Download Distribution of Prime Numbers 6CCM320A / CM320X

Distribution of Prime Numbers
6CCM320A / CM320X
Manuel Breuning
Department of Mathematics
King’s College London
January 2010
Contents
Chapter 1. Divisibility
1. Basic notations and definitions
2. Euclid’s algorithm
3. The fundamental theorem of arithmetic
4. Computational problems
5. Addendum: Proofs of results in §2 and §3
3
3
4
5
7
8
Chapter 2. Basic distribution results
1. Elementary observations
2. The function π
3. Chebyshev’s theorem
4. Bertrand’s postulate
5. Addendum: Proof of Bertrand’s postulate
11
11
12
14
15
16
Chapter 3. The prime number theorem
1. Statement of the prime number theorem
2. Addendum: Error terms and Riemann hypothesis
19
19
21
Chapter 4. Arithmetic functions and Dirichlet series
1. The Riemann zeta function
2. Arithmetic functions
3. Dirichlet series
4. Addendum: The Möbius function
23
23
26
27
29
Chapter 5. Primes and arithmetic progressions
1. Primes in arithmetic progressions
2. Addendum: Arithmetic progressions in the sequence of primes
31
31
32
Appendices
Notation
Further reading
The first 400 prime numbers
35
35
36
37
1
CHAPTER 1
Divisibility
In this chapter we discuss divisibility of integers. The most important result
is the fundamental theorem of arithmetic: every integer greater than 1 can be
expressed as a product of prime numbers and apart from the order of the factors
this product representation is unique. We also look at some computational problems
related to prime numbers.
1. Basic notations and definitions
In this course we are mainly interested in properties of positive integers (also
called natural numbers). We write N for the set of all positive integers, so
N = {1, 2, 3, . . . }.
We will use many well-known properties of N without further explanation, for
example we will use that N is ordered (i.e. we know what a ≤ b means for a, b ∈
N) and that certain algebraic operations (addition, multiplication) are defined for
positive integers. Furthermore we will prove some statements by induction, so we
assume familiarity with the principle of induction.
To study properties of N we must sometimes consider larger sets, in particular
the set of all non-negative integers N ∪ {0} = {0, 1, 2, 3, . . . } and the set of all
integers Z = {. . . , −2, −1, 0, 1, 2, . . . }. Later we will also require the real numbers
R and the complex numbers C.
Definition 1.1. Let a, b ∈ Z. We say that b divides a (or equivalently that b
is a divisor of a, or that b is a factor of a, or that a is divisible by b, or that a is a
multiple of b) if there exists an integer q ∈ Z such that a = bq. We write b | a if b
divides a, and b - a if b doesn’t divide a.
Most of the time we will be interested in positive divisors of positive integers.
Example 1.2.
(1) 3 | 12 because 12 = 3 · 4. Also −3 | 12 because 12 =
(−3) · (−4). But 3 - 10 because 10 6= 3 · q for all q ∈ Z.
(2) Let n be any integer. Then 1 | n and n | n.
(3) The number 1 has only one positive divisor, namely 1 itself.
(4) If a and b are positive integers and if b | a then b ≤ a. Indeed, b | a implies
that there exists q ∈ Z with a = bq, and since a and b are positive so is q.
Therefore q ≥ 1 and it follows that b ≤ bq = a.
(5) If b | a1 and b | a2 then b | a1 + a2 .
Definition 1.3. A positive integer n is called a prime number (or simply a
prime) if n 6= 1 and if the only positive divisors of n are 1 and n. A positive integer
n is called composite if n has a positive divisor different from 1 and n.
3
4
1. DIVISIBILITY
The number 1 is neither prime nor composite. Every positive integer n other
than 1 is either prime or composite (but not both).
Example 1.4.
(1) The first 10 prime numbers are 2, 3, 5, 7, 11, 13, 17, 19,
23, 29.
(2) The first 10 composite numbers are 4, 6, 8, 9, 10, 12, 14, 15, 16, 18.
2. Euclid’s algorithm
Lemma 1.5 (Division algorithm). Given a ∈ Z and b ∈ N, there exist unique
q, r ∈ Z such that a = qb + r and 0 ≤ r < b.
The proof of Lemma 1.5 is in Chapter 1, §5 (this proof is not examinable). If
a = qb + r and 0 ≤ r < b as in Lemma 1.5, then r is called the remainder of a when
divided by b.
Definition 1.6. Let a, b ∈ N. The greatest common divisor of a and b is the
greatest number d ∈ N with the property that d | a and d | b. We write (a, b) for the
greatest common divisor of a and b. The numbers a and b are said to be coprime
(or relatively prime) if (a, b) = 1.
We remark that any two natural numbers a, b have a greatest common divisor.
Indeed, the set {d ∈ N : d | a and d | b} is non-empty (because it contains 1) and
bounded above (a is an upper bound) and therefore contains a greatest element.
Example 1.7. We want to find the greatest common divisor of 8 and 12. The
set of positive divisors of 8 is {1, 2, 4, 8} and the set of positive divisors of 12 is
{1, 2, 3, 4, 6, 12}. Therefore the set of positive common divisors is {1, 2, 4} and thus
4 is the greatest common divisor of 8 and 12.
We can compute the greatest common divisor of a, b ∈ N using Euclid’s algorithm. This works as follows. We apply the division algorithm to a and b to get
a = q1 b + r1 with 0 ≤ r1 < b. If r1 6= 0 then we apply the division algorithm to b
and r1 and get b = q2 r1 + r2 with 0 ≤ r2 < r1 . If r2 6= 0 then we apply the division
algorithm to r1 and r2 and get r1 = q3 r2 + r3 with 0 ≤ r3 < r2 . We repeat this
until we find a remainder which is zero (this must happen at some point because
b > r1 > r2 > · · · ≥ 0). Then we have a system of equations
a = q1 b + r1 ,
b = q2 r1 + r2 ,
r1 = q3 r2 + r3 ,
..
.
rn−3 = qn−1 rn−2 + rn−1 ,
rn−2 = qn rn−1 + rn ,
rn−1 = qn+1 rn + 0.
0 < r1 < b
0 < r2 < r1
0 < r3 < r2
0 < rn−1 < rn−2
0 < rn < rn−1
To show that this algorithm really computes the greatest common divisor of a
and b we need the following lemma.
Lemma 1.8. If a = qb + r then (a, b) = (b, r).
Proof. If d is a common divisor of a and b, then d | a − qb = r which shows
that d is also a common divisor of b and r. Conversely, if d is a common divisor of
b and r, then d | qb + r = a which shows that d is also a common divisor of a and b.
3. THE FUNDAMENTAL THEOREM OF ARITHMETIC
5
Therefore the set of common divisors of a and b is equal to the set of common
divisors of b and r, and thus the greatest common divisors are equal.
We apply this lemma repeatedly to the equations in Euclid’s algorithm and
obtain (a, b) = (b, r1 ) = (r1 , r2 ) = (r2 , r3 ) = · · · = (rn−2 , rn−1 ) = (rn−1 , rn ). But
the equation rn−1 = qn+1 rn + 0 shows that rn | rn−1 and therefore (rn−1 , rn ) = rn .
We have shown the following result.
Theorem 1.9. The last non-zero remainder rn in Euclid’s algorithm is the
greatest common divisor of a and b.
Example 1.10. We want to compute the greatest common divisor of 99 and
42. Euclid’s algorithm gives
99 = 2 · 42 + 15,
42 = 2 · 15 + 12,
15 = 1 · 12 + 3,
12 = 4 · 3 + 0.
The last non-zero remainder is 3, thus (99, 42) = 3.
3. The fundamental theorem of arithmetic
Note: The proofs of Lemmas 1.11, 1.12, 1.16 and of Theorem 1.13 are in
Chapter 1, §5 (these four proofs are not examinable).
Lemma 1.11. Let p be a prime number and let a1 , a2 ∈ N. If p | a1 a2 then
p | a1 or p | a2 . More generally, if a prime number p divides a product a1 a2 · · · an
then p divides ai for some i.
By a prime factor (or prime divisor ) of a positive integer a we mean a prime
number p which is a factor of a.
Lemma 1.12. Every integer a > 1 has a prime factor.
Theorem 1.13 (Fundamental theorem of arithmetic). Every integer a > 1 is
a product of prime numbers, i.e. a = p1 p2 · · · pn where n ≥ 1 and the pi are prime
numbers (not necessarily distinct). Apart from the order of the factors this product
representation is unique.
It is often convenient to write the prime factorisation of a number a > 1 in
the form a = pk11 · · · pkr r where r ≥ 1, the pi are distinct prime numbers and the
exponents ki are positive integers. One often allows ki = 0; since p0i = 1, a prime
number pi with exponent ki = 0 is not a prime factor of a. One advantage of
allowing the exponent 0 is that then a = 1 can be written in the same form:
1 = p01 · · · p0r .
Example 1.14. 200 = 2 · 5 · 2 · 5 · 2 = 23 · 52 = 23 · 30 · 52 .
If we know the prime factorisation of a positive integer a then we can immediately write down all positive divisors of a: if a = pk11 · · · pkr r then b | a if and only
if b = pl11 · · · plrr with 0 ≤ li ≤ ki for all i. We apply this observation to prove the
following lemma (which is needed in Chapter 4, §2).
Lemma 1.15. If m, n ∈ N are coprime then every d | mn can be written uniquely
as d = d1 d2 where d1 | m and d2 | n (we assume that d, d1 , d2 are all positive).
