Exploring the Primes: A Mathematical Approach with a Million Dollar Question

[Canute]

* Win a Million Dollars Here! *

Over the last couple of years I've spent some time on a pseudo-mathematical investigation the primes. I feel the results of this investigation are important, but I cannot be sure of this since I know so little about mathematics. I have written, emailed and phoned a number of senior mathematicians and called in a few favours in the hope of interesting someone in my notes but none have even replied. Recently I've been trying to tidy up these notes with a view to submitting them for publication. Not because they are likely to be published, I've no idea how to write on this topic, but because this would at least get my notes read by someone competent and establish some sort of priority.

Just this week, however, I was horrified to discover that a new website is under construction called 'The Orbits of the Primes'. It claims that the content, once the site is up and running, will revolutionise number theory. I suspect, from the few clues given, that this site will present the same conclusions about the primes that I have reached. This is very annoying, to put it mildly. A year ago I constructed a virtual 'observatory' in Excel showing the orbits of the primes and how these orbits allowed us to predict the postion and distribution of the primes. I think I'm onto something but do not know how to express it in a way a mathematician would immediately grasp, and am trapped in complete (mathematical) obscurity.

So, I'd like to ask for some help and advice. I need to reframe my ideas in mathematical language, and I need to find out whether or not my approach to the primes is mathematically interesting.

Using my approach it is possible, I think, to prove that there are infinitely many twin primes. Also, this approach seems to shed some light on why there is a correlation between the pattern of the primes and the pattern of energy levels in a random quantum drum. I already have a rough function for pi(x). More importantly, for I need the money, I think that at least in theory this approach would allow a proof of the Riemann hypothesis, or at least a proof of its decidability.

Whoever proves that Riemann's hypothesis is true (or false) would win fame, a bright future in mathematics and a million dollars. If I am right about the primes then such a proof need not be beyond the average graduate mathematician, and not beyond some of the mathematicians here. I know how unlikely this will seem, but it seems likely to me that a proof, if one is possible, will come from an unusual source. After all, all the obvious methods have been tried and have so far failed.

I am not interested in fame or a bright future in mathematics, but I am interested in a share of a million dollars. This gave me an idea. Suppose I post my ideas here. Then you could rip them to shreds if that is what they deserve and I can forget all about number theory. However, if by some miracle I am right and some interesting new proofs can be derived from these ideas, then as mathemticians you would be in a position to develop and exploit them. As you would be able to express them properly you might even be able to create something publishable out of them. If so, I hope I'll get a credit.

Also if, by some miracle, I am not mad after all but have found a not-too-difficult route to a proof of RH (or of its decidability) then someone here might be able to construct such a proof, or pass on some ideas to someone capable of doing so. This is almost certainly not going to happen, but sometimes truth is stranger than fiction. You will all have connections in the mathematics world, I have none.

If anyone here can make use of these ideas then feel free to do so. All I ask is for a mention, or for a mention of this discussion. If they lead to a proof of RH then all I ask for is for 25% of the prize.

What I would like to do is post chunks of notes and ask for your comments, for some help in expressing them properly, and for your opinion on whether they are interesting or trivial. I'll explain all the principles and work towards my thoughts on RH. The notes are quite long so I wanted to generate some interest before inflicting them on you. Shall I start posting them or shall I go back to philosophy?

The million dollar question would be: Can you prove the Riemman hypothesis by using my methods? If so, you've won $1,000,000 less my cut.

To start things off here is one 'result'. All numbers at 6n+/-1 are prime unless they can be written in the form 6np+/-p. From this 6np+/-p rule the behaviour of the primes is entirely predictable.

Shall I post more? My notes are long and incompetently expressed so I don't want to start posting them unless someone's interested in trying to understand them.

Regards
Peter Jones (Canute)

 

[shmoe]

I've just skimmed, so only a couple things for now.

* Whoever proves that Riemann's hypothesis is true (or false) would win fame, a bright future in mathematics and a million dollars.

Proving it false would net zero cash from the clay institute.

To start things off here is one 'result'. All numbers at 6n+/-1 are prime unless they can be written in the form 6np+/-p. From this 6np+/-p rule the behaviour of the primes is entirely predictable.

This looks to me like you're saying every number of the form 6n+/-1 is prime unless it has a prime divisor? Or are you saying it's a special thing that a composite 6n+/-1 has a divisor of the form 6m+/-1?

