THE DIFFERENCE BETWEEN CONSECUTIVE PRIMES, II
R. C. BAKER, G. HARMAN and J. PINTZ
[Received 3 May 2000; revised 15 November 2000]
1. Introduction
Beginning with Hoheisel [8], many authors have found shorter and shorter
intervals x ÿ x
v
; x that must contain a prime number. The most recent result is
v 0:535: see Baker and Harman [1], where the history of the problem is
discussed. In the present paper we prove:
Theorem 1. For all x > x
0
, the interval x ÿ x
0:525
; x contains prime numbers.
With enough effort, the value of x
0
could be determined effectively.
The paper has much in common with [1]; in particular we use the sieve method
of Harman [4, 5]. We no longer use zero density estimates, however, but rather
mean value results on Dirichlet polynomials similar to those that give rise to such
estimates. Compare, for example, work of Iwaniec and Pintz [9] and Baker,
Harman and Pintz [2]. Much of the improvement over [1] arises from the use of
Watt's theorem [11] on a particular kind of mean value. More accurate estimates
for six-dimensional integrals are also used to good effect. There is in addition a
device which uses a two-dimensional sieve to get an asymptotic formula for a
`one-dimensionally sieved' set; see Lemmas 16, 17. Unfortunately, these lemmas,
which would be of great signi®cance for v 0:53, are not very numerically
signi®cant when v drops to 0.525; the same applies to the `ro
Ã
le reversals'
discussed below.
Let us introduce enough notation to permit an outline of the proof. When E is a
®nite sequence of positive integers, counted with multiplicity, we write jEj for the
number of terms of E, and
E
d
fm: dm 2 Eg:
Let
Pz
Y
p < z
p;
where the symbol p is reserved for a prime variable; and let
SE; zjfm 2 E: m; Pz 1gj:
Let v be a positive number,
0:524 < v < 0:535: 1:1
Research of the ®rst author was supported in part by the National Security Agency and the National
Science Foundation.
2000 Mathematics Subject Classi®cation: 11N05.
Proc. London Math. Soc. (3) 83 (2001) 532±562. q London Mathematical Society 2001.
Let L log x; y
1
x expÿ3L
1 = 3
; y x
v «
,
A x ÿ y; x Ç Z and B x ÿ y
1
; x Ç Z;
where « is a suf®ciently small positive number.
Buchstab's identity is the equation
SE; zSE; wÿ
X
w < p < z
SE
p
; p;
where 2 < w < z; SA; x
1 = 2
counts the primes we are looking for. Our
philosophy is to use Buchstab's identity to produce parallel decompositions of
SA; x
1 = 2
and SB; x
1 = 2
:
SA; x
1 = 2
X
k
j 1
S
j
ÿ
X
l
j k 1
S
j
;
SB; x
1 = 2
X
k
j 1
S
j
ÿ
X
l
j k 1
S
j
:
Here S
j
> 0, S
j
> 0 and for j < t < k or j > k we have
S
j
y
y
1
S
j
1 o1
as x ! 1. Thus
SA; x
1 = 2
>
y
y
1
SB; x
1 = 2
ÿ
X
k
j t 1
S
j
1 o1:
We must thus ensure that not too many sums are discarded, that is, fall into the
category t < j < k.
