SPACINGS BETWEEN INTEGERS HAVING TYPICALLY MANY
PRIME FACTORS
RIZWANUR KHAN
Abstract. We show that the sequence of integers which have nearly the typical number of distinct prime factors forms a Poisson process. More precisely,
for δ arbitrarily small and positive, the nearest neighbor spacings between
integers n with |ω(n) − log2 n| ≤ (log2 n)δ obey the Poisson distribution law.
1. Introduction
Consider n random variables independently and uniformly taking real values
in the interval [0, n]. Let Y1 < ... < Yn denote the order statistics obtained by
arranging these random variables in increasing order. Setting Y0 = 0 and Yn+1 = n,
let Di = Yi+1 − Yi for 0 ≤ i ≤ n denote the nearest neighbor spacings of the order
statistics. Thus D1 + . . . + Dn = n and by symmetry it follows that for 0 < λ < n
a real number,
n − λ n
Prob(Di > λ) = Prob(D1 > λ) =
.
n
Thus Prob(Di > λ) ∼ e−λ as n → ∞. This is the exponential or Poisson distribution.
We are interested in the spacing distributions of arithmetic sequences. An example of such a sequence is the sequence of prime numbers less than x, which form
a sparse subset of the integers of density 1/ log x by the prime number theorem.
This is similar to Y1 , . . . , Yn being sparse in the interval [0, n]. If pi denotes the
i-th prime less than x, we rescale to consider instead the sequence pei = pi / log x
of ‘normalized’ primes so that that the average spacing between consecutive normalized primes is 1 as x → ∞. This matches the expected value of Di above as
n → ∞. Gallagher [4] showed that assuming the validity of the Hardy-Littlewood
prime k-tuple conjectures, we have for λ > 0 real that
1
(1.1)
#{i ≤ x : pei+1 − pei > λ} ∼ e−λ
x
as x → ∞. Thus conditionally we see that the spacings between primes obey the
Poisson distribution law, as in the prototypical situation of randomly dispersed
objects mentioned at the start. More recently Kurlberg and Rudnick [9] showed
that the spacings between quadratic residues modulo q , as the number of distinct
prime divisors of q tends to infinity, follow the Poisson distribution. There are many
other interesting arithmetic sequences that are conjectured to be Poisson processes,
but only few examples exist with proof. For example, √
it is an open problem to show
that the spacings between the fractional parts of n2 2 for n ≤ x, as x → ∞ are
Poisson distributed (see [11]). The reader may find a few more examples of such
work listed in the references section (see [1, 6, 7]). Of course there are important
1
2
RIZWANUR KHAN
arithmetic sequences which are not expected to behave like randomly dispersed
elements in this sense, such as the non-trivial zeros of the Riemann Zeta function.
In this paper we are interested in the spacings between integers with not only one
prime factor as in Gallagher’s work, but with the typical number of distinct prime
factors. We first explain what is meant by ‘typical’.
Let ω(n) denote the number of distinct prime factors of n. It is easy to see that
integers n ≤ x have log log x distinct prime factors on average:
1 X jxk
1 XX
1 XX
1X
(1.2)
1=
= log2 x + O(1),
ω(n) =
1=
x
x
x
x
p
n≤x
n≤x p|n
p≤x n≤x
p|n
p≤x
where we write log2 x for log log x, and similarly for logj x. Also, throughout this
paper p and q will be used to denote primes. The variance can be shown to be
1X
(1.3)
(ω(n) − log2 x)2 ∼ log2 x.
x
n≤x
Note that (1.2) and (1.3) imply that ω(n) ∼ log2 x for all most all n ≤ x. Erdős
and Kac [2] further p
showed that ω(n) is normally distributed with mean log2 x and
standard deviation log2 x. Rényi and Turán [10] proved this result with a sharp
error term. The following theorem can also be found in Tenenbaum’s book [14].
Theorem 1.1. Given a real number C > 0 we have for 0 < c < C that the number
2x
√
of integers n ≤ x for which −c < ω(n)−log
< c is
log2 x
1
x√
2π
Z
c
exp(−u2 /2)du + OC (x/
p
log2 x).
