y
i=1
5
U(pess ; a).
For a prime power pe , it is easy to see that U(pe ; a) = pe−1 U(p; a). Indeed,
all solutions to
u1 + u2 + a ≡ 0
(mod pe ),
1 ≤ u 1 , u 2 ≤ pe ,
gcd(u1 u2 , p) = 1
are of the form u1 ≡ v1 + wp (mod pe ), u2 ≡ v2 − wp (mod pe ), where
w = 0, . . . , pe−1 , and (v1 , v2 ) is a solution to
v1 + v2 + a ≡ 0 (mod p),
1 ≤ v1 , v2 ≤ p,
gcd(v1 v2 , p) = 1.
Clearly, the last congruence has at least p − 2 solutions because for each
v1 = 1, . . . , p − 1, v1 6≡ −a (mod p), the value of v2 is uniquely determined
with v2 6≡ 0 (mod p). Hence, U(pe ; a) ≥ pe−1 (p − 2), which finishes the
proof.
Lemma 3. For some absolute constant κ > 0,
X Y
2
= κx + O(log2 x).
1−
p
1≤ν≤x
p|ν
Proof. We have
Y
p|ν
2
1−
p
X
Y
τ (d)µ(d)
τ (p)
=
1−
=
,
p
d
p|ν
d|ν
where µ(d) denotes the Möbius function; we recall that µ(1) = 1, µ(d) = 0
if d ≥ 2 is not square-free, and µ(d) = (−1)ω(d) otherwise. Therefore,
X Y
X X τ (d)µ(d)
X
X τ (d)µ(d)
2
1−
=
1
=
p
d
d
1≤ν≤x
1≤ν≤x
1≤ν≤x
1≤d≤x
p|ν
d|ν
ν≡0
X τ (d)µ(d) x
+ O(1)
=
d
d
1≤d≤x
X τ (d)µ(d)
= x
+O
2
d
1≤d≤x
Using the well-known fact that
X
1≤d≤y
τ (d) ≪ y log y
6
X τ (d)
d
1≤d≤x
!
.
(mod d)
(see Theorem 3.3 in [1], or see Theorem 320 in [9] for a much more precise
statement), and using partial summation, we have the estimates
X τ (d)
= O(log2 x)
d
1≤d≤x
and
X τ (d)
d>x
d2
= O(x−1 log x).
The latter estimate implies
∞
X τ (d)µ(d) X
τ (d)µ(d)
=
+ O(x−1 log x) = κ + O(x−1 log x),
2
d
d2
1≤d≤x
d=1
where κ is the constant
Y
2
κ=
1 − 2 = 0.3226 . . . ,
p
p
and the product is taken over all primes p. The result follows.
3
Exponential Sums with ρ(n)
Let e(x) = exp(2πix) for all x ∈ R. For any integers a, and N with N ≥ 1,
let Sa (N) be the exponential sum given by
Sa (N) =
N
X
e(aρ(n)).
n=1
Theorem 1. For every integer a 6= 0, the following inequality holds:
Sa (N) ≪
|a| N
.
log2 N
Proof. Let P (n) denote the largest prime divisor of n ≥ 2, and put P (1) = 1.
As usual, we say that an integer n ≥ 1 is y-smooth if and only if P (n) ≤ y,
and we put
ψ(x, y) = #{1 ≤ n ≤ x : n is y-smooth}.
We now choose Q = N 1/u , where
u=
2 log3 N
,
log4 N
7
and we denote by E1 the set of Q-smooth positive integers n ≤ N. According
to Corollary 1.3 of [11] (see also [3]), we have the bound
#E1 = ψ(N, Q) ≤ Nu−u+o(u) ≪
N
.
log2 N
Next, we denote by E2 the set of the positive integers n ≤ N not in E1
such that P (n)2 | n. Clearly,
#E2 ≤
XN
p>Q
p2
≪ N/Q ≪
N
.
log2 N
Now let
k = ⌊(1 − δ) log2 N⌋
and
K = ⌊(1 + δ) log2 N⌋ ,
where δ > 0 is a sufficiently small absolute constant to be chosen later, and
let E3 denote the set of positive integers n ≤ N such that either ω(n) < k or
ω(n) > K. By the Turán-Kubilius inequality (see [14, 24]), it follows that
#E3 ≪
N
.