Just as in [1] we use Buchstab's identity twice to reach the decomposition
SA; x
1 = 2
SA; x
n0
ÿ
X
n0 < a
1
< 1 = 2
SA
p
1
; x
na
1
X
n0 < a
1
< 1 = 2
na
1
< a
2
< mina
1
; 1 ÿ a
1
= 2
SA
p
1
p
2
; p
2
S
1
ÿ S
2
S
3
; say. 1:2
(Here p
j
x
a
j
.) We give asymptotic formulae for S
1
and S
2
. The piecewise
linear function n ... is larger (for given v) than its counterpart in [1]. From this
point on, ro
Ã
le reversals are employed. To illustrate this, note that
SA
p
1
p
2
; p
2
jfp
1
p
2
h 2 A: p j h ) p > p
2
gj:
If K is a region in which a
1
> 1 ÿ a
1
ÿ a
2
, we note that
X
a
1
; a
2
2K
SA
p
1
p
2
; p
2
fhp
2
h
1
2 A: L
ÿ1
log h
1
; L
ÿ1
log p
2
2K;
p j h
1
) p > h
1 = 2
1
; p j h ) p > p
2
g;
533the difference between consecutive primes
leading readily to the formula (in which h x
b
3
)
X
a
1
; a
2
2K
SA
p
1
p
2
; p
2
1 o1
X
1 ÿ a
2
ÿ b
3
; a
2
2K
p j h ) p > p
2
S
A
hp
2
;
x
hp
2
1 = 2
which we term a ro
Ã
le-reversal. The point here is that our asymptotic formulae
X
m , M
X
n , N
a
m
b
m
SA
mn
; x
n
y
y
1
1 o1
X
m , M
X
n , N
a
m
b
n
SB
mn
; x
n
1:3
require certain upper bounds on M and N; see Lemmas 12 and 13. Here m , M
means M < m < 2M; m } M means B
ÿ1
M < m < BM; B is a positive absolute
constant, which need not have the same value at each occurrence.
It will generally be bene®cial to attempt as many decompositions as possible.
There are two reasons for this. First, if there are several variables, there should often
be a combination of variables which satisfy one of our criteria for obtaining an
asymptotic formula. Second, if there are many variables, the contribution is already
quite small. To see this, note that if represents x
n
< p
n
< p
n ÿ 1
< ... < p
1
< x
l
, then
X
SB
p
1
... p
n
; p
n
y
1
L
1 o1
Z
l
a
1
n
Z
a
1
a
2
n
...
Z
a
n ÿ 1
a
n
n
q
1 ÿ a
1
ÿ ...ÿ a
n
a
n
da
1
...da
n
a
1
...a
2
n
(compare [1]). Moreover,
q
1 ÿ a
1
ÿ ...ÿ a
n
a
n
< 1 and
Z
a
1
...
Z
a
n
da
1
a
1
...
da
n
a
2
n
<
logl=n
n
n!n
:
For v 0:525 we shall have n > 0:05. Hence the contribution from p
1
< x
1 = 10
(for which one can take n 8) is at most
y
L
log 2
8
8!0:05
1 o1 < 0:000002yL
ÿ1
:
(If `asymptotic formula regions', in the sense of (1.4) below, are not discarded, we
get a better estimate still.)
However, when ro
Ã
le-reversals are used it may not always be bene®cial to perform
as many decompositions as possible. The reason for this is that with ro
Ã
le-reversals, a
sum may be replaced by the difference of two sums, each substantially larger than
the original one. If not enough combinations of variables lie in `asymptotic formula
regions', we have made matters worse. For example, when decomposing in
straightforward fashion we count
p
1
...p
n
m; p j m ) p > p
n
:
When ro
Ã
le-reversals are used we may have
p
1
...p
n
klm; p j k ) p > p
r
; p j l ) p > p
s
; p j m ) p > p
n
:
The ®rst expression gives rise to a term
q
1 ÿ a
1
ÿ ...ÿ a
n
a
n
1
a
1
...a
n ÿ 1
a
2
n
;
534 r . c. baker, g. harman and j. pintz
while the second leads to a term
q
f
1
a
r
q
f
2
a
s
q
f
3
a
n
1
a
1
...a
n
a
r
a
s
a
n
for certain expressions f
1
, f
2
and f
3
. The corresponding integral can then be larger
than the original term under consideration.
The ®nal decomposition of S
2
, given in § 6, arises from Lemmas 12 and 13,
together with formulae of the type
X
a
1
; ... ; a
r
2K
SA
p
1
... p
r
; p
r
y
y
1
1 o1
X
a
1
; ... ; a
r
2K
SB
p
1
... p
r
; p
r
1:4
discussed in § 5.
