Turkish Journal of Analysis and Number Theory, 2015, Vol. 3, No. 4, 97-103
Available online at http://pubs.sciepub.com/tjant/3/4/2
© Science and Education Publishing
DOI:10.12691/tjant-3-4-2
Study of Some Sequences of Prime Numbers Defined by
Iteration
Idir Sadani*
Department of Mathematics, University of Mouloud Mammeri, Tizi-Ouzou. Algeria
*Corresponding author: sadani.idir@yahoo.fr
Abstract We study the properties of prime number sequences obtained using a well-defined equivalence relation
. It will be seen that the elements of each class ̇ of are all prime numbers which constitute the fundamental
object of our study. The number of prime numbers of each class less than or equal to a given quantity , the number
of the different equivalence classes and some other results will be deduced.
Keywords: distribution of prime numbers, equivalence relation, iteration, asymptotic formula
Cite This Article: Idir Sadani, “Study of Some Sequences of Prime Numbers Defined by Iteration.” Turkish
Journal of Analysis and Number Theory, vol. 3, no. 4 (2015): 97-103. doi: 10.12691/tjant-3-4-2.
1. Introduction
Let be the set of prime numbers, and for all
,
let
denote the number of prime numbers less than or
equal to . The prime number theorem which was shown
independently by de la Vallée Poussin [1], and Hadamard
[2] in
, states that:
  x ~
x
, as x  
ln x
(1)
or
 ( x)  Li( x), as x   ,
where
is the logarithmic integral of
defined by:
x dy 
 1 dy
Li  x   lim  ‍

.
1 ln y 
 0  0 ln y


1
 n   pn  Li  n  ~ n ln n
as n   .
p x
this function count the number of primes less than .
2) We define the following functions:
We can give an equivalent statement for this theorem as,
for example, let
denote the n'th prime number. Then
1
conjectures, such as Goldbach's Conjecture and Twin
Prime Conjecture.
The structure of this paper is as follows. In section 2,
we begin with the definition of our equivalence relation
and its equivalence classes ̇ where is a prime number,
which constitute the fundamental object of study, and we
propose some preliminary results. In section 3, we exhibit
the main results of this work by introducing and studying
the functions
and
. The methods used
here are part of elementary number theory and we have
attempted to present the ideas in as elementary a way as
possible. Finally, in section 4, we give some open
questions related this subject.
Notation
1) We set
  x    1,
n times
 ( x)      ( x),
n
(2)
One of our objectives here is to use a restriction of the
function
to
to study intrinsic properties of some
sequences of primes defined by iterations. The point of
departure for this study is the construction of an
equivalence relation that we denote by . The purpose of
this equivalence relation is to show first, that there is a
recurrence relation between prime numbers which can be
arranged in an infinity of well-defined classes dependent
on the initial value. Second, the use of its properties with
the famous prime number theorem is one way to find and
prove other results for the possible applications in number
theory. Part of our motivation came from the prime
number theorem and the Riemann hypothesis. Also, it was
an attempt to establish the relationship between these
prime numbers classes and several as-yet unproved
n times
 n ( y)   1  1   1 ( y),
where is the composition operator.
3) In many situations, we search to estimate ∑
where
. Then, we use the following formula:
 f ( p) 
p x

n x
,
x
f ( n)
f (t )

dt.
ln n
ln t
2
2. Preliminary results
2.1. The Classes ̇ and Its Elements
We start with the following lemma:
(3)
Turkish Journal of Analysis and Number Theory
Lemma 2.1. Let
be the restriction of to . Then,
is a bijection and its inverse function is
.
Notation. Throughout this paper, we simply use the
notation to designate the restriction of to the set
instead of using
.
The proof of the following theorem is obvious.
Theorem 2.1. We define the relation on the set of prime
numbers defined by: if and are two prime numbers,
if and only if:
1)
for any prime number ,
2) there exists
such that
.
Then, is an equivalence relation.
Notation. The elements of the equivalence class ̇ are
defined by:


p  ,  2  p  ,   p  , p,  1  p  ,  2  p  , .
The smallest element of ̇ , which we denote by , is
called the origin of the class ̇ . Then, in this case, we note
that the sets ̇ and ̇ have the same elements.
Example 1.
2  2,3,5,11,31,127, ,7  7,17,59,.
0
. The
 {2,7,13,19, 23, 29,37, 43},
and we denote by
the set of all origins less than or
equal to .
Theorem 2.2. Let be a prime number. Then,
is not
a prime number implies that, for all
, there is no an
integer such that
.
Proof. Let
. By the prime number theorem,
represents the -th prime number which we denote by
.
Next,
, implies that
which is the
-th prime number, so we obtain the
formula:


