Chapter
1
Number-theoretic Functions and the
Distribution of Primes
Let N denote the set of positive integers and let Z be the set of integers.
1.1 Multiplicative Functions
Definition. A real- or complex-valued function defined on the positive integers is called an
arithmetic function or a number-theoretic function.
Throughout this chapter, variables occurring as arguments of number-theoretic functions are
understood to be positive. The same applies to their divisors.
Examples 1.1.1. The following functions are arithmetic functions.
1. φ(n) = |{r ∈ Z : 0 ≤ r < n and gcd(r, n) = 1}|.
X
2. τ (n) = the number of positive divisors of n =
1.
d|n
3. σ(n) = the sum of positive divisors of n =
X
d.
d|n
Here,
X
f (d) means the sum of the values f (d) as d runs over all positive divisors of the positive
d|n
integer n. E.g.,
X
f (d) = f (1) + f (2) + f (3) + f (4) + f (6) + f (12).
d|12
Theorem 1.1.1. Let p be a prime and k ∈ N ∪ {0}. Then
τ (pk ) = |{1, p, p2 , . . . , pk }| = k + 1
and
σ(pk ) = 1 + p + p2 + · · · + pk =
1
pk+1 − 1
.
p−1
2 Number-theoretic Functions and the Distribution of Primes
Y. Meemark
Definition. A number-theoretic function f which is not identically zero is said to be multiplicative
if ∀m, n ∈ N, gcd(m, n) = 1 ⇒ f (mn) = f (m)f (n).
Remark. Let f be a multiplicative function. Then f (1) = f (1 · 1) = f (1)f (1), so f (1) = 0 or 1.
If f (1) = 0, then f (n) = f (1 · n) = f (1)f (n) = 0, so f is the zero function. Hence, if f is
multiplicative, then f (1) = 1.
Lemma 1.1.2. f is multiplicative ⇔ f (1) = 1 and f (pk11 pk22 . . . pkr r ) = f (pk11 )f (pk22 ) . . . f (pkr r ) for
all distinct primes pi and r, ki ∈ N.
Remarks.
1. From the above lemma, to compute the values of a multiplicative function f , it
suffices to know only the values of f (pk ) for all primes p and k ∈ N.
2. If f and g are multiplicative functions and f (pk ) = g(pk ) for all primes p and k ∈ N, then
f = g.
Definition. A number-theoretic function f which is not identically zero is said to be completely
multiplicative if f (mn) = f (m)f (n) for all m, n ∈ N.
E.g., (i) U (n) = 1, for all n ∈ N, and (ii) N (n) = n, for all n ∈ N, are completely multiplicative.
Remark. If f is completely multiplicative, then
f (pk11 pk22 . . . pkr r ) = f (p1 )k1 f (p2 )k2 . . . f (pr )kr .
Thus, to determine the values of a completely multiplicative function f , it suffices to know only
the values of f (p) for all primes p.
Lemma 1.1.3. Let n > 1 be factored as n = pk11 pk22 . . . pkr r for some primes pi and r, ki ∈ N for all
i ∈ {1, 2, . . . , r}. Then for d ∈ N,
d | n ⇔ d = pa11 pa22 . . . par r , where 0 ≤ ai ≤ ki for all i ∈ {1, 2, . . . , r}.
Hence, {d ∈ N : d | n} = {pa11 pa22 . . . par r : 0 ≤ ai ≤ ki for all i ∈ {1, 2, . . . , r}}.
Theorem 1.1.4. If n = pk11 pk22 . . . pkr r is the prime factorization of n > 1, then
τ (n) = (k1 + 1)(k2 + 1) . . . (kr + 1) = τ (pk11 )τ (pk22 ) . . . τ (pkr r ).
Moreover, τ is multiplicative.
Definition. A positive integer n is a perfect square number if ∃a ∈ Z, n = a2 .
Remarks.
1. If n is a perfect square number, then n ≡ 0 or 1 (mod 4).