6
1. DIVISIBILITY
Proof. Since m and n are coprime, they don’t have any prime factors in
kr+s
kr+1
where all the pi are distinct. If
· · · pr+s
common. So m = pk11 · · · pkr r and n = pr+1
lr+s
l1
d | mn then d = p1 · · · pr+s with 0 ≤ li ≤ ki for 1 ≤ i ≤ r + s. Let d1 = pl11 · · · plrr
lr+s
lr+1
. Then obviously d = d1 d2 , d1 | m and d2 | n.
· · · pr+s
and d2 = pr+1
l0
l0
Conversely if d = d01 d02 , d01 | m and d02 | n then we must have d01 = p11 · · · prr
l0
l0
r+1
r+s
and d02 = pr+1
· · · pr+s
. From d = d01 d02 it follows that li = li0 for 1 ≤ i ≤ r + s and
0
therefore d1 = d1 and d2 = d02 . This shows uniqueness.
The following result is used in Chapter 2, §3.
Lemma 1.16. Let p1 , p2 , . . . , pr be distinct prime numbers and let n be any
integer. If pi | n for all i then p1 p2 · · · pr | n.
Finally we want to give an explicit formula for the prime factorisation of N !.
QN
Recall that for N ∈ N one defines N ! (read: N factorial) by N ! = i=1 i =
1 · 2 · · · (N − 1) · N . Thus 1! = 1, 2! = 2, 3! = 6, 4! = 24, etc. First note that
if p is a prime factor of N ! then p must divide one of the numbers 1, 2, . . . , N (by
Lemma 1.11) and therefore p ≤ N . On the other hand every prime number p ≤ N
is obviously a prime factor of N !.
To simplify the notation we will use the convention that an index
P p in a sum
or product denotes a prime number. For example in the expression p p1 (considered in Corollary 4.7) the sum is over all prime numbers p, and in the expression
Q
kp
(in Lemma 1.17) the product is over all prime numbers p not greater
p≤N p
than N .
Q
P∞ h i
Lemma 1.17. N ! = p≤N pkp where kp = m=1 pNm .
Here we have used the following standard notation: for any x ∈ R, [x] is the
greatest integer that is less
than or equal to x, e.g. [2.6] = 2, [5] = 5, [−1.5] = −2.
P∞ h N i
Note that the sum m=1 pm has only finitely many non-zero summands because
h i
if pm > N then pNm = 0.
Proof of Lemma 1.17. Consider a prime number p ≤ Nh. We
i must count
N
how often p appears in the product N ! = 1 · 2 · · · · · N . Clearly p of the factors
h i
1, 2, . . . , N are multiples of p; pN2 of these factors are multiples of p2 ; etc. Hence
h i h i
in kp = Np + pN2 + . . . we have counted once the number of factors which are
divisible by p but not by p2 , we have counted twice the number of factors which
are divisible by p2 but not by p3 , etc. This shows that the prime factor p appears
kp times in N !.
Example 1.18. We want to compute the largest integer k such that 7k | 50!.
By Lemma 1.17
∞ X
50
50
50
k=
= 1 + 2 = 7 + 1 = 8.
m
7
7
7
m=1
Note that only the summands for
50m
= 1 and m = 2 are non-zero because for m ≥ 3
= 0.
one has 750
m < 1 and therefore
m
7
4. COMPUTATIONAL PROBLEMS
7
Using Lemma 1.17 one can easily deduce formulas for the prime factorisation
n!
of other expressions formed with factorials, e.g. binomial coefficients nk = k!(n−k)!
.
This is used in Chapter 2, §5.
4. Computational problems
Lemma 1.19. A positive
integer n is composite if and only if n has a prime
√
divisor p satisfying p ≤ n.
Proof. If n is composite then n = ab with 1 < a < n and 1 < b < n. We can
assume that a ≤ b. By Lemma
1.12
a prime factor p. Then p is also a prime
√ a has √
√
factor of n and p ≤ a = a · a ≤ a · b = n.
√
Conversely suppose that n has a prime divisor
p ≤ n. Then n 6= 1 (because
√
1 doesn’t have a prime divisor) and therefore n < n. Hence 1 < p < n and p | n,
i.e. n is composite.
An algorithm that determines whether an integer n > 1 is prime or composite
is called a primality test. The easiest primality test is √
the following method which is
known as trial division. For every prime number p ≤ n test whether√n is divisible
by p or not. By Lemma 1.19 we know that if p | n for some p ≤ n then n is
composite, otherwise n is prime. Trial division is only useful to test primality of
small numbers.
Remark 1.20. In 2002, Agrawal, Kayal and Saxena developed the first polynomial time primality test (now known as the AKS primality test). Polynomial
time means that there exist constants C, k such that for every integer n > 1 the
algorithm needs at most C · (log n)k many steps to decide whether n is prime or
composite. If n is composite then the AKS primality test will confirm this without
finding a factor of n, so this primality test can’t be used to find prime factorisations.
The sieve of Eratosthenes is a method to compute a list of all prime numbers
not greater than n (where n ≥ 2 is given). We explain it for the example n = 50.
Write down all integers from 2 to 50. We will cross out more and more composite
numbers from this list until in the end only the prime numbers remain. We cross
out all multiples of 2 except 2 itself, i.e. 4, 6, 8, . . . , 50. The first remaining integer
after 2 is 3. We cross out all multiples of 3 except 3 itself, i.e. 6, 9, 12, . . . , 48 (note
that some of these numbers were already crossed out because they are multiples of
2). The first remaining integer after 3 is 5. We cross out all multiples of 5 except
5 itself, i.e. 10, 15, 20, . . . , 50 (again some of these numbers have been crossed out
earlier). The first remaining integer after 5 is 7. We cross out all multiples of
7 except 7 itself, i.e. 14, 21, . . . , 49 (some of these numbers have been crossed out
earlier). Now the list looks as follows.
11
21
31
41
2
1
2
2
2
3
2
4
2
3
13
23
3
3
43
4
14
24
34
44
5
1
5
2
5
3
5
4
5
6
1
6
2
6
3
6
4
6
7
17
27
37
47
8
1
8
2
8
3
8
4
8
9
19
29
3
9
4
9
10
20
30
40
50
√
The first remaining integer after 7 is 11. But 11 is greater than 50 and we are
therefore finished. Indeed,
we have crossed out all multiples ip with i ≥ 2 of all
√
prime numbers p ≤ 50. Since by Lemma 1.19 every composite number ≤ 50 has
8
1. DIVISIBILITY
√
a prime factor ≤ 50, this implies that we have crossed out all composite numbers
≤ 50. Hence the remaining numbers 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47
are precisely the primes not greater than 50.
A variation also works if one wants to find all primes p with a ≤ p ≤ b for given
√
2 ≤ a ≤ b. Write down all integers from a to b and for all prime numbers p ≤ b
cross out all multiples of p (except for p itself if it is in the list). The remaining
numbers are precisely the prime numbers in the given range.
5. Addendum: Proofs of results in §2 and §3
Note: The material in this section is not examinable.
In the following proof we use the fact that every non-empty set S ⊆ N ∪ {0}
has a smallest element. This is known as the well-ordering principle.
Proof of Lemma 1.5 (Division algorithm). Let S = {a − tb : t ∈ Z, a − tb ≥
0}. Then S ⊆ N ∪ {0} and S is non-empty (because if t is any integer with t ≤ a/b
then a − tb ∈ S). By the well-ordering principle it follows that S contains a smallest
element, say r. Since r ∈ S, there exists q ∈ Z such that r = a − qb and r ≥ 0.
Also r < b because otherwise r − b = a − (q + 1)b would be an element of S which
is smaller than r. This shows the existence of q and r with the stated properties.
Suppose that there are two representations a = qb + r = q 0 b + r0 as in the
statement of the lemma. Then (q − q 0 )b = r0 − r. Since −b < r0 − r < b it follows
that |q − q 0 | · b = |r0 − r| < b and therefore |q − q 0 | = 0. Hence q = q 0 and r = r0 .
This shows the uniqueness of q and r.
From Euclid’s algorithm in §2 we can also deduce the following.
Theorem 1.21. There exist integers s, t ∈ Z such that (a, b) = sa + tb.
Proof. By solving the second last equation in Euclid’s algorithm for rn we
obtain rn = rn−2 − qn rn−1 , i.e. we have expressed rn as a linear combination of
rn−2 and rn−1 . Using rn−1 = rn−3 − qn−1 rn−2 (which follows from the third last
equation) we can eliminate rn−1 in the linear combination,
rn = rn−2 − qn rn−1
= rn−2 − qn (rn−3 − qn−1 rn−2 )
= (−qn )rn−3 + (1 + qn qn−1 )rn−2 ,
i.e. we have expressed rn as a linear combination of rn−3 and rn−2 . Continuing
like this we obtain an expression of the form rn = sa + tb. From this we obtain
(a, b) = sa + tb because rn = (a, b) by Theorem 1.9.
Remark 1.22. The proof of Theorem 1.21 shows not only that integers s, t
with (a, b) = sa + tb exist, but also how to find such integers by using Euclid’s
algorithm. Note that s and t are in Z and not necessarily in N. We remark that s
and t are not uniquely determined.
Proof of Lemma 1.11. The case of a product with n factors follows easily
from the case with two factors.
Assume that p | a1 a2 . If p | a1 we are finished. If p - a1 then p and a1
are coprime, and by Theorem 1.21 there exist s, t ∈ Z such that 1 = sp + ta1 .
Multiplying this by a2 gives a2 = spa2 + ta1 a2 which shows that a2 is divisible by
p (because obviously p | spa2 and by assumption p | ta1 a2 ).