To be bluntly honest "pseudo-mathematical investigation" scares the willies out of me. Please tell me you've at least familiarized yourself with some elementary number theory, like Hardy and Wright's intro book. Too many times have I seen people claiming great ideas about primes yet having apparently no clue about even the most trivial number theory techniques and the only parts of their attempt that were remotely correct were already very well known.

 

[RandallB]

Proving it false would net zero cash from the clay institute.

NO, I think they would pay off on what they describe as a acceptable “counterexample”.

If, in the opinion of the SAB, the counterexample effectively resolves the problem then the SAB may recommend the award of the Prize. If the counterexample shows that the original problem survives after reformulation or elimination of some special case, then the SAB may recommend that a small prize be awarded to the author. The money for this prize will not be taken from the Millennium Prize Problem fund, but from other CMI funds.

In this case it theoretical could be much easier to prove false than true. As always it is easier to falsify something; as it only takes one solid example of a proper “interesting solution” that does not fall on the expected line would be all it takes! (1,500,000,000 solutions already checked)

The problem with hunting for that one solution that provides falsification is that it very likely just doesn’t exist, because the idea is true. Making the real problem, just as they have framed the problem, is providing an acceptable and adequate real positive PROOF. Including a two year acceptance in proper journals, etc.

Picking one of the other six “Millennium Problems” (Total $7 Million Prize Pool) may be “easier” (The term easy being relative).

 

[shmoe]

Interesting, the interpretation of the wording of the prize that I've known did not allow for a counterexample to get the prize, though they still seem to have the leeway to not award in this case. The http://www.zetagrid.net/zeta/prizes.html" people for example allowed a paltry $1,000 award for a counterexample found with their software. This was always the reason no prize was reasonable to me in this case, finding a counterexample could be just a matter of running software for a long enough time (assuming it's false of course).

They've gone much furthur than 1,500,000,000 zeros. The zetagrid people claimed over 100 billion I believe (*my belief must be old, see below).

edit- http://mathworld.wolfram.com/RiemannHypothesis.html apparently believed as I did about a disproof. They mention ZetaGrid claims 10^12 zeros and Gourdon claims 10^13:

http://numbers.computation.free.fr/Constants/Miscellaneous/zetazeros1e13-1e24.pdf

 

[RandallB]

They've gone much furthur than 1,500,000,000 zeros.

Over 1,500,000,000 Solutions not zeros

 

[shmoe]

Over 1,500,000,000 Solutions not zeros

What do you mean by a "solution" then? Solution to what?

Is it a coincidence that van de Lune, te Riele, and Winter verified the first 1,500,000,001 zeros where on the critical line? You do know that the Riemann Hypothesis is all about the location of the non-trivial zeros of the zeta function right?

 

[RandallB]

What do you mean by a "solution" then? Solution to what?

I am by no means a expert on the Riemann Hypothesis just sharing some of the details from "Clay Inst." I’d rather work on Yangs-Mills, than Riemann. I figure it can be solved before Rienman will be. If you need some details on Riemann and the prize rules just go to their web site. Google “Millennium Problems”, there is a book by the same name as well.

 

[shmoe]

I am by no means a expert on the Riemann Hypothesis just sharing some of the details from "Clay Inst." I’d rather work on Yangs-Mills, than Riemann. I figure it can be solved before Rienman will be. If you need some details on Riemann and the prize rules just go to their web site. Google “Millennium Problems”, there is a book by the same name as well.

No thank you, I do know about RH and I'm quite certain that whatever you read talking about "solutions" was talking about "zeros" as they are almost always called.

 

[Hurkyl]

To start things off here is one 'result'. All numbers at 6n+/-1 are prime unless they can be written in the form 6np+/-p. From this 6np+/-p rule the behaviour of the primes is entirely predictable.

I also think you're just using this trivial fact in disguise:

If mn is relatively prime to k, then m and n are both relatively prime to k.

"m is relatively prime to 6" means the exact same thing as "m can be written either as 6n+1 or 6n-1 for some integer n".

You will find similar theorems, such as:

Any number of the form 12n+1 / 12n+5 / 12n+7 / 12n+11 is prime, unless it could be written in the form p(12m+1) / p(12m+5) / p(12m+7) / p(12m+11).

In fact you can say slightly more:

If m = 6n+1 or m = 6n-1 (for the integer n), and p is a prime divisor of m, then p = 6k+1 or p = 6k-1 for some integer k.