2. n is a perfect square number if and only if τ (n) is odd.
Let n = pk11 pk22 . . . pkr r is the prime factorization of n > 1. Consider the product
(1 + p1 + p21 + · · · + pk11 )(1 + p2 + p22 + · · · + pk22 ) . . . (1 + pr + p2r · · · + pkr r )
X
=
{pa11 pa22 . . . par r : 0 ≤ ai ≤ ki for all i ∈ {1, 2, . . . , r}}
X
=
{d ∈ N : d | n} = σ(n).
Y. Meemark
1.1 Multiplicative Functions
3
Theorem 1.1.5. If n = pk11 pk22 . . . pkr r is the prime factorization of n > 1, then
σ(n) = (1 + p1 + p21 + · · · + pk11 )(1 + p2 + p22 + · · · + pk22 ) . . . (1 + pr + p2r · · · + pkr r )
pk11 +1 − 1 pk22 +1 − 1
pkr +1 − 1
··· r
p1 − 1
p2 − 1
pr − 1
=
= σ(pk11 )σ(pk22 ) . . . σ(pkr r ).
Moreover, σ is multiplicative.
Lemma 1.1.6. Assume that gcd(m, n) = 1. Then
{d ∈ N : d | mn} = {d1 d2 : d1 , d2 ∈ N, d1 | m, d2 | n and gcd(d1 , d2 ) = 1}.
Proof. The result is clear when m or n is 1. Assume that m, n > 1 and gcd(m, n) = 1. Let
mr and n = q n1 . . . q ns , where p and q are all distinct primes and m , n ∈ N for
1
m = pm
i
j
i j
s
1 . . . pr
1
all i ∈ {1, . . . , r} and j ∈ {1, . . . , s}.
Suppose that d | mn. By Lemma 1.1.3, d = pa11 . . . par r q1b1 . . . qsbs for some 0 ≤ ai ≤ mi and
0 ≤ bj ≤ nj for all i, j. Thus d = d1 d2 where d1 = pa11 . . . par r , d2 = q1b1 . . . qsbs , so d1 | m, d2 | n and
gcd(d1 , d2 ) = 1. The converse is clear.
Remark. If gcd(m, n) = 1, then the above lemma gives
X
X
f (d) =
f (d1 d2 ).
d|mn
d1 |m,d2 |n,
gcd(d1 ,d2 )=1
Theorem 1.1.7. If f is a multiplicative function and F is defined by
X
F (n) =
f (d),
d|n
then F is also multiplicative.
Proof. Let m, n ∈ N be such that gcd(m, n) = 1. Then
X
X
XX
F (mn) =
f (d) =
f (d1 d2 ) =
f (d1 )f (d2 ) (since gcd(d1 , d2 ) = 1)
d1 |m,d2 |n,
gcd(d1 ,d2 )=1
d|mn
=
X
f (d1 )
d1 |m
X
d1 |m d2 |n
f (d2 ) = F (m)F (n).
d2 |n
Hence, F is multiplicative.
Recall that U (n) = 1 for all n ∈ N and N (n) = n for all n ∈ N are multiplicative. The above
theorem gives another proof of the following result.
X
X
Corollary 1.1.8. τ (n) =
1 and σ(n) =
d are multiplicative.
d|n
Theorem 1.1.9.
X
d|n
φ(d) = n.
d|n
4 Number-theoretic Functions and the Distribution of Primes
Y. Meemark
Proof. We first observe that
[n
o
1 2
a
n−1 n
=
, ,...,
,
: 1 ≤ a ≤ d and gcd(a, d) = 1 .
n n
n
n
d
d|n
Moreover, for d | n, each set in the union is of cardinality φ(d). Assume that d1 | n, d2 | n and
a
b
=
for some 1 ≤ a ≤ d1 , gcd(a, d1 ) = 1 and 1 ≤ b ≤ d2 , gcd(b, d2 ) = 1. Then ad2 = bd1
d1
d2
which implies d1 | ad2 and d2 | bd1 . Since gcd(a, d1 ) = 1 = gcd(b, d2 ), d1 | d2 and d2 | d1 , so
d1 = d2 and a = b. This shows that the union on the right hand side is a disjoint union. Hence,
[ n
o
1 2
a
n−1 n n=
, ,...,
,
:
1
≤
a
≤
d
and
gcd(a,
d)
=
1
=
n n
n
n d
d|n
o X
X n a
φ(d)
: 1 ≤ a ≤ d and gcd(a, d) = 1 =
=
d
d|n
d|n
as desired.
Definition. An integer n is said to be square-free if it is not divisible by the square of any prime.
Remark. Every positive integer n can be written uniquely in the form n = ab2 , where a, b ∈ N
and a is square-free.
Definition. [Möbius, 1832]
rules



