Centre de Recherches Mathématiques
CRM Proceedings and Lecture Notes
Volume 46, 2008
On the Distribution of ω(n)
Rizwanur Khan
Abstract. We give yet another proof of the normal distribution of ω(n).
1. Introduction
Let ω(n) denote the number of distinct prime factors of n. That is,
ω(n) =
X
1.
p|n
In this article we revisit the old question of how the values of w(n) are distributed.
The average value of this additive function is easily found to be
1X
1 XX
1 XX
ω(n) =
1=
1
x
x
x
(1.1)
n≤x
n≤x p|n
=
p≤x n≤x
p|n
¹ º
1X x
= log2 x + C + O(1/ log x),
x
p
p≤x
where we write log2 x for log log x (and similarly for logj x) and C is an absolute
constant. Next let us follow Turán [11] in finding the variance of ω(n). We have
(1.2)
1X
1X
1X
bxc
(ω(n) − log2 x)2 =
ω(n)2 − 2 log2 x
ω(n) + (log2 x)2
x
x
x
x
n≤x
n≤x
n≤x
1X
=
ω(n)2 − (log2 x)2 − 2C log2 x + O(1).
x
n≤x
2000 Mathematics Subject Classification. Primary: ??; Secondary: ??
This is the final form of the paper.
c
°2008
American Mathematical Society
1
2
R. KHAN
For the first term we have
¹ º
¹ º
1X
1X x
1X X
1 X x
ω(n)2 =
1=
+
x
x
x
pq
x
p
n≤x
p≤x n≤x
q≤x p|n,q|n
=
pq≤x
p6=q
µX ¶2
X 1
1
−
−
2
p
√ p
p≤x
p≤ x
p≤x
X
p≤x,q≤x
pq>x
1
+ log2 x + O(1)
pq
= (log2 x)2 + (2C + 1) log2 x + O(log3 x).
Thus (1.2) equals
(1.3)
1X
(ω(n) − log2 x)2 = log2 x + O(log3 x),
x
n≤x
after a delicate cancellation of the (log2 x)2 terms! Note that (1.3) implies that that
w(n) ∼ log2 n for almost all n.
P
With this method, computation of the higher moments x1 n≤x (ω(n)−log2 x)k
for integers k > 2 would quickly become difficult, as cancellations and errors
would become arduous to keep track of. It is in fact not a good idea to separate
ω(n) and log2 x as we did when we expanded (ω(n) − log2 x)2 to find the second
moment above. Granville and SoundararajanP[3] show how powerful this simple
point is by computing very high moments: x1 n≤x (ω(n) − log2 x)k uniformly for
k ≤ (log2 x)1/3 . Let us follow them in defining for a prime p,
(
1 − p1 if p | n,
fp (n) =
− p1
if p - n.
Q
Q
αi
i
Extend this definition multiplicatively: if m = i pα
i , define fm (n) =
i fpi (n) .
(ThusP
f1 (n) = 1.) Also, instead of log2 x let us work with the more exact mean
value p≤x p1 . Then we have
ω(n) −
X1 X
=
fp (n).
p
p≤x
p≤x
What is the advantage of this? If we think of a prime p dividing n with probability
1/p independently ofP
other primes, then we have E(fm )=0 for squarefree m. So we
have written ω(n) − p≤x 1/p as a sum of independent random variables of mean
0. When working with this expression, the delicate cancellations mentioned above
take care of themselves in moment calculations.
P Further, this model would predict
by the central limit theorem that ω(n) − p≤x 1/p is normally distributed with
mean 0.
The prediction above is in fact true. Erdős and Kac [1] were the first to show
that ω(n) is normally distributed with mean log2 x and variance log2 x. In [6] Rényi
and Turán proved this with a sharp error term, using results on πk (x), the number
of integers up to x with exactly k prime factors (see [8, 9]). The following theorem
can also be found in Tenenbaum’s book [10].
ON THE DISTRIBUTION OF ω(n)
3
p
Theorem 1.1. The number of integers n ≤ x for which |(ω(n)−log2 x)/
< W for some W > 0 is
µ
¶
Z W
x
x
2
√
exp(−u /2) du + O p
,
2π −W
log2 x
log2 x|
where the implied constant depends on W .
We show how a smooth version of the theorem above, with an error term nearly
as good, can be proven with little effort.
Theorem 1.2. Fix a smooth (C ∞ ) and compactly supported real function ψ.
We have
µ
¶
µ
¶
Z ∞
1X
ω(n) − log2 x
1
log x
p
ψ
=√
ψ(u) exp(−u2 /2) du + O p 3
,
x
2π −∞
log2 x
log2 x
n≤x
where the implied constant depends on ψ.
