Twin Prime Numbers
Bounded gaps between primes
Author: I.F.M.M. Nelen
Supervisor: Prof. Dr. F. Beukers
28 januari 2015
Table of Contents
Twin Prime Numbers
Abstract . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1
2
Introduction
Prime Numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Twin Prime Numbers . . . . . . . . . . . . . . . . . . . . . . . . .
Heuristic Approach To Twin Prime Numbers . . . . . . . . . . . .
2
2
2
3
Bounding The Gaps
Admissible Tuples and Sets . . . . . . . . . . . . . . . . . . . . . .
4
5
Outline of Maynard’s Proof
6
Estimating The Sums
Technical Lemmas . . . .
Changing the Variables .
Rewriting the second sum
Relating the Variables . .
Choosing Suitable y . . .
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9
10
12
18
21
22
Comparing the sums
29
Obtaining a lower bound of Mk for small k . . . . . . . . . . . . . 31
Length of Bounded Gaps . . . . . . . . . . . . . . . . . . . . . . . 35
1
Twin Prime Numbers
Abstract
In this thesis an in-depth explanation will be given of the proof Maynard
gave in his article. In which the steps of the proofs will be expanded. He
used a refinement on the GPY sieve to study k-tuples and small gaps between primes. This will show that lim inf(pn+1 − pn ) ≤ 600, and, by assuming the Elliott-Halberstam conjecture, that lim inf(pn+1 − pn ) ≤ 12 and
lim inf(pn+2 − pn ) ≤ 600.
Introduction
Prime Numbers
For thousands of years people have been fascinated by numbers. First the
positive integers where used to describe quantities. Then these numbers
were extended by the negative integers and fractions. But the fascination
with positive integers or natural numbers stayed. The building blocks of
natural numbers are prime numbers, every natural number greater than
two is a prime number or a unique product of prime numbers. The greeks
started with examining prime numbers and found there are infinitely many.
The proof of this theorem is based on the divisibility of numbers p and
p + 1. Because there are infinitely many prime numbers one may wonder
about the distribution of those prime numbers. For x > 0 let π(x) denote the
number of prime numbers not exceeding x. Because there are infinitely many
prime numbers π(x) → ∞ as x → ∞. Gauss (1792) and Legendre(1798)
proposed the distribution of prime numbers around a number x is asymptotic
to x/ log x. This means heuristically there is a 1 in log x chance of a number
close to x to be a prime number[9]. It is known as the prime number theorem.
2
Twin Prime Numbers
When one looks at the first prime numbers one can see some patterns emerge.
For example 17, 19 or 41, 43 and 107, 109. These are all prime numbers which
differ by exactly 2. One may wonder if there are infintely many of such pairs.
This can be phrased as there are infinitely many pairs of natural numbers
(p, p + 2) with p and p + 2 both prime. This conjecture is known as the
twin prime conjecture. One may wonder if there are more patterns in prime
numbers instead of 2 one may take a difference of 4 or 6 between pairs of
primes. These prime pairs are called respectively cousin and sexy primes.
Heuristic Approach To Twin Prime Numbers
To estimate the number of twin primes up to a natural number x one can
use the distribution of the prime numbers and the prime number theory.
This states that a number smaller than x has at least probability 1/ log x of
being a prime number. This means if we pick two numbers smaller than x
the probability of both of them being a prime numbers is at least 1/(log x)2
but only when the event of ”p is prime” is independent of the event ”p+2
is prime”. This isn’t true if p ≡ 1 mod 3 and prime then p + 2 ≡ 0 mod 3
thus p + 2 isn’t prime.
One needs to correct for this dependence by some correction factor. The
probability for an arbitrary number to be divisible by a number q is 1/q. So
the probability for two arbitrary numbers not to be divisible by a number
q is (1 − 1/q)2 . For two numbers p and p + 2 this is different because we
need p 6≡ 0 mod q and p 6≡ −2 mod q which is in 2/q of the cases. The ratio
between these factors is the correction factor. Thus the correction factor for
any number q > 2 becomes:
(1 − 2q )
(1 − 1q )2
For q = 2 one has 1 modulo class which is restricted for p. This correction
factor becomes:
(1 − 12 )
=2
(1 − 12 )2
Being divisible by a small prime number is independent of the other small
primes. Thus one may multiply the correction factors of the small prime
numbers. In fact one may multiply over all prime numbers because when q
is large the correction factor converges to 1. This suggests a definition of a
3
twin prime constant of:
C=2
Y
q prime
q≥3
(1 − 2q )
(1 − 1q )2
≈ 1, 3203236316
This is the total correction factor over all primes q. One may guess the
estimate of the number of twin prime pairs smaller than an integer x is:
C
x
(log x)2
This would mean by a heuristic approach there would be infinitely many
pairs of primes which differ 2. Because the formula for the estimate of the
number of pairs under x goes to infinity when x goes to infinity.
4
Bounding The Gaps
There are two main ways to attempt to prove the twin-prime conjecture.
One can try to find the difference between the nth prime pn and the next
prime pn+1 . And prove for infinitely many n the difference between them is
2. Or one can use ’GPY method’ which takes sets of numbers of the same
length and proves at least two of them have to be prime.
Admissible Tuples and Sets
We are interested in sets which when you add n all of them could, in theory,
be prime. If you take the tuple 0, 2, 4 then it is known when you add a
arbitrary integer n. One of the n, n + 2, n + 4 is always divisible by 3. This
is a restriction on the tuple. Thus when one needs to find prime numbers
this tuple is not used.
Definition 1. A k-tuple H Q
= h1 , ..., hk is called an admissible set when there
is no integer q such that q| ki=1 (n + hi ) for all n ∈ Z
The tuples used here are all tuples of natural numbers but this is not
needed. One may take k−tuples of lineair forms such as m, m+n, m+4n for
two natural numbers m, n. The next conjecture isn’t proven for any k > 1.
It is proven for any admissible set of linear forms provided no two satisfy
a linear equation over the integers. Unfortunatly most of the questions
mathematicians are interested in do not satisfy these conditions. Such as
the twin prime conjecture which satisfies the linear equation q − p = 2.
Conjecture 1. (Prime k-tuples conjecture).
Let H = {h1 , ..., hk } be admissible. Then there are infinitely many integers
n such that all of n + h1 , ..., n + hk are prime.
If one wants to prove the twin prime conjecture the prime k-tuples conjecture has to be proven for k = 2 and h2 = h1 + 2. When a bound needs
to be proven a less strict conjecture is needed.
5
Conjecture 2. Let H = {h1 , ..., hk } be admissilbe. Then there are infinitely
many integers n such that at least two of n + h1 , ..., n + hk are prime.
When this conjecture was proven for some H we have lim inf n pn+1 −pn ≤
maxi,j |hi − hj |. The breaktrough of finding a bounded gaps between primes
is made by Zhang. Who in his paper proved lim inf n pn+1 − pn = B for a
B < ∞ and even gave an upper bound of B ≤ 70000000. As well in his
paper he stated ”This result is not optimal ... to replace the [upper bound
of B] by a value as small as possible is an open problem that will not be
discussed in this paper”[10]. The mathematical community worked together
in the Polymath8 project to show this bound could indeed be improved.
This project features different mathematicians all over the world who with
modern technology try to make a breaktrough in a certain mathematical
problem. In Polymath8 new ways to calculate the bounds were found. The
results followed each other in a span of weeks.1 In less than a year the bound
has been lowered to less than 600. This step was a major breaktrough in the
project because a different method was used to find this bound. All other
improvements where made by small improvements in the proof by Zhang.
This improvement uses a different method. In my thesis I will give the proof
of Maynard [6] in a extended and simplified way. In the next chapter an
outline of the proof will be given.
1
For the complete timeline look at http://michaelnielsen.org/polymath1/index.php?
title=Timeline of prime gap bounds
6
Outline of Maynard’s Proof
The proof of Maynard is based on the ’GPY method’ named after Goldston,
Pintz and Yildrim[2]. This method uses the distribution of primes to study
prime tuples and small gaps between primes. Given θ > 0 we say the primes
have ”level of distribution θ’ if, for any A > 0
X
π(x) x
max π(x; q, a) −
A
(1)
ϕ(x)
(log
x)A
(a,q)=1
θ
q≤x
where π(x; q, a) is the number of primes up to x of the form a mod q. We
have the following results.
Theorem 1. (Bombieri-Vinogradov Theorem)
The primes have a level of distribution θ for any θ < 1/2.
Conjecture 3. (Elliott-Halberstam Conjecture)
The primes have a level of distribution θ for any θ < 1.
Now the basic idea behind the approach of Maynard is when H =
{h1 , ..., hk } is a fixed admissible set one considers the sum
S(N, ρ) =
X
k
X
(
χP (n + hi ) − ρ)wn
(2)
N ≤n≤2N i=1
where χP is the characteristic function of the primes. Thus χP (p) = 1 when p
is prime and 0 otherwise. ρ > 0 and wn are non negative weights. If one can
show S(N, ρ) > 0. Then at least one term in (2) has a positive contribution.
This means there exists some integer n ∈ [N, 2N ] such that at least bρ + 1c
of the n + hi are prime. Thus if S(N, ρ) > 0 for all large N , then there are
infinitely many integers n for which at least bρ + 1c of the n + hi are prime.
Thus there are infinitely many bounded intervals containing bρ + 1c primes.
The hard part in this is choosing the weights such that this happens. The
7
difference between these sieves and the GPY sieve is in the λ’s. In the
standard Sielberg sieve the following weights are chosen.
P
wn = ( d| Qk (n+hi ) λd )2 , λd = µ(d)(log R/D)k
(3)
i=1
d<R
But Goldston, Pintz and Yildrim used different weights of the form
λd = µ(d)F (log R/D)
(4)
for a suitable smooth function F and µ the Möbius function. They chose
F (x) = xk+l for suitable l ∈ N, which has been shown to be essentially
optimal when k is large. This will prove the existence of bounded gaps
between primes when the level of distribution of primes θ > 1/2 but not
when θ < 1/2. Thus the existence of bounded gaps between primes then
relies on the the Elliott-Halberstam Conjecture (conjecture 3) which has not
been proven. Thus a weaker condition is needed to prove the existence of
bounded gaps. Zhang used in his paper a modified form of the distribution
of primes. Maynard uses in his paper a different
Q weight.
Maynard’s weights aren’t only based on a d| ki+1 (n + hi ) but uses different
di |n + hi ∀i. Which gives sieve weights of the form
X
wn = (
λd1 ,...,dk )2
(5)
di |n+hi ∀i
In Maynard’s proof he choses his weightsQto be zero unless n lies in a fixed
residue class v0 ( mod W ), where W = p≤D0 p the product of all primes
numbers smaller than D0 . Thus will remove some minor complications when
dealing with small prime factors. He choses D0 to be D0 = log log log N to
be sure that W (log log N )2 by the prime number theorem.
To estimate the sum (2) it is split in two parts.