1,
µ(n) = 0,



(−1)r ,
For a positive integer n, we define the Möbius function, µ, by the
if n = 1,
if ∃ a prime p, p2 | n, i.e., n is not square-free,
if n = p1 p2 . . . pr where p1 , p2 , . . . , pr are distinct primes.
Theorem 1.1.10. The Möbius function µ is multiplicative.
Proof. Note that µ(1) = 1. Suppose n > 1 and write n = pk11 pk22 . . . pkr r , where pi are distinct
k
primes and ki ≥ 1 for all i. If kj > 1 for some j ∈ {1, 2, . . . , r}, we have µ(n) = 0 and µ(pj j ) = 0, so
µ(n) = µ(pk11 )µ(pk22 ) . . . µ(pkr r ). Assume that ki = 1 for all i. Then n = p1 p2 . . . pr , so µ(n) = (−1)r .
Since µ(pi ) = −1 for all i, we have µ(p1 )µ(p2 ) · · · µ(pr ) = (−1)r = µ(n). Hence, µ is multiplicative
by Lemma 1.1.2.

1, if n = 1,
X
Theorem 1.1.11. E(n) =
µ(d) =
for all n ∈ N is a multiplicative function.
0, if n > 1,
d|n
Proof. By Theorem 1.1.7, E is multiplicative, and since

1,
if k = 0,
E(pk ) =
1 − 1 + 0 + · · · + 0, if k ≥ 1,
we see that E(n) = 0 if n is divisible by a prime p, if n > 1.
Remark. For n ∈ N, {d ∈ N : d | n} = {n/d : d ∈ N and d | n}.
Y. Meemark
1.1 Multiplicative Functions
5
Lemma 1.1.12. Let f and g be multiplicative functions. Then
1. f g and f /g are multiplicative (whenever the latter function is defined), and
X
X
2. F (n) =
f (d)g(n/d) =
f (n/d)g(d) is a multiplicative function.
d|n
d|n
Proof. Exercises.
Definition. For arithmetic functions f and g, we define the Dirichlet convolution by
X
(f ∗ g)(n) =
f (d)g(n/d)
d|n
for all n ∈ N.
Remarks.
1. Clearly, f ∗ g = g ∗ f and we can verify that f ∗ (g ∗ h) = (f ∗ g) ∗ h.
2. By Lemma 1.1.12, if f and g are multiplicative, then f ∗ g is also multiplicative.
3. The set of multiplicative functions
is an abelian group under the Dirichlet convolution with

1, if n = 1,
identity element E(n) =
0, if n > 1.
Theorem 1.1.13. [Möbius Inversion Formula] Let F and f be two arithmetic functions (not necessarily multiplicative) related by the formula
X
F (n) =
f (d) = (f ∗ U )(n).
d|n
Then f (n) =
X
F (d)µ(n/d) =
d|n
X
F (n/d)µ(d) =
X
d1 d2 =n
d|n
F (d1 )µ(d2 ), i.e., f = F ∗ µ.
Proof. We have
X
X
F (n/d)µ(d) =
F (d1 )µ(d2 ) =
d1 d2 =n
d|n
=
X
X
d1 d2 =n
f (d)µ(d2 ) =
d2 d|n
X
d|n
f (d)


X
d|d1
X

f (d) µ(d2 )
µ(d2 ).
d2 |(n/d)
P
But d2 |(n/d) µ(d2 ) = 0 unless n/d = 1 (that is, unless d = n) when it is 1, so that this last sum
is equal to f (n).
X
Example 1.1.2. We know by Theorem 1.1.9 that
φ(d) = n, i.e., φ ∗ U = N . The Möbius
d|n
inversion formula gives φ = N ∗ µ, i.e., we have
X µ(d)
Xn
µ(d) = n
φ(n) =
d
d
d|n
d|n
for all n ∈ N.
It can also be obtained directly from the product representation of φ(n) by multiplying out:


Y
X
X µ(d)

1
(−1)r 
=n
φ(n) = n
1−
1
+
= n
for all n ∈ N.