Let us adopt the convention that ε always denotes an arbitrary small positive
constant, but not necessarily the same one from one occurrence to the next, and
that all implied constants may depend implicitly on ε and ψ.
The moments computed by Granville and Soundararajan match the moments
of a normally distributed random variable, and so their work implies the result
of Erdős
Pand Kac. We do use the idea of Granville and Soundararajan of writing
ω(n) − p≤x 1/p as a sum of random variables of mean 0, but instead of computing
P
moments we work with the Fourier transform of ω(n) − p≤x 1/p, also called the
characteristic function in the language of probability theory.
2. Set Up
Define y = y(x) = log x and z = z(x) = x1/ log2 x and
X
ω(n; y, z) =
1.
p|n
y<p<z
It is more convenient to work with
P
ω(n; y, z) − y<p<z 1/p
qP
y<p<z 1/p
p
in place of (ω(n) − log2 x)/ log2 x. The following lemma will be used to show that
there is no significant loss in disregarding these small or large primes.
Lemma 2.1. We have
P
¯ µ
¶
µ
¶¯
ω(n; y, z) − y<p<z 1/p ¯
1 X ¯¯
ω(n) − log2 x
log3 x
¯¿ p
p
q
−
ψ
.
ψ
P
¯
¯
x
log
x
log2 x
1/p
2
n≤x
y<p<z
Proof. Let E = E(x) denote the (exceptional) set of integers less than or
equal to x with more than 10 log3 x distinct prime factors less than y or more than
10 log3 x distinct prime factors greater than z. The size of this set is
µX ¶10 log3 x
µ X ¶10 log3 x
x
x
x
1
1
|E| ≤
+
¿
,
b10 log3 xc!
p
b10 log3 xc!
p
log2 x
p≤y
z≤p≤x
4
R. KHAN
P
using that
p≤x 1/p = log2 x + C + O(1/ log x) and Stirling’s estimate n! ∼
√
n+1/2 −n
2πn
e . Therefore we have
P
¯ µ
¶
µ
¶¯
X
¯
ω(n; y, z) − y<p<z 1/p ¯
1
ω(n)
−
log
x
2
¯ ¿ kψk∞ .
¯ψ
p
q
−
ψ
P
¯
¯
1
x
log x
log x
n∈E
2
y<p<z p
2
For n ∈
/ E we have ω(n) − ω(n; y, z) ¿ log3 x, and so it follows that
P
ω(n) − log2 x ω(n; y, z) − y<p<z 1/p
p
qP
−
log2 x
y<p<z 1/p
P
µ
¶
ω(n) − log2 x − ω(n; y, z) + y<p<z 1/p
|ω(n) − log2 x| log3 x
qP
=
+O
log2 x
y<p<z 1/p
log x
|ω(n) − log2 x| log3 x
+
.
¿p 3
log2 x
log2 x
Thus
P
¯ µ
µ
¶
¶¯
ω(n; y, z) − y<p<z 1/p ¯
1 X ¯¯
ω(n) − log2 x
¯
p
q
−
ψ
ψ
P
¯
¯
x
log2 x
n∈E
/
y<p<z 1/p
¿
kψ 0 k∞ log3 x kψ 0 k∞ X |ω(n) − log2 x| log3 x
p
+
.
x
log2 x
log2 x
n≤x
By the Cauchy – Schwarz inequality and (1.3) the second term above is seen to
be ¿ log3 x(log2 x)−1/2 .
¤
Let
Z
b )=
ψ(T
∞
ψ(u)e−iuT du
−∞
b )| ¿ 1 and by integration by parts
denote the Fourier transform of ψ. We have |ψ(T
−B
several times we have |ψ̂(T )| ¿B T
for any B > 0. By Fourier inversion and
these bounds we have,
P
µ
¶
ω(n; y, z) − y<p<z 1/p
1X
qP
(2.1)
ψ
x
n≤x
y<p<z 1/p
P
µ
¶
Z
∞
ω(n; y, z) − y<p<z 1/p
1X 1
b ) exp iT
qP
=
ψ(T
dT
x
2π −∞
1/p
n≤x
y<p<z
!
P
ε
µ
µ
¶
Z
(log
x)
2
ω(n; y, z)− y<p<z 1/p
1
1X 1
b
qP
dT + O
=
ψ(T ) exp iT
,
x
2π −(log2 x)ε
(log2 x)15
1/p
n≤x
y<p<z
for any ε > 0. In the next section we will prove the following theorem.