 1  p p    1  1  p    2  p   p
 1 p 
.
(4)
 1( p)
  2 ( p)   1 (p
 1( p )
that is the
)   1 ( 2 ( p)),
-th prime number. We can write
 1 (p
 1( p)
)   1 ( 2 ( p))  p
 2 ( p)
  3 ( p)
Inductively, we obtain the following general formula:
p
 ( n1) ( p )
  n ( p), n  1.
To each number , we associate the set
is a prime
if and only if
But, for all
is prime,
which proves what we wanted.
Remark. The cardinality of the set
is infinite.
Theorem 2.3. There exists a partition of the set of
prime numbers defined by
P  {Ap1 , Ap2 ,, Apn ,},
such that
, where
and
is not prime number.
Proof. On the one hand, according to the prime number
theorem, the number of prime numbers belonging to
is equal to
. On the other hand, by
the Theorem 2.2, there are
prime
numbers
which
do
not
belong
to
. We denote these numbers
by
, where
. So, we
obtain
of sets
which are defined by:
l p
k
( pk )},
such that ∑
and
. Finally, it is not difficult to see that the sets
constitute a partition of the set of prime
numbers less than or equal to . So, since is arbitrary in
, letting tend to
we obtain an infinity of sets
which then form a partition of the set .
Theorem 2.4. Let be a finite set of prime numbers
defined by
A  { p1, p2 ,, pn },
such that
are consecutive. Then,
there exists at least one set ̇
where
, such that
̇
.
Before giving the proof of this result, we give an
illustrative example.
Example 2. Let be a set which defined by
A  {5,7,11,13,17}.
Now, we use the equation (4) which composed by
gives
p
and we show that
number. Hence, we have,
, which implies that
Apk  { pk ,  1 ( pk ),, 
Notation. We denote by
the set of all origins
set
is given explicitly by:
98
(5)
:
Ap  { n ( p), n  0}  { p,  1 ( p),  2 ( p),}.
Now, suppose that
is not a prime number, we
must prove that for all
, there is no
such that
. To show this, suppose that there exists a
positive integer such that
, i.e.,
• The class of the integer
is ̇
and its
cardinality is greater than .
̇ , therefore the integer constitutes
• We notice that
the origin of a new class which is ̇
and its
cardinality is greater than . It only remains to see that the
prime number
does not belong to the two classes ̇ and
̇ . Thus the prime number
constitutes the origin of a
new class ̇ and clearly its cardinality is .
Proof of theorem 2.4. We suppose that each class
̇
containing at least two elements i.e., the
cardinality of ̇ is equal or greater than , and we suppose
that
is prime. According to the theorem of prime
numbers, the interval
contains
prime numbers. This leads to the two
following cases:
• If
, i.e., there exists only one prime number in
, namely . Therefore, we obtain
and
is not a prime number since
and
are
99
Turkish Journal of Analysis and Number Theory
consecutive. Then ̇ contains only one element in which
is the prime number .
• If
, i.e., there exist at least two prime numbers in
, namely
. We suppose that where
, is a prime number. Then
is not prime
and since
, then, the unique element of ̇
is
.
Finally, in both cases, there exist at least one class ̇
where
, such that is a unique element of this class
in , i.e., ̇
. The proof now is completed.
We have the following definition:
Definition 1. Let be the set defined as in the Theorem
2.4 and is a prime number belonging to . We say that ̇
is an outside class of , if
  p  is not prime in A and 1 ( p)  A i.e.,| p | 1in A.
if
̇
n
x0
x0
 lim
 lim  ln xi
n xn 1 n p  xn  n i 0
lim
 lim ln x0  ln ln p ( xi ).
n
i 0
Consequently,

