Prime Spacing and the Hardy-Littlewood Conjecture B

Prime Spacing and the Hardy-Littlewood Conjecture B
David Schmidt


I. Introduction

For junior seminar I used a computer program to generate large primes
and then tested to see if prime spacings were exponentially distributed, if
the spacings between successive k-spacings for a given k were exponentially
distributed, and to test the Hardy-Littlewood Conjecture B. All three
appeared to hold. Large numbers (on the order of 1015) were tested for
primality using the java math library function IsProbableprime. I chose to
start at a large number (1015) and then consider a smaller block of numbers
(107) around that large number so that normalizing term (1/logn) would be
constant, and also because I felt that while a number of experiments had
considered all primes up to a given number, fewer had considered a block of
numbers around a much larger number.


II. Theoretical Background

There are two major theoretical results that are very related to
these problems of prime spacing and distribution. The first is the Prime
Number Theorem, and the second is the Hardy-Littlewood Conjecture B. For
the Prime Number Theorem I will follow the treatment of A. E. Ingham in his
book "The Distribution of Prime Numbers", and for the Hardy-Littlewood
Conjecture B, I will present a heuristic proof that can also be found in
Michael Rubinstein's paper "A Simple Heuristic Proof of Hardy and
Littlewood's Conjecture B." For a different treatment of the Conjecture,
another good source is the paper in which the Conjecture was first printed,
namely "Some Problems of 'Partitio Numerorum'; III: On the Expression of a
Number as a Sum of Primes" by Hardy and Littlewood.
The Prime Number Theorem states that ((n) is asymptotic to n/log(n)
as n((, where ((n) is defined to be the number of prime numbers less than
or equal to n. To begin the proof we start by defining two other related
functions, Chebychev's functions ((n) and ((n) as
where the first extends over every combination of prime p with a positive
integer m for which pm?n.
We wish to compare the values of these functions with ((n), so we
notice that
with the right hand side only containing a finite number of non-zero terms,
because ((n) = 0 when n0, as a single-valued function having as
its only singularity in this half plane a simple pole with residue 1 at
s=1.
From our background with series we know if X(1


Then writing [x] = x - (x), so that 0(x (, with ( any fixed positive number, and therefore represents a
regular function of s in ( > 0.
Next we need to develop some bounds upon ((s). A denotes a positive
constant.
Theorem 3. (((s)( < Alog(t) (((1, t(2),
(('(s)( < Alog2(t) (((1, t(2),
(((s)( < A(()t1-( ((((, t(1) if 0 0,

for ( > 1. This means that 1+it cannot be a zero of ((s), because if it
were, then since ((s) is regular at 1+it and 1+2it, and has a simple pole
at 1, the left hand side would tend to a finite limit, and the right hand
side to infinity when ((1+0.
Now, in proving the rest of the theorem we may suppose 1 ( ( ( 2,
since when ( ( 2
If 1 ( ( ( 2, t ( 2, then from above we have
So, since log(2t) ( logt2 = 2logt
Now, let 1 < ( < 2, then if 1 ( ( ( (, t ( 2
Now choose ( = ((t) so that
assuming t large enough to ensure ( < 2. Then
Now that we have established enough properties of the ( function, we
look to deduce from them properties of ((n). They are related as
For s>1. Now, to avoid problems of convergence, instead of working
directly with ((n) we will work with
We shall first prove the formula
Where x > 0, c > 1, and the path of integration is the line ( = c.
The proof is based on
Theorem A. If k is a positive integer, c>0, y>0, then
If y ( 1, and 0 if y ( 1. Notice that the integral is absolutely
convergent, so denote by J the infinite integral and by JT the integral
from c-iT to c+iT. Cauchy's theorem of residues allows us to replace the
line of integration in JT by an arc of the circle C with center s = 0 and
passing through the points s = c ( iT. If y(1, we use the arc C1 which
lies to the left of the line ( = c, assuming T so large that R>2k, where R
is the radius of C. This gives JT = S + J(C1), where S is the sum of the
residues of the integrand at its poles 0, -1, -2, ..., -k, and J(C1) is the
integral along C1. Now on C1 we have ( ( c, and so (ys( = y( ( yc, since y
( 1. Also, (s+n( ( R-k > R/2. It follows that
So by JT = S+J(C1), JT(S as T((, so J=S. But,
Which proves the theorem when y ( 1. For the case y ( 1 the proof is
similar except that the right hand arc is used, and no poles are passed
over.
To deduce the full desired result we have for x > 0, by Theorem A, if
c > 0,

If c>1 the order of summation and integration may be interchanged,
yielding the desired result
Finally now we are prepared to take the crucial step in the proof of
the Prime Number Theorem, developing an asymptotic formula for (1(x).
Theorem 5. We have (1(x) asymptotic to x2/2 when x((.
Suppose throughout x>1. We have from before where c > 1, and (c)
denotes the line ( = c,

By theorems 2,3, and 4, g(s) is regular in ( ( 1, except at s = 1, and
Now take ( > 0, and take the infinite broken line L1 + L2 + L3 + L4 + L5
where L1 is the line from 1-i( to 1-iT, L2 from 1-iT to a-iT, L3 from a-iT
to a+iT, L4 from a+iT to 1+iT, and L5 from l+iT to 1+i(, and where T = T(()
is chosen so that
and then a = a(T) = a(() (0 < a < 1) so that the rectangle cut into the
critical strip contains no zeros. The first choice is possible because of
the constraint on g, the second because ((s) has no zeros on the line ( = 1
and (since it is regular) is zero at a finite number of places in the
region ½ < ( < 1.



By Cauchy's theorem we obtain
with the term ½ coming from the pole at 1. Now, by our choice of L the
integrand is regular between and on the lines (c) and L (except at s = 1)
and integrating around a closed contour bounded by portions of (c) and L
and by segments of the lines t = (U, where U>max(t0,T), the integrals along
the latter segments are
and thus tend to 0 when U((.
Now consider J = J1 + J2 + J3 + J4 + J5 where Jn is the integral around Ln.
Since g(s)xs-1 takes conjugate values for conjugate values of s,
And also, since x > 1,
Where M = M(T,a) = M(() is the maximum of (g(s)( on the finite segments L2,
L3, L4. So we have
And thus (1(x)/x2 ( ½ as x((. The final step is to deduce from this, the
corresponding relation for ((n). For this we need
Theorem B. Let c1, c2, ... be a given sequence of numbers, and let
If cn ( 0 (n=0,1,2,...), and C1(x) is asymptotic to Cxc as x((, where C and
c are positive constants, then C(x) is asymptotic to Ccxc-1.
Let 0 < a < 1 < b. Since cn ( 0, C(u) is an increasing function,
hence for x > 0,
Let x((, keeping b fixed, then since C1(y)/yc(C when y((
By considering the interval (ax,x) we similarly arrive at
By taking a and b very near to 1 we can make the two right hand sides very
near to Cc, thus
So then with Theorem B, and Theorem 5, and Theorem 1 we arrive at the
desired result, the Prime Number Theorem, namely that ((n) is asymptotic to
n/log(n).
The other major theoretical topic to be discussed is the Hardy and
Littlewood Conjecture B, which states that there are infinitely many prime
pairs (p, p+m) for every even m. If (m(n) is the number of pairs less than
n, then (m(n) is asymptotic to
Two outside mathematical ideas will be needed in our consideration of
this problem. The first is given a set A = {1,2,...,a}, and two subsets