p
p1 p2 · · · pr 
d
p|n
p prime
p1 p2 ...pr |n
p1 <p2 <···<pr
d|n
6 Number-theoretic Functions and the Distribution of Primes
Y. Meemark
Corollary 1.1.14. Let F and f be two arithmetic functions related by the formula
F (n) =
X
f (d).
d|n
If F is multiplicative, then f is also multiplicative.
Proof. It follows from Theorem 1.1.13 and Theorems 1.1.12, 1.1.10.
Remark. The above corollary and Theorem 1.1.9 imply that φ is multiplicative.
1.2 Dirichlet Series
∞
X
Theorem 1.2.1. Suppose f is a multiplicative function and
f (n) converges absolutely. Then
n=1
∞
X
f (n) =
n=1
Y
Y
1 + f (p) + f (p2 ) + . . . =
p prime
1+
∞
X
!
f (pn )
n=1
p prime
and the product is absolutely convergent. Moreover, if f is completely multiplicative, then
∞
X
f (n) =
n=1
Y
p prime
1
.
1 − f (p)
Granting the above theorem, we have an immediate result.
Corollary 1.2.2. Let f be a multiplicative function.
∞
X
Y
The series
f (n) converges absolutely if and only if
n=1
Proof. Let S =
1+
n=1
p prime
Y
1+
p prime
∞
X
n=1
n
|f (p )|
!
T (m) ≤
and T (m) =
n=1
Y
p≤m
Since T (m) is increasing and bounded,
m
X
∞
X
1+
∞
X
n=1
∞
X
|f (pn )|
!
converges.
|f (n)|. Then
n
|f (p )|
!
≤ S.
f (n) converges absolutely.
n=1
Proof of Theorem 1.2.1. Since f is multiplicative, f (1) = 1. Let P (x) =
Y
p≤x
1+
∞
X
n=1
!
f (pn ) .
Each series in the product converges
X absolutely so that we can rearrange the finite products in
any way we like. Thus, P (x) =
f (n), where N (x) is the set of n ∈ N all of whose prime
n∈N (x)
factors are ≤ x. We have
X
X
P (x) −
≤
f
(n)
|f (n)|.
n>x
n≤x
Y. Meemark
1.2 Dirichlet Series
By hypothesis, the right hand side converges to 0 as x → ∞. Also,
X
n≤x
f (n) →
∞
X
7
f (n) and so
n=1
P (x) has the same limit. Finally, we observe that
!
!
∞
∞
X
X
X
Y
n n f (p )
log 1 + f (p ) =
log
1+
n=1
n=1
p≤x
p≤x
∞
∞
∞
XX
X
X X
n |f (pn )| ≤
|f (m)|,
f (p ) ≤
≤
p≤x n=1
p≤x n=1
m=2
so the product is absolutely convergent.
Definition. A Dirichlet series is a series of the kind
∞
F (s) = f (1) +
X f (n)
f (2) f (3)
+
+
·
·
·
=
,
2s
3s
ns
n=1
where f is an arithmetic function and s is a complex variable usually written as s = σ + it with
Re s = σ and Im s = t. Recall that nσ+it = nσ nit = nσ eit log n . We may write F (s) ↔ f (n).
Dirichlet series serve as a type of generating functions for arithmetic functions, adapted
to the multiplicative structure of the integers, and they play a role similar to that of ordinary
generating functions in combinatorics. For example, just as ordinary generating functions can
be used to prove combinatorial identities, Dirichlet series can be applied to discover and prove
identities among arithmetic functions.
On a more sophisticated level, the analytic properties of a Dirichlet series, regarded as a
function of the complex variable s, can be exploited to obtain information on the behavior of
P
partial sums n≤x f (n) of arithmetic functions. Analytic properties of Dirichlet series may be
listed as follows.
1. Every Dirichlet series has an abscissa of convergence σc with the property that F (s)
converges when σ > σc and diverges when σ < σc . It can be −∞, ∞ or any numbers in