2
X
X

S1 =
λd1 ,...,dk 
(6)
N ≤n≤2N
n≡v0 ( mod W )
S2 =
X
N ≤n≤2N
n≡v0 ( mod W )
k
X
di |n+hi ∀i
2
!
χP (n + hi ) 
i=1
X
λd1 ,...,dk 
(7)
di |n+hi ∀i
So S = S2 − ρS1 with a few minor resrictions added. In the same way
as Zhang’s proof the first part is rewriting both sums untill they are more
8
or less smooth functions. In the sum itself you have some functions which
can’t be specifically calculated. By rewriting both sums, and changing the
variables the sum will be easier to handle. This will be the first part of his
proof.
The second part will be proving the desired result will be achieved when the
sum is calculated. The next part is finding the length of the tuple in which
two prime numbers will be found. To finish the proof a tuple is given and
the bound will be set at 600.
9
Estimating The Sums
We will use a proposition to evaluate the sums.
Proposition 1. Let the primes have exponent of distribution θ > 0, and let
R = N θ/2−δ for some small fixed δ > 0. Let λd1 ,...,dk be defined in terms of
a fixed piecewise differentiable function F by
λd1 ,...,dk
k
Y
= ( µ(di )di )
X
r1 ,...,rk
di |ri ∀i
(ri ,W )=1∀i
i=1
Q
µ( ki=1 ri )2 log r1
log rk
F(
, ...,
)
Qk
log R
log R
i=1 ϕ(ri )
Q
whenever ( ki=1 di , W ) = 1 and let λd1 ,...,dk = 0 otherwise. Let F be supporP
ted on R = {(x1 , ..., xk ) ∈ [0, 1]k : ki=1 xi ≤ 1}. This means if λd1 ,...,dk 6= 0
then d1 ...dk ≤ R and (di , dj ) = 1 when i 6= j. Then we have
S1 =
(1 + o(1))ϕ(W )k N (LogR)k
Ik (F ),
W k+1
k
(1 + o(1))ϕ(W )k N (LogR)k+1 X (m)
Jk (F ),
W k+1 log N
S2 =
j=1
(m)
provided Ik (F ) 6= 0 and Jk
6= 0 for each m where
1
Z
Ik (F ) =
...
0
(m)
Jk
Z
(F ) =
1
Z
...
0
1
Z
(
0
Z
1
F (t1 , ..., tk )dt1 ...dtk
0
1
F (t1 , ..., tk )dtm )2 dt1 ...dtm−1 dtm−2 ...dtk
0
10
Technical Lemmas
The following Lemmas are used troughout the proofs
Lemma 1.
X
τk (d) < R(log R)k−1
(8)
d<R
Proof.
X
d<R
d1 ,...,dk
di <R
1
=R
di
X
1=R
d1 ,...,dk
Q
i di <R
d1 ,...,dk
d1 ...dk <R
X
<R
X
τk (d) =
X1
d
1
R
!k
< R(log R)k−1
(9)
d<R
Lemma 2. (Generalized Möbius inversion)
Q
If Ad1 ,...,dk with support in di ∈ N, i di < R.
Define
X
Br1 ,...,rk =
Ad1 ,...,dk
(10)
ri |di
then
Ad1 ,...,dk =
k
XY
µ(ri )Br1 ,...,rk
(11)
ri |di i=1
Lemma 3.
R
X
µ(a)2
a=1
Proof.
Y
log
(1 +
p≤R
p prime
ϕ(a)
log R
(12)
X
X 1
1
1
)=
log(1 +
)≤
< c + log log R (13)
p−1
p−1
p−1
p≤R
R
X
µ(a)2
a=1
ϕ(a)
<
p≤R
Y
(1 +
p≤R
p prime
11
1
) log R
p−1
(14)
Lemma 4.
µ(u)2
1
2
ϕ(u)
D0
(15)
Y µ(u)2
1
1+
=
ϕ(u)2
(p − 1)2
(16)
X
(u,W )=1
u>D0
Proof.
X
(u,W )=1
u>D0
p>D0
If one looks at the logarithm of the right hand side we get
X
X
X
1
2
2
2
log 1 +
<
<
<
2
2
2
(p − 1)
(p − 1)
(n − 1)
D0
p>D0
p>D0
(17)
n>D0
Changing the Variables
To prove this proposition a change of variables will be introduced. This
approach is based on the elementary combinatorial ideas of Selberg. [7]
We assume that the primes have a fixed level of distribution Q
θ, and R =
θ/2−
N
. The weight λd1 ,...,dk is restricted to tuples with d = ki=1 di and
d < R, (d, W ) = 1 and µ(d)2 = 1. This implies all the di are pairwise
coprime and square-free.
Proposition 2. Let
yr1 ,...,rk =
k
Y
µ(ri )ϕ(ri )
i=1
X λd ,...,d
1
k
Qk
i=1 di
d1 ,..,dk
(18)
ri |di ∀i
Let ymax = sup |yr1 ,...,rk | then
r1 ,...,rk
S1 =
2
N X yr21 ,...,rk
ymax
ϕ(W )k N (LogR)k
+
O(
)
Qk
W r ,...r
W k D0
i=1 ϕ(ri )
1
k
12
(19)
Proof. Expand out the square in S1 and then swap the order of summation