x0 
  ln xi  ln x0  ln p( xi ),
e i 0
i 0
2.2. Study of
The function
Therefore, the sequence
,
, is a contraction.
defined by:
x
xn1  p( xn )  n ,
ln xn
admits a fixed point. Since
decreasing and is bounded
below by
, then it converges to the single
fixed point .
⏞
Notation.
, where
is the
function composition operator.
Theorem 2.5. Let
be an initial value. Then


i 0
i 0
p( x0 )  e ln p( xi )  e ln pi 1 ( x0 ),
Proof. We set
x1  p( x0 ) 
(6)
lim xn  e,
n
the formula (10) is obviously equivalent to

p( x0 )  e ln p( xi ).
i 0
Which is the desired result.
Lemma 2.2. Let
be the sequence which is
decreasing and bounded below by , and let
be a
real number. Then, the number of iterations, which denote
by
, is depend on and the initial value
, and
given by :
niter ( x,  ) ~ 
x0
 ln x1  ln x0  ln 2 x0
ln x0
where
is a bounded value, and
depend on .
Proof. We have
where
is the derivative of the function
x3  x2  p' 2  x2  x1  p'(2 ) p'(1 ) x1  x0 ,
where,
Inductively, we obtain
, such that
n
Or in an equivalent way, since,
,
and
xn  xn1   p'(0 )  .  x0  x1  .
n
We can extract the value of
And combining all these, we obtain
n
(7)
and replacing
i 0
Which is equivalent to
Then, passing to the limit, we obtain
, i.e.,
xn1  xn  p'(0 ) . x1  x0
x
xn1  p( xn )  n  ln xn1  ln xn  ln 2 xn .
ln xn
n
x0
  ln xi .
xn 1 i 0
(11)
x2  x1  p'(1 ) x1  x0 , 1  [ x1, x0 ],
Next, there exists a real number
x
x2  p( x1 )  1  ln x2  ln x1  ln 2 x1
ln x1
n
ln c
ln x

, as x   ,
lnln0 lnln0
 n

xn  xn1    p'(i )  . x1  x0 , n  [ xn , xn 1 ].


 i 1

and we have
ln xn1  ln p( xn )  ln x0   ln 2 xi .
(10)
and since
in , we say that this class is an inside class of
.
Remark. According to the Definition 1, the number is
the smallest element of the class ̇ in , therefore, it is the
origin of this class i.e.,
.
(9)
n 1
niter 
(8)
from the above equality:
ln( xn  xn1 ) ln( x0  x1 )

,
ln p'(0 )
ln p'(0 )
by its value, we get
ln( xn  xn1 )
ln( x0  x1 )

ln(ln 0  1)  2lnln0 ln(ln 0  1)  2lnln0
Now, according to the definition of , the sequences
and
are bounded, then it holds for the difference
Turkish Journal of Analysis and Number Theory
which we denote it by . Then, we have, by
and
:
setting
ln c
niter  x,   
ln  ln 0  1  2lnln0

 ( x, p0 ) 
ln( x  x1 )
ln(ln 0  1)  2lnln0
 ( x, p0 )~lnx  o(ln x).
n
n
i 0
i 0
ln x0  ln  ( xi ) ~ ln x0  ln p  xi  ~
ln(ln 0  1)  2lnln0  lnln0 as x   ,

ln c
ln x

, as x   .
lnln0 lnln0

p0  p  x
p p0
n
i 0
i 0
ln ln p ~ ln x  ln p0 .
Finally, for all fixed
, then
3. Principal Results

3.1. Definition and Estimate of the Functions
and
We set the following definition:
Definition 2. Let
. We define

1
1, l  .
p0  p  x
p  l ( p0 )
Then
counts the number of prime numbers
less than or equal to belonging to the class ̇ .
Lemma 3.1. We have
 iter ( x, p0 )~niter ( x,  )~niter ( x, p0 ), as x  .
Proof. According to the definition of , there exists
such that
. Next, from the prime number
theorem, we have
 ( x)~p( x) 
n
ln ln x0   ln ln  ( xi )  ln ln x   ln ln  ( xi )
then,
p0  p  x
p p0
p0  p  x
p p0
with
3) iter  x, p0  