5. ADDENDUM: PROOFS OF RESULTS IN §2 AND §3
9
Proof of Lemma 1.12. Let S = {b ∈ N : b | a and b > 1}. Then S is nonempty (because a ∈ S) and therefore by the well-ordering principle S has a smallest
element, say p. We claim that p is a prime factor of a. Clearly p is a factor of a
since p ∈ S. If p was composite then p would have a factor r with 1 < r < p;
but then r ∈ S which is a contradiction (because p was chosen to be the smallest
element in S). Hence p must be a prime number.
Proof of Theorem 1.13 (Fundamental theorem of arithmetic). Let a > 1
be an integer. We first show that a is a product of prime numbers. By Lemma 1.12
a has a prime factor, i.e. a = p1 a1 for a prime number p1 and a positive integer a1 .
Since p1 ≥ 2 we have a1 < a. If a1 = 1 we are finished. Otherwise apply Lemma
1.12 to a1 to obtain a1 = p2 a2 with p2 prime and 1 ≤ a2 < a1 . Repeating this
process we must finally find an = 1 and then a = p1 a1 = p1 p2 a2 = · · · = p1 p2 · · · pn .
Next we show uniqueness. If a = p1 · · · pn = q1 · · · qm are two representations of
a as product of primes, then by Lemma 1.11 p1 | qi1 for some i1 . Since qi1 is prime
this implies that p1 = qi1 . Therefore a/p1 = p2 · · · pn = q1 · · · qi1 −1 qi1 +1 · · · qm .
Next we repeat this process for p2 in p2 · · · pn = q1 · · · qi1 −1 qi1 +1 · · · qm , then for p3 ,
etc. After having cancelled all factors p1 , p2 , . . . , pn on the left hand side and the
corresponding qi1 , qi2 , . . . qin on the right hand side, the product of the remaining
factors on the right hand side must be equal to 1. Hence there are no remaining
factors, i.e. n = m. Now p1 = qi1 , p2 = qi2 , . . . shows that the two product
representations a = p1 · · · pn and a = q1 · · · qm only differ by the order of the
factors.
Proof of Lemma 1.16. For every r ≥ 1 we must show the following statement: If p1 , p2 , . . . , pr are distinct primes, n is any integer and pi | n for i =
1, 2, . . . , r, then p1 p2 · · · pr | n. To do this we use induction on r.
First we consider the case r = 1. The statement for r = 1 is: If p1 is a prime,
n is an integer and p1 | n, then p1 | n. This is obviously true.
Now assume that r ≥ 2 and that we have shown the statement for r − 1. Let
p1 , p2 , . . . , pr be distinct primes and n an integer such that pi | n for all i. We
must show that p1 p2 · · · pr | n. Since p1 , p2 , . . . , pr−1 are distinct primes which
divide n, the induction hypothesis implies that p1 p2 · · · pr−1 | n. So there exists
m ∈ Z such that n = p1 p2 · · · pr−1 m. Now it follows from Lemma 1.11 applied to
pr | n = p1 p2 · · · pr−1 m that pr divides one of the pi for 1 ≤ i ≤ r − 1 or that pr
divides m. But pr - pi for 1 ≤ i ≤ r − 1 (because pr and pi are distinct primes),
hence pr | m. It follows that p1 p2 · · · pr−1 pr | p1 p2 · · · pr−1 m = n as required.
(There exist several alternative proofs. For example one can use induction on
r and the general fact that if a | n, b | n and (a, b) = 1 then ab | n. Or one can use
the fundamental theorem of arithmetic.)
CHAPTER 2
Basic distribution results
In this chapter we show that there are infinitely many prime numbers and we
prove various elementary results concerning their distribution among the integers.
In particular we show that for x → ∞ the number of primes not greater than x
has the same order of magnitude as x/ log(x) (Chebyshev’s theorem), and that for
every positive integer n the interval (n, 2n] contains a prime (Bertrand’s postulate).
1. Elementary observations
The following result was already known to Euclid (about 325 BC–about 265
BC).
Theorem 2.1. There exist infinitely many primes.
Proof. Suppose for a contradiction that there exist only finitely many primes,
and let p1 , p2 , Q
. . . , pn be the complete list of all prime numbers. We consider the
n
number N = i=1 pi + 1. Then N is greater than 1 and must therefore have a
prime factor p (by Lemma 1.12). By assumption the list p1 , p2 , . . . , pn contains all
prime numbers,
hence p must be equal toQpi for some 1 ≤ i ≤ n, so in particular p
Qn
n
divides i=1 pi .QSince p divides N and i=1 pi , it follows that p also divides the
n
difference N − i=1 pi = 1. This is a contradiction because 1 is not divisible by
any prime number.
The next result shows that the sequence of primes contains arbitrarily large
gaps.
Proposition 2.2. Let N ∈ N. Then there exists a sequence of N consecutive
composite numbers.
Proof. Consider the N consecutive numbers (N + 1)! + a for 2 ≤ a ≤ N + 1.
Each of these numbers is composite because a | (N + 1)! + a and 1 < a < (N +
1)! + a.
Now we consider small gaps in the sequence of primes. Clearly the only possibility for two consecutive positive integers (p, p + 1) to be both prime is (2, 3).
There are many prime pairs (p, p + 2), i.e. pairs of prime numbers of distance two,
for example (3, 5), (5, 7), (11, 13), (17, 19), . . . , however it is not known if there exist
infinitely many such pairs.
Conjecture 2.3 (Twin prime conjecture). There exist infinitely many prime
pairs (p, p + 2).
How many prime triplets of the form (p, p + 2, p + 4) are there? One of the
three numbers p, p + 2, p + 4 must be divisible by 3 (because if p = 3n then 3 | p;
if p = 3n + 1 then 3 | p + 2; if p = 3n + 2 then 3 | p + 4). The only prime number
11
12
2. BASIC DISTRIBUTION RESULTS
divisible by 3 is 3 itself, therefore one of the numbers p, p + 2, p + 4 must be equal
to 3. Clearly p + 2 = 3 and p + 4 = 3 are impossible, thus the only prime triplet of
the form (p, p + 2, p + 4) is (3, 5, 7).
No simple formula is known to compute the n-th prime number for given n (of
course the sieve of Eratosthenes can be used to compute a list of all prime numbers
and therefore to compute the n-th prime, however that’s certainly not a simple
formula).
2. The function π
We have seen that on the one hand the sequence of prime numbers contains
arbitrarily large gaps. On the other hand it is conjectured that there are infinitely
many pairs of primes of distance two. Thus the distribution of primes in detail is
very irregular. However the average distribution of primes is much more regular.
The following table shows the number of primes in the first ten blocks of 1000
integers.
range number of primes
range number of primes
1 to 1000
168
5001 to 6000
114
1001 to 2000
6001 to 7000
135
117
2001 to 3000
127
7001 to 8000
107
3001 to 4000
120
8001 to 9000
110
4001 to 5000
9001 to 10000
119
112
To study the average distribution of primes it is useful to introduce the function
π. For a real number x we define
π(x) = the number of primes not greater than x.
For example π(7) = 4, π(10) = 4, π(3.14) = 2, π(1) = 0. Figure 1 shows π(x) for
x ≤ 30. Note that if we know the function π then we also know all prime numbers:
an integer n is prime if and only if π(n) > π(n − 1).
Figure 1. π(x) for x ≤ 30
2. THE FUNCTION π
13
Obviously π(x) ≤ x for all x ≥ 0. Theorem 2.1 implies that π(x) → ∞ as
x → ∞. One would like to have more precise statements for the behaviour of π(x)
as x → ∞. Around 1800, several mathematicians conjectured approximations for
π(x) as x → ∞. Legendre suggested in 1798 that π(x) is approximately equal
x
to the function logx x (and more generally to functions of the form log(x)−A
for
A ∈ R). Here log x denotes the natural logarithm of x, i.e. the logarithm having
base e = 2.718 . . . . Figure 2 shows π(x) and logx x for x ≤ 400. Gauss observed
that the logarithmic integral li(x) seems to give a good approximation to π(x) (see
Chapter 3, §2 for the definition of li(x)).
Figure 2. π(x) and x/ log(x) for x ≤ 400
Of course one has to define precisely what is meant by an approximation to
π(x) as x → ∞. A weak notion of approximation is that π(x) and x/ log(x) have
the same order of magnitude (see Definition 2.6), in symbols
x
π(x) as x → ∞.
log x
This was shown by Chebyshev in 1850 and we will give part of the proof in the
next section. A stronger notion of approximation is that π(x) and x/ log(x) are
asymptotically equal (see Definition 3.1), in symbols
x
π(x) ∼
as x → ∞.
log x
This is the statement of the prime number theorem which was proved in 1896
independently by Hadamard and de la Vallée Poussin. We will discuss the prime
number theorem in Chapter 3, §1. Finally Chapter 3, §2 contains a few remarks
about the approximation to π(x) given by li(x).
14
2. BASIC DISTRIBUTION RESULTS
3. Chebyshev’s theorem
To study the function π, Chebyshev (also spelled Tchebychef) introduced an
auxiliary function, the θ-function. It is defined by
X
θ(x) =
log p
p≤x
for real numbers x (summation over all prime numbers p ≤ x). For example
θ(10) = log 2 + log 3 + log 5 + log 7. Chebyshev proved upper and lower estimates
for the function θ, and then deduced upper and lower estimates for the function π.
The following result is the upper estimate for θ.
Lemma 2.4. If x > 0 then θ(x) < log(4) · x.
Proof. The statement is clearly true for 0 < x < 1, so we can assume that
x ≥ 1. Since θ(x) = θ([x]), it is enough to prove the statement θ(n) < log(4) · n for
all n ∈ N.