2
X
X
X
X

λd1 ,...,dk λe1 ,...,ek
1
S1 =
λd1 ,...,dk  =
N 6n<2N
n≡v0 modW
di |n+hi ∀i
N 6n<2N
n≡v0 modW
[di ,ei ]|n+hi ∀i
d1 ,...,dk
e1 ,...,ek
(20)
The inner sum runs over all n such that N ≤ n < 2N with n ≡ v0 mod W
and n+hi ≡ 0 mod [di < ei ] for all i. The W, [di , ei ] are pairwise coprime. If
not it would mean ([di , ei ], [dj , ej ]) = d > 1 thus d|di , dj but the product over
the di is square free. This is a contradiction so they are pairwise coprime.
Because they are coprime there exists,
Qkby the chinese remainder theorem,
a residue class a mod q with q = W i=1 [di , ei ] such that the summation
runs over n ≡ a mod q and N ≤ n < 2N
There is N/q times a n modulo this residue class in an interval with lenght
N. This calculation has an error which approaches 1 when N → ∞. The
inner sum becomes N/q + O(1) when the integers W, [di , ei ] are pairwise
coprime, when they are not the sum is zero. This restriction on the integers
X0
W, [di , ei ] will be denoted by
. This gives
S1 =
X0
N X0 λd1 ,...,dk λe1 ,...,ek
+ O(
|λd1 ,...,dk λe1 ,...,ek |)
Qk
W
i=1 [di , ei ]
(21)
d1 ,...,dk
e1 ,...,ek
d1 ,...,dk
e1 ,...,ek
To ease notation we put ymax = supr1 ,...,rk |yr1 ,...,rk |. The assumptions on
Q
λ state it is only non zero when ki=1 di < R, thus the errorterm contributes
X
|λd1 ,...,dk λe1 ,...,ek | 6
d1 ,...,dk
e1 ,...,ek
X
(λmax )2 λ2max (
d1 ,...,dk
e1 ,...,ek
X
τk (d))2
(22)
d<R
Where τk (d) means in how many ways d can be written as a product of
k integers.. This is bigger than the number of ways d can be written with k
divisors which are square free. This can be estimated in order of magnitude
by:
λ2max (
X
τk (d))2 λ2max R2 (LogR)2k
d<R
13
(23)
In this estimation Lemma 1 is used to estimate the sum. In the main
sum we remove the dependencies between the di and the ej variables. We
use the identity:
1
1
1 X
=
(ei , di ) =
ϕ(ui )
[di , ei ]
di ei
di ei
ui |di ,ei
This means the least common multiple can be rewritten to the product
divided by the greatest common divisor. This is a combination
of all ui
P
dividing the greatest common divisor and the equality n|d ϕ(n) = d. Thus
the sum is the greatest common divisor.
The main term becomes
!
k
Y
X0 λd ,...,d λe ,...,e
N X
1
1
k
k
S1 =
(24)
ϕ(ui )
Qk
Qk
W u ,...,u
d
e
i=1 i
i=1 i
1
k
i=1
d1 ,...,dk
e1 ,...,ek
ui |di ,ei ∀i
Now recall the requirement on the summation which is that W, [d1 , e1 ], ..., [dk , ek ]
are alle pairwise coprime. The weight λ is only supported on integers
d1 , ..., dk which are coprime with W. Thus the requirement of W being
coprime with the least common multiples can be dropped. Similarly the
requirements of (di , dj ) = 1 for all i 6= j and (ei , ej ) = 1 for all i 6= j
may be dropped. The only restriction left from the pairwise coprimality of
W, [d1 , e1 ], ..., [dk , ek ] is (di , ej ) = 1 for all i 6= j.
To be certain the requirement willPbe held without a requirement in
the sum, multiply the main term with si,j |di ,ej µ(si,j ). This works by the
P
equality d|n µ(d) is 1 when n = 1 and 0 otherwise. So the sum for a pair
(i, j) will not be empty only when (di , ej ) = 1. This applies to all i, j with
i 6= j. The main term becomes

S1 =
N
W
X
k
Y
u1 ,...,uk
i=1

!
ϕ(ui )
X
s1,2 ,...,sk,k−1
 Y


µ(si,j )


1≤i,j≤k
i6=j
X
d1 ,...,dk
e1 ,...,ek
ui |di ,ei ∀i
si,j |di ,ej ∀i6=j
λd1 ,...,dk λe1 ,...,ek
Qk
Qk
i=1 di
i=1 ei
(25)
Assume (ui , si,j ) 6= 1 this means si,j > 1 and si,j |di , ei , ej . Thus (ei , ej ) ≥
si,j > 1 and λe1 ,...,ek = 0 unless (ei , ej ) = 1. So there may be a restriction to
si,j by only summing over the ones which are coprime to ui and uj . Because
14
when this is not true λe1 ,...,ek does not support the give tuple. In a similar
way the restriction may be expanded by demanding si,j to be coprime with
si,a and sb,j for all
P∗a 6= j and b 6= i. The summation over these will be
further noted by
. To make our sum S1 more straightforward we will
introduce a change of variables. Let
k
Y
X λd ,...,d
1
k
yr1 ,...,rk = ( µ(ri )ϕ(ri ))
Qk
d
i=1 i
i=1
d1 ,...,dk
(26)
ri |di ∀i
This change of variables has to be invertible. When it is not invertible
the sum will not run over all possible tuples, for d1 , ..., dk square free. We
will prove our change of variables is invertible. This follows by using Lemma
2
X
r1 ,...,rk
di |ri ∀i
λd ,...,dk
yr1 ,...,rk
= Qk 1
Qk
i=1 ϕ(ri )
i=1 µ(di )di
(27)
This means any choice of yr1 ,...,rk supported on r1 , ..., rk when the product
i=1 ri = r is square-free, r < R and (r, W ) = 1 will give a suitable choise
of λd1] ,...,dk . By changing our variable we needPto re-estimate the λmax . Let
ymax = supr1 ,...,rk |yr1 ,...,rk |. Since d/ϕ(d) = e|d 1/ϕ(d) for d square-free.
Q
Take r0 = ki=1 ri /di then by the change of variables and the definition of
λmax
Qk
λmax ≤
sup
d1 ,...,dk
i=1 di square-free
Qk
k
Y
ymax ( di )
X
k
Y
µ(r1 )2
(
)
ϕ(ri )
(28)
r1 ,...,rk
i=1
Qkdi ri ∀i
ri <R
Qk i=1
i=1 ri square-free
i=1
0
variable
Qk by changing the
Qk from di to r the limits of the sum change
0
r < R → r < R/ i=1 di and because di |ri the other limit changes
Qki=1 i
Qk
0
0
i=1 ri square-free → (r ,
i=1 di ) = 1. When r is used instead of a sum
over r1 , .., rk there need to be corrected by the number of ways the r1 , ..., ri
can form a certain product. This correction is smaller than the number of
ways the product can be written by all it’s divisors. So we can estimate this
term true multiplying by τk (r0 ).
15
λmax ≤ ymax
sup
(
k
Y
di
)
ϕ(di )
d1 ,...,dk
i=1
Qk
i=1 d1 square-free
d1 ,...,dk
µ(d)
)
ϕ(d)
X
≤ ymax sup (
d|
Qk
i=1
di
X
Qk
r0 <R/ i=1 di
Q
(r0 , ki=1 di )=1
µ(r0 )2 τk (r0 )
ϕ(r0 )
X
r0 <R/
(r0 ,
Qk
Qk
i=1
µ(r0 )2 τk (r0 )
ϕ(r0 )
(29)
(30)
di
di )=1
i=1
The product we can rewrite by using the equality
Qk stated and rewriting
2
1 = µ(d) to remove the requirement of needing i=1 di to be square-free.
If we let u = dr0 and by using the fact that τk (dr0 ) ≥ τk (r0 ) both sums can
be combined an the estimate becomes
≤ ymax
X µ(u)2 τk (u)
ymax (logR)k
ϕ(u)
(31)
u<R
This is by combining (Lemma 1) and Lemma 3.
By substituting the change of variables and with the estimation of the
error term, we obtain
S1 =
N
W
X
(
k
Y
ϕ(ui ))
u1 ,...,uk i=1
X∗
(
Y
s1,2 ,...,sk,k−1 1≤i,j≤k
i6=j
k
Y
µ(ai µ(bi ))
µ(si,j ))(
)ya ,...,ak yb1 ,...,bk
ϕ(ai )ϕ(bi ) 1
i=1
2
+O(ymax
R2 (LogR)4k )
(32)
Q
where ai = ui i6=j si,j and bj = uj i6=j si,j The functions ϕ and
Q µ are
multiplicative and all factors are pairwise coprime thus µ(ai ) = µ(ui ) i6=j µ(si,j )
the same for µ(bi ), ϕ(ai ) and ϕ(bi ) but only when all si,j are coprime with
all other terms in the ai and bi . All terms with these not square free do not
contribute to the sum and µ(si,j )3 = µ(si,j ) hence the sum can be rewritten
to as
Q
N
S1 =
W
X
(
k
Y
µ(ui )2
u1 ,...,uk i=1
ϕ(ui )
)
X∗
Y
s1,2 ,...,sk,k−1 1≤i,j≤k
i6=j
2
+O(ymax
R2 (LogR)4k )
16
µ(si,j )2
ya ,...,ak yb1 ,...,bk
ϕ(si,j )2 1
(33)
Again there is no contribution from si,j with (si,j , W ) 6= 1 because of
the restricted support of y. Thus only si,j = 1 and si,j > D0 need to be
considered. When si,j > D0 the contribution to the sum is
k