log2 N
Let E4 denote the set of positive integers n ≤ N/ log2 N. Of course,
#E4 ≪
N
.
log2 N
Now let n ≤ N be a positive integer not in ∪4i=1 Ei . This integer n has a
unique representation of the form n = mp, where m is such that m < N/Q
and P (m)m > N/ log2 N, and p = P (n) is a prime number in the half-open
interval p ∈ Lm , where
N
Lm = max Q, P (m),
and
Lm = (Lm , N/m].
m log2 N
Let E5 be the set of those n such that Lm = Q. In this case,
N
N
<m< .
Q log2 N
Q
8
When m is fixed, p ≤ N/m can take at most π(N/m) values. Thus, the
number of elements n ∈ E5 is
X
N
#E5 ≪
π
m
N/(Q log2 N )<m<N/Q
≪
X
N/(Q log2 N )<m<N/Q
N
m log(N/m)
X
1
m
N/(Q log2 N )<m<N/Q
Nu
N
N
≪
log
− log
log N
Q
Q log2 N
2
Nu log3 N
N log3 N
N
≪
≪
≪
.
log N
log N log4 N
log2 N
≪
N
log Q
Finally, let E6 be the set of those positive integers n ≤ N which are not
in ∪5i=1 Ei and such that Lm = P (m). In this case,
P (m) ≥
N
,
m log2 N
so we see immediately that p = P (n) ≤ P (m) log2 N. Thus, E6 is contained in
the set of all those positive integers n ≤ N which are divisible by two primes
q < p, such that p ≤ q log2 N and p > Q. In particular, q ≥ Q/ log2 N >
Q1/2 . Fix q and p. The number of such n ≤ N is O(N/pq). Hence, using (1),
we derive that for each q the total number Tq (N) of such n ≤ N with some
prime p in the interval (q, q log2 N] can be estimated from above as
X
N
1
N
N
Tq (N) ≪
=
(log2 (q log2 N) − log2 q) + O
q q<p≤q log N p
q
q log q
2
N
N log3 N
log3 N
N
=
+O
≪
log 1 +
.
q
log q
q log q
q log q
Summing the above inequality over all q > Q1/2 , we get that
X
X
N log3 N
1
≪
#E6 ≤
Tq (N) ≪ N log3 N
q log q
log Q
1/2
1/2
q>Q
≪
q>Q
log23
N
N
N
Nu log3 N
≪
≪
.
log N
log N log4 N
log2 N
9
Thus for each i = 1, . . . , 6, we have the estimate
#Ei ≪
N
.
log2 N
(6)
We now let N be the set of positive integers n ≤ N which do not belong
to any of the sets Ei for i = 1, . . . , 6. We see from (6) that
X
N
Sa (x) =
e(aρ(n)) + O
.
(7)
log2 N
n∈N
Each n ∈ N admits a unique representation of the form n = pm, where
p > P (m). Moreover, in this case, p ∈ Lm = (N/(m log2 N), N/m]. Let M
be the set of permissible values for m.
Let Nν be the subset of n ∈ N with ω(n) = ν. Note that each such n is
of the form n = pm, where m ∈ M has ω(m) = ν − 1 and p ∈ Lm . We write
Mν−1 for the set of those m ∈ M with ω(m) = ν − 1. Note that
X
#N =
π(Lm ),
m∈M
and for k ≤ ν ≤ K, we have
#Nν =
X
π(Lm ),
(8)
m∈Mν−1
where we have used π(Lm ) to denote the number of primes in the interval Lm .
With the above notations, we can write
X
e(aρ(n)) =
K
X
X
X
e(aρ(mp))
ν=k m∈Mν−1 p∈Lm
n∈N
=
K
X
X
ν=k m∈Mν−1
e(aρ(m)(ν − 1)/ν)
X
e(ap/ν).
p∈Lm
Therefore, by the Page
√ bound (4), for any integers b and d with gcd(b, d) = 1,
and with 1 ≤ d ≤ log x, we have
X
X
p
x
e(bp/d) =
e(bc/d) + O xd exp −A log x .