−c
We [8] proved a slightly weaker version of the Theorem 1.1 by methods similar to
those in this paper.
p πWe conjecture that the spacings between integers n ≤ x with |ω(n) − log2 x| ≤
2 (that is, integers with more or less exactly log2 x distinct prime factors) obey
the Poisson distribution law but we are unable to prove it. Instead we look at an
easier question. For any fixed 0 < δ < 1/2, let us say an integer less than x2 is
‘δ-normal’ if
r
π
|ω(n) − log2 x| ≤
(log2 x)δ .
2
We study the sequence of δ-normal numbers. These are integers having nearly the
expected
number of prime factors, as (log2 x)δ is smaller than the standard deviation
p
log2 x of ω(n). Denote the sequence of δ-normal numbers in increasing order by
N1 , N2 , N3 . . .. Up to x, there are x(log2 x)−1/2+δ such integers by Theorem 1.1,
p
√
2 x
since an integer is δ-normal if and only if ω(n)−log
≤ π2 (log2 x)−1/2+δ . Thus we
log2 x
ei = Ni (log2 x)−1/2+δ . Our main theorem
should rescale these integers by setting N
is
Theorem 1.2. For any fixed real number λ > 0 we have
1
ei+1 − N
ei > λ} ∼ e−λ .
#{i ≤ x : N
x
Throughout this paper, all implicit constants may depend on δ and λ.
SPACINGS BETWEEN INTEGERS HAVING TYPICALLY MANY PRIME FACTORS
3
2. Independence between additive shifts of the ω(n) function
In this section we show how Theorem 1.2 can be reduced to studying correlations
between the additive shifts of the function ω(n). We will show for example that
ω(n) − log2 x, ω(n + 1) − log2 x, and ω(n + 2) − log2 x behave independently. Define
N (x) to be the number of δ-normal integers less than x. The left hand side of
Theorem 1.2 is asymptotic to
1
#{i ≤ x : Ni+1 − Ni > λ(log2 x)1/2−δ }
x
1
(2.1)
∼ #{N ≤ x(log2 x)1/2−δ : N N + λ(log2 x)1/2−δ − N N = 0},
x
where N denotes a δ-normal number. Define Nb1 ,...,br (x) to be the number of
integers n ≤ x for which n + bi is δ-normal for all 1 ≤ i ≤ r and let σ(m, r)
denote the number of maps from the set
{1, ...,m} onto {1, ..., r}. We have the
m-th moment of N N + λ(log2 x)1/2−δ − N N :
X
m
1
N N + λ(log2 x)1/2−δ − N N
x
1/2−δ
N ≤x(log2 x)
m
(2.2)
=
1X
σ(m, r)
x r=1
X
N0,b1 ,...,br x(log2 x)1/2−δ .
1≤b1 <...<br ≤λ(log2 x)1/2−δ
We will prove
Theorem 2.1. For a fixed integer r and any integers 0 ≤ b1 < ... < br ≤
λ(log2 x)1/2−δ , we have
1
Nb ,...,br (x) ∼ (log2 x)(−1/2+δ)r .
x 1
Throughout this paper all implicit constants may depend r. Since a randomly
chosen integer less than x is δ-normal with probability (log2 x)−1/2+δ , the theorem
above says that n + b1 , . . . , n + br are independently likely to be δ-normal. Theorem
2.1 implies that for fixed m we have that (2.2) is asymptotic to
(2.3)
∼
m
X
r=1
∞
σ(m, r)
X
e−λ λj
λr
=
jm
,
r!
j!
j=0
which is the m-th moment of the Poisson distribution (the identity above is known
as Dobinski’s formula). The Poisson distribution can be recovered from these moments. Thus Theorem 1.2 follows from Theorem 2.1. Next we discuss the demonstration of Theorem 2.1.
The characteristic function of a random variable with a normal distribution
√ i )−log2 x for 1 ≤ i ≤ r by
is exp(−T 2 /2). We show the independence of ω(n+b
log2 x
Qr
showing that their joint characteristic function equals essentially i=1 exp(−Ti2 /2).