1
p0  p  x
p p0

1~
p0  p  x
p  l ( p0 )

  x, p0  
not a
l 1

1~
1
  p x
  p x
p p0
p ~p l (  )
as
Definition 3. Let
. We define the functions
and
as follows:

p0  p  x
p p0
ln ln p.
ln ln x
 ln x 
 O
 , x  .
 ln ln x 

ln ln p  ln ln x
p0  p  x
p p0
 iter  x, p0  
1  niter ( x,  )~niter ( x, p0 ),
  x, p0  
  x, p0 

1
p0  p  x
p p0
Then we obtain
  x, p0 
ln ln x
.
On the other hand, for all
, we have
  x, p0   ln x
n ( x,  )

ln ln p ~ ln x  o  ln x  .
  iter  x, p0  ln ln x.
And,

tend to infinity, we have
Proof.
1) For the proof of the first formula, we have, on the one
hand
( p0   n ( x))~( pn ( x)   ), x   .
 iter ( x, p0 ) 
and
i.e.,
Theorem 3.2. We have,
  x, p0 
1) iter  x, p0  ~
, x 
ln ln x
ln x
 ln x 
2) iter  x, p0  ~ iter  x  ~
 o
, x 
ln ln x
 ln ln x 
x
, x     n ( x)~p n ( x), x   .
ln( x)
Moreover, supposing that
prime number, it follows that
x0 x0

,
p0 p0
which is equivalent to
x  x1  x as x   ,

p  x, p p0
lnlnx
ln ln p.
lnlnp
Proof. In view of the proof of Lemma 3.1, we have,
0   as x   ,
 iter ( x, p0 ) 

Theorem 3.1. We have
(12)
Finally, we have the following limits:
niter ( x,  ) ~ 
100

with
1   ln   n ln x 
x  p  x

p p0


  ln   ln ln x   iter  x, p0    iter x , p0

  ln x  
Then
 ( x, p0 )
ln   ln ln x
1
p0  p  x
p p0

101
Turkish Journal of Analysis and Number Theory

 iter  x, p0    ln x  
  x, p0 
ln δ  ln ln x
1   1   iter t, p0    iter 
.
p t
Now,
according
to
Lemma
3.1,
(
) , and then for x sufficiently
large (depending on  ),
,
and thus
 iter  x, p0  
Now, for all
For this , so that
we have
  x, p0 
  ln δ  ln ln x 
thus
 iter  x, p0  ln ln x    x 
.







  x, p0 
ln ln x
  x, p0 
ln ln x
~
~  iter  x  , x  .
 iter  t , p0 
x
x , p0  
dt
t ln ln t
x
2 ln t
x , p0  
dt
x t ln t ln ln t
x
2
x , p0  
dt.
x t ln ln t
x


By using
, since
  x, p0    iter  x, p0  ln ln x 
.
ln x
ln ln x  ln x.
ln ln x
we obtain
2) According to Theorem 3.1, we have
 iter  x, p0  ~
p x
, we can choose more near to 1.
, and for sufficiently large,
 iter  x, p0   1   

x , p0   iter  t , p0  ,
ln x
 ln x 
 o

ln ln x
 ln ln x 


(
)
 ln x 
 iter  x, p0  ln ln x    x, p0   O ln x  O 

 ln ln x 
Then,
3) Concerning the third equality, we have
3.2. Definition and Estimate of
  x, p0    iter  x, p0  ln ln x.
3.2.1. Definition and Estimate of
Evaluate now the difference
 lnlnx 
 iter  x, p0  ln ln x   p x, p p 
ln ln p.
0  lnlnp 
Then
 lnlnx 
 iter  x, p0  ln ln x   
 ln ln p
lnlnp 
p  x, p p0 


