Thursday, December 21, 2006

Primes: Randomness and Prime Twin Proof

Author: Martin Winer


Randomness and Prime Twin Proof

Martin C. Winer

Referring sites: I’m greatly appreciative of sites that have found my work interesting and have linked to me: Most Notably, I appreciate:

Google Directory Google Prime Directory DMOZ Open Directory Project DMOZ Open Directory H. Peter Aleff @ Recovered Science

Introduction Overview The purpose of this work is to look into some long pondered questions. First, is the distribution of primes across the number line random? Next, what is random anyways? Finally the theories and axioms derived are used to solve the long discussed “Prime Twin Problem” to show possible applications of the understanding of what it means to be random.

Sieves and Patterns Consider all odd numbers starting at 3.

Mark a 1 on the number line where the number is a product of 3, (including 3x1), 0 otherwise. We get a pattern (sieve) such as:

1 0 0 1 0 0

3 5 7 9 11 13

1)The pattern is 100...

2)Note that the numbers corresponding to the zeros between 3 and 3^2=9 are also prime (5 and 7).

3)The length of the pattern is 3

Consider the pattern formed by 3 and 5:

1) the pattern is 110100100101100...

2) Note that the numbers that correspond to zeros between 5 and 5^2=25 are also prime (7,11,13,17,19,23)

3) The length of the pattern is 3x5=15

Definition of P(x) [The Xth Prime] In general

If we let P(x) be the xth prime starting from 3 such that

P(1)=3, P(2)=5, P(3)=7 and so on, we can consider the patterns on a larger scale.

Definition of Pat(n) Suppose we define a function Pat(n) which will produce the string of ones and zeros as defined above from P(1) to P(n).

I.e. Pat(1) = 100…

Pat(2) = 110100100101100… (That’s the pattern or sieve of 3 and 5)

In such a case,

(1) The pattern will consist of 1's and zeros corresponding to the products and non-products of the n composing prime factors,

(2) The numbers corresponding to zeros between P(n) and P(n)^2 are guaranteed to be prime

[Why? because a number is either a prime or a product of primes. A zero means that it's not the product of any prime below it. The first unique contribution a prime factor gives to the number line occurs at P(n)^2 = P(n)xP(n) because below that at say P(n)xP(n-1) can be rewritten as P(n-1)xP(n) and thus is already accounted for in the number line. Thus a zero between P(n) and P(n)^2 is not a product of primes and must therefore be prime.]

(3) The length of the pattern will be:

P(n) x P(n-1) x P(n-2) x ... x P(1)

Unique Contributions of P(x) Description As we build iteratively build Pat(n)’s over time, each successive prime adds to our knowledge of which numbers are prime and which numbers are not. In the case of the prime number 3, we know that 3 is prime and that all (other) multiples of 3 are not prime. However, when we come to the prime number 5, 5 does not ADD to our knowledge that all multiples of 5 are not prime. For example 15 is a multiple of 5, but we already knew that this number was not prime because of the prime number 3. Therefore, the unique contribution to our knowledge as we build Pat(n)’s that a given prime (P(x)) provides us is given below:

Definition uniqueContribution(P(x)) For any prime, P(x), define

UniqueContribution(P(x)) = {P(x)*k; k is odd, k>=P(x), primeFactorization(P(x) contains no primes

inf) (1/x) = 0 Examine the model of:

let f(x) = 1/x

At no x, is f(x) = 0, however


and the

lim(x->inf)f(x) = 0

Likewise, for no x is Pat(x) a random pattern, however

The mr(Pat(n+1))>mr(Pat(n)), and then the

Important Identities (4a) lim(n->inf) mr(Pat(n)) = inf (i.e. grows infinitely complex)

(4b) lim(n->inf) uniqueContribution(P(n)) = random set

(4c) lim(n->inf) Pat(n) = random binary pattern (i.e. absolute random)

Definition of Random in English Pat(n) always produces patterns in the lowest reducible form (ask me for a proof if you like). Pat(n) has n smallest repeating units (we know this because the units are prime). Therefore as you create Pat(n) with greater and greater n, you produce lowest reducible patterns of greater and greater complexity (higher mr). Each individual P(n) as n increases, has a more and more complicated uniqueContribution, leading to more complexity in the resulting Pat(n)’s, hence more randomness. As you do this without bound, you create complexity based on previous complexity, resulting in infinite complexity = random.

So what can we do with this knowledge?