Actually it is more convenient to work with ω(n; y, z) in place of ω(n), where we
set
y = y(x) = (log x)3r
and
−3r
z = z(x) = x((log2 x)
)
4
RIZWANUR KHAN
and define
ω(n; y, z) =
X
1.
p|n
y<p<z
P
Accordingly we work with ω(n; y, z) − y<p<z p1 in place of ω(n) − log2 x. We will
soon see that there is not much loss in disregarding the primes less than y or greater
than z. In an imprecise sense, the reason for this is that on average integers have
few small prime factors and few large prime factors. In the next section we will
prove the following theorem.
P
P
1
( y<p<z p1 )1/2 ,
Theorem 2.2. Let ti = Ti ( y<p<z p1 )−1/2 be real. For |Ti | ≤ 1000
we have
P
1
r
r
ω(n + bi ; y, z) −
X 1
Y
1 XY
y<p<z p
q
exp iTi
=
exp eiti − 1 − iti
P
1
x
p
y<p<z
i=1
i=1
n≤x
y<p<z p
+ O(1/ log x).
Observe that for Ti ≤ (log2 x) for small enough > 0 we have
−t2
X 1
X 1
i
= exp
+ O(t3i )
exp eiti − 1 − iti
p
2
p
y<p<z
y<p<z
−T 2 T 3 i
= exp
1+O p i
(2.4)
,
2
log2 x
and for (log2 x) < Ti ≤ (log2 x)1/2− we have
X 1
(2.5)
exp(−Ti2 /4),
exp eiti − 1 − iti
p
y<p<z
where the implied constants depend on .
To see how Theorem 2.2 implies Theorem 2.1 we will use the following lemmas.
Lemma 2.3. Let ψ(x) be a real function differentiable b4r/δc times and satisfying
0 ≤ ψ(x) ≤ 1 for x ∈ R,
ψ(x) = 0 for |x| ≥ 2(log2 x)−1/2+δ
Z ∞
√
ψ(x)dx ∼ 2π(log2 x)−1/2+δ ,
−∞
(2.6)
|ψ (j) (x)| (log2 x)j(1−δ)/2 for any positive integer j ≤ b4r/δc.
We have
r
1 X Y ω(n + bi ; y, z) −
qP
ψ
x
i=1
n≤x
1
y<p<z p
P
1
y<p<z p
∼ (log2 x)(−1/2+δ)r .
Proof. Let
Z
∞
ψ̂(T ) =
−∞
ψ(u)e−iuT du
SPACINGS BETWEEN INTEGERS HAVING TYPICALLY MANY PRIME FACTORS
5
denote the Fourier transform of ψ. By Fourier inversion we have
P
1
r
1 X Y ω(n + bi ; y, z) − y<p<z p qP
ψ
1
x
n≤x i=1
(2.7)
y<p<z p
r
Y
1X
1
=
x
2π
i=1
n≤x
Z
∞
−∞
ω(n + bi ; y, z) −
qP
ψ̂(Ti ) exp iTi
P
1
y<p<z p
1
y<p<z p
dTi .
We have that |ψ̂(Ti )| (log2 x)j(1−δ)/2 |Ti |−j , by integrating by parts j times and
using (2.6). Thus (2.7) equals
P
Z (log2 x)1/2−δ/4
1
r
ω(n + bi ; y, z) −
1 XY 1
y<p<z p
qP
(2.8)
ψ̂(Ti ) exp iTi
dTi
1
x
2π −(log2 x)1/2−δ/4
n≤x i=1
y<p<z p
+ O (log2 x)−r .
Now by Theorem 2.2 and observations (2.4) and (2.5), the main term above equals
!
Z (log2 x)
r
−T 2 Y
1
1
i
ψ̂(Ti ) exp
dTi + O p
2π −(log2 x)
2
log2 x
i=1
!
Z ∞
−T 2 r
1
1
dT + O p
(2.9)
=
ψ̂(T ) exp
.