P
P
Theorem 2.2. Let t = T ( y<p<z 1/p)−1/2 . For |T | ≤ ( y<p<z 1/p)1/2 /1000,
we have
P
¶
µ
¶
µ
¶
µ
X
ω(n; y, z) − y<p<z 1/p
1
1X
it
qP
1/p +O
.
exp iT
= exp (e −1−it)
x
log x
y<p<z
n≤x
y<p<z 1/p
ON THE DISTRIBUTION OF ω(n)
5
Certainly the theorem is true with 10−3 replaced by a larger constant, but we
have not tried to improve it. For |T | ≤ (log2 x)ε the main term above equals
õ
!
¶
µ
¶ X
X
−t2
it
3
(2.2)
exp (e − 1 − it)
1/p = exp
+ O(t )
1/p
2
y<p<z
y<p<z
¶!
µ
¶Ã
µ
−T 2
T3
= exp
1+O p
.
2
log2 x
The function exp(−T 2 /2) is the characteristic function of the normal distribution.
Using Theorem 2.2 and the observation (2.2) we get that the main term of (2.1)
equals
µ
¶
µ
¶
Z (log2 x)ε
2
1
1
b ) exp −T
(2.3)
ψ(T
dT + O p
2π −(log2 x)ε
2
log2 x
¶
µ
µ
¶
Z ∞
1
1
−T 2
b
dT + O p
=
ψ(T ) exp
.
2π −∞
2
log2 x
√
Recall that the Fourier transform of (1/ 2π) exp(−u2 /2) is exp(−T 2 /2). By the
Plancherel formula the main term of (2.3) equals
µ 2¶
Z ∞
1
−u
√
ψ(u) exp
du.
2
2π −∞
Thus Theorem 1.2 follows from Lemma 2.1 and Theorem 2.2.
3. Proof of Theorem 2.2
For the range of T under consideration in Theorem 2.2 we have that |t| ≤ 10−3 .
The left-hand side of Theorem 2.2 is
µ X
¶
¡
¢
1X
1X Y
exp it
fp (n) =
exp itfp (n) .
x
x
y<p<z
y<p<z
n≤x
n≤x
Expanding the exponential as a Taylor series this equals
µ
¶
1X Y
1
1
(3.1)
1 + itfp (n) − t2 fp2 (n) − it3 fp3 (n) + · · ·
x
2!
3!
y<p<z
n≤x
=
X
Ka tΩ(a)
a≥1
p|a⇒y<p<z
1X
fa (n),
x
n≤x
where Ω(a) is the number of prime factors of a counted with multiplicity and
Y iα
Ka =
.
α!
α
p ka
α
α
α+1
Above p k a means that p | a but p
result of Granville and Soundararajan,
- a. To evaluate (3.1) we use the following
Lemma 3.1. Let A denote the square-free
divisors of A. We have
¶α µ
Y µ1µ
1
1X
1−
+ 1−
fa (n) =
x
p
p
α
n≤x
p ka
part of a and d(A) the number of
1
p
¶µ
−1
p
¶α ¶
µ
+O
¶
d(A)
.
x
6
R. KHAN
The main term is just what we would expect from the heuristic that p divides
n with probability 1/p, independently of other primes. Note that the main term is
zero unless a is 1 or square-full (a is square-full if p | a implies p2 | a). Before using
this lemma, we make a couple of observations which will be needed later.
Lemma 3.2. We have
X
(3.2)
|Ka | |t|Ω(a)
a≥1
p|a⇒y<p<z
ω(a)≥(log2 x)/2
1X
1
.
|fa (n)| ¿
x
log x
n≤x
Proof. Let r ≥ 12 log2 x be an integer. We first bound absolutely the contribution of the terms of (3.2) with ω(a) = r. Note that |Ka | ≤ 1 and |fpα (n)| ≤ |fp (n)|.
Therefore
X
X
1X
1X
(3.3)
|Ka ||t|Ω(a)
|fa (n)| ¿
|t|Ω(a)
|fA (n)|,
x
x
a≥1
ω(a)=r
p|a⇒y<p<z
n≤x
a≥1
ω(a)=r
p|a⇒y<p<z
n≤x
where A denotes the square-free part of a. Note that for a fixed square-free integer
A with ω(A) = r we have
X
|t|Ω(a) ¿ |2t|r ,
a≥1
A=square-free part of a
as |t| ≤ 10−3 . Thus (3.3) is bounded by
µ X
¶r
1X 1
(3.4)
¿ |2t|r
|fp (n)| .
x
r! y<p<z
n≤x
Since |fp (n)| ≤ 1 if p | n and |fp (n)| ≤ 1/p if p - n, this is bounded by
µ
¶r
1X 1
¿ |2t|r
ω(n; y, z) + log2 x
(3.5)
x
r!
