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On the prime k-tuple conjecture Liu Fengsui Feb 12, 2002 Tel: 0086 10 65887888-7731# Fax: 0086 10 64951424, 64201375 Address: 6/F Full Link Plaza 18 Chao Yang Men Wai Avenue Chao Yang District, Beijing, China 100020 Email: [email protected].

Abstract: In this paper we defined some operations on the sets of natural numbers P(N) , given a

recursive formulaT'n of the set sequences, which approximates the set of prime k-tuples, and use

the limit of the set sequences limT'n show the patterns of prime k-tuples k  41 infinitely often.

Key words: second order arithmetic, model, prime formula, strict constructive proof, twin prime conjecture, triple prime conjecture, k-tuple prime conjecture, infinite, limit of set sequences, recursive formula, Dirichlet's Theorem, prime constellation.

MBC2000: 11B37, 03C62.

After all give some operations on the sets of natural numbers P(N)

Let A = < a1 ,a2 ,...,ai ,...,an >, B = be the finite sets of natural

number, we define

A + B = < a1  b1 ,a2  b1 ,...,ai  b j ,...,an1  bm ,an  bm >,

A * B = < a1 *b1 ,a2 *b1 ,...,ai *b j ,...,an1 *bm ,an *bm >. A \ B be the set subtraction. Define the solution of the system of congruencies

X = < a1 ,a2 ,...,ai ,...,an > mod a ,

X = mod b

be X = D = < d11 ,d 21 ,...,dij ,...,d n1m ,d nm > mod a *b .

1 Where x = dij mod a *b is the solution of the system of congruencies x = ai mod a , x =b j mod b .

Now, we had founded a model of the second order arithmetic

 P(N),,,,0,1, .

Where N is the set of natural numbers and P(N) is the power set of N .

We identify the set < a >, which has only single element a , with the number a < a > = a .

So that, we had isomorphic embedded the model  N,,,,0,1  in the model

 P(N),,,,0,1, .

In the theory PA + ZF of the second order arithmetic  P(N),,,,0,1, , we shall

show that the prime k-tuple conjecture with the limit of the set sequences limT 'n . Perhaps, it is undecidable in the first order Peano theory PA .

The prime k-tuple conjecture states that every admissible pattern for a prime constellation occurs infinitely often.

To describe easily, we discuss the conjecture: There are infinity a such that all

x 2  x  a are prime for x  0,1,2,...,k . [1]

Example: k = 2, this is the Twin Prime Conjecture, k =3, this is the Triple Prime Conjecture, and so on.

We use R(k,a) denote that all x 2  x  a are prime for x  0,1,2,...,k , and say the number a be a prime k-tuple.

Let pn be n -th prime, p0 = 2.

For any prime pi > 2, we consider the set

2 Bi = { a : pi | x  x  a for some 0  x  k }

2 = { a : x  x  a = 0 mod pi for some 0  x  k }.

2 2 2 By ( pi 1 x)  ( pi 1 x)  ( pi 1 x) *( pi  x)  x  x mod pi ,

2 it is easy to prove that, in the congruence x  x  a = 0 mod pi , when x runs through the

complete system mod pi , the number a runs through the classes of residues

2 <0,-2,-6,..., ( pi 1) / 2  ( pi 1) / 4 > mod pi .

Thus the set Bi are the classes of residues:

2 Bi = <0,-2,-6,..., ( pi 1) / 2  ( pi 1) / 4 > mod pi , ( pi 1) / 2  k .

2 Bi = <0,-2,-6,..., k  k > mod pi , ( pi 1) / 2 > k

For example, let k  4 , the first few terms of the sets Bi are

B1 = <0,-2> = <0,1> mod 3,

B2 =<0,-2,-6> =<0,3,4> mod 5,

B3 = <0,-2,-6,-12> = <0,5,1,2> mod 7,

B4 = < 0,-2,-6,-12> = <0,9,5,10> mod 11.

Let mn1 = p0 * p1 *...* pn , from the set of all odd numbers X = <1> mod 2 we

cancel the classes B1 , B2 ,..., Bn successively, and obtain the class Tn1 mod mn1 such that

2 Tn1 = {a: x  x  a =/= 0 mod pi for x  0,1,2,...,k and pi  p1 , p2 ,..., pn }.

Then the recursive formula of Tn1 , which is the set of non negative representatives

mod mn1 is as follows:

T1 = <1>,

Tn1 = (Tn   mn  *  0,1,2,..., pn 1 ) \ Dn .

Where X = Dn mod mn1 is the solution of the system of congruencies.

X = Tn mod mn ,

X = Bn mod pn

3 For example, let k  4 , the first few terms of Tn1 are

T1 = <1>,

T2 = (<1>+<0,2,4>) \ <1,3> = <5>,

T3 = (<5> + <0,6,12,18,24>) \ <5,23,29> = <11,17>,

T4 = (<11,17>+<0,30,60,...,180>) \ <47,71,77,107,131,,161,191,197>

= <11,17,41,101,137,167>.

Given any integer k  1 , take an integer s such that

( ps 1) / 2  s  ( ps1 1) / 2 ,

then the number of all elements of the set Tn1 is

Tn1 = ( p1 1) / 2*( p2 1) / 2*...*( ps 1) / 2*( ps1  k) / 2*...*( pn  k) / 2 .

