Distribution It

#A* INTEGERS **B (****/**): Integers Conference **11 Proceedings

ON THE DIFFERENCES BETWEEN CONSECUTIVE PRIME

NUMBERS, I

D. A. Goldston1

Department of Mathematics, San Jos State University, San Jos, California

e e

abqb1q@r.postjobfree.com

A. H. Ledoan

Department of Mathematics, University of Tennessee at Chattanooga,

Chattanooga, Tennessee

abqb1q@r.postjobfree.com

Received: 2/1/12, Accepted: 6/4/12, Published: 10/26/12

Abstract

We show by an inclusion-exclusion argument that the prime k-tuple conjecture of

Hardy and Littlewood provides an asymptotic formula for the number of consecutive

prime numbers which are a speci ed distance apart. This re nes one aspect of a

theorem of Gallagher that the prime k-tuple conjecture implies that the prime

numbers are distributed in a Poisson distribution around their average spacing.

1. Introduction and Statement of Results

In 1976, Gallagher [5], [6] showed that a uniform version of the prime k-tuple

conjecture of Hardy and Littlewood implies that the prime numbers are distributed

in a Poisson distribution around their average spacing. Speci cally, let Pr (h, N )

denote the number of positive integers n less than or equal to N such that the

interval (n, n + h] contains exactly r prime numbers. Gallagher then proved that an

appropriate form of the prime k-tuple conjecture implies, for any positive constant

and h log N as N, that

r

Pr (h, N ) e N.

r!

In particular, if r = 0, then we obtain by an argument using the prime number

1 During the preparation of this work, the rst author received support from the National

Science Foundation Grant DMS-1104434.

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theorem that, as N,

N

1 e (1)

.

log N

pn+1 N

pn+1 pn log n

Here, pn is used to denote the nth prime number. The purpose of the present paper

is to obtain a re nement of (1), which shows that the Poisson distribution of the

prime numbers in short intervals extends down to the individual di erences between

consecutive prime numbers. To obtain this result, we employ a version of the prime

k-tuple conjecture formulated as Conjecture H in Section 2, which is equivalent to

the form of the conjecture used by Gallagher.

Theorem. Assume Conjecture H. Let d be any positive integer, and let p be a prime

number. Let, further,

p 1

2C2, if d is even;

p 2

S(d) = p d

p>2

0, if d is odd;

where

1

C2 = 1 = 0.66016 . . .,

(p 1)2

p>2

and de ne

N (x, d) = 1,

pn+1 x

pn+1 pn =d

where pn denotes the nth prime number. Then for any positive constant and d

even with d log x as x, we have

x

N (x, d) e S(d) (2)

.

(log x)2

Here, we note that S(d) is the singular series in the conjectured asymptotic

formula for the number of prime pairs di ering by d. Our theorem shows that, for

consecutive prime numbers, the Poisson density is superimposed onto this formula

for prime pairs.

Our theorem as well as its proof are implicitly contained in the 1999 paper of

Odlyzko, Rubinstein and Wolf [11] on jumping champions.2 Without claiming any

originality, we think it is worthwhile to explicitly state and prove (2). More precise

results when d/ log x 0 will be addressed in a second paper.

2 An integer d is called a jumping champion for a given x if d is the most frequently o ccurring

di erence between consecutive prime numbers up to x.

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2. The Hardy-Littlewood Prime k-Tuple Conjectures

Let H = {h1, . . ., hk } be a set of k distinct integers. Let (x; H) denote the number

of positive integers n less than or equal to x for which n + h1, . . ., n + hk are

simultaneously prime numbers. Then the prime k-tuple conjecture of Hardy and

Littlewood [8] is that, for x,

(x; H) S(H) lik (x), (3)

where

k

1 H (p)

S(H) = 1 1,

p p

p

H (p) denotes the number of distinct residue classes modulo p occupied by the

elements of H, and x

dt

lik (x) = (4)

.

2 (log t)

k

Note in particular that, if H (p) = p for some prime number p, then S(H) = 0.

However, if H (p)

set H is called admissible. In (3), H is assumed to be admissible, since otherwise

(x; H) is equal to 0 or 1.

The prime k-tuple conjecture has been veri ed only for the prime number the-

orem. That is to say, for the case of k = 1. It has been asserted that, in its

strongest form, the conjecture holds true for any xed integer k with an error term

that is Ok (x1/2+ ) at most and uniformly for H [1, x]. (See Montgomery and

Soundarara jan [9], [10].) However, we do not need such strong conjectures here.

Using

x kx

lik (x) = +O (5),

(log x)k (log x)k+1

obtained from integration by parts, we replace lik (x) by its main term and make

the following conjecture.

Conjecture H. For each xed integer k 2 and admissible set H, we have

x

(x; H) = S(H) (1 + ok (1)),

(log x)k

uniformly for H [1, h], where h log x as x and is a positive constant.

3. Inclusion Exclusion for Consecutive Prime Numbers

The prime k-tuple conjecture for the case when k = 2 provides an asymptotic

formula for the number of prime numbers with a given di erence d. We need to

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nd a corresponding formula where we restrict the count to prime numbers that

are consecutive, and for this one can use the prime k-tuple conjecture with k =

3, 4, . . . and inclusion-exclusion to obtain upper and lower bounds for the number of

consecutive prime numbers with di erence d. This method has appeared in a series

of papers of Brent [1], [2], [3] and was used by Erd s and Strauss [4] and Odlyzko,

o

Rubinstein and Wolf [11] in their study of jumping champions.

We consider a special type of tuple Dk for which

D2 = {0, d}

and, for k 3,

Dk = {0, d1, . . ., dk 2, d}.

Here, we require that d is even. We want to count the number of consecutive prime

numbers which do not exceed x and have di erence d, namely N (x, d), and for this

we do inclusion-exclusion with

2 (x, d) = 1,

p x

p p =d

where p is also a prime number and, for k 3,

k (x, d1, . . ., dk 2, d) = 1.

p x

p p =d

p pj =dj, 1 j k 2

Inserting the expected main term, we obtain

2 (x, d) = S(d) li2 (x) + R2 (x, d) (6)

and, for k 3,

k (x, d1, . . ., dk 2, d) = S(Dk ) lik (x) + Rk (x, Dk ). (7)

We now carry out the inclusion-exclusion. We trivially have

N (x, d) 2 (x, d).

The consecutive prime numbers that di er by d are those prime numbers p and p

satisfying p p = d such that there is no third prime number p with p 0,

M

1 e

e x = ( x)k + ( x)M +1,

k! (M + 1)!

k=0

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where lies in the open interval joining 0 and x. Hence, we have

x N 2 k

x

1 d d

dt dt

= exp

k! log t (log t)2 log t (log t)2

2 2

k=0

N 1

1

x

d dt

+O

2 (N 1)! log t (log t)2

x

d dt

= exp

log t (log t)2

2

N 1

1 3d x

+O,

N N log x (log x)2

by the Stirling formula

1

(M 1)! = 2 M 1+O

M 1/2 M

e .

M

Therefore, we have proved the following lemma.

Lemma. For N 2, we have

N

x

dt

QN (x, d) = S(d)I (x, d) + ( 1)k Ek (d) + R2 (x, d)

(log t)k

2 k=3

N

+ ( 1)k Rk (x, Dk )

1 d1

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