 

[RandallB]

No thank you, I do know about RH and I'm quite certain that whatever you read talking about "solutions" was talking about "zeros" as they are almost always called.

I’m sure ClayMath doesn’t intend for such an error to be misleading in their description of the problem, you could let them know if their wrong.

Do you think Sebastian Wedeniwski / ZetaGrid is on track to providing a ClayMath acceptable Proof or acceptable falsification by counterexample?
(First I’ve seen of ZetaGrid – interesting)

 

[shmoe]

I’m sure ClayMath doesn’t intend for such an error to be misleading in their description of the problem, you could let them know if their wrong.

Do you think Sebastian Wedeniwski / ZetaGrid is on track to providing a ClayMath acceptable Proof or acceptable falsification by counterexample?
(First I’ve seen of ZetaGrid – interesting)

It could be that they've altered in the past couple years to allow payment for a counter-example, or that there was a misleading phrase somewhere. It's not really a serious issue to me as I have no delusions of my ability to resolve it one way or another.

Assuming RH is false, all the computations are on the right track to provide a counterexample. Even at the rate of finding 1 zero a week it would only be a matter of time before a counterexample was found (of course it may be an extremely long time). In any case these computations do provide some insight into the problem. It was computations like this (Odlyzko and others work in in the 80's) that gave enough data to strongly suggest Montgomery's pair correlation conjecture may be true and it can be useful in some problems like finding explicit bounds for the nth prime for example (there was a thread about this https://www.physicsforums.com/showthread.php?t=100265").

 

[Canute]

Hang on. I don't think I'm some genius who has proved RH. I just have some idea and wondered if they would be interesting to anybody here. I think they might be, but then I'm not a mathematician. All I know is that I haven't come across anybody discussing the primes in this way, and I was very surprised to discover anything at all about the primes seeing as how I'm such an idiot at mathematics.

Hurkyl - I think you've misinterpreted me, but in any case what follows will make it clear one way or the other.

I apologise for the length of what follows but I'm stuck with natural language and not much skill at using even that. This first section is a more or less stream of consciousness preamble to some notes on twin primes and Riemann. They may seem to lead nowhere but I think I can prove from just these simple observations that, to put it Euclid's way, twin primes are more than any assigned multitude of twin primes.

Got to eat but I'll come back and post the notes a bit later.

 

[-Job-]

All numbers at 6n+/-1 are prime unless they can be written in the form 6np+/-p.

From what i can see 6np+/-p is always a multiple of p. Because it is a multiple of p, it is never a prime. Saying that a number can be decomposed into np+/-p is equivalent to saying it is not a prime. Saying that it can be decomposed into 6x is the same as saying it is not a prime. Saying that it can be decomposed into 6np+/-p means that it is not a prime.
So what you're saying is equivalent to:
6n+/-1 is always a prime, unless it is not a prime.

In my opinion you're committing a logical error similar to "begging the question". But i can appreciate the fact that you've tried to solve this problem.

 

[shmoe]

Hang on. I don't think I'm some genius who has proved RH. I just have some idea and wondered if they would be interesting to anybody here. I think they might be, but then I'm not a mathematician. All I know is that I haven't come across anybody discussing the primes in this way, and I was very surprised to discover anything at all about the primes seeing as how I'm such an idiot at mathematics.

It was clear you weren't claiming a proof. Search this forum and you'll find more than one example of someone claiming a different way of looking at primes that turned out to be a trivial and well known observation to anyone familiar with basic number theory. This is why I *hope* you are at least familiar with the elementary stuff.

That said, you do seem open to the possibility that what you've done is trivial, and I hope you will keep an open mind that it may be just wrong. In return I'll keep an open mind that what you have may be correct and won't pass judgement until I've given an honest attempt to understand it (provided it's comprehensible and provided I have time).

 

[Canute]

Shmoe - Thanks. That's what I was hoping someone would say. If what follows turns out to be trivial I wouldn't be completely suprised.

The Behaviour of the Multiples of the Primes

The numbers 2 and 3 generate a combination wave of multiples such that every sixth number from 6 is divisible by 6 and only 2 numbers in 6 can be prime. The two null points in this combination wave occur at 6n-1 and 6n+1, so except for the numbers 2 & 3 only a number at 6n+/-1 can be prime. Mathematical rules determine that a number at 6n+/-1 cannot be a multiple of a non-prime number. Therefore, if a number at 6n+/-1 is not a multiple of a prime it is a prime.