2π −∞
2
log2 x
2
2
is exp −T2 . By the Plancherel
Recall that the Fourier transform of √12π exp −u
2
formula, (2.9) equals
!
Z ∞
r
−u2 1
1
√
(2.10)
du + O p
∼ (log2 x)(−1/2+δ)r .
ψ(u) exp
2
2π −∞
log2 x
To prove Theorem 2.1 we need to show
r
1 X Y ω(n + bi ) − log2 x p
ψ
∼ (log2 x)(−1/2+δ)r ,
x
log
x
2
i=1
n≤x
where ψ is a suitable
p smooth function
p approximating the characteristic function
of the interval [− π2 (log2 x)−1/2+δ , π2 (log2 x)−1/2+δ ]. This is accomplished by
Lemma 2.3, provided that we can show that we may neglect prime factors smaller
than y or larger than z without significant loss. This is the purpose of the next
lemma.
Lemma 2.4. Except for O x(log2 x)−r integers less than x we have
1
ω(n) − log x ω(n; y, z) − P
y<p<z p 2
qP
−
p
(log2 x)−1/2+δ/2 .
1
log2 x
y<p<z p
Proof. Let E(x) denote the set of integers less than or equal to x with more than
(log2 x)δ/2 distinct prime factors less than y or more than (log2 x)δ/2 distinct prime
6
RIZWANUR KHAN
factors greater than z. The size of this set is
δ/2
|E(x)| ≤
X 1 (log2 x)
x
δ/2
b(log2 x) c! p≤y p
+
X 1 (log2 x)δ/2
x
b(log2 x)δ/2 c! z≤p≤x p
x
,
(log2 x)r
√
P
1
using that p≤x p1 = log2 x+C+O(1/ log x) and Stirling’s estimate n! ∼ 2πnn+ 2 e−n .
For n ∈
/ E(x) we have ω(n) − ω(n; y, z) (log2 x)δ/2 , and so it follows that
ω(n) − log2 x ω(n; y, z) −
p
qP
−
log2 x
P
1
y<p<z p
1
y<p<z p
ω(n) − log2 x − ω(n; y, z) +
qP
=
1
1
y<p<z p
P
y<p<z p
|ω(n) − log2 x| log3 x
+O
log2 x
!
δ/2
|ω(n) − log2 x| log3 x
(log x)
+
p2
.
log2 x
log2 x
The proof is complete by noting that except for O x/(log2 x)r integers less than
2x
√
x we have |ω(n) − log2 x| (log2 x)1/2+δ/4 . This is because ω(n)−log
has normal
log2 x
8br/δc
P
2x
√
1, where the implied
moments and in particular x1 n≤x ω(n)−log
log2 x
constant depends on r/δ.
3. Proof of Theorem 2.2
Define for a prime p
(
1−
fp (n) =
− p1
1
p
if p|n
,
if p - n
Q
Q
αi
i
and if m = i pα
i , define fm (n) =
i fpi (n) . (Thus f1 (n) = 1.) If we think of a
prime p dividing n with probability 1/p independently of other
P we have
Pprimes, then
E(fm (n)) = 0 for square-free m. So we have written ω(n) − p≤x p1 = p≤x fp (n)
as a sum of independent random variables
P of mean 0, which already suggests by the
Central Limit Theorem that ω(n) − p≤x p1 is normally distributed. This simple
idea is actually very powerful. It is borrowed from Granville and Soundararajan
P
[5], who use it to efficiently compute very high moments of ω(n) − p≤x p1 and
provide a new proof of the Erdős-Kac theorem.