2
ymax
N
W

k2 −k−1
X µ(si,j )2
X µ(s)2
 


(34)
ϕ(u) 
ϕ(si,j )2
ϕ(s)2
µ(u)2 
 X


u<R
(u,W )=1

si,j >D0
s≥1
The first sum can be rewritten in much the same way as in Lemma 3
to ϕ(W )k (LogR)k /W k . Because one needs to correct for the terms u which
are not coprime with W this correction term is ϕ(W )/W . Thus the ratio
between W and the numbers which are coprime to it. The last sum is
convergent thus can be rewritten as a contstant. For the middle sum we use
Lemma 4.
If we combine the Lemma with the other estimations then the equation
of (34) is estimated by
2
ymax
ϕ(W )k N (LogR)k
W k+1 D0
(35)
When si,j = 1 and all other terms are in our error term we get
yu21 ,...,uk
y 2 ϕ(W )k N (LogR)k
2
+ O( max
+ ymax
R2 (LogR)4k )
Qk
k+1 D
W
0
ϕ(u
)
i
u1 ,...,uk
i=1
(36)
To prove the lemma it now suffices to show the first error term dominates
the second. Recall R2 = N θ−2δ ≤ N 1−2δ , W N δ and R (LogR)3k .
Thus N/W D0 R2 (LogR)3k hence N/W D0 N 1−2δ ≥ R2 and the first
term dominates. This ends the proof and
S1 =
N
W
X
yu21 ,...,uk
N X
y 2 ϕ(W )k N (LogR)k
S1 =
+ O( max
)
Qk
W u ,...u
W k D0
i=1 ϕ(ui )
1
k
17
(37)
Rewriting the second sum
In a similar way we estimate S2 =
(m)
S2
X
=
(m)
m=1 S2
Pk
χP (n + hm )(
N ≤n<2N
n≡v0 (modW )
where
X
λd1 ,...,dk )2
(38)
d1,...,dk
di |n+hi ∀i
(m)
In the next lemma we estimate S2
in a similar way to S1
Proposition 3. Let
yr(m)
=
1 ,...,rk
k
Y
X
µ(ri )g(ri )
i=1
d1 ,..,dk
ri |di ∀i
dm =1
λd1 ,...,dk
Qk
i=1 ϕ(di )
(39)
where g is the totally multiplicative function defined on primes by g(p) =
(m)
p − 2. Let ymax = sup |yr(m)
| then
1 ,...,rk
r1 ,...,rk
(m)
S2
=
(m)
2
2
X (yr(m)
(ymax )2 ϕ(W )N (LogR)k−2
N
ymax
N
1 ,...,rk )
+O(
)+O(
)
Pk
k−1
ϕ(W )LogN r ,...r
(LogN )A
W
D0
i=1 g(ri )
1
k
(40)
Proof.By interchanging the order of summation and expanding out the
squares we get a resemblence with S1 in equation (20).
(m)
S2
=
X
X
λd1 ,...dk λe1 ,...ek
χp (n + hm )
(41)
N ≤n<2N
n≡v0 mod W
[ei ,di ]|n+hi
d1 ,...,dk
e1 ,...,ek
The inner sum is over k + 1 residue classes which canQbe written as a
sum over a single residue. This residue class is q = W ki=1 [di , ei ] with
W, [d1 , e1 ], ..., [dk , ek ] are pairwise coprime. Because of the χ-prime function
the inner sum for S2m is only non zero when dm = em = 1. By the same
reason the inner sum will contribute XN /ϕ(q) and an error term. Where
XN =
X
N ≤n<2N
18
χP (n)
(42)
It means we have at least the sum of prime numbers between N and 2N
divide by the quantity of numbers which are coprime by q. Let
X
X
1
E(N, q) = sup |
χP (n) −
χP (n)|
(43)
ϕ(q)
(a,q)=1
N ≤n<2N
n≡a(modq)
N ≤n<2N
Thus the size of the contribution to the error term becomes
O(E(N, q))
(44)
It’s easy to see by summing the two we get the desired result. Thus by the
same notation of the restriction the sum becomes
(m)
S2
=
XN
ϕ(W )
X0
d1 ,...,dk
e1 ,...,ek
em =dm =1
X
λd1 ,...,dk λe1 ,...,ek
|λd1 ,...,dk λe1 ,...,ek |E(N, q))
+ O(
Qk
ϕ([d
,
e
])
i
i
i=1
d1 ,...,dk
e1 ,...,ek
(45)
In a similar way to the first sum we estimate the error term. The support
of λd1 ,...,dk restricts q to be only square free and smaller than R2 W . Given a
square free integer
r there are only τ3k (r) possibilities of d1 , ..., dk , e1 , ..., ek
Qk
for which W i=1 [di , ei ] = r. Recall λmax ymax (LogR)k . Hence the error
term becomes
2
ymax
(LogR)2k
X
µ(r)2 E(N, r)τ3k (r)
(46)
r<R2 W
Rewrite the sum
2
= ymax
(LogR)2k
X τ3k (r)µ(r)E(N, r)1/2
µ(r)E(N, r)1/2
(47)
r<R2 W
Then you will get by using Cauchy Schwartz on both parts between
brackets and E(N, r) N/ϕ(r)
1/2 
1/2

