Topics in Analytic Number Theory, Second lecture
Monday February 13
Gaps between consecutive primes
For any prime p we denote by p0 the next prime p0 > p and let
d(p) := p0 − p.
The prime number theorem states that
π(x) := #{p 6 x} ∼
x
, as x → +∞,
log x
hence
d(p) ∼ log x
on average for primes p of size x. Cramér inspired by gave a heuristic model for predicting
more precisely the distribution of d(p). Assuming that
0<α<β
are fixed constants and that we are interested in finding how many primes p < x satisfy
(1)
α log x 6 d(p) 6 β log x.
Then for h in the interval above the probability that for a prime p of size x the integers
p + 1, . . . , p + h − 1 are not primes and p + h is prime should be
h−1
1 h−1 1
1
≈ 1−
≈ e− log x
.
log x
log x
log x
Therefore the probability that (1) occurs heuristically equals
Z β
X
h−1
1
− log
x
≈
e−t dt,
e
log x
α
h∈N
h
α6 log
6β
x
by interpreting the last integral as a Riemann sum. This means that the sequence d(p)/ log p
follows a Poisson distribution; for example it suggests that
#{p 6 x : p0 6 p + log p} ∼
e−1 x
1 x
>
.
e log x
2 log x
This model should be applied with precaution; for example, if fully stretched it could
predict that there are x/(log x)2 primes p 6 x such that p + 1 is also a prime.
1
2
Note that the tail of the Poisson distribution converges to zero exponentially fast, hence
we should expect that small gaps are easier to find than large gaps. For example, for every
0 < δ < 1/100 we would have that
x
#{p 6 x : d(p) 6 δ log p} ∼ δ
,
log x
which is a positive probability. Taking δ → 0, this would imply that
lim inf
p→+∞
d(p)
= 0,
log p
the last statement is also valid if there were infinitely many twin primes. The prime number theorem immediately yields that the limit above is 6 1 and the first to prove that it is
strictly less than 1 was Erdős in 1940. It follows from the succeeding stronger result.
Theorem (Erdős): There exist positive constants , δ and x0 ∈ R such that for all
x > x0 we have
#{p 6 x : d(p) 6 log p(1 − )} > δπ(x).
To prove it we shall use a theorem proved in the first set of Exercises. It states that there
exists an absolute constant c1 > 0 such that for all x, y > 3 and even integer r in the range
r 6 y 1/5 we have
Y
y
1
πr (x + y) − πr (x) 6 c1
.
1+
(log y)2
p
p|r
For some ∈ (0,
1
)
2
we partition all primes p ∈ (x, 2x] in the following sets,
A+ := {x < p 6 2x : d(p) > (1 + ) log x},
A0 := {x < p 6 2x : (1 − ) log x < d(p) 6 (1 + ) log x},
A− := {x < p 6 2x : 2 log x < d(p) 6 (1 − ) log x},
E := {x < p 6 2x : d(p) 6 2 log x}.
Now for any constants 0 6 A < B 6 2 we have
X
#{p ∈ (x, 2x] : A log x < d(p) 6 B log x} =
r∈N
A log x<r6B log x
which by the theorem above is
6 c1
x
(log x)2
X
X1
r∈N
d|r
A log x<r6B log x
d
.
π2 (r),
3
The last sum is
6 log x
1
6 (B − A)(log x),
2
d
d6B log x
X
therefore
#{p ∈ (x, 2x] : A log x < d(p) 6 B log x} 6 ζ(2)(B − A)c1
x
.
log x
We obtain directly that
x
x
and #E 6 2 ζ(2)c1
.
log x
log x
Observe that for large enough x the prime number theorem implies that there is a prime
between 2x and (2 + 2 )x, therefore
X
(1 + 2 )x >
d(p) > (1 + )(log x)#A+ + (1 − )(log x)#A0 + 2 (log x)#A− .
(2)
#A0 6 2ζ(2)c1
x<p62x
Again by the prime number, the number of primes in (x, 2x] is at least (1 − 2 ) logx x , hence
x
.
#A+ + #A0 + #A− + #E > (1 − 2 )
log x
Combining the two inequalities with (2) yields
(1 + − 2 )#A− > K(, 2 )
x
,
log x
where
K(, 2 ) := − 22 − 3 − 2 ζ(2)c1 (1 + ) − 42 c1 ζ(2).
Choosing 2 = /n for a large enough positive integer n we see that the resulting function
K() remains strictly positive for all small enough enough .