2. Application of Watt's theorem
Let T x
1 ÿ v ÿ « = 2
and T
0
expL
1 = 3
. In this section we seek a result of
the type
Z
1 = 2 iT
1 = 2 iT
0
jMsNsKsj jdsj p x
1 = 2
L
ÿ A
2:1
where Ms and Ns are Dirichlet polynomials,
Ms
X
m , M
a
m
m
ÿ s
; Ns
X
n , N
b
n
n
ÿ s
;
and Ks is a `zeta factor', that is,
Ks
X
k , K
k
ÿ s
or
X
k , K
log kk
ÿ s
:
Note the convention of the same symbol for the polynomial and its `length'. Of
course, 1 is a Dirichlet polynomial of length 1. We shall assume without comment
that each Dirichlet polynomial that appears has length at most x and coef®cients
bounded by a power of the divisor function t: thus, whenever a sequence
a
m
m , M
is mentioned, we assume that
ja
m
j < tm
B
:
(This property may be readily veri®ed for the particular polynomials employed
later.) The bound (2.1), and any bound in which A appears, is intended to hold for
every positive A; the constant implied by the `p'or`O' notation may depend on
A, B and «.
It is not a long step from (2.1) to a `fundamental lemma' of the type
X
m , M
a
m
X
n , N
b
n
SA
mn
; w
y
y
1
1 o1
X
m , M
a
m
X
n , N
b
n
SB
mn
; w2:2
with
w exp
L
log L
: 2:3
This will be demonstrated in § 3.
535the difference between consecutive primes
Lemma 1. Let
Ns
X
p
i
, P
i
p
1
...p
u
ÿ s
2:4
where u < B, P
i
> w and P
1
...P
u
< x. Then, for Re s
1
2
,
jNsj < g
1
s... g
r
s; with r < L
B
; 2:5
where each g
i
is of the form
L
B
Y
h
i 1
jN
i
sj; with h < B; N
1
...N
h
< x; 2:6
and among the Dirichlet polynomials N
1
; ...; N
h
the only polynomials of length
greater than T
1 = 2
are zeta factors.
Proof. It clearly suf®ces to prove (2.5) for
Ns
X
n , N
Lnn
ÿ s
where L is von Mangoldt's function. We now obtain the desired result by the
identity of Heath-Brown [6].
We shall refer to polynomials Ns `of type (2.4)' to indicate that the
hypothesis of Lemma 1 holds for Ns.
Lemma 2. If Ks is a zeta factor, 1 < U < T, K < 4U and M < T, then
Z
1 = 2 iU
1 = 2 iU= 2
jMsj
2
jKsj
4
jdsj p U
1 «
1 M
2
U
ÿ 1 = 2
: 2:7
Proof.ForK < U
1 = 2
and M < U
1 = 2
this is proved in all essentials by Watt
[11] in the course of the proof of his main theorem. For K < U
1 = 2
and M > U
1 = 2
we have
Z
1 = 2 iU
1 = 2 iU= 2
jMsj
2
jKsj
4
jdsj p kMk
2
1
Z
jKsj
4
jdsj
p M
1 «
U p M
2
U
1 = 2 «
:
Now suppose that U
1 = 2
< K < 4U. Using a re¯ection principle based on [10,
Theorem 4.13], we may replace K by a zeta factor of length K
0
< U
1 = 2
with
error E O1. Thus jKj
4
p jK
0
j
4
jEj
4
. Since
Z
1 = 2 iU
1 = 2 iU= 2
jMsj
2
jEj
4
jdsj p
Z
1 = 2 iU
1 = 2 iU= 2
jMsj
2
jdsj
p M UU
«
;
the general case of Lemma 2 now follows.
536 r . c. baker, g. harman and j. pintz