X
X
N
2
2
 
ymax
(LogR)2k 
µ(r)2 τ3k
(r)
µ(r)2 E(N, r)
ϕ(r)
2
2
r<R W
r<R W
(48)
Remember the primes have level of distribution θ thus the last sum 0
(N/LogN A )1/2 for any A0 > 0. In the first sum they are smaller than
N 1/2 (LogN )B for a B. Hence this can be rewritten to
19
2
ymax
N
(49)
(LogN )A
As in the treatment of the first sum the conditions
of all (di , ej ) = 1 can
P
be rewritten by multiplying the expression by si,j |di ,ej µ(si,j ). Again the
P
same requirements may restrict si,j and will be denoted by ∗ . In the same
way the ϕ([di , ei ]) term can be split, because they are square-free, by the
equation
X
1
1
=
g(ui )
ϕ([di , ei ])
ϕ(di )ϕ(ei )
ui |di ,ei
Where g is the totally multiplicative function defined on the primes by
g(p) = p − 2. This transforms the main term to
k
X Y
XN
( g(ui ))
ϕ(W ) u ,...,u
s
1
k
i=1
X∗
(
1,2 ,...,sk,k−1
Y
µ(si,j ))
1≤i.j≤k
X
λd1 ,...,dk λe1 ,...,ek
Qk
i=1 ϕ(ei )
i=1 ϕ(di )
Qk
d1 ,...,dk
e1 ,...,ek
ui |di ,ei ∀i
si,j |di ,ej ∀i6=j
dm =em =1
(50)
A similar substition may be used in this situation we have only one extra
(m)
demand rm = 1. When rm 6= 0 then yr1 ,...,rk is 0. Thus let
yr(m)
=
1 ,...,rk
k
Y
µ(ri )g(ri )
i=1
X
d1 ,..,dk
ri |di ∀i
dm =1
λd1 ,...,dk
Qk
i=1 ϕ(di )
(51)
Hence the main term becomes in a similar way as equations (??) to (32)
k
X Y
XN
µ(ui )2
( (
)
ϕ(W ) u ,...,u
g(ui ) s
1
k
i=1
X
(
Y
1,2 ,...,sk,k−1 1≤i.j≤k
i6=j
(
µ(si,j ) (m)
(m)
)y
y
g(si,j ) a1 ,...,ak b1 ,...,bk
(52)
Again there are two different cases for si,j is 1 or > D0 . When si,j isn’t
1 the contribution is by the same calculation as S1
(m)
X
(ymax )2 N
(
ϕ(W )LogN
u<R
(u,W )=1
µ(u)2 k−1 X µ(s)2 k(k−1)−1 X µ(si,j )2
) (
)
g(u)
g(s)2
g(si,j )2
s
si,j >D0
20
(m)
(ymax )2 ϕ(W )k−2 N (logR)k−1
W k−1 D0 LogN
(53)
And by the prime number theorem XN = N/LogN + O(N/(LogN )2 )
which contributes to the error term by
(m)
X
(ymax )2 N
(
ϕ(W )(LogN )2
u<R
(u,W )=1
(m)
µ(u)2 k−1
(ymax )2 ϕ(W )k−2 N (logR)k−3
)
g(u)
W k−1
(54)
Which will be absorbed by the first error term. With this we can rewrite
the sum
(m)
S2
(m)
(m)
2
X (yr(m)
N
(ymax )2 ϕ(W )N (LogR)k−2
ymax N
1 ,...,rk )
=
+O(
)+O(
)
Pk
ϕ(W )LogN r ,...r
(LogN )A
W k−1 D0
i=1 g(ri )
1
k
(55)
Which ends the proof.
Relating the Variables
Next the new variables of S1 will be related to S2m by the following lemma.
Lemma 5. if rm = 1 then
yr(m)
=
1 ,...,rk
X yr1 ,...,rm−1 ,am ,rm+1 ,...,r
am
ϕ(am )
k
+ O(
ymax ϕ(W )LogR
)
W D0
Proof. Substitue the expression from (??) to (27) in definition (51) we
get
k
k
Y
X Y
ya1 ,...,ak
µ(di )di X
yr(m)
=
(
µ(r
)g(r
))
(
)
Q
i
i
1 ,...,rk
ϕ(di ) a1 ,...,ak ki=1 ϕ(ai )
i=1
d1 ,...,dk i=1
ri |di ∀i
dm =1
(56)
di |ai ∀i
By switching the summation of d and a variables the last sum is over d. This
sum we can calculate explicitly. Because all the di , ai and ri are square-free
21
and the functions are multiplicative
X
k
Y
µ(di )di
d1 ,...,dk i=1
di |ai ,ri |di ∀i
dm =1
ϕ(di )
=
Y X µ(di )di µ(ri )ri
Y µ(ai /ri ) µ(ri )ri
=
ϕ(di ) ϕ(ri )
ϕ(ai /ri ) ϕ(ri )
i6=m
di
di |ai /ri
i6=m
(57)
Hence we get
k
Y
X
ya1 ,...,ak Y µ(ai )ri
yr(m)
=
(
µ(ri )g(ri ))
Qk
1 ,...,rk
ϕ(ai )
a1 ,...,ak
i=1 ϕ(ai ) i6=m
i=1
(58)
ri |ai ∀i
(m)
Looking at the support of yr1 ,...,rk we can restrict the summation over
ai to (ai , W ) = 1. Thus either ai = ri or ai > D0 ri . First look at the
contribution of ai 6= ri for j 6= m. Split the last sum in three parts. The
part of aj > D0 rj . The part of am and the rest. The first and last sums
converge thus restricted to a constant. The middle sum is estimated by (14).
k
Y
X
ymax ( g(ri )ri )(
i=1
(
aj >D0 rj
µ(aj )2
)(
ϕ(aj )2
X
am <R
(am ,W )=1
µ(am )2 Y X µ(ai )2
)
(
)
ϕ(am )
ϕ(ai )2
1≤i≤k ri |ai
i6=j,m
k
Y
g(ri )ri ymax ϕ(W )LogR
ymax ϕ(W )LogR
)
2
ϕ(ri )
W D0
W D0
(59)
i=1
Thus estimate the main contribution when aj = rj for all j 6= m
k
Y
ymax ϕ(W )LogR
ri g(ri ) X yr1 ,...,rm−1 ,am ,rm+1 ,...,rk
yr(m)
=
(
)
+ O(
)
1 ,...,rk
2
ϕ(ri ) a
ϕ(am )
W D0
i=1
m
(60)
2
−2
Note that g(p)p/ϕ(p) = 1 + O(p ). Since the contribution is zero unless
the product of the ri is coprime to W every ri > D0 thus the product in the
expression may be replaced by 1 + O(D−1 ). This will be dominated by the
error term we already have. Thus this gives the result
yr(m)
=
1 ,...,rk
X yr1 ,...,rm−1 ,am ,rm+1 ,...,r
am
ϕ(am )
k
+ O(
ymax ϕ(W )LogR
)
W D0
This will end the proof and gives a way te relate both new variables.
22
(61)
Choosing Suitable y
To complete the proof of proposition 1 we need a suitable choice of y.
The choice of y is such that the ratio between the main terms of S1 and S2
is maximized. A second demand is that the y are smooth. Such thatQit has
no dependence on the prime factorisation of the ri . Remember r = ki=1 ri
satisfies (r, W ) = 1 and µ(r)2 = 1. Choose
yr1 ,...,rk = F (
log ri
log rk
, ...,
)
log R
log R
(62)
for some piecewise differentiable
function F : Rk → R supported on Rk =
Pk
k
{(x1 , ..., xk ) ∈ [0, 1] : i=1 xi ≤ 1}. When r is nog coprime to W or not
square free set yr1 ,...,rk = 1. With this choice suitable asymptotic estimates
for S1 and S2 can be made.
Lemma 6. Let κ, A1 , A2 , L > 0, Let γ be a multiplicative function satisfying
0≤
γ(p)
≤ 1 − A1
p
and
−L ≤
X γ(p) log p
− κ log z/w ≤ A2
p
w≤p≤z
for any 2 ≤ w ≤ z.Let g be the totally multiplicative function defined on
primes by g(p) = γ(p)/(p − γ(p)). Finally, let G : [0, 1] → R be a piecewise
differentiable function and let Gmax = supt∈[0,1] (|G(t)| + |G0 (t)|). Then
X
d<z
(log z)κ
log d
µ(d) g(d)G(
)=G
log z
Γ(κ)
2
Z1
G(x)xκ−1 dx+OA1 ,A2 ,κ (G LGmax (log z)κ−1 )
0
where
G =
Y
1
γ(p) −1
(q −
) (q − )κ
p
p
p
The constant implied by O is independent of L and G
23
Proof. This proof is divided in two lemmas in [3]. Lemmas two and three
prove this result with a slight notation difference. Which again is divided in
multiple lemmas and based on explicit estimates of selbergs upper bounds.
Lemmas 5.3 and 5.4 [5]
The next two lemmas will finish the estimation of S1 and S2m this will conclude the proof of Proposition 1. The next lemma will estimate S1
Lemma 7. Let yr1 ,...,rk be given in terms of a piecewise differentiable function F supported on Rk = {(x1 , .., xk ) ∈ [0, 1]k : |sumki=1 xi ≤ 1} by (62).
Let
k
X
δF
Fmax =
sup
|F (t1 , ..., tk )| +
| (t1 , ..., tk )|
(63)
δti
(t1 ,...,tk )∈[0,1]k
i=1
Then
S1 =
2 ϕ(W )k N (log N )k
ϕ(W )k N (log R)k
Fmax
I
(F
)
+
O(
)
k
W k+1
W k+1 D0
where
Z
1
Z
1
...
Ik (F ) =
0
(64)
F (t1 , ..., tk )2 dt1 ...dtk
(65)
0
Proof. We substitute the choice of y (62) into the expression of S1 which
was rewritten in lemma (2). There are no restrictions given for the k-tuples
which are supported on y so we will add the restrictions to the sum. This
gives
S1 =
N
W
X
k
Y
log ui
log uk 2
y 2 N (LogR)k ϕ(W )k
µ(ui )2
)F (
, ...,
) +O( max
(
)
ϕ(ui )
log R
log R
W k D0
u1 ,...uk
i=1
(ui ,uj )=1∀i6=j
(ui ,W )=1∀i
(66)
Note that the requirement of (ui , uj ) = 1 can be dropped at the cost of
an reasonably sized error. This is because every prime divider they have in
common has to be bigger than D0 for they both are coprime with W . This
error is of size
2 N X
Fmax
W
p>D0
X
k
Y
µ(ri )2
u1 ,...uk i=1
p|ui ,uj
(ui ,W )=1∀i
24
ϕ(ui )
2 N X
X
Fmax
1
(
2
W
(p − 1)
p>D0
u<R
(u,W )=1
µ(u)2 k
F 2 N (log R)k ϕ(W )k
) max
ϕ(u)
W k+1 D0
(67)
The second sum is rewritten in the same way as in the estimations of
2 to ϕk (log R)k /W k while in the first sum ”hier moet nog een afschatting”
and rewrite it to 1/D0 . Which will give the desired result.
Thus the sum that needs to be evaluated is
X
k
Y
µ(u1 )2
log uk 2
log ui
, ...,
)
(
)F (
ϕ(ui )
log R
log R
(68)
ui ,...,uk
i=1
(ui ,W )=1∀i
By k-fold application of Lemma.6 the sum can be estimated. Take for
each application κ = 1 and
1,
p 6 |W
(69)
γ(p) =
0, otherwise
X log p
log D0
(70)
L1+
p
p|W
and A1 and A2 constants of suitable size.
Lemma 8. If γ(p), κ stated as above then the lower limit L stated in Lemma
6 will be log D0
Proof. One chooses κ = 1 and no terms are counted when p < D0
and use the maximum of log z/w. The prime counting function π(x) will
be used. But not x/ log x
is needed. A stricter
R xa stricter approximation
A
approxomation is π(x) = 2 dt/ log t + O(x/(log x) [1]
Z x
t x
d
1
π(x) =
−
t
dt + O(x/(log x)A )
log t 2
dt
log
t
2
x
x
+
+ O(x/(log x)3 )
(71)
log x (log x)2
P
Look at D0 <p<R log p/p over primes.
X
D0 <p<R
Z
R
log p/p =
D0
Z R
log x
log x R
d log x
dπ(x) = π(x)
−
π(x)
x
x D0
dx
x
D0
25
By the prime number theorem [4, p. 352] the first part is O(1) for the second
part the above estimation of π(x) is used. This gives.
Z
R
= O(1) +
D0
R
Z
= O(1) +
D0
x
x
+
log x (log x)2
log x
1
− 2
2
x
x
dx
1
1
−
dx = C + log R/D0
x x(log x)2
This constant C > 0 thus our estimation becomes
− log D0 ≤ C + log R/D0 − log R ≤ C
(72)
Thus L = log D0
It will be shown how this works when k = 2. When κ = 1 and because
W is square-free
Y
1
ϕ(W )
G =
1−
=
(73)
p
W
p|W
Thus we get
X µ(u1 )2 X µ(u2 )2 log u1 log u2
F(
,
)2
ϕ(u1 )
ϕ(u2 )
log R log R
u1 <R
u2 <R