between.
P∞
P∞ 2n
1
Example 1.2.1. (a)
n=1 2n ns has σc = −∞.
n=1 ns has σc = ∞ and
P∞
1
(b)
n=1 ns−a , a ∈ R, has σc = a + 1.
P∞
1
(c)
n=1 ns log2 n has σc = 1 and converges for all s with Re s = 1.
P∞ 1
(d)
n=1 ns has σc = 1 and diverges for all s with Re s = 1.
2. When the abscissa of convergence is finite, the series may converge everywhere on the
line σc + it, it may converge at some but not all points on the line, or nowhere on the line.
3. If s0 is a point with σ0 > σc , then there is a neighborhood of s0 (an open disc) in which
F (s) converges uniformly.
4. Unlike power series, the region of convergence of the Dirichlet series does not coincide
with the region of absolute convergence. For example, the series
∞
X
(−1)n−1
n=1
ns
8 Number-theoretic Functions and the Distribution of Primes
Y. Meemark
converges for σ > 0, but it is absolutely convergent only for σ > 1. In general, we let
P∞ |f (n)|
< ∞. Then σa , the abscissa
σa denote the infimum of those σ for which
n=1 nσ
P
|f (n)|
of absolutely convergence, is the abscissa of convergence of the series ∞
n=1 ns , and
P
f (n)
we see that ∞
n=1 ns is absolutely convergent if σ > σa , but not if σ < σa . Moreover,
0 ≤ σa − σc ≤ 1.
P
P∞ g(n)
f (n)
5. If ∞
n=1 ns =
n=1 ns for all s with σ ≥ σc , then f (n) = g(n) for all n ∈ N.
Next, we study algebraic properties of Dirichlet series.
Theorem 1.2.3.
1. If F (s) ↔ f (n) and G(s) ↔ g(n), then F (s)G(s) ↔ (f ∗ g)(n).
2. If f is multiplicative, then
F (s) =
∞
X
f (n)
n=1
ns
=
Y p prime
f (p) f (p2 ) f (p3 )
1 + s + 2s + 3s + · · ·
p
p
p
.
3. If F (s) ↔ f (n), then F (s − k) ↔ nk f (n) for all k ∈ Z.
Example 1.2.2.
1. 1 ↔ E(n).
2. Let
∞
X
1
ζ(s) =
ns
for all σ > 1.
n=1
It is called the Riemann zeta function. Then ζ(s) ↔ U (n).
P
Since (U ∗ U )(n) = d|n 1 = τ (n), we have ζ 2 (s) ↔ τ (n). Moreover,
Y
Y 1
1
1
1
ζ(s) =
1 + s + 2s + · · · =
=Q
−s
−s
p
p
1−p
p (1 − p )
p
p prime
and
∞
X
µ(n)
n=1
ns
Y
Y µ(p) µ(p2 )
1
1
=
1 + s + 2s + · · · =
1− s =
.
p
p
p
ζ(s)
p
p prime
Suppose that f and g be two arithmetic functions (not necessarily multiplicative) related by
the formula
X
g(n) =
f (d) = (f ∗ U )(n).
d|n
1
Then G(s) = F (s)ζ(s), so F (s) = G(s) ζ(s)
which implies f = g ∗ µ and gives another proof of the
Möbius inversion formula (Theorem 1.1.13).
1.3 The Greatest Integer Function
Let x ∈ R. By Archimedean property and well-ordering principle, we can prove that there exists
an nx ∈ Z such that
nx ≤ x < nx + 1.
This leads to the following definition.
Y. Meemark
1.3 The Greatest Integer Function
9
Definition. For each real number x, [x] is the unique integer such that
x − 1 < [x] ≤ x < [x] + 1.
That is, [x] is the largest integer ≤ x. Sometimes, [x] is called the floor of x. Note that
[x] = max((−∞, x] ∩ Z). The greastest integer function is the map x 7→ [x] for all x ∈ R.
Some properties of [x] are listed in the following theorem.
Theorem 1.3.1. Let x, x1 and x2 be real numbers.
(1) x = [x] + {x}, where 0 ≤ {x} < 1. {x} is called the fractional part of x.
(2) [x] = x if and only if x is an integer.
(3) [x + a] = [x] + a, if a ∈ Z.