SPACINGS BETWEEN INTEGERS HAVING TYPICALLY MANY PRIME FACTORS
7
We have that the left hand side of Theorem 2.2 equals
r
ω(n + bi ; y, z) −
1 XY
qP
exp iTi
x
i=1
1
y<p<z p
n≤x
=
1
y<p<z p
P
r
X
1 XY
exp
iti fp (n + bi )
x
y<p<z
i=1
n≤x
=
r
X
1X Y
exp
iti fp (n + bi )
x
y<p<z
i=1
n≤x
(3.1)
=
r
r
X
1 X
2
1X Y 1+
iti fp (n + bi ) +
iti fp (n + bi ) + . . . .
x
2! i=1
y<p<z
i=1
n≤x
Now upon expansion of the product, (3.1) equals
1X
x
(3.2)
n≤x
X
Ka1 ,...,ar
ai ≥1
p|ai ⇒y<p<z
r
Y
Ω(ai )
ti
fai (n + bi ),
i=1
for some constants Ka1 ,...,ar of modulus bounded by 1, where Ω(a) is the number
of prime factors of a counted with multiplicity. Note that when the integers ai are
pairwise coprime we have that
(3.3)
Ka1 ,...,ar =
r Y α
Y
i
.
α!
i=1 α
p |ai
We will evaluate (3.2) using the following results. The first generalizes a result from
[5].
Lemma 3.1. Let ai be pairwise coprime integers for 1 ≤ i ≤ r. Denote the squarefree part of ai by Ai . We have
r
r Y Y
1 XY
1
1 α −1 α 1 fai (n + bi ) =
1−
+
1−
x
p
p
p
p
i=1
i=1 α
n≤x
p kai
+O
r
1 Y
x i=1
τ (Ai )2 ,
where τ (A) denotes the number of divisors of A and pα k a means that pα |a and
pα+1 - a. Note that the main term is zero unless each ai is square-full (that is, p|ai
implies p2 |ai ).
8
RIZWANUR KHAN
Proof. For a fixed integer a the value fa (n) depends only on the common prime
factors of a and n, so fa (n) = fa ((A, n)). Thus we can group terms this way:
r
1 XY
1 X
fai (n + bi ) =
x
x
i=1
n≤x
X
di |Ai
=
1 X
x
(3.4)
=
fai (di )
i=1
n≤x
(Ai ,n+bi )=di
X
di |Ai
r
Y
r
Y
X
µ(ei )fai (di )
(A ,n+b )
n≤x
ei | i d i i=1
i
di |(n+bi )
1 X X
x
A
di |Ai ei | i
d
i
X
r
Y
1
µ(ei )fai (di ).
i=1
n≤x
ei di |n+bi
By
Theorem, since the integers ai are pairwise coprime,
P the Chinese Remainder
1 = Qr x ei di + O(1). Therefore the above sum is
n≤x
i=1
ei di |n+bi
r
r X
1 Y
Y
fai (di ) X µ(ei )
τ (Ai )2
+O
di
ei
x i=1
A
i=1
di |Ai
ei | d i
i
r X
r
A
1 Y
Y
fai (di ) φ( dii )
2
=
τ
(A
)
+
O
i
Ai
di
x i=1
d
i=1
di |Ai
=
(3.5)
r X
Y
i=1 di |Ai
i
r
1 Y
1 Ai fai (di ) φ
τ (Ai )2 .
+O
Ai
di
x i=1
Now it is easily verified (by multiplicativity in ai ) that the main term of the last
line above equals
r Y Y
1
1 α −1 α 1 1−
+
1−
.
p
p
p
p
i=1 α
p kai
In the case that a1 , . . . , ar are not pairwise coprime we will need the following.
Lemma 3.2. Let r ≥ 2 and 0 ≤ b1 < ... < br ≤ λ(log2 x)1/2−δ . Suppose that for
some prime y < q < z, we have that q|a1 and q|a2 . Let q αi k ai and let a0i = ai q −αi .
Let Ai denote the squarefree part of ai and let A0i denote the square free part of a0i .