ϕ(d) log x 1≤c≤d
p≤x
gcd(c,d)=1
10
The sum over c is the classical Ramanujan sum and (since gcd(b, d) = 1) is
equal to µ(d), where, as before, µ(d) denotes the Möbius function (see, for
example, Theorem 272 of [9]). Consequently, writing νa = ν/ gcd(a, ν), and
noting that
Q
N
νa ≤ ν ≤ 2 log2 N < log
< log
log2 N
m log2 N
holds for all m ∈ M and all sufficiently large N, we derive that
X
n∈N
e(aρ(n)) ≪
K
X
X
ν=k m∈Mν−1
We have
K
X
X
ν=k m∈Mν−1
Nνa
1
√
π(Lm ) +
ϕ(νa )
m exp 0.5A log Q
!
.
X 1
Nνa
N log2 N
≪
√
√
m exp 0.5A log Q
exp 0.5A log Q m<N m
≪
In particular, by (8),
X
n∈N
e(aρ(n)) ≪
K
X
ν=k
N
N log N log2 N
≪
√
.
log2 N
exp 0.5A log Q
1
N
#Nν +
.
ϕ(νa )
log2 N
(9)
We now substitute the inequality
ϕ(ν)
ν
ϕ(ν)
≥
ϕ(νa ) = ϕ
≥
gcd(ν, a)
gcd(ν, a)
|a|
in (9), and use the trivial inequality #Nν ≤ πν (N) in combination with (5),
to get that
X
n∈N
K
|a|N X (log2 N)ν−1
N
e(aρ(n)) ≪
+
.
log N
ϕ(ν)(ν − 1)! log2 N
(10)
ν=k
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:
1 2
xν
−1/2
3
≤x
exp x 1 − λ + O(λ )
,
ν!
2
11
where λ = |1 − ν/x|. In particular, we see that if δ is sufficiently small, then
(log2 N)ν−1
|ν − log2 N|2
log N
exp −
.
(11)
≪p
(ν − 1)!
3 log2 N
log2 N
For an integer ℓ, let Iℓ be the interval defined by
h
i
p
p
Iℓ = log2 N + ℓ log2 N , log2 N + (ℓ + 1) log2 N .
Using (11), we derive that
K
X
(log2 N)ν−1
ϕ(ν)(ν − 1)!
ν=k
log N
≤p
log2 N
−δ
√
X
log2 N −1≤ℓ≤δ
√
exp(−ℓ2 /3)
log2 N +1
X
ν∈Iℓ
1
.
ϕ(ν)
Furthermore, by (2), we have
X
ν∈Iℓ
1
= β log
ϕ(ν)
!
p
log2 N + (ℓ + 1) log2 N
log3 N
p
+O
log2 N
log2 N + ℓ log2 N
1
≪p
.
log2 N
(12)
Combining the preceding two estimates and substituting into (10), the proof
is completed.
4
Uniformity of Distribution of ρ(n)
For a sequence of N points in the half-open interval [0, 1), the discrepancy
of the sequence is defined as
A(u)
∆ = sup − u ,
N
0≤u<1
where A(u) is the number of points of the sequence in the interval [0, u).
Let ∆N denote the discrepancy of the sequence consisting of the fractional
parts of the numbers ρ(n) for n = 1, . . . , N. In this section, we give an upper
bound for ∆N .
12
Theorem 2. For N → ∞, the following bound holds:
∆N ≪
log23 N
.
log2 N
Proof. According to the well-known Erdős-Turán relation between the discrepancy and the appropriate exponential sums (see [6, 15]), we find that for
any H ≥ 1:
1 X 1
1
+
|Sa (N)|.
∆N ≪
H N 1≤a≤H a
We now choose u, Q, k and K as in the proof of Theorem 1. Using the
bound (9), we derive
K
X 1X
1 X 1
1
N log H
|Sa (N)| ≪
#Nν +
.