n≤x
¿ |4t|r
(log2 x)r
1 X ω(n; y, z)r
+ |4t|r
.
r!
x
r!
n≤x
Since r ≥ 12 log2 x and |t| ≤ 10−3 we have that the first term on the right-hand side
of (3.5) is ¿ 10−r . For the second term we have
µ
¶r
X ω(n; y, z)r
|4t|r 1 X X
r1
(3.6)
|4t|
=
1
x
r!
r! x
n≤x
n≤x
¿
|4t|r
r!
p|n
y<p<z
X
y<p1 ,...,pr <z
1
x
X
1,
n≤x
[p1 ,...,pr ]|n
where [p1 , . . . , pr ] is the least common multiple of the primes p1 , . . . , pr . We have
that (3.6) is less than or equal to
X
|4t|r
1
(3.7)
.
r! y<p ,...,p <z [p1 , . . . , pr ]
1
r
ON THE DISTRIBUTION OF ω(n)
Now for 0 ≤ k ≤ r − 1 we have
X
y<p1 ,...,pr <z
{p1 , . . . , pr } is a set of
r − k distinct primes
1
¿
[p1 , . . . , pr ]
7
µ ¶µ X ¶r−k
r
1
.
k
p
y<p<z
So by the binomial theorem we get that (3.7) is
P
|4t|r (1 + y<p<z 1/p)r
|4t|r (log2 x)r
¿
¿
.
r!
r!
Since r ≥ 12 log2 x and |t| ≤ 10−3 , this is ¿ 10−r .
Finally, summing 10−r over the integers r ≥ 12 log2 x, we get the lemma.
¤
Lemma 3.3. We have
¶α µ
¶µ
¶α ¯
X
Y ¯¯ 1 µ
1
1
−1 ¯¯
1
Ω(a)
¯
(3.8)
|Ka ||t|
+ 1−
¯p 1 − p
¯ ¿ log x .
p
p
α
a≥1
p|a⇒y<p<z
ω(a)≥(log2 x)/2
p ka
Proof. Let r ≥ 12 log2 x be an integer. We first bound absolutely the contribution of the terms of (3.8) with ω(a) = r. We have
¯
¶α µ
¶µ
¶α ¯
X
Y ¯1µ
1
1
−1 ¯¯
Ω(a)
¯
(3.9)
|Ka ||t|
+ 1−
¯p 1 − p
¯
p
p
a≥1
pα ka
p|a⇒y<p<z
µ X ¶r
1
1
1
1
ω(a)=r
¿ |4t|r (log2 x)r ¿ r ,
¿ |4t|r
r! y<p<z p
r!
10
since r ≥ 12 log2 x and |t| ≤ 10−3 . Summing 10−r over the integers r ≥
get the lemma.
1
2
log2 x, we
¤
Now we are ready to to find the main term of Theorem 2.2. By Lemma 3.2 we
have that (3.1) equals, up to an error of O(1/ log x), the sum
X
1X
(3.10)
Ka tΩ(a)
fa (n).
x
a≥1
p|a⇒y<p<z
ω(a)≤(log2 x)/2
n≤x
We use Lemma 3.1 to evaluate this. The error incurred from the use of this lemma
is bounded by
X
X
d(A)
2log2 x
log x
2log2 x
|t|Ω(a)
¿
¿ z (log2 x)/2
¿ √ .
x
x
x
x
a≥1
p|a⇒y<p<z
ω(a)≤(log2 x)/2
a square-free
p|a⇒p<z
ω(a)≤(log2 x)/2
The main term of (3.10) equals
¶α µ
¶µ
¶α ¶
X
Y iα µ 1 µ
1
−1
1
(3.11)
tα
1−
.
+ 1−
α!
p
p
p
p
α
a≥1
p ka
p|a⇒y<p<z
ω(a)≤(log2 x)/2
The expression above is zero if a is not 1 or square-full. Thus for a 6= 1 we may
further impose the implication pα k a ⇒ α ≥ 2. By Lemma 3.3, we may also
8
R. KHAN
extend, up to an error of O(1/ log x), the sum in (3.11) to all a ≥ 1 whose prime
factors lie between y and z. Thus (3.11) equals, up to this error,
µ
¶α µ
¶
µ
¶α ¶
Y ³
1 X iα α
1
1
1 X iα α
(3.12)
1+
t 1−
+ 1−
t −
p
α!
p
p
α!
p
y<p<z
α≥2
α≥2
Ã
µ ¶!
Y
1 it
1
=
1 + (e − 1 − it) + O 2
p
p
y<p<z
Ã
!