0,
if x ∈ Z,
(4) [x] + [−x] =
−1, otherwise.
(5) [x1 ] + [x2 ] ≤ [x1 + x2 ] ≤ [x1 ] + [x2 ] + 1.
(6) [x/n] = [[x]/n] if n ∈ N.
(7) −[−x] is the least integer ≥ x and [x + 12 ] is the nearest integer to x.
(8) 0 ≤ [x] − 2[x/2] ≤ 1.
(9) If x1 < x2 , then |(x1 , x2 ] ∩ Z| = [x2 ] − [x1 ].
(10) For d ∈ N and x > 0, |{n ∈ N : d | n and n ≤ x}| = [x/d], so
n
X
τ (k) =
k=1
n
X
[x/k].
k=1
Proof. Exercises
Theorem 1.3.2. If a ∈ Z and m ∈ N, then
and 0 ≤ m{a/m} < m.
a = m[a/m] + m{a/m}
That is, [a/m] and m{a/m} are the quotient and the remainder in the division of a by m.
Proof. By the division algorithm, there exist q, r ∈ Z such that
a = mq + r,
where 0 ≤ r < m.
Then
a/m = q + r/m,
where 0 ≤ r/m < 1.
Hence, [a/m] = q and r/m = {a/m} as desired.
Remark. In order to simplify the notation, sometimes we write
x
X
k=a
f (k) for
[x]
X
k=a
f (k) and
X
k≤x
f (k) for
[x]
X
k=1
f (k).
10 Number-theoretic Functions and the Distribution of Primes
Y. Meemark
We write pe kn if pe | n and pe+1 - n, i.e., e is the highest exponent of p that divides n.
Theorem 1.3.3. [de Polignac’s Formula] If n is a positive integer and p is a prime, then the highest
exponent of p that divides n! is
∞
X
n
n
n
[n/pj ].
e=
+ 2 + 3 + ··· =
p
p
p
j=1
That is, pe kn!.
Proof. The sum has only finitely many nonzero terms, since [n/pk ] = 0 for pk > n. Note that if
P
j
p > n, then p - n! and ∞
j=1 [n/p ] = 0. If p ≤ n, then [n/p] integers in {1, 2, . . . , n} are divisible
by p, namely,
p, 2p, 3p, . . . , [n/p]p.
Of these integers, [n/p2 ] are again divisible by p2 :
p2 , 2p2 , . . . , [n/p2 ]p2 .
By the same idea, [n/p3 ] of these are divisible by p3 :
p3 , 2p3 , . . . , [n/p3 ]p3 .
After finitely many repetitions of this argument, the total number of times p divides number in
{1, 2, . . . , n} is precisely
n
n
n
+ 2 + 3 + ··· .
p
p
p
Hence, this sum is the exponent of p appearing in the prime factorization of n!.
Remark. Recall that [x/k] = [[x]/k] if k ∈ N, this shortens the computation for e as follows:
[n/p]
[[n/p]/p]
n
e=
+
+
+ ··· .
p
p
p
Example 1.3.1. Find the highest power of 7 that divides 1000!.
Proof. We compute [1000/7] = 142, [142/7] = 20, [20/7] = 2 and [2/7] = 0. Thus e = 142 + 20 +
2 + 0 = 164 is the highest power of 7 divides 1000!.
n
n!
=
is an integer.
Theorem 1.3.4. If 0 ≤ k ≤ n, then the binomial coefficient
k
k!(n − k)!
Proof. This follows from the fact that
[n/pj ] = [(n − k + k)/pj ] ≥ [(n − k)/pj ] + [k/pj ]
for all j ∈ N.
Corollary 1.3.5. For k ∈ N, k! divides the product of k consecutive integers.
Y. Meemark
1.4 The Symbols “O”, “o”, “” and “∼”
11
1.4 The Symbols “O”, “o”, “” and “∼”
Let g(x) be defined and positive for all x in some unbounded set S of positive numbers (which
will usually be either the set of positive integers of the set of positive real numbers but might,
for example, be the set of primes). Then if f (x) is defined on S, and if there is a constant M
such that
|f (x)|
<M
g(x)
for all sufficiently large x ∈ S, then we either write f (x) = O(g(x)) (read ‘f (x) is big oh of g(x)’)
or f (x) g(x) (read ‘f (x) is less-than-less-than g(x)’). If
lim
x→∞
x∈S
f (x)
= 0,
g(x)
we write f (x) = o(g(x)) (read ‘f (x) is small oh of g(x)’), and if
lim
x→∞
x∈S
f (x)
= 1,
g(x)
we write f (x) ∼ g(x), and say that f (x) is asymptotically equal to g(x).
Example 1.4.1. For x ∈ R and n ∈ N,
1. sin x x, sin x = o(x), sin x = O(1),
2. φ(n) = O(n) (note that O(n) could not be replaced by o(n) since φ(p) = p − 1 ∼ p),
√
3. [x] ∼ x, x = o(x), xk = o(ex ) for every constant k,
x
4. (The prime number theorem) π(x) ∼
, where π(x) = |{p : p prime ≤ x}|.
log x
Theorem 1.4.1.
1. O(O(g(x))) = O(g(x)),
i.e., if f (x) = O(g(x)) and h(x) = O(f (x)), then h(x) = O(g(x)).
2. O(o(g(x))), o(O(g(x))) and o(o(g(x))) = o(g(x)).
3. O(g(x)) ± O(g(x)) = O(g(x)) ± o(g(x)) = O(g(x)).
4. o(g(x)) ± o(g(x)) = o(g(x)).
5. (O(g(x)))2 = O(g 2 (x)).
6. O(g(x))o(g(x)) = (o(g(x)))2 = o(g 2 (x)).
Theorem 1.4.2. O(f (x)) ± O(g(x)) = O(max{f (x), g(x)}).
Remarks.
1. The implication
“f (x) = O(g(x)) implies h(f (x)) = O(h(g(x)))”
does not hold in general; a sufficient condition is that h be monotonic and that h(kx) =
O(h(x)) for every k > 0, if h(x) and f (x) → ∞ as x → ∞. Thus, if f (x) is larger than some
positive constant for every x > 0, then
f (x) = O(g(x)) implies log f (x) = O(log(g(x))),
12 Number-theoretic Functions and the Distribution of Primes
Y. Meemark
but it does not imply that ef (x) = O(eg(x) ), since, for example, log x = O(log
√
x 6= O( x).
√
x) but
2. However, we have a different situation for “o”. If f (x) = o(g(x)), then ef (x) = o(eg(x) )
if f (x) → ∞ as x → ∞. But the relation log f (x) = o(log g(x)) may be false, e.g., if
√
f (x) = x and g(x) = x.
3. “= O of” and “= o of” are not equivalence relations.
Theorem 1.4.3. f (x) ∼ g(x) if and only if f (x) = g(x) + o(g(x)). Hence, if g(x) → ∞ as x
increases indefinitely, the difference f (x) − g(x) need not remain bounded. Thus, f (x) ∼ g(x) is
asserted that f (x) − g(x) is of smaller order of magnitude than g(x) itself.