We use induction on n. The cases n = 1 and n = 2 are obviously true. Now
assume that n ≥ 3 and that θ(m) < log(4) · m for all m < n. We must distinguish
the cases n even and n odd.
If n is even then θ(n) = θ(n − 1) < log(4) · (n − 1) < log(4) · n, as required.
If n is odd then n = 2m + 1 with m ≥ 1. We will show below that θ(2m + 1) −
θ(m + 1) < log(4) · m. It then follows that
θ(n) = (θ(2m + 1) − θ(m + 1)) + θ(m + 1)
< log(4) · m + log(4) · (m + 1) = log(4) · (2m + 1) = log(4) · n,
as required.
It remains to show that θ(2m + 1) − θ(m + 1) < log(4) · m for every m ≥ 1.
. If p is a prime number with m + 2 ≤
Consider M = 2m+1
= (2m+1)(2m)···(m+2)
m!
m
p ≤ 2m + 1 then p divides M (because p divides the
Q numerator of M but not the
denominator). Hence by Lemma 1.16 the product m+2≤p≤2m+1 p divides M . On
the other hand M < 22m because 2m+1
= 2m+1
m
m+1 and these are two of the terms
in the binomial expansion of (1 + 1)2m+1 . It follows that
!
X
Y
θ(2m + 1) − θ(m + 1) =
log p = log
p
m+2≤p≤2m+1
m+2≤p≤2m+1
≤ log M < log(22m ) = log(4) · m.
The next lemma gives a weak form of the lower estimate for θ. We omit its
proof.
Lemma 2.5. There exists a constant c > 0 such that θ(x) > cx for all sufficiently large x.
Definition 2.6. Let f and g be functions which are defined for all sufficiently
large real numbers, and assume that f (x) and g(x) are positive for all large x. We
say that f and g are of the same order of magnitude as x → ∞, in symbols
f g
as x → ∞,
if there exist positive constants a and A such that af (x) < g(x) < Af (x) for all
sufficiently large x.
4. BERTRAND’S POSTULATE
15
Example 2.7. Consider the functions f (x) = x2 +x and g(x) = 2x2 . For x > 1
one has f (x) < g(x) < 2f (x), i.e. the condition in Definition 2.6 is satisfied with
a = 1 and A = 2. Hence f g as x → ∞.
We remark that the order of magnitude relation is symmetric: f g as x → ∞
if and only if g f as x → ∞. From Lemmas 2.4 and 2.5 we can now deduce the
order of magnitude of the functions θ(x) and π(x).
Theorem 2.8. θ(x) x as x → ∞.
Proof. This is immediate from Lemmas 2.4 and 2.5.
Theorem 2.9. π(x) x
log x
as x → ∞.
Proof. Since
θ(x) =
X
log p ≤
p≤x
X
log x = log(x) ·
p≤x
X
1 = log(x) · π(x),
p≤x
θ(x)
x
we have π(x) ≥ log
x for x > 1. Hence Lemma 2.5 implies that π(x) > c log x for all
sufficiently large x.
To get the other inequality we note that for x > 0
X
X
X
√
θ(x) =
log p ≥
log p ≥
log( x)
p≤x
√
X
1
= · log(x) ·
2
√
x<p≤x
√
√
x<p≤x
x<p≤x
√ 1
1 = · log(x) · π(x) − π( x) .
2
√
√
Now π( x) ≤ x and from Lemma 2.10 below it follows that x < logx x for all
sufficiently large x. Using Lemma 2.4 we therefore find that for all sufficiently large
x
√
x
θ(x)
log(4) · x
x
+ π( x) < 1
+
π(x) ≤ 1
= (2 log(4) + 1)
.
log
x
log
x
·
log
x
·
log
x
2
2
In the proof of Theorem 2.9 we used the following well-known result from
analysis.
Lemma 2.10. For any δ > 0, limx→∞
log x
xδ
= 0.
So Lemma 2.10 says that the function log x tends to infinity more slowly than
any power xδ (with δ > 0). The proof of Lemma 2.10 can be found in most books
on analysis.
4. Bertrand’s postulate
The following theorem is known as Bertrand’s postulate. It was conjectured by
Bertrand in 1845 and proved by Chebyshev in 1850.
Theorem 2.11. For every integer n ≥ 1 there is a prime number p satisfying
n < p ≤ 2n.
Example 2.12.
(1) For n = 1 Bertrand’s postulate states that there is
a prime number p such that 1 < p ≤ 2. Clearly p = 2 is the only prime
with this property.
16
2. BASIC DISTRIBUTION RESULTS
(2) For n = 4 Bertrand’s postulate states that there is a prime number p such
that 4 < p ≤ 8. Clearly p = 5 and p = 7 have this property. Hence ‘there
is a prime number p’ means ‘there is at least one prime number p’.
8
(3) We know by Bertrand’s postulate applied to n = 1010 −1 = 1099,999,999
99,999,999
that there exists a prime p such that 10
< p ≤ 2 · 1099,999,999 .
However at the moment no prime p with this property is known. In fact
the largest known prime has only 12,978,189 decimal digits and is therefore
less than 1099,999,999 . There is a prize of $150,000 for the discovery of
a prime greater than 1099,999,999 (i.e. a prime with at least 100 million
decimal digits), see http://www.eff.org/awards/coop for details.
Corollary 2.13. Let pn denote the n-th prime number. Then pn ≤ 2n .
Proof. By Bertrand’s postulate we know that each of the intervals (1, 2], (2, 4],
(4, 8], etc. contains at least one prime number. Hence the interval (1, 2n ] contains
at least n prime numbers. Thus the n-th prime number must be contained in this
interval, i.e. pn ≤ 2n .
5. Addendum: Proof of Bertrand’s postulate
Note: The material in this section is not examinable.
Proof of Theorem 2.11. Assume for a contradiction that we have an n such
that there is no prime p in the range n < p ≤ 2n. Let N = 2n
n . We will find
an upper estimate and a lower estimate for log N and obtain a contradiction by
showing that these estimates are incompatible (at least for sufficiently large n).
This is done in the following three lemmas.
Lemma 2.14 (Upper estimate for log N ).
√
2
log N ≤ log(4) · n + 2n · log(2n)
3
for n ≥ 5.
Proof. By Lemma 1.17 the prime factorisation of N =
2n
n
=
(2n)!
n!·n!
is given
by
N=
Y
pkp
where
kp =
p≤2n
∞ X
2n
n
−
2
.
pm
pm
m=1
So
(1)
log N =
X
kp log p =
p≤2n
X
p:kp =1
log p +
X
kp log p
p:kp ≥2
P
P
where p:kp =1 (resp. p:kp ≥2 ) denotes the sum over all prime numbers p for which
kp = 1 (resp. kp ≥ 2). We will estimate these two sums separately.
By our assumption there are no prime numbers p in the range n < p ≤ 2n,
2
so every prime divisorh p of
i N satisfies
h ip ≤ n. Now if 3 n < p ≤ n, then we have
2p ≤ 2n < 3p and so
2n
p
= 2 and
n
p
= 1, hence kp = 0 (in the formula for kp
5. ADDENDUM: PROOF OF BERTRAND’S POSTULATE
17
we only need to consider the summand for m = 1, because p > 23 n implies p2 > 2n
for n ≥ 5). Therefore, if p | N then p ≤ 23 n, and so
X
X
X
2
2
(2)
log p ≤
log p ≤
log p = θ
n ≤ log(4) · n
3
3
2
p:kp =1
p|N
p≤ 3 n
(the last inequality follows from Lemma 2.4).
One easily sees that in the formula for kp each summand
h
2n
pm
i
−2
h
n
pm
i
is either
0 or 1. Moreover it is clearly equal to 0 for all m > log(2n)/ log(p), so there are
at most log(2n)/ log(p) non-zero summands.
Thus kp ≤ log(2n)/√
log(p). If kp ≥ 2
√
then 2 ≤ log(2n)/ log(p), i.e. p ≤ 2n. Hence there are at most 2n many prime
numbers p with kp ≥ 2, and for each such p we have kp log p ≤ log(2n). This implies
X
√
(3)
kp log p ≤ 2n · log(2n).
p:kp ≥2
Combining the equality (1) with the inequalities (2) and (3) shows the upper
estimate for log N .
Lemma 2.15 (Lower estimate for log N ).
log N ≥ 2n log 2 − log(2n)
for n ≥ 1.
Proof. N is the largest term in the binomial expansion of 22n = (1 + 1)2n ,
and so 22n ≤ 2nN , since this expansion has 2n + 1 terms, two of which are equal
to 1. Hence 2n log 2 ≤ log(2n) + log N .
Lemma 2.16. If n > 29 = 512, then
√
2
log(4) · n + 2n · log(2n) < 2n log 2 − log(2n).
3
Proof.
It
is
not
difficult
to see that the required inequality is equivalent to
√
3(1 + 2n) log(2n) < 2n log 2.
We need the following two inequalities, valid for n > 512:
√
33 √
1 + 2n <
· 2n,
32
√
10
log(2n) <
· log(2) · 2n.
32
The first of these inequalities is easy to show.
The second one can be seen by
√
10
observing that the function x 7→ 32
· log(2) · 2x − log(2x) is 0 for x = 512 and has
positive derivative for all x ≥ 512.
Using these inequalities we find that
√
√
33 √
10
· 2n ·
· log(2) · 2n < log(2) · 2n,
3(1 + 2n) log(2n) < 3 ·
32
32
the last inequality because
990
1024
< 1.