 lnlnx 

  ln ln x   lnlnp  ln ln p 



p  x, p p0 


 lnlnx 
 ln ln x  
 ln ln p 
lnln
p



p  x , p p0 
+


 lnlnx 
 ln ln x  
 ln ln p 
 lnlnp 

x  p  x , p p0 

 lnlnx 

 ln ln p
lnlnp 
p  x , p p0 









x , p0  
However, for
x  p  x, p p0
x
[√
1 pt 

x

dt
t ln t
p
ln p0 .
 .
(13)
i 0 i
dt
1 p t
t ln t
1 pt 
1
p
Next, let
obtain
]
p x, p p0

p0  x
Proof. Suppose that the number of different classes is
finite as
. We know that
x  p  x, p p0

the function defined by
lim  c ( x)   .
x
x  p  x, p p0
0.
x
 ln ln x  ln ln p 
x , p0 
 1, p0 
p0  x
2. We denote by

x  p  x , p p0
Then,
 c ( x) :
.
Example 3. In
, the value represents the origin
̇
of the class but the values
do not, since they
belong to the same class ̇ . The value represent the
origin of the class ̇ . Thus in this case, we have
.
We have the following result:
Theorem 3.3. We have,


Definition 4.
1. Let be a positive real number. We denote by
the number of different classes ̇ such that
Precisely,
0 ( x) 


and

1 pt
p  x , p p0
̇ , where
is finite by hypothesis. We


1
p

i 0 i
k i 0 i

Therefore, we have
1
  pk .
1
 pk
i 0
  , and since the second
i
sum has a finite number of terms, we deduce
Turkish Journal of Analysis and Number Theory


1
k
k  i 0 pi
n p0 ~ ( x),
 ,
p0
i.e., the number of elements of the form
be equal to number of prime numbers
According to the formula (3), we have
which contradicts formula (13).
Theorem 3.4. Let
. Then
 c ( x)   ( x)   ( ( x)).
(14)
Proof. To find the value of
means that we estimate
the number of origins . Clearly, the prime number is
an origin that means
is not a prime number. Then,
let
be between
and , and let
be the
greatest prime number in
, therefore
is not a
prime number and for all integer
,
. So, we only have to search the numbers which
are not primes and less than
. Thus, we have
1°The number of the even numbers less than or equal
to
equal to
.
2°The number of the odd numbers less than or equal to
equal to
such that
is
the number of the prime numbers less than or equal to
.
Next, we add the two quantities, we obtain, since
, the quantity
which is equal to
 c ( x)   ( x)   ( ( x)).
xlnlnx

 p0  xlnp0
ln x
ln x
p x



p0 ,n
ln x  ln p0
~
lnlnp0
  ln x   iter  x  

p0  x
is
obtained
directly
by
(16)
0 ( x)~x as x   ..
Proof.
1. We have
 ( x)   ( ( x)) 
xlnlnx
ln 2 x

 p0  xlnp0
ln x
 lnp0  ( ( x)   ( ( x))) ln x 
p0  x
xlnlnx
.
ln x
ln x ln p0
lnlnp0
p0  x
And recalling that
 ( x)~
x
.
ln x
We obtain

,
lnlnp0
  ln x  ln p0 
p0  x
ln p0 .
1
lnlnx 

 x 1 

.
 ln x  lnlnx ln x 
The second inequality is obtained in the same way,
 x

x
ln p0  ln x 


ln
x
ln
x
(ln
x

lnln
x
)


p0  x

1


 x 1 
.
 ln x  lnlnx 
2. It is enough to tend
In the expression:

 lnp0  ln x  1   c ( x) ln x.
p0  x
 x

x
xlnlnx
ln p0  

 ln x 
ln x
 ln x ln x(ln x  lnlnx) 
p0  x
xlnlnx 0 ( x)

,
ln x
ln 2 x
l‍nlnp0 
p0 ln n p0  x,n
 lnlnp0
p0  x

xlnlnx
,
ln x
1
lnlnx 

x 1 


 ln x  lnlnx ln x 
1


 0 ( x)  x 1 

ln
x

lnln
x


(15)
where represents the origin of the classes ̇ .
Proof.. Since for all
, we have
then
 lnlnp 
therefore, the inequality
substitution.
Proposition 3.1. We have,
must
.
Moreover, since
It is not difficult to see that, this sequence is stationary
for
and increasing divergent to infinite for
and as
. It is clear that, inductively,
, then we have the
following consequence:
Lemma 3.2. We have,
2
p x
, we define the following
yn1  yn ln yn .
 c ( x) 
 lnlnp 

