Download The Prime Number Double Product

International Mathematical Forum, Vol. 10, 2015, no. 2, 57 - 67
HIKARI Ltd, www.m-hikari.com
http://dx.doi.org/10.12988/imf.2015.412199
The Prime Number Double Product
Raphael Schumacher
Department of Mathematics
ETH Zürich, Switzerland
Copyright © 2014 Raphael Schumacher. This is an open access article distributed under the
Creative Commons Attribution License, which permits unrestricted use, distribution, and
reproduction in any medium, provided the original work is properly cited.
Abstract
The paper presents a new formula for the characteristic function of the prime
numbers in form of a finite double product using only roots of unity.
Keywords: characteristic function of prime numbers, double product, exponential
sums, logarithm, elementary group theory, roots of unity
1 Introduction
There exist formulas for the Möbius function [2], the divisor functions or the
Ramanujan sums involving only roots of unity. There are also many formulas for
the characteristic function of the prime numbers using the floor function and
Wilson’s theorem among other things [1].
However, to the best of the author’s knowledge, there isn’t a formula for the
characteristic function of the prime numbers involving only roots of unity without
using the floor function and Wilson’s theorem.
In this paper, we present such a formula similar to these for the Möbius function
or the divisor functions. This formula has the form of a finite double product of
simple terms of roots of unity.
The only proof found in the literature uses infinite methods, namely an infinite
series, even if the two products are finite and there is no proof based only on finite
methods.
2 Definitions
Let n  N be a natural number.
Let   (n) be the characteristic function of 1 and the prime numbers, that is:


58
Raphael Schumacher
1
if n = 1

  (n)  1
if n = prime .

0 if n = composite
x
Define  (x)     (n) as the prime counting function. (the number of prime

numbers  x .)
n2
Let also be pn  n - th prime number .


3The Formula
Theorem: (The Prime Number Double Product)
Let n  N be a natural number. Then one has:

n1 n1 
2 ikl 
1
  (n)  n1   1 e n .
n k1 l1 

This formula is interesting because it is a product of (n 1)2 1 complex
numbers giving as result a real number, moreover a natural number.
It is surprisingly that the natural number nn1 is exactly equal to the double
product term if n is a prime number.

Now it is time for the proof.

Proof:

In the proof we distinguish three cases:
1.) n 1
2.) n  composite

3.) n  prime

Let us start with the first.

The prime number double product
59
1.) n 1.
In this case the formula takes the form:

0
0
1
1    (1)  0   1 e2 ikl  1,
1 k1 l1
which is already true.

2.) n  composite k1 *l1 for some k1,l1  N and k1,l1  n -1.

Then we have:
 2 iknl 
n1 n1 
2 ikl 
2 ikl n1 
2 ik1l* 
11

1
1


0    (n)  n1   1 e n  n1   1 e n  1 e n 1 e n ,
n k1 l1 
 n k1 l1 
l*1 




kk1
l*l1
n1 n1
0
which is again true, because

 2 ik1l1   2 in 
2 i
1 e n  1 e n  1 e  11  0 and 0 * x  0 x  C.


 
3.) n  prime.


This is the hardest part of the proof, but also the most beautiful.
In this case we should have:
n1 n1 
2 ikl 
1
1    (n)  n1   1 e n .
n k1 l1 

Since, if n is a prime, every term of the double product is not contained in
,0 , we can take the logarithm on both sides to get:




 1 n1 n1  2 ikl 
0  ln(1)  ln   (n)  ln n1   1 e n 

n k1 l1 
60
Raphael Schumacher
 2 ikl 
n
.
 0  (n 1)ln(n)  ln1 e


k1 l1
n1 n1


xa
Now it is time to use the Taylor expansion of ln(1 x)  
a
a1
(because everything is well-defined) to get:
n1 n1 
1 2 niakl
0  (n 1)ln(n)   e
.
a
k1 l1 a1
Next we interchange the sums and obtain:

1 n1 n1 2 niakl 
0  (n 1)ln(n)   e
.
a


a1
k1 l1

1 n1 n1 2 niakl 
 (n 1)ln(n) .
   e

a1 a k1 l1


n1 n1

Define now
fn (a)   e
2 iakl
n
.
k1 l1
Now we study the behaviour of fn (a) for every a  N and n  prime.
First, let a  n *b for some b  N .

Then we have:


n1 n1
fn (a)  fn (n *b)   e
k1 l1

2 inbkl
n
n1 n1
ibkl
  e1223
 (n 1)2 .
k1 l1
1
Second, let a  n *b for all b  N .
In this case we have to use two well-known facts from algebra.