Lemmas 2.14, 2.15 and 2.16 imply that
√
2
log N ≤ log(4) · n + 2n · log(2n) < 2n log 2 − log(2n) ≤ log N
3
18
2. BASIC DISTRIBUTION RESULTS
for n > 512. This is a contradiction, therefore our assumption that there are no
primes p satisfying n < p ≤ 2n was wrong. Hence Theorem 2.11 is true for all
n > 512.
It remains to consider the case n ≤ 512. Of course we could simply try for
each n = 1, 2, . . . , 512 to find a prime p such that n < p ≤ 2n. The method in the
following lemma is a bit shorter.
Lemma 2.17. For all n ≤ 630 there is a prime p satisfying n < p ≤ 2n.
p+1 Proof. If a prime p is given, then n < p ≤ 2n for n = hp+1
, i 2 +1, . . . , p−
2
pk+1 +1
for each k will
1. Hence a list p1 , . . . , pr of primes where p1 = 2 and pk ≥
2
verify the statement for n = 1, 2, . . . , pr − 1. Such a list is
2, 3, 5, 7, 13, 23, 43, 83, 163, 317, 631
which verifies the statement for all n up to n = 630.
We have shown that the statement in the theorem holds for all n > 512 and
for all n ≤ 630, thus it holds for all n ∈ N. This completes the proof of Theorem
2.11.
CHAPTER 3
The prime number theorem
In the previous chapter we saw that the functions π(x) and x/ log(x) have
the same order of magnitude. A stronger result is true: π(x) and x/ log(x) are
asymptotically equal. This is the statement of the prime number theorem which
we discuss in this chapter. We also consider the approximation to π(x) given by
the logarithmic integral li(x). The Riemann hypothesis can be formulated as a
conjecture about the error term |π(x) − li(x)| in this approximation.
1. Statement of the prime number theorem
Definition 3.1. Let f and g be functions which are defined for all sufficiently
large real numbers, and assume that f (x) and g(x) are positive for all large x. We
say that f and g are asymptotically equal as x → ∞, in symbols
f ∼g
as
x → ∞,
if f (x)/g(x) → 1 for x → ∞, i.e. the limit limx→∞ f (x)/g(x) exists and is equal to
1. (Some authors use the expression asymptotically equivalent instead of asymptotically equal).
Example 3.2.
(1) x + 1 ∼ x as x → ∞.
(2) x2 +x ∼ x2 as x → ∞. Note that in this example for x → ∞ the difference
(x2 + x) − (x2 ) tends to infinity even though the quotient (x2 + x)/(x2 )
tends to 1.
Lemma 3.3. If f ∼ g as x → ∞ then f g as x → ∞.
Proof. f ∼ g as x → ∞ means that limx→∞ f (x)/g(x) = 1. Hence for all
sufficiently large x we have 21 < f (x)/g(x) < 23 and thus 12 g(x) < f (x) < 23 g(x).
This shows that g f as x → ∞ and therefore (since the order of magnitude
relation is symmetric) f g as x → ∞.
Recall that for a real number x we defined π(x) by π(x) = #{p ≤ x : p prime}.
Theorem 3.4 (Prime number theorem).
x
π(x) ∼
as
log x
x → ∞.
The prime number theorem was proved in 1896 independently by Hadamard
and by de la Vallée Poussin. Both proofs use methods from complex analysis
to deduce the prime number theorem from certain properties of the Riemann zeta
function. In 1948 Selberg and Erdös gave a new proof of the prime number theorem
which didn’t require any complex analysis. This new proof is now known as the
elementary proof of the prime number theorem.
19
20
3. THE PRIME NUMBER THEOREM
Remark 3.5. It follows from Lemma 3.3 that the prime number theorem implies Chebyshev’s theorem π(x) x/ log(x) as x → ∞ (Theorem 2.9).
We now deduce two corollaries from the prime number theorem.
Corollary 3.6. Let pn denote the n-th prime number. Then
pn ∼ n log n
as n → ∞.
Here pn ∼ n log n as n → ∞ means limn→∞
3.1 for the variable n ∈ N.
pn
n log n
= 1, i.e. we use Definition
Proof of Corollary 3.6. By the prime number theorem we have
π(x) log x
(4)
lim
= 1.
x→∞
x
Taking the logarithm gives limx→∞
log π(x) + log(log x)− log x = 0, and dividing
this by log x shows that limx→∞
(5)
lim
x→∞
log π(x)
log x
+
log(log x)
log x
− 1 = 0. Therefore
log π(x)
log(log x)
= 1 − lim
= 1,
x→∞
log x
log x
log(log x)
log x
= 0 by Lemma 2.10. Using (4) and (5) we obtain
π(x) log π(x)
π(x) log x log π(x)
lim
= lim
·
= 1.
x→∞
x→∞
x
x
log x
because limx→∞
Now take x = pn . Then π(x) = n, so
lim
n→∞
n log n
= 1.
pn
By Bertrand’s postulate (Theorem 2.11) the interval (x, 2x] contains a prime
number for every integer x ≥ 1. This is not true if we replace the factor 2 by
a smaller number (e.g. (1, δ · 1] doesn’t contain a prime if δ < 2). However if we
restrict to large enough x then the statement remains true as the following corollary
shows.
Corollary 3.7. Let δ > 1. Then for all large enough x, the interval (x, δx]
contains a prime number.
Proof. The interval (x, δx] contains a prime number if and only if π(δx) −
π(x) ≥ 1. Thus we must show that π(δx) − π(x) ≥ 1 for all sufficiently large x.
We have
π(δx)
x/ log(x) δx/ log(δx)
π(δx)
=
·
·
.
π(x)
δx/ log(δx)
π(x)
x/ log(x)
By the prime number theorem limx→∞ δx/π(δx)
log(δx) = 1 and limx→∞
For the third factor we find
δx/ log(δx)
log(x)
=δ·
→δ
as x → ∞.
x/ log(x)
log(δ) + log(x)
Hence limx→∞
π(δx)
π(x)
x/ log(x)
π(x)
= 1.
= δ.
Now fix any γ with 1 < γ < δ. From limx→∞ π(δx)
π(x) = δ it follows that π(δx) ≥
γπ(x) for all sufficiently large x, and from π(x) → ∞ as x → ∞ it follows that
(γ − 1)π(x) ≥ 1 for all sufficiently large x. Thus for all large enough x we obtain
π(δx) − π(x) ≥ (γ − 1)π(x) ≥ 1 as required.
2. ADDENDUM: ERROR TERMS AND RIEMANN HYPOTHESIS
21
2. Addendum: Error terms and Riemann hypothesis
Note: The material in this section is not examinable.
The prime number theorem states that π(x) ∼ logx x as x → ∞. We can therefore consider logx x as a reasonable approximation to π(x). A different approximation
is given by the function li(x) which is defined as follows.
Definition 3.8. For x ≥ 2 define
Z
li(x) =
2
x
1
dt.
log t
The function li(x) is called the logarithmic integral. (Some books use the notation
Li(x) for this function.)
One can show that li(x) ∼
theorem is equivalent to
x
log x
as x → ∞ and that therefore the prime number
π(x) ∼ li(x)
as
x → ∞.
So li(x) can also be considered as an approximation to π(x). The following figure
shows π(x) and li(x) for x ≤ 400.
To measure the quality of the two approximations to π(x) we must estimate
the error terms |π(x) − logx x | and |π(x) − li(x)|.
Example 3.9. The following table gives some values of the functions π(x),
x/ log(x) and li(x). The values of x/ log(x) and li(x) are rounded to the nearest
22
3. THE PRIME NUMBER THEOREM
integer.
x
π(x)
x/ log(x)
li(x)
103
168
145
177
106
78498
72382
78627
109
50847534
48254942
50849233
From Example 3.9 one gets the impression that li(x) is a better approximation
to π(x) than logx x . This is indeed the case: one can show that for large x the
difference |π(x) − li(x)| is smaller than the difference |π(x) − logx x |.
It is a very difficult problem to find good upper bounds for |π(x) − li(x)|. The
Riemann hypothesis is the following conjecture about the existence of such upper
bounds.
Conjecture 3.10 (Riemann hypothesis). Let ε > 0. Then
|π(x) − li(x)| < x1/2+ε
for all sufficiently large x.
Remark 3.11.
(1) Usually the Riemann hypothesis is formulated as a
conjecture about the zeros of the Riemann zeta function. The Riemann
zeta function, denoted by ζ(s), is a meromorphic function of a complex
variable s, and the Riemann hypothesis states that if ζ(s) = 0 and 0 <
Re(s) < 1, then Re(s) = 21 . In this form the Riemann hypothesis was
conjectured by Riemann in 1859. One can show that this formulation is
equivalent to the formulation given in Conjecture 3.10.
(2) There is a prize of $1 million for the solution of the Riemann hypothesis.
See http://www.claymath.org/millennium/ for details.
CHAPTER 4
Arithmetic functions and Dirichlet series
An arithmetic function is a function from the positive integers to the complex
numbers. To such a function we can associate a Dirichlet series, which is a differentiable function of a real (or complex) variable. The most important example of
a Dirichlet series is the Riemann zeta function which we consider first.
1. The Riemann zeta function
Definition 4.1. The Riemann zeta function, denoted by ζ(s), is the function
of a real variable s > 1 defined by the series
∞
X
1
.
ζ(s) =
ns
n=1
We must check that the series converges for s > 1. Since all summands in the
PN
series are positive, it suffices to show that the partial sums n=1 n−s are bounded
above. These partial sums can be estimated as follows.