Now we had proved an elementary result: for every k and n there are number a such that

x 2  x  a

have not any prime factor p  pn . This is the Ludovicus conjecture..[2]

Obviously the criterion of prime k-tuple is

R(k,a) iff a = pn  mimTn  k 1

Where mim Tn is the smallest number of the set Tn . This criterion R(k,a) recursively enumerates all prime k-tuples.

The recursive formula Tn expresses a process of sieve method. We can perform this sieve method infinite times on the entire set of natural numbers N in the second order arithmetic. This is difference from traditional sieve method.

Let k  1 , then Bi =< 0> mod pi , we simplify the recursive formula Tn to be

T1 = <1>,

Tn1 = (Tn   mn  *  0,1,2,..., pn 1 )\  pn  *Tn .

4 From this formula we obtain a prime formula

p1  3

pn  S(Tn ) .

Where the function S(Tn ) takes the second element of the set Tn .

S(Tn ) = S  t1 ,t2 ,......  t2 .

This formula provides a direct strict constructive proof, that the number of primes is infinite. [3] [4].

If we canceled all classes B1 , B2 ,..., Bi ,...... , from entire set of natural numbers N , will

obtain what result? In other word, we consider the limit of the sequences of sets T1 ,T2 ,…,Tn ,.....

… In the classes of residues there is include relation

N  T1  T2  ...  Ti  ...... Thus the sequences of sets T1 ,T2 ,...,Tn ,...... have lim Tn .[3]

It is easy to prove lim Tn is empty, since that when we canceled classes B1 , B2 ,..., Bi ,...... to sift the prime k-tuple a , we removed the prime k-tuple a also by

pn = a and pn | a .

We modified the set B j to be B' j

2 2 B' j = { a : p j | x  x  a for some 0  x  k but p j =/= x  x  a }

2 2 = { a : x  x  a = 0 mod p j for some 0  x  k but p j =/= x  x  a }.

And we modified the set Tn1 to be T 'n1

2 2 T 'n1 = { a : x  x  a =/= 0 mod p j but p j = x  x  a for

x  0,1,2,...,k and p j  p1 , p2 ,..., pn ,}.

Except reserving the prime k-tuples a , Tn and T'n are same, so that | Tn || T'n | .

Now the recursive formula T'n will approximates the set of all prime k-tuples lim T'n if n go to infinity.If we have known that the set of prime k-tuples is not empty, example k  41, we

5 try prove the pattern R(k,a) occurs infinitely often with the lim T'n .

Proof:

Suppose that the number of patterns R(k,a) is finite and not 0, then there is a maximum number

a0 , such that for every a > a0 , R(k,a) is false. From the set of all odd numbers X  1  mod 2 ,

We cancel the classes B1 , B2 ,..., Br1 successively, and obtain the class Tr such that

a0 = mim Tr = pr ,

Then for every j  r , in the class T j there is not any a such that R(k,a) , and for every a  a0

there is not any a such that R(k,a) .

From the class Tr we continue cancel the set Br , B'r1 ,..., B' j ,...... Now as n goes to

infinity, the recursive formula T'n approximated all prime k-tuple a  a0 . Because lim | Tn | is

infinite, and| Tn || T'n | , so that lim | T 'n | is infinite and lim T'n is not empty. Take a number

e belongs lim T'n , e  limT 'n , then

x 2  x  e

does not contain any prime pi or p j as factor except itself for x  0,1,2,...,k Thus R(k,a) hold by definition and

e > a0 .

This is a contradiction, so we have proved the prime k-tuple conjecture:

The pattern R(k,a) , k  41, occurs infinitely often. #

Let k  1 , as a practical example of this method, we provided an abstract proof that the

number of primes is infinite with lim T'n . Thus we can prove that the prime and prime k-tuple

both are infinite with same method.

Change the sets Bi , we can prove the infinity of various patterns with above method.

6 Example: the primes in arithmetic sequence ax  b , (a,b)  1 is infinite.

Give any prime pi , except pi | a , let

Bi = { x : ax  b  0 mod pi },

| Tn | = ( p1 1) *( p2 1) *...*( pn1 1) , except pi | a .

Repeat above proof, we obtained the Dirichlet's Theorem again.

If the set of prime k-tuples is empty for some patterns, it is no meaning to discuss if they are infinite, to extend above method to empty set, we describe the empty set with the sentence.

There is no prime k-tuple a  a0 or there is no prime k-tuple a  a0 .

Assume the component of disjunction "there is no prime k-tuple a  a0 ", run above proof, if we do not obtain contradiction, we can not prove anything, if we obtain a contradiction, then we have proved the component of disjunction " there is no prime k-tuple a  a0 " is true, namely we have proved the set of prime k-tuples is empty again.

There is an Euler's formula

x 2  x  41 all are primes for x  0,1,2,...,k . So that, for k  41, the set of prime k-tuples is not empty, by above proof, the pattern R(k,a) , k  41, occurs infinitely often. For k  41, we do not know if there is a number a such that

x 2  x  a all are primes for x  0,1,2,...,k , we do not know if they are infinite. If we found one a , such that R(k,a) , then they would infinite.

Reference : [1] 华罗庚，堆垒素数论（中国科学院 1953） 204-205。 [2] http://www.primepuzzles.net/conjectures/conj_017.htm [3] http://www.primepuzzles.net/problems/prob_037.htm [4] P.Ribenboim, The new book of prime number records, 3rd edition, Springer- Verlag, New Tork. NY,(1995)213. [5] K. Kuratowsky and A. Mostowsky , Set Theory, With an introduction to descriptive set theory, North-Holland Publishing com. (1976) 118-120.