1
2
X µ(u1 )2 ϕ(W )(log R) Z
log u2 2
ϕ(W )Fmax (log D0 ) 

F (x,
) dx + O(
)
=
ϕ(u1 )
W
log R
W

u1 <R
0

=
ϕ(W )(log R) 
W
Z1
ϕ(W )(log R)
W
0
Z1
F (x, y)2 dx + O(
2 (log D )
ϕ(W )Fmax
0
W

)dy +O(
2 (log D )
ϕ(W )Fmax
0
)
W
0
ϕ(W )2 (log R)2
=
I2 (F )+
W2
1
Z
O(
0
2 (log D ) log R
2 (log D )
ϕ(W )2 Fmax
ϕ(W )Fmax
0
0
)
)dy+O(
W2
W
2 (log D ) log R
ϕ(W )2 Fmax
ϕ(W )2 (log R)2
0
I2 (F ) + O(
)
2
W
W2
If this is applied k times to the sum in equation (68) we get
=
X
k
Y
µ(u1 )2
log ri
log rk 2
(
)F (
, ...,
)
ϕ(ui )
log R
log R
u1 ,...,uk
i=1
(ui ,W )=1∀i
26
(74)
k (log D )(log R)k−1
ϕ(W )k (log R)k
ϕ(W )k Fmax
0
I
(F
)
+
O(
)
(75)
k
Wk
Wk
And by combining (75), (67) and (66) this ends the proof of lemma
=
(7)
To end the proof of the proposition the following lemma needs to be proved.
Lemma 9. Let yr1 ,...,rk , F and Fmax be as defined in Lemma 7. Then we
have
(m)
S2
=
2 ϕ(W )k N (log N )k
Fmax
ϕ(W )k N (log R)k+1 (m)
J
(F
)
+
O(
)
k
W k+1 log N
W k+1 D0
(76)
where
(m)
Jk (F )
1
Z
=
Z
1
...
0
1
Z
F (t1 , ..., tk )dtm )2 dt1 ...dtm−1 dtm+1 ...dtk
(
0
(77)
0
Proof. The proof is similar to the proof of Lemma 7. First estimate
Q
(m)
Recall yr1 ,...,rk = 0 unless rm = 1 and r = ki=1 ri satisfies (r, W ) =
(m)
1 and µ(r)2 = 1. Then yr1 ,...,rk = 0 is given by Lemma 5. First the case
(m)
when yr1 ,...,rk 6= 0 is checked. Substitute (62) in the expression of Lemma 5
and this gives
(m)
yr1 ,...,rk .
log rm−1 log u log rm+1
log rk
µ(u)2 log ri
F(
, ...,
,
,
, ...,
)
ϕ(u)
log R
log R log R log R
log R
X
yr(m)
=
1 ,...,rk
(u,W
Qk
i=1 )ri )=1
+O(
Fmax ϕ(W ) log R
)
W D0
(78)
(m)
Which makes ymax ϕ(W )Fmax (log R)/W . Now estimate the sum over u.
Lemma 6 is used with κ = 1,
Q
1, p 6 |W ki=1 ri
γ(p) =
(79)
0,
otherwise
X
L1+
p|W
Qk
i=1 ri
X log p
log p
+
p
p
p<log R
X
Q
p|W ki=1 ri
p>log R
log log R
log log N
log R
(80)
and with A1 , A2 suitable Q
fixed constants.
In
the
same
way
as
in
lemma
7 it
Qk
k
is easy to see G = ϕ(W ) i=1 ϕ(ri )/W i=1 ri which gives
27
k
yr(m)
= (log R)
1 ,...,rk
ϕ(W ) Y ϕ(ri ) (m)
Fmax ϕ(W ) log R
)Fr1 ,...,rk + O(
)
(
W
ri
W D0
(81)
i=1
where
Fr(m)
1 ,...,rk
Z
1
F
=
0
logrm+1
log r1
log rm−1
log rk
, ...,
, tm ,
, ...,
log R
log R
log R
log R
dtm
(82)
If this is substituted in the expression (40). The term of ymax is to the po2 ϕ(W )k N (log R)k )/(W k+1 D ).
wer of two. Thus the error term becomes (Fmax
0
The sum obtained is
(m)
S2
2
ϕ(W )N (log R)2 Fmax
=
W 2 log N
O
X
r1 ,...,rk
(ri ,W )=1∀i
(ri ,rj )=1∀i6=j
rm =1
k
Y
µ(ri )2 ϕ(ri )
g(ri )ri
i=1
2 ϕ(W )k N (log R)k
Fmax
W k+1 D0
!
(Fr(m)
)2 +
1 ,...,rk
(83)
In the same way as in the first sum the condition (ri , rj ) = 1 can be removed.
This will introduce an error of size