N
X
n−s = 1 +
N
X
n−s
n=2
N
n=1
Z
<1+
x−s dx < 1 +
1
PN
−s
Z
1
RN
∞
x−s dx = 1 +
1
.
s−1
−s
Here the inequality
< 1 x dx can be seen by comparing the area
n=2 n
under the curve x−s for 1 ≤ x ≤ N to the sum of the areas of the rectangles of
width 1 and height 2−s , 3−s , . . . , N −s under this curve, as illustrated in the following
picture (in the picture s = 1.5, but for any s > 1 one obtains a similar picture).
Remark 4.2.
ferentiable.
(1) One can show that the Riemann zeta function is dif23
24
4. ARITHMETIC FUNCTIONS AND DIRICHLET SERIES
(2) Riemann (in 1859) studied the function s 7→ ζ(s) as a function of a complex variable s (the series in Definition 4.1 converges for all complex numbers s with real part greater than 1). To consider complex s is important
for many questions in analytic number theory (including the prime number theorem), but we will not discuss it in this course.
The following theorem shows that the Riemann zeta function is closely related
to prime numbers.
Theorem 4.3 (Euler product). For all s > 1,
Y
1
.
ζ(s) =
1 − p−s
p
Proof. Recall the following formula for the geometric series (valid for |x| < 1)
1
= 1 + x + x2 + . . . .
1−x
In this formula we let x = p−s (which is valid because |p−s | < 1 for s > 1) and
obtain
1
= 1 + p−s + p−2s + . . . .
1 − p−s
If we take p = 2, 3, 5, . . . , M (where M is a prime) and multiply the series together,
we obtain
∞
∞
∞ X
X
Y
X
−s
1
·
·
·
2a2 3a3 · · · M aM
.
=
−s
1−p
a =0
a =0 a =0
p≤M
2
3
M
From the fundamental theorem of arithmetic (Theorem 1.13) we know that a number n ∈ N has a representation of the form 2a2 3a3 · · · M aM if and only if n doesn’t
have any prime factors greater than M . Moreover every such number n has only
one such representation. Hence
Y
X
1
=
n−s ,
1 − p−s
p≤M
n∈N (M )
where N (M ) = {n ∈ N : if p | n then p ≤ M }. The set N (M ) includes all numbers
up to M , so that
∞
∞
X
X
X
0≤
n−s −
n−s ≤
n−s ,
n=1
n∈N (M )
n=M +1
and the last sum tends to 0 as M → ∞. Hence
∞
X
X
Y
n−s = lim
n−s = lim
n=1
M →∞
n∈N (M )
M →∞
p≤M
1
,
1 − p−s
as required.
Remark 4.4. The essential step in the proof was the existence of a unique
prime factorisation for every positive integer. Therefore the Euler product can
be considered as an analytic expression of the fundamental theorem of arithmetic
(Theorem 1.13).
Proposition 4.5. ζ(s) → ∞ as s → 1+.
Here s → 1+ means that s tends to 1 from the right, i.e. s → 1 and s > 1.
1. THE RIEMANN ZETA FUNCTION
25
R∞
Proof of Proposition 4.5. For every s > 1 the integral 1 x−s dx is smaller
P∞
the sum n=1 n−s (this can be seen from the following picture: the integral
Rthan
P∞
∞ −s
x dx is the area under the curve x−s for 1 ≤ x < ∞ and the sum n=1 n−s
1
is the sum of the areas of the rectangles).
Thus for every s > 1 we have
ζ(s) =
∞
X
n
−s
n=1
Since
1
s−1
Z
>
∞
x−s dx =
1
1
.
s−1
→ ∞ as s → 1+, it follows that ζ(s) → ∞ as s → 1+.
Remark 4.6. We can use the Euler product for the Riemann zeta function and
Proposition 4.5 to give a new proof of the fact that there are infinitely many prime
numbers. SupposeQfor a contradiction that there are only finitely many primes.
Then the product p (1 − p−s )−1 has only finitely many factors. It follows that
lim ζ(s) = lim
s→1+
s→1+
Y
Y
Y
(1 − p−s )−1 =
lim (1 − p−s )−1 =
(1 − p−1 )−1 ,
p
p
s→1+
p
contradicting ζ(s) → ∞ as s → 1+.
The following result was shown by Euler in 1737.
P
Corollary 4.7. The series p p1 is divergent.
P
Proof. For every s > 1 the series p p−s is convergent because it is a subseries
P∞
P
of n=1 n−s . We first show that p p−s → ∞ as s → 1+. Note that
log
1
x2
x3
=x+
+
+ ...
1−x
2
3
for |x| < 1 and so
log
1
p−2s
p−3s
−s
=
p
+
+
+ ....
1 − p−s
2
3
26
4. ARITHMETIC FUNCTIONS AND DIRICHLET SERIES
Therefore taking the logarithm of the Euler product gives
!
Y
1
log ζ(s) = log
1 − p−s
p
X
1
=
log
1 − p−s
p
X
p−2s
p−3s
=
p−s +
+
+ ...
2
3
p
=
X
p−s +
p
X X p−ks
p k≥2
k
.
The second summand is bounded by 1 because
X X p−ks
XX
X
X
X p−2
<
p−k =
p−2
p−k =
k
1 − p−1
p
p
p
p
k≥2
k≥2
<
∞
X
k≥0
−2
∞ X
n
1
1
=
−
= 1.
1 − n−1
n−1 n
n=2
n=2
P −s
Hence
> log ζ(s) − 1 for all s > 1. Since ζ(s) → ∞ as s → 1+ (by
pp
P
Proposition 4.5), this implies that p p−s → ∞ as s → 1+.
P
P
P
If p p−1 was convergent then p p−s < p p−1 for all s > 1, which is imP
possible because the left hand side tends to infinity as s → 1+. Hence p p−1 is
divergent.
2. Arithmetic functions
Definition 4.8. An arithmetic function is a function from the set of positive
integers to the complex numbers. (Some authors call this an arithmetical function.)
So if f is an arithmetic function then for every n ∈ N we have a number
f (n) ∈ C. The idea is that for an arithmetic function f the value f (n) contains
some interesting arithmetical information about n. In all our examples the values of
f will be real numbers (but arithmetic functions with complex values are important
in more advanced applications).
P
In the following we will consider only positive divisors, e.g. in d|n the sum is
over all positive divisors d of n.
Example 4.9. The following functions are arithmetic functions.
(1) u(n) = 1 for all n ∈ N.
(2) d(n) = #{d
P ∈ N : d | n}, so d(n) is the number of positive divisors of n.
(3) σ(n) = d|n d, so σ(n) is the sum of all positive divisors of n.
Definition 4.10. An arithmetic function f is called multiplicative if f (mn) =
f (m)f (n) whenever (m, n) = 1.
Note that for a multiplicative function f the equality f (mn) = f (m)f (n) holds
for all coprime m, n ∈ N; if m, n are not coprime then f (mn) and f (m)f (n) may
or may not be equal.
3. DIRICHLET SERIES
27
If f is a multiplicative function then f (1) = f (1 · 1) = f (1)f (1) and therefore
either f (1) = 0 (in which case f (n) = f (1 · n) = f (1)f (n) = 0 for all n ∈ N)
or f (1) = 1. A multiplicative function f is uniquely determined by its values on
powers of primes:
if n ∈ N then by the fundamental theorem of arithmetic
we can
Q
Q
write n = p|n pkp , and since f is multiplicative we have f (n) = p|n f (pkp ).
Example 4.11.
(1) The arithmetic function n 7→ u(n) = 1 is multiplicative, because if (m, n) = 1 then u(mn) = 1 = u(m)u(n) (in fact this is
true for all m, n, the condition (m, n) = 1 is not needed in this example).
(2) The arithmetic function n 7→ log n is not multiplicative. For example
log(2 · 3) 6= log(2) · log(3).
Lemma 4.12. Let f : N →PC be an arithmetic function. Define an arithmetic
function g : N → C by g(n) = d|n f (d). If f is multiplicative then g is multiplicative.
Proof. Let m, n ∈ N with (m, n) = 1. We must show that g(mn) = g(m)g(n).
Now d | mn if and only if d = d1 d2 with d1 | m and d2 | n, and in this case d1 and
d2 are uniquely determined (see Lemma 1.15). Hence
X
X
X
X
g(mn) =
f (d) =
f (d1 d2 ) =
f (d1 ) ·
f (d2 ) = g(m)g(n),
d|mn
d1 |m,d2 |n
d1 |m
d2 |n
as required.
P
P
Example 4.13. We have d(n) = d|n 1 = d|n u(d) where d(n) and u(n) are
defined in Example 4.9. Furthermore u is multiplicative by Example 4.11. Lemma
4.12 implies that d is multiplicative. Note that for the function d we really need the
condition (m, n) = 1 to ensure that d(mn) = d(m)d(n). For example if m = n = 2
then d(2 · 2) = 3 but d(2) · d(2) = 4.
3. Dirichlet series
Definition 4.14. A Dirichlet series is a series of the form
∞
X
a(n)
F (s) =
ns
n=1
where a(n) ∈ C.
P∞
We say that the Dirichlet series F (s) = n=1 a(n)
ns is associated to the arithmetic function a(n). The Dirichlet series associated to an arithmetic function a(n)
allows us to study a(n) with methods from analysis.
P∞
Example 4.15. The Riemann zeta function ζ(s) = n=1 n1s (see §1) is the
Dirichlet series associated to the arithmetic function u(n) = 1 for all n ∈ N.
In a Dirichlet series, s is a real variable (or more generally a complex variable).
Of course, for F (s) to be defined one must check that the series converges. And
to study Dirichlet series in more detail one also needs to know for which s the
series is absolutely convergent, whether it is uniformly convergent, etc. We showed
convergence in the case of the Riemann zeta function in §1 (for s > 1), but from
now on we will ignore all convergence questions.