µ
³ 1 ´¶
X
1 it
log 1 + (e − 1 − it) + O 2
= exp
.
p
p
y<p<z
Now since
µ ¶!
µ ¶
1 it
1 it
1
1
= (e − 1 − it) + O 2 ,
log 1 + (e − 1 − it) + O 2
p
p
p
p
Ã
we have that (3.12) equals
µ
µ ¶
X 1¶
1
it
(3.13)
exp (e − 1 − it)
+O
.
p
y
y<p<z
4. Concluding Remarks
In [4], the characteristic function of Theorem 2.2 is used to detect when
P
¯
¯
¯ ω(n) − y<p<z 1/p ¯
1
¯ q
¯≤
P
¯
¯ (log x)1/2−ε ,
2
y<p<z 1/p
or when |ω(n) − log2 x| ≤ (log2 x)ε . These are integers with very nearly the average
number of prime factors. Let us call them ε-normal numbers. How are these
numbers distributed amongst the integers?
p
The number of ε-normal numbers up to x is ∼ x/( π/2(log2 x)1/2−ε ), by
Theorem 1.1. Let us write them
in order as N1 , N2 , . . .. The distance between
p
consecutive elements is then π/2(log2 x)1/2−ε on average. How does the spacing
e
distance
p vary from the average? Rescale the ε-normal numbers by defining Ni =
Ni /( π/2(log2 x)1/2−ε ), so that the consecutive spacing between the rescaled εnormal numbers is 1 on average. In [4] we prove that given 0 < α < β, we have
Z β
ei+1 − N
ei < β} ∼ x
(4.1)
#{i ≤ x : α < N
e−u du,
α
as x → ∞.
The distribution function e−u is called the Poisson or exponential distribution.
It is significant because it predicts how randomly scattered objects are spaced. More
precisely, suppose that we have bxc random variables independently and uniformly
taking real values in the interval (0, bxc) and written in order as Y1 < · · · < Ybxc .
Then for α > 0 we have that Prob(Yi+1 − Yi > α) ∼ e−α as x → ∞ for any i.
So (4.1) implies that ε-normal numbers are somewhat randomly dispersed amongst
the integers.
The study of spacings in arithmetic sequences is of great interest to number
theorists. We name just a few examples. Gallagher [2] showed that conditional on
the truth of the Hardy – Littlewood prime k-tuple conjectures, the primes less than
ON THE DISTRIBUTION OF ω(n)
9
x, as x → ∞, form a Poisson process (that is, their spacings follow the Poisson
distribution law as described above). Kurlberg and Rudnick [5] proved that the
Error: it should be
quadratic residues modulo q, as
√ ω(q) → ∞, form a Poisson process. It is conjectured
that the fractional parts of n 2 for n ≤ x, as x → ∞, form a Poisson process (see $n^2 \sqrt{2}$
[7]). Of course, some arithmetic sequences do not behave randomly, such as the
zeros of the Riemann zeta function. We conjecture that integers with more or less
exactly the average number of prime factors, that is for |ω(n) − log2 x| ≤ 12 , form a
Poisson process.
The main strategy of our proof is to show that additive shifts of ω(n) behave
independently. For example we find the joint characteristic function
P
P
µ
¶
µ
¶
ω(n; y, z) − y<p<z 1/p
ω(n + 2; y, z) − y<p<z 1/p
1X
q
q
exp iT1
exp iT2
P
P
x
n≤x
y<p<z 1/p
y<p<z 1/p
P
µ
¶
ω(n + 5; y, z) − y<p<z 1/p
qP
× exp iT3
,
y<p<z 1/p
and show that it essentially equals exp(−T12 /2) exp(−T22 /2) exp(−T32 /2).
References
1. P. Erdős and M. Kac, The Gaussian law of errors in the theory of additive number theoretic
functions, Amer. J. Math. 62 (1940), 738 – 742.
2. P. X. Gallagher, On the distribution of primes in short intervals, Mathematika 23 (1976),
no. 1, 4 – 9.
3. A. Granville and K. Soundararajan, Sieving and the Erdős – Kac theorem, Equidistribution
in Number Theory, an Introduction (Montréal, 2005) (A. Granville and Z. Rudnick, eds.),
NATO Sci. Ser. II Math. Phys. Chem., vol. 237, Springer, Dordrecht, 2007, pp. 15 – 27.
4. R. Khan, On the distribution of normal numbers, preprint.
5. P. Kurlberg and Z. Rudnick, The distribution of spacings between quadratic residues, Duke
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Department of Mathematics, University of Michigan, 530 Church St., Ann Arbor,
MI 48109, USA.
E-mail address: rrkhan@umich.edu