Solution to prime twin, triple, quadruplet problem Well it solves the prime twin, triplet, and quadruplet problems in a shot...

From 2 above, we know that the zero's between P(n) and P(n)^2 are prime. A prime twin will occur in this region whenever you see the pattern 00 (two adjacent prime candidates). Can one predictably say that there exists a certain Prime P(K) after which, there will never be a 00 in the pattern between P(k+q) and P(k+q)^2?

An examination of pattern combinatorics reveals that there is a 00 in the base case P(1) (100..). As we combine patterns, there will always be a 00 somewhere in the pattern Pat(n) (ask me for the proof if you like). The trick is, will it be between P(n) and P(n)^2.

Well the pattern subtended by P(n) and P(n)^2 is a subset of the pattern Pat(n) and grows without bound as n does. You can tell me less and less about it as n grows without bound. By (4) I can let n grow without bound until it is a truly random pattern at which time you can no longer tell me that you can predictably state that there won't be a 00 in the pattern between P(n) and P(n)^2. At the time that there is a 00 between P(n) and P(n)^2, a new prime twin will occur.

This is true of all prime triples, quadruplets etc that are allowable.

Why do they keep finding patterns in primes? This becomes evident knowing that we only have a finite list of primes in our knowledge. The patterns produced by a finite list of prime factors are never absolutely random, just relatively random, or ‘sufficiently complex to avoid simple categorization’. Statistical tools, depending on their power, will find a pattern in those patterns not produced by an infinite number of prime factors (the number line).

Interesting Patterns in Non-Primes There are some interesting patterns in non-primes that emerge from this work.


LowMarker(n) = 3 + 2(P(1)xP(2)x…x(P(n)),

HighMarker(n) = 3 + 2(3x5x…xP(n)), and

Offset(n) = P(n) – 3

We can say conclusively, thanks to pattern combinatorics that numbers in the ranges:

LowRepeater(n) = [LowMarker(n),LowMarker(n)+Offset(n)] and

HighRepeater(n) = [HighMarker(n),HighMarker(n)+Offset(n)] are non-prime (product of primes)

Moreover they follow a similar pattern to the base pattern that spawned them.

Example Let’s work an example:

Examine Pat(4) at the start of the pattern Examine the numbers


3 is a product of 3,

5 is a product of 5,

7 is a product of 7,

9 is a product of both 9 and 3,

11 is a product of 11

Recall P(4) = 11

Let’s examine the numbers in the ranges, LowRepeater(4) and HighRepeater(4)

Examine LowRepeater(4) LowRepeater(4) = {2313,2315,2317,2319,2321}

2313 is a product of 3,

2315 is a product of 5,

2317 is a product of 7,

2319 is a product 3, and necessarily, not a product of 9 (this could only occur in HighRepeater)

2321 is a product of 11

Examine HighRepeater(4) HighRepeater(4) = {20793, 20795, 20797,20799, 20801}

20793 is a product of 3,

20795 is a product of 5,

20797 is a product of 7,

20799 is a product of both 9 and 3,

20801 is a product of 11

LowRepeater(n,k) and HighRepeater(n,k) LowRepeater and HighRepeater repeat over the number line

Adding a factor k to the previous functions we get:

LowMarker(n,k) = 3 + 2k(P(1)xP(2)x…x(P(n)),

HighMarker(n,k) = 3 + 2k(3x5x…xP(n)), and

Offset(n) = P(n) – 3


LowRepeater(n,k) = [LowMarker(n,k),LowMarker(n,k)+Offset(n)] and

HighRepeater(n,k) = [HighMarker(n,k),HighMarker(n,k)+Offset(n)] are non-prime (product of primes)

All, where k>=0 and k is an integer.

Interesting observation about the difference between LowMarker(n,k) and HighMarker(n,1) For a fixed n, and HighMarker’s k=1, LowMarker’s k = the product of the non-primes between P(1) and P(n).

The following table will clarify:

LowMarker(3,1) = HighMarker(3,1) LowMarker(4,9) = HighMarker(4,1) LowMarker(5,9) = HighMarker(5,1) LowMarker(6,9*15) = HighMarker(6,1) LowMarker(7,9*15) = HighMarker(7,1) LowMarker(8,9*15*21) = HighMarker(8,1)

Questions? Comments?

© Martin C. Winer, 2004

Posted to the web on Mar 16, 2004 after years of being ignored :(

About the author: Martin Winer is a computer scientist by day, working on and an amateur mathematician by night. For a formatted version of this article, please view:


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