In this way it is the behaviour of the multiples of the primes that precisely determines the behaviour of the primes. The primes are the gaps or null points in the combination wave of multiples generated by previous primes. If you like, to jump ahead to some ideas about the connection between Riemann’s non-trivial zero’s and the pattern of energy levels asociated with random quantum drums, the primes are missing frequencies in the noise created by the multiples of the primes, the opposite of spectral lines.

Take the wave formed by the multiples of 5. As this wave of multiples propogates up the number line it beats against the 6/8 rhythm of the multiples of 2 and 3 in such a way that in every 30 numbers above 30 there are exactly two multiples of 5 at 6n+/-1. These multiples occur at 6n5+/-5 for all values of n. This entails that except for the number 5 itself (and -5) if a number is at 6n5+/-5 it is not prime. These two multiples of 5 in every 30 numbers are the only contribution made by 5 to the combination wave of multiples of primes as it develops, the null points in which are the primes.

When we add the multiples of 7 to those of 2, 3 and 5 the combination wave changes in such a way that from 7 onwards to infinity there cannot be a prime, a null point, at 6n7+/-7. From 7 onwards there are 2 less null points in the combination wave in every 42 numbers.

In the same way, as we go up the number line, each prime p contributes two multiples of itself in every 6p numbers to the evolving wave. Each prime transmits multiples on its own unique frequency and so, as we we travel up the number line, every time we pass a prime and set it vibrating the frequency of the combination wave becomes longer.

[The mathematics of all this is too difficult for me, but as a musician and ex-studio engineer this feels like familiar territory. It would be very easy to programme a music sequencer to play the music of the primes if it were hooked up to Mathematica, and give a concert. The time signature would 6/8 and the 6n numbers would be the first beat of each bar. The backbeat would be the underlying two against three rhythm. (To hear this tap out the rhythm of the words ‘nice cup of tea’ using your hands in the pattern: Together - Right, Left, Right - T - RLR - T - RLR …. You’re then playing all the quavers in the bar except those at Tn+/-1. The sequencer would play the primes in these gaps, triggering some samples. It might be quite funky.]

The frequency of the combination wave for the multiples of 2 and 3 is 6. Adding 5 it becomes 30, adding 7 it becomes 210, adding 11 it becomes 2310, and so on. The frequency of the wave generated by each individual prime p >3 is 6p, and for a consecutive sequence of primes it is 6(p1 x p2 x p3 … x pn) where p1 is greater than 3. (I expect I’m not writing these things in the proper way, but I’m hoping you’ll see what I mean and can tell me how it should be written.)

The 6n numbers are like stepping stones running up the number line. They provide the steady pulse against which the ‘wave’ emitted by each prime ‘beats’, a metric by which their wavelength can be measured, a frequency analyser marked off in units of six.

Piano tuners tune pianos by listening to the way different frequencies beat aginst each other. If two notes an interval of a fifth apart are in tune then their combination wave will have an audible beat as it cycles. If the two notes being tuned are supposed to cycle 2 against 3, then the tuner will listen out for the 6n numbers, the downbeat of the bar. If he cannot hear them then the two notes are not in tune. What I’m suggesting here is that we can know most of what can be known about the primes by using the same methods. Fourier analysis would be the obvious choice of method but I can’t do that.

The wave of multiples generated by a prime p greater than 3 is always locked into a cyclic relationship with the 6n numbers in such a way that it takes 6p numbers for the two of them to return to the same position in respect of each other.

For example, where p = 7 then f = 42. Every 42 numbers the wave of multiples of 7 returns to the same position in relation to the 6n numbers. It is for this reason that 42 +/-7 are both multiples of 7 at 6n+/-1, 84 +/-7are both multiples of 7 at 6n+/-1, 126 likewise and so on to infinity. These are the only multiples of 7 that occur at 6n+/-1.

If we were sieving for primes then we could start by crossing out every number not at 6n+/-1 in the range. To cross out any remaining non-primes we can instruct an algorithm to cross off 6np+/-p for all values of n greater than 1 and all values of p greater than 3. (This need not require a list of primes. For values of p we could step through all values of 6n+/-1 rather than just the primes. A list of primes would be more efficient but the result would be same.) .