3.2.2. Estimate of
For all initial value
sequence:
102
to
in the inequality (16).
4 Conclusion and Future Work
ln ln p0 ,
p0 ln n p0  x,n
the value of obviously depends on
by
, then, we have
and if we denote
There is a lot of results on properties of prime numbers.
There are innumerable ideas in this field regarding the
randomness of prime numbers. However, it turns out that
prime numbers do not appear absolutely randomly,
meaning, it is not entirely true that there is no way
whatsoever to see some relations and find some functions
103
Turkish Journal of Analysis and Number Theory
to generate a lenght of few prime numbers. Patterns do
emerge in the distribution of primes over varied ranges of
number sets. In this paper we have investigated sequences
of primes obtained by the equivalence relation , from
which we have associated various arithmetic functions. As
we have seen in the sections above, many classical results
stated in elementary number theory can be restated with
our sequences. Also, there are many other asymptotic
formulas can be deduced, we have just presented several
of them as examples.
As far as motivation, the concern relates to the
developing of new ideas for solving great unsolved
problems and conjectures encountered in this field.
A highly intriguing area in primes is the concept of twin
primes. These are prime numbers which differ by the
number for example: and , and ,
and , etc.
There is an attempt being made to prove that there are
infinitely many twin primes that exist in the natural
number system. These are the concepts that come to mind.
Others are pretty much minor results.
Yitang Zhang, in his paper [3], attacked the problem by
proving that the number of primes that are less than
million units apart is infinite. While
million is a long,
long way away from , Zhang's work marked the first
time anyone was able to assign any specific proven
number to the gaps between primes.
In November
, James Maynard, in [4] , introduced
a new refinement of the GPY sieve, allowing him to
reduce the bound to
and show that for any there
exists a bounded interval containing prime numbers.
At the present time, let me explain and share some
general ideas and questions about my future work. Firstly,
the questions raised are very broad in scope and cannot be
addressed directly. This means that, it is preferable to
resort to a methodology (plan in stages) to tackle the great
problems in a structured manner. Secondly, for that reason,
we have proposed and developed the results of this paper
(see the problem 1 and 2).
Finally, here are just a few questions and conjectures a
little more direct, I think are important.
Problem 1 The twin prime conjecture is equivalent to
conjecturing the translates of the
(the
values of the pair
) are simultaneously prime
values infinitely often. The question is: show that if the
are simultaneously prime values
infinitely often then the
are
simultaneously twin primes infinitely often, and then the
twin prime conjecture is true. We have posed this question
from the perspective of finding recurrence relation
between primes.
Problem 2 Let and
be twin primes. Show that
 i n 1
   ( pi  2 )   1 ( pi )

 i 1


2

 ~O(n3 ln 2 n).


Problem 3
1. Does the set
contain an infinity of twin primes?
2. Let
fixed. Are there an infinity of primes of
the form
such that
is prime?
3. Study of

1
p 0 1 
1
, s is a complex number ,
ps
There is continuing research to prove these conjectures
and questions rigorously, using the results of this paper
and advanced techniques in number theory in the next
work.
Acknowledgement
We sincerely thank the editor and the referees for their
valuable comments and suggestions.
References
[1]
[2]
[3]
[4]
C. J. de la Vallée Poussin, “Recherche analytique sur la théorie
des nombres”, Ann. Soc. Sci. Bruxelle, 20. 183-256. 1896.
J. Hadamard, “Sur la distribution des zéros de la fonction
et
ses conséquences arithmétiques”, Bull. SOC. Math. France, 24.
199-220. 1896.
Y. Zhang, “Bounded gaps between primes”, Ann. of Math, 179,
1121-1174. 2014.
J. Maynard, “Small gaps between primes”, Ann. of Math, 181.
383-414. 2015.