X1
Theorem 1.4.4. The limit lim 
− log x exists and is denoted by γ. It satisfies
x→∞
n
n≤x
1
X1
= log x + γ + O
.
n
x
(1.4.1)
n≤x
γ is called the Euler’s constant and γ = 0.5772156649 . . . .
Remark. The relation
X1
− log x ∼ γ
n
n≤x
or


X1
lim 
− log x = γ
x→∞
n
(1.4.2)
n≤x
is weaker than (1.4.1), since it says nothing about the error except that it approaches zero. Note
that (1.4.2) is not equivalent to
X1
∼ log x + γ.
n
n≤x
1.5 The sieve of Eratosthenes
√
One method of estimating π(x) is based upon that if n ≤ x and is not divisible by prime ≤ x,
then it is prime. We then cross out the integers between 1 and x first all multiples of 2, then
√
multiples of 3, then multiples of 5, and so on, until all multiples of all primes less than x have
been eliminated, thus the number remaining are primes. This method is known as the sieve of
Eratosthenes.
Consider the first r primes, p1 , p2 , . . . , pr , retaining r as an independent variable. Let
A(x, r) = # of integers remaining after all multiples of p1 , p2 , . . . , pr (including p1 , p2 , . . . , pr
themselves) have been removed from integers between 1 and x.
Then
π(x) := (# of primes less than or equal to x) ≤ r + A(x, r).
√
Of course, pr may not be the largest prime ≤ x. In order to estimate A(x, r) we use the
principle of cross-classification of the principle of inclusion-exclusion.
Y. Meemark
1.5 The sieve of Eratosthenes
13
Theorem 1.5.1. [PIE] Let S be a set of N distinct elements, and let S1 , . . . , Sr be arbitrary subsets
of S containing N1 , . . . , Nr elements, respectively. For 1 ≤ i < j < · · · < l ≤ r, let
Sij...l = Si ∩ Sj ∩ · · · ∩ Sl
Nij...l = |Sij...l |.
and
Then
r
X
[
Ni +
Si = N −
S −
i=1
1≤i≤r
X
1≤i<j≤r
Nij −
X
1≤i<j<k≤r
Nijk + · · · + (−1)r N12...r .
Take N = [x] and S = {1, 2, . . . , N }. For 1 ≤ k ≤ r, we define
Sk = {a ∈ S : pk |a}.
Then
N1 = [x/2], N2 = [x/3], . . . , N12 = [x/(2 · 3)], N13 = [x/(3 · 5)], . . .
and so
A(x, r) = [x] −
r X
x
i=1
pi
+
X
1≤i<j≤r
x
−
pi pj
X
1≤i<j<k≤r
x
x
r
+ · · · + (−1)
.
pi pj pk
p1 p2 . . . pr
Note that


