Primes: Randomness and Prime Twin Proof
Primes:Randomness and Prime Twin ProofMartin C.
[email protected] sites:Im greatly appreciative of sites that have found my work interesting and have linked to me: Most Notably, ... Primes:Randomness and Prime Twin ProofMartin C.
[email protected] sites:Im greatly appreciative of sites that have found my work interesting and have linked to me: Most Notably, I appreciate:Google Directory Google Prime DirectoryDMOZ Open Directory Project DMOZ Open DirectoryH. Peter Aleff @ recoveredscience.com Recovered ScienceIntroductionOverviewThe 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 PatternsConsider 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 03 5 7 9 11 131)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 3Consider 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 generalIf 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
(Thats 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)DescriptionAs 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), defineUniqueContribution(P(x)) = {P(x)*k; k is odd, k>=P(x), primeFactorization(P(x) contains no primes In English
The uniqueContribution a prime number (P(x)) gives us as to which numbers are not prime while building successive Pat(n)s is a function of all odd multiples of P(x) such that the odd multiples have no primes less than P(x) in their prime factorizationsExamples:Consider the prime 5.5*5 = 25 is a unique contribution of 55* 15 = 5*3*5 = 75 is NOT a unique contribution because it has 3 in its prime factorization.
Ie, we already knew that 75 was not prime thanks to the prime number 3.5*5*5 = 125 is a unique contribution.Powers of a primeIt turns out that powers of primes (greater than the first power) are unique contributions.Important Notes on uniqueContribution(P(x))For larger P(x), uniqueContribution(P(x)) becomes increasingly difficult to calculate and more complicated. The unique contribution becomes more random as P(x) increases.General Notes on RandomnessAxioms of Randomness1) All truly random patterns must be infinite length2) A pattern is said to be random if there is an infinite supply of complexityBlack Box Pattern ParadoxIt can only ever be said that an infinite length pattern follows a pattern for a certain finite length.
Suppose you have a machine that spits out 1s and 0s and it spits out 1010101010
for a certain number of times you make the attempt. You can only say that it follows the patter 10
for the number of attempts you made because on the next attempt, the pattern may change.
Thus it is impossible to ever say that an infinite length pattern follows a certain pattern unless you are aware of the algorithm that generates it.On Randomness of PrimesMeasure of Randomness in a Binary Pattern Lets define a measure of randomness (mr) for a binary pattern to be the number of smallest repeating units in the lowest reducible pattern. Definition of Lowest Reducibility:A pattern is reducible if it can be rewritten in a simpler, shorter form, such as:11111111
is reducible to 1
10101010101010
is reducible to 10
Definition of Smallest Repeating UnitsTake the pattern:Pat(2) = 110100100101100
This repeats every 3rd and every 5th.
Note it repeats every 9th as well but thats not the smallest repeating unit because the every 3rd subsumes the every 9th. Thus the mr of this pattern is 2.Some Examples for ClaritySo, 100000000
is no more or less random than100
, or100000000000000000000000
(Because in all cases mr = 1)However,110100100101100
(every 3rd and 5th) is more random than those above since mr=2.Some other interesting examples for clarity:110
has mr = 2 because it has two repeating units of size 3, (the second offset by 1)101
has mr = 2 because it has two repeating units of size 3, (the second offset by 2)111
has mr = 1 because this is a reducible pattern, reducing to 1
The latter is an important example because one might be tempted to say this pattern repeats every 3rd, every 3rd offset by 1 and every 3rd offset by 2, but this pattern reduces to 1
therefore, the mr is calculated on the lowest reducible pattern.Examining Pat(n) re: Randomness with increasing nSo as we take higher n in Pat(n), the number of smallest repeating units increases.
In fact it exactly equals n. For any given n, Pat(n) isnt absolutely random, but P(n+1) is more random than Pat(n).Model of Lim(x->inf) (1/x) = 0Examine the model of:let f(x) = 1/xAt no x, is f(x) = 0, however closenesstozero(f(x+1))>closenesstozero(f(x))and thelim(x->inf)f(x) = 0 Likewise, for no x is Pat(x) a random pattern, howeverThe mr(Pat(n+1))>mr(Pat(n)), and then theImportant 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 EnglishPat(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 problemWell 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-PrimesThere are some interesting patterns in non-primes that emerge from this work.Define:LowMarker(n) = 3 + 2(P(1)xP(2)x
x(P(n)),HighMarker(n) = 3 + 2(3x5x
xP(n)), andOffset(n) = P(n) 3We can say conclusively, thanks to pattern combinatorics that numbers in the ranges:LowRepeater(n) = [LowMarker(n),LowMarker(n)+Offset(n)] andHighRepeater(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.ExampleLets work an example:Examine Pat(4) at the start of the patternExamine the numbers{3,5,7,9,11}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 11Recall P(4) = 11Lets 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 11Examine 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 11LowRepeater(n,k) and HighRepeater(n,k)LowRepeater and HighRepeater repeat over the number lineAdding 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)), andOffset(n) = P(n) 3Then,LowRepeater(n,k) = [LowMarker(n,k),LowMarker(n,k)+Offset(n)] andHighRepeater(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 HighMarkers k=1, LowMarkers 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?
[email protected]© Martin C.
Winer, 2004Posted to the web on Mar 16, 2004 after years of being ignored :( Source: Free Articles from ArticlesFactory.com .