The multiples of a prime p occur at 6np+/-p. We can analyse this waveform into two sine waves, one responsible for the sequence of multiples (p+p)+6p+6p+6p …, the other the sequence (p-p)+6p+6p+6p… . (E.g., where p = 5 these sequences are 35, 65, 95, 125 …, and 25, 55, 85,115, …). Each prime is like a planet orbiting 6n as it travels up the number line, with an eccentric orbit 6p in length. Or, more simply, two moons in perfectly circular orbits around a planet, each with a frequency of 6p but offset.

As we travel up the number line the perceived outcome of the beating together of all these unique sine waves, two for each prime, if we imagine we could hear it, is simple to start with but very quickly becomes too complex to be considered as anything other than pink noise. More like raindrops on a tin roof than music. Euclid proved that however many raindrops we add it never becomes white noise, there will always be moments of silence however many raindrops there are.

I picture the primes as being generated by a collection of tuning forks, each emitting two unique sine waves. As the peaks and troughs of all these sine waves travel up the number line their beating creates a near cacophony, but a cacophony so strictly ordered that however many frequencies are added, even if it is an infinite number, there will always be null points in it, numbers that are at 6n+/-1 but not at 6np+/-p relative to any prime and so are prime.

End
Notes

By using only the 6np+/-p rule for multiples of primes and a few very basic Excel functions I’ve constructed a prime sieve that will sieve a range that can be placed anywhere on the number line, in principle, a prime checker and a function for calculating pi(x), all of which work fine. The first two are reliable and quick up to the available system limit of about twelve digits. The calculation of pi(x) is a very crude approximation but whether by luck or good judgement the output correlates with the true figure for values of x up to about twelve digits. None of these programmes are at all efficient. It would take a mathematician to figure out how efficient they could be made. The point here is simply that they work, and all I’m using is the 6np+/-p rule.

Shall I go on? Am I wasting your time and mine? Is anyone still there? Is any of it unclear? Am I an idiot? Is any of this mathematically interesting? I really have no idea. Perhaps it will only become mathematically interesting if I can prove something interesting by the use of just these simpleminded methods. For this reason I’ve had a go at proving that there are infinitely many twin primes or, rather, to paraphrase Euclid, that there are more twin primes than any finite number of them. I think I’ve succeeded, but I won’t know until I’ve discussed it here. The next bit is twin primes, then Riemann.

By the way, in case it’s not a common analogy, the idea of tuning forks I picked up from Marcus du Sautoy’s book on the primes. He likens Riemann’s non-trivial zeros to tuning forks, suggesting that they emit waves that somehow encode for the distribution of primes, or something like that. I probably misunderstood what he actually said, but it was this analogy that first got me thinking about RH.

Thank you for reading this far. I’m sure it was a struggle.

Canute

 

[Hurkyl]

I think you mentioned this before, and it still sounds just like a wheel sieve.

We start knowing 2*3, and that 1 and 5 are the only numbers less than 6 relatively prime to both.

So we take 5 as being prime. We then go up to 6*5 = 30 by adding multiples of 6:
1 | 7, 13, 19, 25
5 | 11, 17, 23, 29

we have to take out the multiples of 5 (only 5*5 = 25), allowing us to add to our putative list:
7, 11, 13, 17, 19, 23, 29

Then, we take 7 and go up to 30*7 = 210 by adding multiples of 30.
1 | 31, 61, 91, 121, 151, 181
7 | 37, 67, 97, 127, 157, 187
11 | 41, 71, 101, 131, 161, 191
13 | 43, 73, 103, 133, 163, 193
17 | 47, 77, 107, 137, 167, 197
19 | 49, 79, 109, 139, 169, 199
23 | 53, 83, 113, 143, 173, 203
29 | 59, 89, 119, 149, 179, 209

We then get rid of the multiples of 7
1*7, 7*7, 7*11, 7*13, 7*19, 7*19, 7*23, 7*29

Leaving us with only those numbers not having 2, 3, 5, or 7 as a prime factor.

Then, we go up to 210*11 = 2310 by starting with each of the 48 remaining numbers, and generating a row by going up in steps of 210.


Once we get as far as we want to go, we continue sieving in the ordinary fashion. Maybe we wanted to stop at 210. So then we could sieve the remaining 48 numbers by 11, and then 13, and what's left are the primes.