We can perform various operations on Dirichlet series, e.g. differentiate or multiply them. The result will again be a Dirichlet series as the following two lemmas
show.
28
4. ARITHMETIC FUNCTIONS AND DIRICHLET SERIES
P∞
P∞ b(n)
Lemma 4.16. Let F (s) = n=1 a(n)
n=1 ns be two Dirichlet
ns and G(s) =
series. Then their product is given by a Dirichlet series, more precisely
F (s)G(s) =
∞
X
c(n)
ns
n=1
with
c(n) =
X
a(d)b(n/d).
d|n
Proof. The product is
F (s)G(s) =
∞ X
∞
X
a(d)b(i)
ds is
d=1 i=1
=
∞ X
X
a(d)b(n/d)
ns
n=1 d|n
=
∞
X
n=1
P
d|n
a(d)b(n/d)
ns
where we set n = di. (Analytic remark: To justify the rearrangement of the infinite
sums requires some convergence arguments.)
P∞
Lemma 4.17. Let F (s) =
n=1
P∞ b(n)
with
b(n)
=
−
log(n)
·
a(n).
n=1 ns
Proof.
d
−s
)
ds (n
F 0 (s) =
d
ds
a(n)
ns
be a Dirichlet series. Then F 0 (s) =
= − log(n) · n−s and therefore
X
∞
a(n)
ns
n=1
=
(Analytic remark: We used that
vergence arguments.)
d
ds
X
∞
∞
X
d a(n)
− log(n) · a(n)
.
=
s
ds
n
ns
n=1
n=1
P
=
P
d
ds ;
to justify this requires some con
The following examples show that many Dirichlet series associated to interesting arithmetic functions can be expressed in terms of the Riemann zeta function.
Example 4.18.
(1) The
Dirichlet series associated to the identity funcP∞
tion N → N, n 7→ n, is n=1 nns . We have
∞
∞
X
X
n
1
=
= ζ(s − 1).
s
s−1
n
n
n=1
n=1
P∞
a
More generally, for any a ∈ R we have n=1 nns = ζ(s − a).
P
∞
(2) ζ(s)2 = n=1 d(n)
ns where d(n) is as in Example 4.9. This holds because
P
∞
∞
∞
∞
X
X
X
d(n)
1 X 1
d|n 1 · 1
ζ(s) · ζ(s) =
·
=
=
.
s
s
s
n
n
n
ns
n=1
n=1
n=1
n=1
(3) ζ(s − 1)ζ(s) =
because
P∞
n=1
σ(n)
ns
where σ(n) is as in Example 4.9. This holds
∞
∞
∞
X
X
n X 1
ζ(s − 1)ζ(s) =
·
=
ns n=1 ns
n=1
n=1
(4) ζ 0 (s) = −
P∞
n=1
log(n)
ns
P
d|n d
ns
·1
=
∞
X
σ(n)
.
ns
n=1
4. ADDENDUM: THE MÖBIUS FUNCTION
29
4. Addendum: The Möbius function
Note: The material in this section is not examinable.
Definition 4.19. The Möbius function is the arithmetic function µ defined by


if n = 1,
1
µ(n) = 0
if p2 | n for some prime p,


(−1)k if n = p1 p2 · · · pk where the pi are distinct primes.
P
Lemma 4.20. The arithmetic function n 7→ d|n µ(d) is given by
(
X
1 if n = 1,
µ(d) =
0 if n > 1.
d|n
Proof. It is easy to check
P that µ is multiplicative. From Lemma 4.12 it
follows that the function n 7→ d|n µ(d) is also multiplicative, and it is therefore
determined by its values on prime powers. For n = pk we find
(
X
1
if k = 0,
µ(d) =
1 − 1 + 0 · · · + 0 = 0 if k > 0.
k
d|p
This shows the result.
P∞
Lemma 4.21.
n=1
µ(n)
ns
−1
= ζ(s)
.
P∞
Proof. We compute the product of the two Dirichlet series n=1
P∞ 1
ζ(s) = n=1 ns . Using Lemmas 4.16 and 4.20 we find
P
∞
∞
∞
∞
X
X
X
µ(n)
µ(n) X 1
d|n µ(d)
·
ζ(s)
=
·
=
= 1.
ns
ns n=1 ns
ns
n=1
n=1
n=1
P∞
−1
Hence n=1 µ(n)
.
ns = ζ(s)
µ(n)
ns
and
Theorem 4.22 (Möbius inversion formula). For arithmetic functions f and g,
the following statements are equivalent:
P
(1) g(n) = d|n f (d) for all n ∈ N,
P
(2) f (n) = d|n µ(d)g(n/d) for all n ∈ N.
Proof. Assume that (1) holds. Then
X
X
X
X
X
µ(d)g(n/d) =
µ(d)
f (c) =
f (c)
µ(d) = f (n)
d|n
d|n
c| n
d
c|n
d| n
c
where for the last equality we used Lemma 4.20. Furthermore we used that a pair
(c, d) of positive integers satisfies d | n and c | nd if and only if it satisfies c | n and
d | nc .
Assume that (2) holds. Then
X
XX
XX
XX
f (d) =
µ(c)g(d/c) =
µ(d/c)g(c) =
µ(d0 )g(c) = g(n)
d|n
d|n c|d
d|n c|d
c|n d0 | n
c
where we used that a pair (c, d) of positive integers satisfies d | n and c | d if and
only if the pair (c, d0 ) = (c, d/c) satisfies c | n and d0 | nc .
CHAPTER 5
Primes and arithmetic progressions
In this chapter we discuss (without proofs) two deep results relating prime
numbers and arithmetic progressions. The first result is Dirichlet’s theorem on
primes in arithmetic progressions which states that an arithmetic progression a, a +
b, a + 2b, a + 3b, . . . contains infinitely many prime numbers if a and b are coprime.
The second result is a recent theorem by Green and Tao showing that there are
arbitrarily long (but finite) arithmetic progressions a, a + b, a + 2b, . . . , a + (k − 1)b
consisting solely of primes.
1. Primes in arithmetic progressions
Let a and b be positive integers. Does the arithmetic progression a, a + b, a +
2b, a + 3b, . . . contain infinitely many prime numbers, i.e. are there infinitely many
prime numbers which are of the form a + nb for some integer n ≥ 0?
Example 5.1. We fix b = 4 and look what happens for different values of a.
a = 1: The arithmetic progression 1 + n · 4 is 1, 5, 9, 13, 17, 21, 25, 29, 33, 37, . . .
which contains many prime numbers.
a = 2: The arithmetic progression 2 + n · 4 is 2, 6, 10, 14, . . . which contains only
the prime number 2. All numbers 2 + n · 4 with n ≥ 1 are composite
because 2 | 2 + n · 4 and 1 < 2 < 2 + n · 4 for n ≥ 1.
a = 3: The arithmetic progression 3+n·4 is 3, 7, 11, 15, 19, 23, 27, 31, 35, . . . which
contains many prime numbers.
a = 4: The arithmetic progression 4+n·4 is 4, 8, 12, 16, . . . which doesn’t contain
any prime numbers, because 2 | 4 + n · 4 and 1 < 2 < 4 + n · 4 for n ≥ 0.
We observe that every arithmetic progression in Example 5.1 either contains
many primes or it contains at most one prime. This observation is explained by
the following two results.
Lemma 5.2. Let a and b be positive integers. If a and b are not coprime then the
arithmetic progression a + nb, n = 0, 1, 2, . . . , contains at most one prime number.
Proof. Let d = (a, b). By assumption a and b are not coprime and therefore
d > 1. If n > 0 then a + nb is composite because d | a + nb and 1 < d < a + nb.
Hence the arithmetic progression a + nb contains at most one prime, more precisely
if a + 0 · b is composite then it contains no prime, and if a + 0 · b is prime then it
contains precisely one prime.
Theorem 5.3. Let a and b be positive integers. If a and b are coprime then
the arithmetic progression a + nb, n = 0, 1, 2, . . . , contains infinitely many prime
numbers.
31
32
5. PRIMES AND ARITHMETIC PROGRESSIONS
Theorem 5.3 is known as ‘Dirichlet’s theorem on primes in arithmetic progressions’. It was proved by Dirichlet in 1837 using analytic methods (more precisely
a certain kind of Dirichlet series), but for certain small values of a and b one can
also give elementary proofs.
Example 5.4.
(1) There exist infinitely many prime numbers which are
of the form 3 + 8n for some integer n ≥ 0. Indeed since (3, 8) = 1 this
follows directly from Theorem 5.3.
(2) Consider the arithmetic progression 26 + 39n, n = 0, 1, 2, . . . . Since
(26, 39) = 13 > 1 we can apply Lemma 5.2 which shows that there exists at most one prime in this arithmetic progression. In fact one easily
sees that there is no prime in this arithmetic progression: for all n ≥ 0,
26 + 39n is composite because 13 | 26 + 39n and 1 < 13 < 26 + 39n.
(3) There exist infinitely many prime numbers whose decimal expansion ends
with the digit 9. To see this we first observe that the decimal expansion
of a number p ends with the digit 9 if and only if p is of the form 9 + 10n
for some integer n ≥ 0. By Theorem 5.3 there are infinitely many primes
of the form 9 + 10n (note that (9, 10) = 1 and that therefore Theorem 5.3
can be applied), hence there are infinitely many primes ending with the
digit 9.
2. Addendum: Arithmetic progressions in the sequence of primes
Note: The material in this section is not examinable.
In the previous section we considered an arithmetic progression a, a + b, a +
2b, a + 3b, . . . and asked how many prime numbers it contains. Now we consider the
sequence of all prime numbers 2, 3, 5, 7, 11, . . . and look for arithmetic progressions
contained in it. In other words, we look for arithmetic progressions consisting solely
of primes.