We have
r
r
r
1 1 XY
1 Y
1 XY
fai (n + bi ) = O 2
fa0i (n + bi ) + O
τ (Ai )2 .
x
q x
qx i=1
i=1
i=1
n≤x
n≤x
Proof. From (3.4) we have
(3.6)
r
1 XY
1 X X
fai (n + bi ) =
x
x
A
i=1
n≤x
di |Ai ei | i
d
i
The sum
P
n≤x
ei di |n+bi
X
n≤x
ei di |n+bi
1
r
Y
µ(ei )fai (di ).
i=1
1 is zero for large enough x if q divides more than one integer
ei di . This is because if q|n + bi and q|n + bj then q|bi − bj and hence i = j since
SPACINGS BETWEEN INTEGERS HAVING TYPICALLY MANY PRIME FACTORS
9
q > y and the integers bi are distinct and bounded by λ(log2 x)1/2−δ . Thus we can
suppose that q divides at most one integer ei di . First consider the terms of (3.6)
with q - ei di for all i. These terms contribute
1 X
x
0
X
X
di |Ai ei |A0i /di
=
1
r
Y
n≤x
ei di |n+bi
r
−1 α1 +...+αr 1 X Y
q
x
µ(ei )fa0i (di )fqαi (di )
i=1
fa0i (n + bi ),
n≤x i=1
using that fqαi (di ) = (−1/q)αi and the identity of (3.6). Since α1 + α2 ≥ 2 we get
the desired factor of 1/q 2 . Now say q|e1 d1 and q - ei di for i 6= 1. The contribution
of this case is,
(3.7)
1 X
x
0
X
di |Ai ei |A0i /di
+
1 X
x
0
r
Y
1 µ(e1 )fa01 (qd1 )fqα1 (qd1 )
µ(ei )fa0i (di )fqαi (di )
X
i=2
n≤x
qe1 d1 |n+b1
ei di |n+bi ,i6=1
X
r
Y
µ(ei )fa0i (di )fqαi (di ),
1 µ(qe1 )fa01 (d1 )fqα1 (d1 )
X
di |Ai ei |A0i /d0i
i=2
n≤x
qe1 d1 |n+b1
ei di |n+bi ,i6=1
where the first line corresponds to q|d1 and the second to q|e1 . We have by the
Chinese Remainder Theorem,
X
1 X
1=
1 + O(1).
q
n≤x
qe1 d1 |n+b1
ei di |n+bi ,i6=1
n≤x
ei di |n+bi
Thus the contribution of the sums of (3.7) is
(3.8)
1
q
1−
r
1 α1 −1 α2 +...+αr 1 −1 α1 +...+αr 1 X Y
fa0i (n + bi )
−
q
q
q q
x
i=1
n≤x
+O
1
x
|fqα2 (d2 )|
r
Y
X 1 .
i=1 ei di |A0i
Again we have aQ
factor of 1/q 2 in the first line above since α2 ≥ 1. The second line
r
1
of (3.8) is O( qx i=1 τ (Ai )2 ) since α2 ≥ 1 and q - d2 . This completes the proof as
terms with q|ej dj and q - ei di for i 6= j are dealt with similarly.
We will also need the following observations.
Lemma 3.3. We have
X
1X
x
n≤x
ai ≥1
p|ai ⇒y<p<z
ω(a1 )≥(log2 x)2r
|Ka1 ,...,ar |
r
Y
i=1
|fai (n + bi )||ti |Ω(ai ) 1
.
log x
10
RIZWANUR KHAN
Proof. We first bound the contribution of terms with ω(ai ) = wi for some positive
integers wi . Recall that |Ka1 ,...,ar | ≤ 1 and note that |fpα (n)| ≤ |fp (n)|. Thus
(3.9)
1X
x
X
n≤x
|Ka1 ,...,ar |
ai ≥1
p|ai ⇒y<p<z
ω(ai )=wi
1X
x
n≤x
X
r
Y
|fai (n + bi )||ti |Ω(ai )
i=1
r
Y
|fAi (n + bi )||ti |Ω(ai )
i=1
ai ≥1
p|ai ⇒y<p<z
ω(ai )=wi
For a fixed square-free integer Ai with ω(Ai ) = wi we have
X
|ti |Ω(ai ) ≤ 1,
ai ≥1
Ai =square-free part of ai
since |ti | ≤
1
1000 .