k−1


2
X ϕ(p)2
 X µ(r)2 ϕ(r) 
ϕ(W )N (log R)2 Fmax



W 2 log N
g(p)2 p2 
g(r)r 
p>D0
r<R
(r,W )=1
The first sum will converge to a constant of 1/D0 . The second sum can
be estimated by ϕ(W )k−1 (log R)k−1 /W k−1 ≤ ϕ(W )k−1 (log N )k−1 /W k−1
which give an error term of
2 ϕ(W )k N (log N )k
Fmax
W k+1 D0
(84)
Now the last part of the sum needs to be evaluated. In the same way as the
first sum we will use Lemma 6. Thus the sum we evaluate is


X
Y µ(ri )2 ϕ(ri )

 (Fr(m)
)2
(85)
1 ,...,rk
g(ri )ri
r1 ,...,rm−1,rm+1 ,...,rk
(ri ,W )=1∀i
1≤i≤ki6=j
28
By applying Lemma 6 k − 1 times with κ = 1 and
1
1 + p2 −p−1
,
p 6 |W
γ(p) =
0,
otherwise
L1+
X log p
p|W
p
log D0
(86)
(87)
and A1 , A2 suitable fixed constants. It is easy to see that error term (84)
dominates the dominant error term we get when using this lemma k − 1
times. Thus this gives
(m)
S2
=
2 ϕ(W )k N (log N )k
ϕ(W )k N (log R)k+1 (m)
Fmax
J
(F
)
+
O(
)
k
W k+1 log N
W k+1 D0
(88)
where
(m)
Jk
Z
(F ) =
1
Z
...
0
1
Z
(
0
1
F (t1 , ..., tk )dtm )2 dt1 ...dtm−1 dtm+1 ...dtk
0
as required.
29
(89)
Comparing the sums
In the previous chapter we estimated the sums. Now it needs to be proven
that by taking these estimations and a suitable k-tuple that there are infinitely many integers n such that several, but at least two, of the n + hi are
prime. In the next proposition we will prove this and we will introduce a
way to calculate the bound.
Proposition 4. Let the primes have level of distribution θ > 0. Let δ > 0
(m)
and H = h1 , ..., hk an admissible set. Let Ik (F ) and Jk (F ) be given as
in Proposition 1, let Sk denote the set of piecewise differentiable
functions
Pk
k
k
F : [0, 1] → R supported on Rk = {(x1 , ..., xk ) ∈ [0, 1] : i=1 xi ≤ 1} with
(m)
Ik (F ) 6= 0 and Jk 6= 0 for each m. Let
Pk
(m)
Jk
Ik (F )
m=1
Mk = sup
F ∈Sk
k
, rk = d θM
2 e
(90)
Then there are infinitely many integers n such that at least rk of the
n + hi (1 ≤ i ≤ k) are prime. In particular, lim inf n (pn+rk −1 − pn ) ≤
max1≤i,j≤j (hi − hj ).
This proposition will tell when we have an upper bound θMk > 2 for a
k then we will have infinitely many n where n + hi with hi in a admissible
set of order k. Where at least two will be prime.
Proof.Let S = S2 − ρS1 . If S > 0 for all large N and ρ > 1 then there are
infinitely many integers n such that at least two of the n + hi are prime.
Put R = N θ/2− for a small > 0. By the definition of Mk , a F0 ∈ Sk can
P
P
(m)
(m)
be chosen such that km=1 Jk (F0 ) > (Mk − )Ik (F0 ) = km=1 Jk (F0 ) −
Ik (F0 ). By using Proposition 1, we can rewrite both sums with given
λd1 ,...,dk such that
ϕ(W )k N (log R)k
S=
W k+1
!
k
log R X (m)
Jk (F0 ) − ρIk (F0 ) + o(1)
log N
m=1
30
ϕ(W )k N (log R)k Ik (F0 )
≤
W k+1
θ
( − )(Mk − ) − ρ + o(1)
2
(91)
When ρ = θMk /2 − δ by choosing sufficiently small then S > 0 for all large
N. Thus there are infinitely many integers n for which at least bρ + 1c of the
n + hi are prime. Since bρ + 1c = dθMk /2e when δ is sufficiently small the
result is proven.
To get the result of bounded gaps between primes an esimation of Mk and
θ is needed. We need θ to be as large as possible. Remember if the primes
have ’level of distribution’ θ if for any A > 0, we have
X
π(x) x
max π(x; q, a) −
(92)
A (log x)A
ϕ(q)
(a,q)=1
θ
q≤x
By the Bombieri-Vinogradov theorem it is proven when θ < 1/2. If we find
Mk > 4 for a k and an admissible set Hk with k elements there are infinitely
many n ∈ N such that two of the n + hi are prime.
Obtaining a lower bound of Mk for small k
Let Sk denote the set of piecewiseP
differentiable function F : [0, 1]k → R
supported on Rk = {(x1 , ..., xk ) : ki=1 xi ≤ 1} such that Ik (F ) 6= 0 and
(m)
JK (F ) 6= 0 for each m. If a lower bound is obtained, with these requirement, this will be
Pk
(m)
m=1 Jk (F )
(93)
Mk = sup
Ik (F )
F ∈Sk
To obtain this lower bound we will consider approximations to the optimal
function F of the form
P (t1 , ..., tk ), if (t1 , ..., tk ) ∈ Rk
(94)
F (t1 , ..., tk ) =
0,
otherwise
for polynomials P .
Lemma 10. Let Pj = sumki=1 tji denote the j th symmetric power sum polynomial. Then we have
Z
a!
(1 − P1 )a Pjb dt1 ...dtk =
Gb,j (k)
(95)
(k + jb + a)!
Rk
31
where
Gb,j (x) = b!
b X
k
r
r=1
r
Y
(jb1 )!
X
(96)
bi !
b1 ,...,br ≥1 i=1
P
r
i=1 bi =b
Proof. First by induction on k it follows
Z
(1 −
k
X
ti )
a
i=1
Rk
k
Y
tai i dt1 , ..., dtk
=
i=1
Qk
a!
(k + a
i=1 ai !
P
+ ki=1 ai )!
(97)
First consider the
P integration with respect to t1 . The limits of the integration
are 0 and 1 − ki=2 ti for (t2 , ..., tk ) ∈ Rk−1 . By substituing v = t1 /(1 −
Pk
i=2 ti ) we find
P
1− ki=2 ti
Z
0
Z 1
k
k
k
k
X
Y
Y
X
ai
ai
a
a+ai +1
(1−
ti ) ( ti )dt1 = ( ti ))(1−
ti )
(1−v)a v a1 dv
i=1
i=1
i=2
0
i=2
k
k
Y
X
a!a1 !
ai
( ti )(1 −
ti )a+ai +1
(98)
=
(a + a1 + 1)!
i=2
i=2
R1
Where the beta function identity is used 0 ta (1 − t)b dt = a!b!/(a + b + 1)!.
The next step will give a fraction before the main term of (a+a1 +1)!a2 !/(a+
a1 + a2 + 2)! thus the identity follows by induction.
By the multinomial theorem,
k
X
Pjb = (
tji )b =
i=1
k
Y
b!
X
Qk
b ,...,bk
Pk1
i=1 bi =b
i=1 bi ! i=1
i
tjb
i
(99)
Thus by applying this the following result will be obtained
Z
(1 − P1 )
a
Pjb dt1 ...dtk
=
b!
X
=
(1 −
Qk
b ,...,bk
Pk1
i=1 bi =b
Rk
Z
b!a!
(k + a + jb)!
i=1 bi !
X
Rk
i=1
k
Y
(jbi )!
b ,...,bk i=1
Pk1
i=1 bi =b
32
k
X
bi !
a
ti )
k
Y
tbi dt1 , ..., dtk
i=1
(100)