We first note that there is no infinite arithmetic progression a, a + b, a + 2b, . . .
(with a, b ∈ N) which consists solely of prime numbers. Indeed, if a is a prime
number then a + ab is composite (because a | a + ab and 1 < a < a + ab).
Next we consider finite arithmetic progressions. A sequence of the form a, a +
b, a + 2b, . . . , a + (k − 1)b with a, b ∈ N is called an arithmetic progression of length
k. What can we say about arithmetic progressions of length k consisting solely of
prime numbers?
Example 5.5.
(1) The sequence 5, 11, 17, 23, 29 (that is 5 + i · 6 for 0 ≤
i ≤ 4) is an arithmetic progression of length 5 consisting solely of prime
numbers.
(2) The sequence 199, 409, 619, 829, 1039, 1249, 1459, 1669, 1879, 2089 (that is
199 + i · 210 for 0 ≤ i ≤ 9) is an arithmetic progression of length 10
consisting solely of prime numbers.
If all the numbers a, a + b, a + 2b, . . . , a + (k − 1)b are prime, then both a and
b must satisfy certain lower bounds depending on k. More precisely we have the
following two results.
Lemma 5.6. If a, a + b, a + 2b, . . . , a + (k − 1)b is an arithmetic progression of
length k consisting solely of prime numbers then a ≥ k.
2. ADDENDUM: ARITHMETIC PROGRESSIONS IN THE SEQUENCE OF PRIMES
33
Proof. The number a+ab is composite (it’s divisible by a, and 1 < a < a+ab).
Therefore a + ab can’t be in the given arithmetic progression, i.e. a ≥ k.
Proposition 5.7. If a, a+b, a+2b, . . . , a+(k −1)b is an arithmetic progression
of length k consisting solely of prime numbers then b is divisible by every prime
p < k.
Proof. Let p < k be a prime number. The first p terms of the progression are
a, a + b, a + 2b, . . . , a + (p − 1)b.
We consider the remainder of these p numbers when divided by p. Assume for a
contradiction that all these p remainders are distinct. Then all possible remainders
0, 1, . . . , p − 1 must occur. So in particular there is a t satisfying 0 ≤ t ≤ p − 1 such
that a + tb has remainder 0, that is p | a + tb. But a + tb ≥ a ≥ k > p (the inequality
a ≥ k is Lemma 5.6), hence a + tb is composite contradicting our assumption that
all numbers in the given arithmetic progression are prime.
We have shown that two of the remainders must be equal. In other words, there
exist integers i, j with 0 ≤ i < j ≤ p − 1 such that a + ib and a + jb have the same
remainder when divided by p. It follows that the difference (a+jb)−(a+ib) = (j−i)b
is divisible by p, and therefore that p divides j −i or b. However p | j −i is impossible
(because 0 < j − i < p) and thus we must have p | b.
Example 5.8.
(1) We consider arithmetic progressions of length k = 5.
Then we must have a ≥ 5 (by Lemma 5.6) and 6 | b (because 2 | b and
3 | b by Proposition 5.7). So the arithmetic progression of length 5 in
Example 5.5.(1) is the smallest possible example.
(2) If k = 10 then a ≥ 10 and 210 = 2 · 3 · 5 · 7 | b. Hence the arithmetic
progression of length 10 in Example 5.5.(2) has the smallest possible b.
Lemma 5.6 and Proposition 5.7 show that if a, a + b, a + 2b, . . . , a + (k − 1)b
is an arithmetic progression of length k consisting solely of primes then a and b
satisfy certain properties. However these results do not imply that such arithmetic
progressions exist for every k, and until recently it had been an open question if
the prime numbers contain arbitrarily long arithmetic progressions. The problem
was solved in 2004 when Green and Tao proved the following theorem.
Theorem 5.9. For every k ≥ 1, the prime numbers contain infinitely many
arithmetic progressions of length k.
Appendices
Notation
Notation
N
Z
R
C
#S
(a, b]
Explanation
set of all positive integers
set of all integers
set of all real numbers
set of all complex numbers
number of elements of a set S
interval {x : a < x ≤ b}
Page
3
3
3
3
b|a
b-a
(a, b)
N !
k
P
Qp
b divides a
b doesn’t divide a
greatest common divisor of a and b
N factorial
binomial coefficient
sum indexed by all primes p
product indexed by all primes p ≤ N
3
3
4
6
7
6
6
[x]
π(x)
θ(x)
ζ(s)
s → 1+
log(x)
li(x)
greatest integer ≤ x
number of primes ≤ x
Chebyshev’s theta function
Riemann zeta function
s tends to 1 from the right
natural logarithm
logarithmic integral
6
12
14
23
24
13
21
n
p≤N
as x → ∞ same order of magnitude as x → ∞
∼ as x → ∞ asymptotically equal as x → ∞
14
19
u(n)
d(n)
σ(n)
µ(n)
26
26
26
29
constant function 1
number of positive divisors of n
sum of positive divisors of n
Möbius function
35
36
APPENDICES
Further reading
The topics in this course belong to the areas of elementary number theory
and analytic number theory. There exist many introductory textbooks which cover
some or most of the material, for example the books in the following list. All of
these books also discuss various topics which were not mentioned in this course.
Books on analytic number theory assume some knowledge of complex analysis.
[1] T. M. Apostol, Introduction to analytic number theory, Springer-Verlag,
1976. (Covers all relevant elementary and analytic number theory.)
[2] D. M. Burton, Elementary number theory, McGraw-Hill Education, 6th
ed., 2005. (A very detailed introduction to elementary number theory.)
[3] G. H. Hardy, E. M. Wright, An introduction to the theory of numbers,
Oxford University Press, 6th ed., 2008. (A classical introduction to all
areas of number theory.)
[4] A. E. Ingham, The distribution of prime numbers, Cambridge University
Press, 1934, reprinted 1990. (A classical, very analytic book about the
prime number theorem, error estimates, etc.)
[5] G. J. O. Jameson, The prime number theorem, Cambridge University
Press, 2003. (Everything about the prime number theorem (including
analytic and elementary proofs) and related material.)
A proof of the prime number theorem (Theorem 3.4) is contained in [1], [3],
[4] and [5]. For more information on the material in Chapter 3, §2 see [4] and [5].
Dirichlet’s theorem on primes in arithmetic progressions (Theorem 5.3) is proved
in [1] and [5]. The two most recent results mentioned in this course are not yet
contained in textbooks. For the AKS primality test (mentioned in Remark 1.20)
see [6], and for Green and Tao’s theorem (Theorem 5.9) see [7]. (Warning: [6] and
[7] are research papers, they are much more difficult than the textbooks [1]–[5].)
[6] M. Agrawal, N. Kayal, N. Saxena, PRIMES is in P, Annals of Mathematics 160 (2004), 781–793.
[7] B. Green, T. Tao, The primes contain arbitrarily long arithmetic progressions, Annals of Mathematics 167 (2008), 481–547.
APPENDICES
37
The first 400 prime numbers
2
31
73
127
179
233
283
353
419
467
3
37
79
131
181
239
293
359
421
479
5
41
83
137
191
241
307
367
431
487
7
43
89
139
193
251
311
373
433
491
11
47
97
149
197
257
313
379
439
499
13
53
101
151
199
263
317
383
443
503
17
59
103
157
211
269
331
389
449
509
19
61
107
163
223
271
337
397
457
521
23
67
109
167
227
277
347
401
461
523
29
71
113
173
229
281
349
409
463
541
547
557
563
569 571 577 587 593 599 601
607
613
617
619 631 641 643 647 653 659
661
673
677
683 691 701 709 719 727 733
739
743
751
757 761 769 773 787 797 809
811
821
823
827 829 839 853 857 859 863
877
881
883
887 907 911 919 929 937 941
947
953
967
971 977 983 991 997 1009 1013
1019 1021 1031 1033 1039 1049 1051 1061 1063 1069
1087 1091 1093 1097 1103 1109 1117 1123 1129 1151
1153 1163 1171 1181 1187 1193 1201 1213 1217 1223
1229
1297
1381
1453
1523
1597
1663
1741
1823
1901
1231
1301
1399
1459
1531
1601
1667
1747
1831
1907
1237
1303
1409
1471
1543
1607
1669
1753
1847
1913
1249
1307
1423
1481
1549
1609
1693
1759
1861
1931
1259
1319
1427
1483
1553
1613
1697
1777
1867
1933
1277
1321
1429
1487
1559
1619
1699
1783
1871
1949
1279
1327
1433
1489
1567
1621
1709
1787
1873
1951
1283
1361
1439
1493
1571
1627
1721
1789
1877
1973
1289
1367
1447
1499
1579
1637
1723
1801
1879
1979
1291
1373
1451
1511
1583
1657
1733
1811
1889
1987
1993
2063
2131
2221
2293
2371
2437
2539
2621
2689
1997
2069
2137
2237
2297
2377
2441
2543
2633
2693
1999
2081
2141
2239
2309
2381
2447
2549
2647
2699
2003
2083
2143
2243
2311
2383
2459
2551
2657
2707
2011
2087
2153
2251
2333
2389
2467
2557
2659
2711
2017
2089
2161
2267
2339
2393
2473
2579
2663
2713
2027
2099
2179
2269
2341
2399
2477
2591
2671
2719
2029
2111
2203
2273
2347
2411
2503
2593
2677
2729
2039
2113
2207
2281
2351
2417
2521
2609
2683
2731
2053
2129
2213
2287
2357
2423
2531
2617
2687
2741