Thus (3.9) is bounded by
(3.10)
r
wi
1 XY 1 X
|fp (n + bi )|
.
x
w ! y<p<z
i=1 i
n≤x
Since |fp (n)| ≤ 1 if p|n and |fp (n)| ≤
(3.11)
1
p
if p - n, we have that (3.10) is bounded by
r
wi
1 XY 1 ω(n + bi ; y, z) + log2 x
x
w!
i=1 i
n≤x
r
r
Y
1 1 X X rwi 2
ω(n + bi ; y, z)rwi + (log2 x)rwi .
w! x
i=1
i=1 i
n≤x
We have
(3.12)
X
1X
1
ω(n + bi ; y, z)rwi x
x y<p ,...,p
n≤x
1
rwi <z
X
1,
n≤x
[p1 ,...,prwi ]|n+bi
where [p1 , . . . , prwi ] denotes the lowest common multiple of p1 , . . . , prwi . Now (3.12)
is bounded by
X
1
(3.13)
2rwi (log2 x)rwi .
[p
,
.
.
.
,
p
]
1
rw
i
y<p ,...,p
<z
1
rwi
Thus we have that (3.11) is bounded by
(3.14)
r
r
Y
1 X rwi
4 (log2 x)rwi .
w
!
i=1 i
i=1
Summing over integers wi ≥ 1 for i ≥ 2 this is bounded by
(3.15)
(4 log2 x)rw1 exp((4 log2 x)r )
.
w1 !
Finally the sum of (3.15) over integers w1 ≥ (log2 x)2r is 1/ log x.
SPACINGS BETWEEN INTEGERS HAVING TYPICALLY MANY PRIME FACTORS
11
Lemma 3.4. We have
r
X
Y
Y 1 1
1 −1 α 1 α + 1−
|ti |Ω(ai )
.
1−
p
p
p
p
log x
α
i=1
ai ≥1
p|ai ⇒y<p<z
ω(a1 )≥(log2 x)2r
p kai
Proof. We first bound the contribution of terms with ω(ai ) = wi for some positive
integers wi . We have
r
X
Y
Y 1 1 α 1 −1 α (3.16)
+ 1−
|ti |Ω(ai )
1−
p
p
p
p
α
i=1
ai ≥1
p|ai ⇒y<p<z
ω(ai )=wi
p kai
r
r
Y
Y
1 X 1 wi
(log2 x)wi
.
w ! y<p<z p
wi !
i=1 i
i=1
Summing over integers wi ≥ 1 for i ≥ 2 this is bounded by
(3.17)
(log2 x)w1 exp(r log2 x)
.
w1 !
The sum of (3.15) over integers w1 ≥ (log2 x)2r is 1/ log x.
Back to the proof. By Lemma 3.3 we see that (3.2) equals, up to an error of
O(1/ log x),
1X
x
(3.18)
X
n≤x
Ka1 ,...,ar
ai ≥1
p|ai ⇒y<p<z
ω(ai )≤(log2 x)2r
r
Y
Ω(ai )
fai (n + bi )ti
.
i=1
Let us first treat the terms of (3.18) with a1 , . . . , ar not pairwise coprime. Applying
Lemma 3.2 repeatedly, these terms contribute an amount bounded by
r
1 X Y
X 1
X
Ω(ai ) (3.19)
f
(n
+
b
)t
a
i
i
i
q2
x
y<q<z
i=1
ai pairwise coprime
p|ai ⇒y<p<z
ω(ai )≤(log2 x)2r
X
+
y<q<z
1
qx
X
r
Y
n≤x
|ti |Ω(ai ) τ (Ai )2 .
i=1
ai ≥1
p|ai ⇒y<p<z
ω(ai )≤(log2 x)2r
For the second line of (3.19), we have
r
(3.20)
1Y
x i=1
r
X
ai ≥1
p|ai ⇒y<p<z
ω(ai )≤(log2 x)2r
r
|ti |Ω(ai ) τ (Ai )2 1 Y (log2 x)2r (log2 x)2r
z
4
x−1/2 .