Though I admit I still have no idea why you focus exclusively on the numbers relatively prime to 6. (i.e. those of the form 6n+1 or 6n+5)

But as I mentioned the last time you brought it up, such a restricted view is simply a line sieve: if you know all the primes are of the form 6n+1 or 6n+5, then you can simply sieve over those sets instead of the whole thing. But, as I mentioned, you could do even better by sieving over the lines:

30n+1, 30n+7, 30n+11, 30n+13, 30n+17, 30n+19, 30n+23, 30n+29

and so forth.

 

[shmoe]

This looks to be essentially what's known as 'pre-sieving' with a sieve of erathosthenes. The most primitive version has you consider only odd numbers, essentially wiping off the multiples of two before you start. To pre-sieve by 2 and 3 you'd only consider the numbers of the form 6n+/-1, and you can carry on.

What you are noticing as repeating waves gets turned into the language of residue classes. Take a product of primes, n=p1p2...pk. mod n what you call 'null points', the only places where primes may occur after removing multples of p1 to pk are exactly the residue classes that are relatively prime to n. (Note there's nothing compelling us to take the primes in order for p1, p2, etc. Everything here will work if you wanted to consider say mod n=11*17*37 or even things like n=5*5, though repeating primes isn't an efficient way to build a sieve.)

For example, if you are working mod 6 then the residue classes are represented by 0, 1, 2, 3, 4, 5, only 1 and 5 are relatively prime to 6, so all primes except 2 and 3 are congruent to either 1 or 5 mod 6 (that is they have the form 6m+/-1 if you like).

If you look mod 30 the null points will be in one of the eight residue classes 1, 7, 11, 13, 17, 19, 23, and 29.

This looks to be basic stuff to me, and quite well studied, and also essentially what someone has mentioned here before. To give a taste of what can be said, mod 30 you know the 8 spots that primes _may_ land in. We can actually guarantee that each residue class will have infinitely many primes in it, moreover they are evenly distributed in an asymptotic sense in these classes. For example, the number of primes less than x that are of the form 30n+1 will approach pi(x)/8 as x goes to infinity (this is a very non-trivial theorem).

The ideas here are certainly not idiotic, but like I say my understanding of your post is that it is ideas that are already very well known. I can't urge you enough to pick up an elementary number theory text and learn about residue classes (look for "modular arithmetic") and also to look up the sieve of erathosthenes in it's various forms (including the wheel sieve Hurkyl mentioned). You should also take a gander a Dirichlets theorem on primes in arithmetic progressions (this is different terminology for residue classes). It doesn't take a mathematician to understand these ideas (the proof of Dirichlet's theorem is difficult but it's statement isn't), and you definitely want to learn about modular arithmetic to understand the terminology involved to see how it relates to your thinking.

By the way, this is rather different from the "waves" from zeta's zeros that I was talking about in the other thread.

 

[Hurkyl]

An example of the use of a line sieve:

I want to find primes in the range 1000-2000.

The numbers in this range of the form 6n+1 have n = 167, 168, ..., 333.

The first multiple of 5 in this list is 1015 = 6*169 + 1. So, I can remove 1015, 1045, 1075, ...

Actually, it's much more convenient to sieve on the n values: that is, instead of sieving the numbers 1000-2000 of the form 6n+1, I'm going to sieve the range of n values 167-333. To sieve by 5, I cross off:
169, 174, 179, 184, ..., 329

The first multiple of 7 occurs at n = 169. So I continue and sieve 176, 183, 190, 197, ...

I figured that out by considering:
7k = 6n + 1
0 = 6n + 1 (mod 7)
6n = 6 (mod 7)
n = 1 (mod 7)

and n=169 is the first number in our range that satisfies this equation.

Anyways, once I've sieved by all my primes, what's left are the n values corresponding to the primes in 1000-2000 of the form 6n+1. I just repeat with 6n+5, and I'm done.

 

[Canute]

Ok, I see that there's nothing new in what I've said so far. But I'll carry on and see if the next bit seems more interesting to you. Just one thing to mention here. Of course sieving is easy and I wasn't suggesting that my way was much better than that of Aritosthenes, except for one small difference. In your example Hurkyl you say 'The first multiple of 7 occurs at n = 169. So I continue and sieve 176, 183, 190, 197, ...'. In regard to sieving, my point was that it is not necessary to sieve all these numbers. For 7 one needs to cross off just two multiples in every 42 numbers, not all the products of 7, since four out of six of these do not fall at 6n+/-1. I mentioned this not because it is a new way of sieving but simply to illustrate that knowing how the multiples of primes >3 behave in relation to the 6n numbers allows us to make some useful predictions, the simplest of which is this 2 multiples in 6 rule.