N 1≤a≤H a
a
ϕ(ν
)
log
N
a
2
1≤a≤H
ν=k
Changing the order of summation and then collecting together the values of
a with the same value of gcd(a, ν) = d for each divisor d of ν, and making
use of the inequality ϕ(νa ) ≥ ϕ(ν)/ gcd(ν, a), we deduce that
K
K
X
X
X 1X
1
1
#Nν ≪
#Nν
d
a
ϕ(ν
ϕ(ν)
a)
1≤a≤H
ν=k
ν=k
≪
d|ν
X
1≤a≤H
gcd(a,ν)=d
K
X
X
X
1
#Nν
d
ϕ(ν)
ν=k
d|ν
≪ log H
K
X
ν=k
1≤b≤H/d
1
a
1
db
τ (ν)
#Nν .
ϕ(ν)
We now proceed as in the proof of Theorem 1, and using Lemma 1 instead
of (12), we obtain
K
X
N log3 N
τ (ν)
#Nν ≪
.
ϕ(ν)
log
N
2
ν=k
Therefore,
1 X 1
log3 N log H
|Sa (N)| ≪
.
N 1≤a≤H a
log2 N
13
Putting everything together, gives the estimate
∆N ≪
1
log3 N log H
+
.
H
log2 N
Choosing H = ⌈log2 N⌉, we obtain the stated result.
5
Integrality of ρ(n)
Let R(N) denote the number of positive integers n ≤ N with ρ(n) ∈ Z.
Theorem 3. The following bound holds
R(N) ≍
N
.
log2 N
Proof. For an integer m, we denote
a(m) =
X
q.
q|m
The proof of the upper bound follows easily from the proof of Theorem 1.
Indeed, with the notations of Theorem 1, we may assume that n ∈ N , since
the number of n ≤ N not in N is at most O(N/ log2 N). Write n = mp and
fix m. Then, in order for ρ(n) to be an integer it is necessary that the prime
p ∈ Lm satisfies p ≡ −a(m) (mod ν), where ν = ω(m) + 1. Since Lm is large
enough with respect to ν = O(log2 N), we can apply the Page bound (4) as
in the proof of Theorem 1, and we see that the number Sm of such choices
for p is bounded by
p
1
N
Sm ≤
π(Lm ) + O
exp −0.5A log Q
ϕ(ν)
m
(in fact, there are no such primes when gcd(a(m), ν) > 1). Summing up the
above bound up over all the possible choices of m ∈ M, and using (8), we
find that the estimate
K
X
N
1
#Nν +
#R(N) ≤
Sm ≪
ϕ(ν)
log2 N
m∈M
ν=k
X
14
holds, and this leads to the desired upper bound using arguments towards
the end of the proof of Theorem 1.
We now turn to the proof of the lower bound .
Suppose
Let us fix ν ∈ J where J is the
that N is sufficiently large.
interval 0.5 log2 (N 1/4 ), 2 log2 (N 1/4 ) .
Choose a positive integer m ≤ N 1/4 with ω(m) = ν − 2.
Now take a solution (u1 , u2 ) to the congruence
u1 + u2 + a(m) ≡ 0 (mod ν)
1 ≤ u1 , u2 ≤ ν,
gcd(u1u2 , ν) = 1. (13)
We now select a prime p1 in the interval N 1/4 < p1 < N 1/3 such that
p1 ≡ u1 (mod ν) and a prime p2 in the interval N 1/3 < p2 < N/mp1 such that
p1 ≡ u1 (mod ν). We remark that N/mp1 > N 1−1/4−1/3 = N 5/12 > N 1/3 ; in
particular, p2 > p1 .
It is clear from the above construction that n = p1 p2 m is in R(N), and
that different choices of m, p1 , p2 lead to different values for n ∈ R(N).
It remains to count how many numbers n can be obtained by this process.
Assume that m, u1 , u2 , p1 are fixed. Because ν = ω(m) + 2 = O(log2 N), the
Page bound (4) applies. Therefore, there are
!
π(N/mp1 ) − π(N 1/3 )
N
N
≫
√
+O
ϕ(ν)
mp1 ν log N
mp1 exp 0.5A log N
possible values for p2 (here we have used the trivial estimate ϕ(ν) ≤ ν).