r
X
X
X
x
x
x
x
r
A(x, r) − [x] −

+
−
+
·
·
·
+
(−1)
pi
pi pj
pi pj pk
p1 p2 . . . pr i=1
1≤i<j≤r
1≤i<j<k≤r
r
r
r
≤1+
+
+ ··· +
= 2r
1
2
r
and consequently
r Y
1
π(x) ≤ r + x
1−
+ 2r .
pi
i=1
This result would be completely useless if we choose r so that the rth prime is the largest one
√
1
< x, for then as we can show later that r > x 2 −ε and 2r > x whereas obviously π(x) < x.
To obtain something useful we need an estimate for
Y
1
1
Theorem 1.5.2. If x ≥ 2, then
1−
<
.
p
log x
p≤x
Theorem 1.5.3. π(x) x
.
log log x
Theorem 1.5.4. The series
X 1
X1
1
diverges. In fact,
> log log x.
p
p
2
p prime
Hence, there are infinitely many primes.
X log p
log x
.
= log x + O
Theorem 1.5.5.
p
log log x
p≤x
p≤x
14 Number-theoretic Functions and the Distribution of Primes
Y. Meemark
Theorem 1.5.6. [Formula for partial summation] Suppose that λ1 , λ2 , . . . is a nondecreasing
sequence of real numbers with limit infinity, c1 , c2 , . . . is an arbitrary sequence of real or complex
numbers and f (x) has continuous derivative for x ≥ λ1 . Put
X
C(x) =
cn ,
n,λn ≤x
where the summation is over all n for which λn ≤ x. Then for x ≥ λ1 ,
Z x
X
cn f (λn ) = C(x)f (x) −
C(t)f 0 (t) dt.
λ1
n,λn ≤x
Theorem 1.5.7.
X1
p≤x
p
∼ log log x.
1.6 Orders of magnitude of arithmetic functions
Theorem 1.6.1.
1. The relation τ (n) logc n is false for every constant c.
That is, τ (n) is sometimes larger than any fixed power of log n.
2. The relation τ (n) nε is true for every fixed ε > 0.
Theorem 1.6.2. [Dirichlet 1849] As x → ∞,
X
√
τ (n) = x log x + (2γ − 1)x + O( x),
n≤x
where γ is the Euler’s constant.
Theorem 1.6.3. Let ε > 0. Then n σ(n) n1+ε .
Theorem 1.6.4. For n ≥ 2, we have
σ(n)φ(n)
1
<
< 1.
2
n2
Corollary 1.6.5. n φ(n) n1−ε .
We say that an arithmetic function F (n) has a mean value c if
N
1 X
lim
F (n) = c.
N →∞ N
n=1
If two arithmetic functions f and F are related by
X
F (n) =
f (d),
(1.6.1)
d|n
then we can write
f (n) =
X
µ(d)F (n/d).
(1.6.2)
d|n
This is the Möbius inversion formula. Conversely, if (1.6.2) holds for all n, then so does (1.6.1). If
f is generally small, then F has an asymptotic mean value. To see this, observe that
X
XX
X
X
X
F (n) =
f (d) =
f (d)
1=
f (d)[x/d].
n≤x
n≤x d|n
d≤x
n≤x
d|n
d≤x
Y. Meemark
1.7 Elementary theorems on distribution of primes
15
Since [y] = y + O(1), we have
X
F (n) = x
n≤x
Thus F has the mean value
d≤x
∞
X
f (d)
X
d
+O
X
d≤x
|f (d)| .
if this series converges and if
d
d=1
Theorem 1.6.6. If F (n) =
X f (d)
(1.6.3)
X
d≤x
|f (d)| = o(x).
f (d), then
d|n
X
F (n) =
n≤x
X
f (d)[x/d] = x
d≤x
Theorem 1.6.7. For x ≥ 2,
X f (d)
d
d≤x
+O
X
d≤x
|f (d)| .
X φ(n)
6
= 2 x + O(log x).
n
π
n≤x
Theorem 1.6.8. For x ≥ 2,
X
φ(n) =
n≤x
3 2
x + O(x log x).
π2
Corollary 1.6.9. The probability that two numbers a and b are relatively prime is
6
.
π2
Let Q(x) denote the number of square-free integers ≤ x. Then
X
X
Q(x) =
µ(n)2 =
|µ(n)|
n≤x
Theorem 1.6.10. For x ≥ 1,
Q(x) =
n≤x
√
6
x + O( x).
2
π
Theorem 1.6.11. For x ≥ 1,
X
σ(n) =
n≤x
ζ(2) 2
π2 2
x + O(x log x) =
x + O(x log x).
2
12
1.7 Elementary theorems on distribution of primes
We define the von Margoldt function as follows. If there exists a prime p and a positive integer
r such that pr = n, then we take Λ(n) = log p. Otherwise, Λ(n) = 0. E.g., Λ(1) = 0, Λ(2) = log 2,
Λ(p) = log p, Λ(4) = log 2, Λ(6) = 0, Λ(8) = log 2, Λ(9) = log 3 and Λ(pr ) = log p.
X
Lemma 1.7.1. We have
Λ(m) = log n.
m|n
Lemma 1.7.2. Suppose x ≥ 1. Then
Theorem 1.7.3.
X
n≤x
f (n) = f (1) +
Z
X
m≤x
Λ(m)[x/m] =
X
n≤x
x
f (t) dt + O(f (x)).
1
log n = log[x]!.
16 Number-theoretic Functions and the Distribution of Primes
Y. Meemark
Theorem 1.7.4. If x ≥ 2, we have
log[x]! = x log x − x + O(log x),
and hence
X
m≤x
Λ(m)[x/m] = x log x − x + O(log x).
√
1
In particular, n! =
2π + O( n1 ) n 2 +n e−n . This is the Stirling’s formula.
X
Let ψ(x) =
Λ(n) for x > 0. This function is called the Chebyshev’s ψ-function.
n≤x
Theorem 1.7.5. [Chebyshev] There are positive numbers C1 and C2 such that
log 2
x, where x ≥ C1
2
Corollary 1.7.6. ψ(x) ≤ 2C2 x.
ψ(x) ≥
and ψ(x) − ψ(x/2) ≤ C2 x, where x > 0.
Theorem 1.7.7. [Chebyshev] There are positive constants C3 and C4 such that
C4 x
C3 x
< π(x) <
.
log x
log x
Y
Lemma 1.7.8.
1. For every positive integer n,
p < 4n .
p≤n
2n
2
.
2. If n ≥ 3 and 3 n < p < n, then p n
Theorem 1.7.9. [Bertrand’s postulate] For every positive integer n, there is a prime p such that
n < p ≤ 2n.
Theorem 1.7.10. [Mertens] As x → ∞,
X Λ(m)
1.
= log x + O(1),
m
m≤x
2.
X log p
p≤x
p
= log x + O(1),
3. there exists a constant C,
X1
p≤x
x
p
= log log x + C + O
1
, and
log x
ψ(u)
du = log x + O(1).
u2
1
X 1
1
+
.
Moreover, C = γ +
log 1 −
p
p
p
X
Theorem 1.7.11. For x ≥ 2, put ϑ(x) =
log p. Then
4.
Z
p≤x
√
ϑ(x) = ψ(x) + O( x) and
Corollary 1.7.12. lim sup
x→∞
ψ(x)
π(x) =
−O
log x
π(x)
π(x)
≥ 1 and lim inf
≤ 1.
x→∞ x/ log x
x/ log x
π(x)
exists, its value must be 1.
Hence, if lim
x→∞ x/ log x
x
log2 x
.