Anyway, thanks for the comments. I'll be back with some more in a short while. If it doesn't get interesting soon then I'll know I'm wasting my time.

 

[Hurkyl]

In regard to sieving, my point was that it is not necessary to sieve all these numbers. For 7 one needs to cross off just two multiples in every 42 numbers, not all the products of 7, since four out of six of these do not fall at 6n+/-1.

I am!

n = 169 corresponds to the number 6*169+1, the first multiple of 7 in my range of the form 6n+1.
n = 176 corresponds to the number 6*176+1, the second multiple of 7 in my range of the form 6n+1. (which is 42 more than the first)
etc
(note that 169 is not a multiple of 7 -- I don't care -- it's that 169 is the first thing I plug into f(n)=6n+1 that gives me a multiple of 7 in my range of interest)

In terms of doing the actual sieving, it's easier to sieve on the values of n, instead of sieving on the values of 6n+1.


If I tried to sieve the entire interval 1000-2000, I'd be wasting a lot of space. I would never look at 67% of the values, and, each of the two phases of the algorithm only look at half of what's left.

However, when sieving the numbers of the form 6n+1, it is much more convenient to sieve on the n's, because there is no wasted memory.

 

[Canute]

Ah. Sorry. Thanks for re-explaining. Now I understand better what you're doing. It seems to be the same thing as me but using a more sensible method. (Mind you, I can sieve two numbers for each value of n so perhaps there's not much in it - assuming I've understood your calculation properly).

 

[Canute]

2. Twin Primes.

To prove that there are infinitely many single primes Euclid asked us to imagine that there is a highest prime. If we multipy together all the primes up to and including this highest prime then the result will be a number at 6n. The number above this, at 6n+1, cannot be a product of any prime in this sequence so it must either be prime or the product of a prime higher than the highest prime. This is absurd, and so there must be infinitely many primes.

This proof also has a bearing on the quantity of twin primes. Call our imaginary highest prime P. Call the product of the primes up to and including this highest prime N(P). The two numbers at N(P)+/-1 have no divisor >1 at or below P. If either of them is not prime then this can only be because it is the product of a prime between P and sqrtN. We can make use of this as the basis for a proof of the infinitude of twin primes if we can show that there are infinitely many values of P for which neither of the numbers at N(P)+/-1 are multiples of primes between P and sqrtN(P).

The quantity of numbers in this range increases as P increases. At the same time the ratio of primes to non-primes in this range decreases as P increases. If we were to attempt to prove that there are infinitely many values of P for which N(P)+/-1 is a twin prime by attempting to create a function for the number of primes between P and sqrtN as P increases then we would have do some very complex mathematics. However, the rules governing the multiples of primes allow us to approach the problem from another direction.

If there are no primes between P and sqrtN(P) then N(P)+/- will be a twin prime. If, therefore, there are infinitely many values of P for which no primes occur between P and sqrtN(P) then there must be infinitely many primes.

From the 6np+/-p rule governing the occurrence of multiples of primes it can be shown that a sequence of numbers containing no primes can be arbitrarily long. (Of course, this is already well known). ‘Arbitrary’ here does not mean finite. It is possible to show that a sequence of non-primes can be infinitely long from this rule. I find this interesting so I’ll explain what I mean.

Note: I’ll write N instead on N(P) because it’s easier and tidier. Wherever I write upper-case N this stands for N(P) or P!

Because N is the product of a consecutive sequence of primes then on each side of N there stretches a consecutive sequence of multiples of these primes. For example, the numbers N+/-5 will be at 6n+/-1 and will not be prime because they are muliples of 5. The numbers at N+/-7 will be at 6n+/-1 and will not be prime because they are multiples of 7 and so on, until we get to N - P and N + P. This entails that between N and N- P only N - 1 can be a prime number, and between N and N + P only N + 1 can be prime number. Thus, for any value of P there will be a maximum of two prime numbers between N - P and N + P.

As there are infinitely primes then eventually P becomes infinitely large and the range N - P to N + P becomes infinitely large. That is, if we multiply together all the primes up to and including P = infinity then N must be divisible by every prime number, and the only prime numbers between N + infinity and N - infinity will be at N +/-1. These two numbers must be prime since they are one number away from a number divisible by every prime. Thus, there are two infinitely long sequences of consecutive non-primes on either side of N+/-1 where P = infinity.