Therefore, if m, u1 , u2 are fixed the above construction gives Rm,u1 ,u2 (N)
integers n ∈ R(N), where
Rm,u1 ,u2 (N) ≫
N
mν log N
X
N 1/4 <p1 ≤N 1/3
p1 ≡u1 (mod ν)
1
.
p1
Using the Page bound (4) again and partial summation, we derive that
Rm,u1 ,u2 (N) ≫
N
N
log2 (N 1/3 ) − log2 (N 1/4 ) ≫
.
mνϕ(ν) log N
mν log N log2 N
Here we have used the estimate ϕ(ν) ≪ log2 N. Recalling Lemma 2, we see
15
that the contribution to R(N) from each fixed value of m is
X
u1 ,u2
N
U(ν; a(m))
mν log N log2 N
Y
N
2
≥
1−
,
m log N log2 N
p
Rm,u1 ,u2 (N) ≥
p|ν
where the sum is taken over all solutions (u1, u2 ) to (13). Thus, we finally
obtain that
X
XY
1
2
N
1−
.
R(N) ≫
log N log2 N ν∈J
p
m
1/4
p|ν
1≤m<N
ω(m)=ν−2
We now apply (5) which, combined with the Stirling formula, yields
#{m ≤ N 1/4 : ω(m) = ν − 2} ≫
N 1/4 (log2 N)ν−2
N 1/4 log N
.
≫ p
(ν − 2)! log N
log2 N
Hence, by partial summation and for an appropriate constant c > 0, we have
X
1≤m<N 1/4
ω(m)=ν−2
Therefore,
1
≥
m
X
N 1/4 (1−c log N/(log2 N )1/2 )≤m<N 1/4
c log N
= log 1 + p
log2 N
N
R(N) ≫
(log2 N)3/2
0.5 log2
!
+O
X
(N 1/4 )≤ν≤2 log
2
1
m
1
log N
(N 1/4 )
log N
≫p
.
log2 N
Y
p|ν
2
1−
p
.
Applying Lemma 3, we complete the proof.
6
Remarks
Clearly, using the bound of Theorem 1 directly in the proof of Theorem 2
leads to a weaker bound (about the square root of the present bound). On
16
the other hand, in the proof of Theorem 1 itself we can probably work much
more carefully with νa and obtain a better dependence on a.
It is easy to see that the results of Theorems 1, 2 and 3 also hold for the
function
1 X
ρe(n) =
p ordp n,
Ω(n)
p|n
where Ω(n) is the number of prime divisors p of n, each one counted with
multiplicity ordp n (one recalls that the Turán-Kubilius inequality [14, 24]
applies to Ω(n) as well). In particular,
N
X
n=1
e(ae
ρ(n)) ≪
|a| N
.
log2 N
Similar but weaker results can be obtained for
1 X s
1 X
ρs (n) =
p
and
ρ(f ; n) =
f (p),
ω(n)
ω(n)
p|n
p|n
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 1 and Theorem 2, replacing
bounds on Ramanujan sums by bounds on sums of the form
X
X
e(bf (c)/d),
e(bcs /d)
and
1≤c≤d
gcd(c,d)=1
1≤c≤d
gcd(c,d)=1
which can be easily derived from the bounds on Gauss sums (leading to the
bound d1/2+o(1) for former sums) and bounds on sums with general polynomials (leading to the bound d1−1/ deg f +o(1) for the latter sums). See [4, 12, 13]
for more details.
The same arguments can also be used to study exponential sums and the
distribution of fractional parts of the functions f (n)/ω(n), f (n)/Ω(n) and
f (n)/τ (n), where f is a polynomial with integer coefficients. We remark that
it has been shown in [16] 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 ≥ 1 for which mρ(n) ∈ Z, where
m ≥ 1 is a fixed integer.
17
It is shown in [18] that only O(x log−2 x log42 x) pairs consecutive integers
(n, n + 1), 1 ≤ n ≤ x, may have the same sum of prime divisors. This
naturally leads to a question how often ρ(n) = ρ(n + 1).
Finally, we remark that it would be very interesting to study the properties of ρ(n) on some special sequences of integers, in particular, on shifted
primes, ρ(p − 1), or on the values of an integer valued polynomial, ρ(F (n)).
These questions appear to be very difficult. On the other hand, it is likely
that our methods can be used to study the values ρ(ϕ(n)), though one first
needs to modify the results of [7] in order to have explicit bounds for the
number of positive integers n ≤ x with atypically small or large values of
ω(ϕ(n)).
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18
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19
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