For computations b will be small, and so it is convenient to split
the sumk
mation over the bi are non-zero. Given an integer r, there are
ways of
r
choosing r of b1 , ..., bk to be non-zero. Thus
k
Y
(jbi )!
X
bi !
b ,...,bk i=1
Pk1
i=1 bi =b
b X
k
=
r=1
r
X
k
Y
(jbi )!
(101)
bi !
b1 ,...,bk ≥1 i=1
P
k
i=1 bi =b
This gives the result.
(m)
Now a lemma is used to write Ik (F ) and Jk (F ) in terms of manageable
expression with this choice of P.
Lemma 11. Let F be given in terms of a polynomial P by (94). Let P be given inP
terms of a polynomial
Pthe symmetric power polynomials
P expression in
P1 = ki=1 ti and P2 = ki=1 t2i by P = di=1 ai (1 − P1 )bi P2ci for constants
ai ∈ R and non-negative integers bi , ci . Then for each 1 ≤ m ≤ k we have
X
Ik (F ) =
ai aj
1≤i,j≤d
(m)
Jk (F )
=
X
ai aj
1≤i,j≤d
(bi + bj )!Gci +cj ,2 (k)
(k + bi + bj + 2ci + 2cj )!
cj ci X
X
ci
ci γbi ,bj ,ci ,cj ,c0 ,c0 Gc0 +c0 (k − 1)
0
0
0
c1 =0 c2 =0
1
0
c1
c1
0
0
2
1
2
(k + bi + bj + 2ci + 2cj + 1)!
where
0
γbi ,bj ,ci ,cj ,c0 ,c0 =
1
0
0
bi !bj !(2ci − 2c1 )!(2cj − 2c2 )!(bi + bj + 2ci + 2cj − 2c1 − 2c2 + 2)!
0
0
(bi + 2ci − 2c1 + 1)!(bj + 2cj − 2c2 + 1)!
2
and G is the polynomial given by Lemma 10
Proof. First consider Ik (F ). Using Lemma 10
Z
ZZ
X
c +c
2
Ik (F ) =
P dt1 ...dtk =
ai aj
(1 − P1 )bi +bj P2 i j dt1 ...dtk
Rk
Rk
1≤i,j≤d
=
X
1≤i,j≤k
ai aj
(bi + bj )!Gci +cj ,2 (k)
(k + bi + bj + 2ci + 2cj )!
(102)
Thus the first part is proven. For the second sum remember that F is
(m)
symmetric in t1 , ..., tk we see that Jk (F ) is independent of m, thus is
(1)
suffices to only consider Jk . The first integral becomes
33
Z
P
1− ki=2
0
Pk
Z 1−
c X
k
X
c
b c
2 c0
(1−P1 ) P2 dt1 =
(
ti )
c0
0
0
i=2
c =0
i=2 ti
(1−
k
X
0
ti )b t2c−2c
dt1
1
i=1
P2c
Where
is split such we have one part with t1 and the rest which can be
handled as a constant.
Z 1
c X
0
c
0 c0
b+2c−2c0 +1
(1 − u)b u2c−c du
=
(P
)
(1
−
P
)
1
2
0
c
0
0
c =0
=
c X
c
c0
c0 =0
Here P10 =
Pk
and P20 =
i=2
1
Z
0
1≤i,j≤d
ai aj
2
i=2 ti .
Z
d
X
F dt1 ) = (
ai
0
=
Pk
2
(
X
0
(P20 )c (1 − P1 )b+2c−2c +1
i=1
cj ci X
X
ci
cj
c01 =0 c02 =0
c01
c02
b!(2c − 2c0 )!
(b + 2c − 2c0 + 1)!
(103)
Hence
P
1− ki=2 ti
0
0
0
(1 − P1 )bi P2ci dt1 )2
0
0
(P20 )c1 +c2 (1 − P10 )bi +bj +2ci +2cj −2c1 −2c2 +2
bi !bj !(2ci − 2c01 )!
(104)
(bi + 2ci − 2c01 + 1)!(bj + 2cj − 2c02 + 1)!
All terms are treated as constants in the other integrals except for the terms
with P10 and P20 by Lemma 10 these terms are
Z
b!
(1 − P10 )b (P20 )c =
Gc,2 (k − 1)
(105)
(k − 1 + b + c)!
Rk
×
with b = c01 + c02 and c = bi + bj + 2ci + 2cj − 2c01 − 2c02 + 2 combining
these, (105) and (104), will give the result. In Lemma 11 it has been proven
(m)
that Ik (F ) and Jk (F ) can be written as quadratic forms of the ai . Thus
both terms can be writen as matrices over coefficients a = (a1 , ..., ad ) of
P . Moreover these will be positive definite real quadratic forms. Thus in
particular Mk can be written as
Pk
(m)
aT M2 a
m=1 Jk (F )
Mk = sup
= sup T
(106)
Ik (F )
F ∈Sk a M1 a
F ∈Sk
for two rational symmetric positive definite matrices M1 , M2 , which can be
calculate explicitly in terms of k for any choice of the exponents bi , ci . This
can be maximized and has a solution
34
Lemma 12. Let M1 , M2 be real, symmetric positive definite matrices. Then
a T M2 a
aT M 1 a
(107)
is maximized when a is an eigenvector of M1−1 M2 corresponding tot the
largest eigenvalue of M1−1 M2 . The value of the ratio at its maximum is this
largest eigenvalue.
Proof. Multiplying a by a non zero scalar doesn’t change the ratio,
so we may assume without loss of generality that aT M1 a = 1. By using
Lagrangian multipliers, aT M2 a is maximized subjest to aT M1 a when
L(a, λ) = aT M2 a − λ(aT M1 a − 1)
(108)
is stationary. This occurs when (using the symmetricity of M1 , M2 )
0=
δL
= ((2M2 − 2M1 )a)i
δai
(109)
for each i. This implies that (recalling M1 is positive definite so invertible)
M1−1 M2 a = λa
(110)
thus aT M1 a = λ−1 aT M2 a
Length of Bounded Gaps
By Lemma 12 a lower bound for Mk can be obtained. If F is chosen in the
same way as in (94) the eigenvalues can be calculated.
Theorem 2.
i) lim inf (pn+1 − pn ) ≤ 600
n
Proof. When k = 105 the largest eigenvalue of M1−1 M2 is
λ ≈ 4.0030697 > 4
(111)
Thus M105 > 4. Because θ = 1/2 − for every > 0 by the Bombierik
Vinogradov Theorem. Proposition 4 can be used. Calculating rk d θM
2 e =
d(1/2)×4.0030697/2e = 2. Hence if an admissible set with k = 105 elements
is given. The longest interval in the admissible set will give a upperbound
for the interval length. The admissible set that can be chosen is of length
600 thus by Proposition 4, Theorem 2 is proven.
35
Theorem 3. Assume that the primes have level of distribution θ for all
θ < 1. Then
lim inf (pn+1 − pn ) ≤ 12
n
lim inf (pn+2 − pn ) ≤ 600
n
Proof. In the same way as in Theorem 2 it can be shown M105 > 4 thus
r105 = 3 when θ < 1 in the same way r5 = 2. By using the admissible
set H = {0, 2, 6, 8, 12} the first part is proven. Because r105 = 3 with this
distribution of primes by Proposition 4 this means there are infinitely many
n such that there are 3 primes in the set n + hi with hi in the admissible set.
This means if the same admissible set is used as in Theorem 2 the second
part is proven.
36
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products of primes. Proc. London Math. Soc., 3:741–774, 2009.
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37