The prime number double product
n
1.)
e
2 im
n
 0 n  N 2 =>
m1

2.)
61
n1
e
2 im
n
 (1) n  N 2 , because e2 i  1.
m1

Z 

 pZ  is a fieldif
p  prime .




Z 
b  Z pZ  and two
=> Every element
 a   pZ  has a unique inverse


different elements have not the same inverse.

  b  n *a1
=> a *b  n








Z n  Z !b  Z with a *b  n
=> a  
pZ 

 pZ 
 pZ 

and for no two"a's" we have the same "b's".
(a1,a2 and a1b  a2b  n <=> a1  a2 )

Now we can compute fn (a) for n  prime and a  n *b.
We have that Z nZ is a field and there is an isomorphism between the sets


 


 2 im n1
n 
e 
and Z nZ .

m0
 
 2 im n1
Therefore, we can think of e n  as a model of

m0

Z nZ  has Z nZ   (n 1)
Z nZ .
*
Furthermore,
identity (=1).
Nowfn (a) 

n1 n1

k1 l1

2 iakl
n

e
 2 ial k
 e n  .

k1 l1 
n1 n1
Because a  n *b and by fact 2.), we have


elements without counting the
62
Raphael Schumacher
 2 ial n1  2 im n1  2 ikm n1  2 iq n1
e n   e n   e n   e n  (without the identity  1).

l1 
m1 
m1 
q1
 2 ial k n1 n1  2 im k n1 n1 2 ikm n1 n1 2 piq
Thus fn (a)  e n    e n     e n   e n .
 k1 m1
 k1 m1
k1 l1 
k1 q1
n1 n1


By fact 1.) we have:
n1
 2 iq  n1
n
fn (a)  
 e 
  (1)  1  (n 1) .
k1 q1
 k1
k1
n1 n1
(1)
In total, we get for n  prime:

(n 1) if a  n *b for all b  N
fn (a)  
.

2
(n 1) if a  n *b for some b  N
To finish the proof, we have to show that:

1 n1 n1 2 niakl   1
 a  e   a fn (a)  (n 1)ln(n) .
 a1
a1 k1 l1


1
For this, define now gn (a)  
if a  n *b for all b  N
(n 1) if a  n *b for some b  N
=> fn (a)  (n 1)gn (a) .



=>

a1


fn (a)
g (a)
 (n 1) n
 (n 1)ln(n)
a
a
a1
.
The prime number double product
63




gn (a)
 ln(n) .
a
a1
In fact, it turns out that this formula is not only valid if n is a prime, moreover it
holds true for all n  N .
The first few cases of this formula are:


(1)a1
1 1 1 1 1
ln(2)  
 1      ....
a
2 3 4 5 6
a1

ln(3)  
g3 (a)
1 2 1 1 2
 1        ....
a
2 3 4 5 6
a1

ln(4)  
g4 (a)
1 1 3 1 1 1 3
 1          ....
a
2 3 4 5 6 7 8
a1
n*k g (a) 
gn (a)
ln(n)  
 lim  n 
k
a
a1 a 
a1

The last general formula has an easy proof.

n*k g (a) 
n*k 1 k n 
n*k 1 k 1 
gn (a)
n

lim

lim


lim
 a k a  k a  n * b  k a   b 
a1

a1

a1
a1
b1
b1 



 lim ln(nk)    onk1  ln(k)    o1k 
k
 lim ln(n)  ln(k)    onk1   ln(k)    o1k 
k
 lim ln(n)  o1k  ln(n).
k
This proves the Prime Number Double Product.


Corollary 1: (The real version of the Prime Number Double Product)
Let n  N be a natural number. Then one has:

64
Raphael Schumacher
2 n1 n1 n1 n1 
2kl 
  (n)   2    1 cos
.
 n 
 n  k1 l1 
Proof:

We must only take the square of the absolute value of the Prime Number Double
Product, to get its real version.
Here we use
z  z * z , z1 * z2  z1 * z2 z, z1, z2  C and Euler' s theorem eix  cos(x) i sin(x).
2

Thus
1 e
2 ikl 2
n
2
 2 kl 
2 kl 
2 kl 
2kl 
 1 cos
 i sin
  1 cos
 i sin

 n 
 n 
 n 
  n 
2

2kl 
2 kl 
2 kl 
2 kl 
 1 cos
 i sin
1 cos
 i sin

 n 
 n 
 n 
 n 

 2kl 
 2kl 
 2kl 
 2kl 
2  2kl 
 2kl 
 1  cos
  i sin
  cos
  cos 
  i sin
 cos

n
n
n
n
n









  n 
 2kl 
 2kl   2kl 
2  2kl 
 i sin
  i sin
 cos
  sin 

n
n
n



 

 n 


 2kl   2  2kl 
 2kl 
 2kl  
2  2kl 
 1  2 cos
  cos 
  sin 
  2  2 cos
  21  cos
 .
n
n
n
n
n 