x i=1
1Y
x i=1
X
ai square-free
p|ai ⇒y<p<z
ω(ai )≤(log2 x)2r
τ (Ai )2
12
RIZWANUR KHAN
Thus the second line of (3.19) falls into the error term of Theorem 2.2. To bound
the first line of (3.19) we use Lemma 3.1 to get that
(3.21)
X
y<q<z
1
y
1
q2
X
ai pairwise coprime
p|ai ⇒y<p<z
ω(ai )≤(log2 x)2r
r
Y
X
r
1 X Y
Ω(a ) fai (n + bi )ti i x
i=1
n≤x
Y 1
+
p
α
|ti |Ω(ai )
i=1
ai ≥1
p|ai ⇒y<p<z
2r
ω(ai )≤(log2 x)
p kai
r
X
ai ≥1
p|ai ⇒y<p<z
ω(ai )≤(log2 x)2r
1 Y Ω(ai )
|ti |
τ (Ai )2
x i=1
r
1Y Y 1
1+
+ x−1/2 ,
y i=1 y<p<z
p
r
where we used the bound of (3.20). Now this is less than (logyx) + x−1/2 1/ log x.
Thus only the terms of (3.18) with a1 , . . . , ar coprime will give a main contribution. Using Lemma 3.1 and (3.3) we get
1X
x
(3.22)
X
Ka1 ,...,ar
r
Y
Ω(ai )
fai (n + bi )ti
i=1
n≤x ai pairwise coprime
p|ai ⇒y<p<z
ω(ai )≤(log2 x)2r
=
r
Y
Y iα 1 1 α 1 −1 α tα
1−
+ 1−
i
α!
p
p
p
p
α
X
i=1 ai pairwise coprime p kai
p|ai ⇒y<p<z
ω(ai )≤(log2 x)2r
+O
r
X
ai ≥1
p|ai ⇒y<p<z
ω(ai )≤(log2 x)2r
1 Y Ω(ai )
|ti |
τ (Ai )2 .
x i=1
We’ve already seen in (3.20) that the error term above is negligible. The main
term of (3.22) is zero if ai is not square-full for all i. Thus we may further impose
the condition p|ai ⇒ p2 |ai . We may also extend by Lemma 3.4 the sum in the
main term of (3.22) to all pairwise coprime and square-full integers ai whose prime
factors lie between y and z, up to an error of O(1/ log x). The contribution of the
terms with ai square-full but not pairwise coprime is
(3.23)
X
y<q<z
1 r
1
1 Y (log z)r
1
+
.
q 2 y<p<z
p
y
log x
SPACINGS BETWEEN INTEGERS HAVING TYPICALLY MANY PRIME FACTORS
13
Thus (3.22) equals up to an error of O(1/ log x),
r
Y
Y 1 α 1 X iα α 1 X iα α 1 α (3.24)
1+
ti 1 −
+ 1−
t −
p
α!
p
p
α! i
p
i=1 y<p<z
α≥2
=
=
r
Y
Y i=1 y<p<z
r
Y
exp
i=1
1+
1
p
α≥2
eiti − 1 − iti
1 +O 2
p
1 1 iti
e − 1 − iti + O 2
.
log 1 +
p
p
y<p<z
X
Now since
1 1 1
1 iti
log 1 +
e − 1 − iti + O 2
=
eiti − 1 − iti + O 2 ,
p
p
p
p
we have that (3.24) equals
r
1 X 1
Y
exp eiti − 1 − iti
+O
(3.25)
p
y
y<p<z
i=1
=
r
Y
i=1
exp
X 1
eiti − 1 − iti
+ O(1/ log x).
p
y<p<z
Acknowledgments. I am grateful to Prof. K. Soundararajan for posing the
problem studied in this paper and for many helpful discussions pertaining to it. I
am also thankful to Prof. A. Granville for numerous useful comments. Part of this
work was done with support through a grant from the NSF (DMS 0500711) and
while I visited the Centre de Recherches Mathématiques, Montréal.
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RIZWANUR KHAN
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University of Michigan, Department of Mathematics, 530 Church St., Ann Arbor,
MI 48109, USA
E-mail address: [email protected]