 
 










1
Thus we obtain:
n1 n1 
2 ikl 
1
1
  (n)    (n)  n1   1 e n   2(n1)
n k1 l1 
 n
2
2


n1 n1
1
n 2(n1)

n1 n1
2 ikl
 2 ikl 
1
  1 e n   n2(n1)   1 e n
k1 l1
k1 l1
n1 n1
2
2
2
n1
2kl  2 (n1) n1 n1 
2kl  2 n1  n1 n1 
2kl 
 2(n1)   1 cos
  2    1 cos
.
 n   n  k1 l1 
 n 
n  n
k1 l1 
  21 cos
k1 l1
This is the claimed formula.

In the next step, we can use the two Prime Number Double Product formulas to
get two formulas for  (x) and pn .


The prime number double product
65
Corollary 2: (Two formulas for  (x) )
Let x  R be a real number. Then one has:

n1 n1
1   2 nikl 
1.)  (x)   n1   1 e .
n k1 l1 

n2
x
2 n1 n1 n1 n1 
2 kl 
2.)  (x)   2    1 cos
.
 n 
n2  n  k1 l1 
x

Proof:

x
Just use  (x)     (n) and insert our two formulas for   (n) .
n2

Corollary 3: (Two formulas for pn )


Let n  N be a natural number. Then one has:


 2 ikl 2 
   t 1 m1m1 2 mikl 2 
 t1 1 m1m1

  1e
  1e m 
n 
n 




n 1 
 
 
 m2 m m1 k1 l1 
 m2 m m1 k1 l1 







1
2
1.) pn   t *1 (1)2
1
(1)

.
4 t1 





2

2 

2
m1

m1 m1m1

t 2 m1 
m1m1
2 kl  
2 kl  

 t1 2 m1 


  1cos 
  1cos 


n   2 
n   2 




m
m




 m2  m 
 
 m2  m 
 
k1 l1
k1 l1





1
2
2.) pn   t *1 (1)2
1 (1)
.
4 t1 





n 2 1


Proof:
It is well-known that pn  n2 +1 n  N.
(because pn  nln(n) nln(ln(n)) n  6 [3])
Let now be n  N and t  N. Note that  pm   m.


0
0
if t  pn
if t  pn 1




Observe that n, (t )  1 if pn  t  pn1 and that n, (k1)  1 if pn 1  t  pn1 1


if t  pn1
if t  pn1 1
0
0


66
Raphael Schumacher
such that 
p ,t
n
0
if t  pn

1
if t  pn 1 if t  pn
 n, (t) 1 n, (t1)  
 
.
0 if pn 1  t  pn1 1 0 if t  pn

if t  pn1 1
0
We
 also have the following formula for i, j :
i, j 



1 if i  j
(i j ) 2
1
1 (1)2
 
.
j
2
0 if i 
Therefore, we have that:
n 2 1
n 2 1
t1
t1
pn   t *  pn ,t   t *n, (t ) 1 n, (t1) 
n 2 1
 t *
t1
n 2 1

 


 






(n  (t ))2 
( n  ( t1))2 
1
1
1 (1)2
1 1 (1)2

 2

2
( n  ( t ))2 
(n  (t1))2 
1
2  1 (1)2

  t * 1 (1)2


4 t1
( n  ( t ))2
( n  ( t1))2
1 n 1
  t * 1 (1)2
2 1 (1)2
4 t1
2

( n  ( t ))2
( n  ( t1))2
1 n 1
  t * 1 (1)2
1 (1)2
.
4 t1
2

Inserting now the two formulas from Corollary 2 for  (x) into this final
expression, we get the two formulas from Corollary 3 for pn .
4 Conclusion



With this, I end my article over the Prime Number Double Product by noting that
this formula, even if it is useless for primality testing, is only true, because three
different areas of mathematics work perfectly together, namely, number theory,
analysis and algebra. For me it was unexpected that a finite region in the complex
plane (complex version) or the real line (real version) determines the distribution
of the prime numbers in a nontrivial way.
The prime number double product
References
[1] http://mathworld.wolfram.com/PrimeFormulas.html
[2] http://en.wikipedia.org/wiki/Moebius_function
[3] http://en.wikipedia.org/wiki/Prime_number_theorem
Received: December 19, 2014; Published: January 23, 2015
67