The only number that is divisible by every prime is zero. In this respect, zero has the precise quality N would have where P = infinity. Further, zero is a 6n number so in this respect also zero qualifies as N. For zero to fully qualify would require that the two numbers at 0 +/-1 are the only two prime numbers between 0 + infinity and 0 - infinity. In this respect zero also qualifies, for -1 and +1 are the only two numbers divisible only by themselves. Thus, er, if P = infinity then N = 0.

All this doesn’t help us much with the twin primes conjecture since it suggests there is only one of them. So, leaving aside the paradoxes to which infinities give rise, what can be said about finite sequences of non-prime numbers?

As the sequence of consecutive primes between 0 and P can be of any length then the sequence of numbers between N + P and N - P can be of any length. There may be a prime at either N+/-1 and so this is not necessarily a sequence of non-primes. However, the sequence of numbers between N - 2 and N - P can also be of any length and there are no prime numbers in this sequence. In this case a sequence of non-primes may be arbitrarily long.

We still do not know whether there are infinitely many occasions on which there are no primes between P and sqrtN, but it now seems more possible. As primes become more scarce as P increases perhaps they become so rare that there are almost always no primes in this range. On the other hand, as P increases, the range of non-primes N - 2 to N - P increases much more slowly than the range P to sqrtN. Perhaps, as a result, beyond a certain value of P there is always one or more primes between P and sqrtN.

Even though a sequence of non-primes may be arbitrarily long the sequence N-2 to N-P is always considerably higher up the number line than P. In this case it is seems likely that byond a certain value of P there may never again be a long enough sequence of non-primes low enough down the number line to ensure that there are no primes between P and sqrtN.

We need not bother with this conjecture. Let's assume that there will always be primes between P and sqrtN but try to make some general predictions for the behaviour of these primes as P increases. This might enable us to predict that there will be infinitely many instances where these primes are so arranged that none of them are divisors of N+/-1. If so, then there are infinitely many twin primes.

We need to show that as the value of P increases there must be infinitely many instances where N+/-1 is a twin prime. Given the astonishing complexity of the combination of wave of multiples of the primes below P where P is of any decent size it would seem impossible to do this by sticking to these simple methods. However, this complexity works in our favour. As there are infinitely many primes there are infinitely many values of P and N, and as the shape of the combination of multiples of primes changes every time a new prime is added to it then it would be odd if it were the case that above some particular value of P one of the numbers at N+/-1 must always be the product of a prime between P and sqrtN.

It seems likely, given that there are infinitely many values of N, that there are infinitely many occasions on which N+/-1 are both prime. That is, if there is no rule that determines that one of the numbers at N+/-1 must always be a product of a prime between P and sqrtN, then now and again, to infinity, N(P)+/-1 will be a twin prime.

We can make this even more certain. Not only are there an infinite quantity of sequences 0 to P, but there are infinitely many instances of N for any given collection of primes. What I mean is, there are infinitely many twin pairs of numbers (x, x+2) at 6n+/-1 that have no divisors >1 at or below P, as well as infinitely many values of P.

This is because the cyclical relationship between the multiples of the primes and the 6n numbers holds also for the product of a collections of primes. If N = 2 x 3 x 5 x 7 x 11 then N+/-1 lie at the center of a 22 number range of non-primes. More generally we can say that for any n, 6n(p1 x p2 x p3 … x P) is always at the centre of a range of 2P numbers in which only N +/-1 can be prime.

This means that every time we multiply N by 6 there will be a sequence of non-primes between N -2 and N - P and between N + 2 and N + P. For example, if we multiply together all the primes below 1000 to get N, then for any n, 6nN will be at the centre of a range of 2000 number in which only 6nN+/-1 can be prime.

The rules ensure then that for any value of P there are an infinite number of possible twin pairs (x,x+2) that have no divisors less than or equal to P. Thus, where P is a prime around 10,000,000 then N will be at the centre of a sequence of 20,000,000 numbers in which N+/-1 are the only two numbers that might be prime, and there are infinitely many instances of this sequence.

[This allows another improvement to Aritosthenes’s sieve. Let P = 1999 and N = P! as usual. Say we sieve the first 500 numbers above N and we find primes at N + 123 and N + 431. We can now sieve an infinite number of similar ranges more easily. We know that for all value of n, between 6nN and 6nN+500 the only two numbers that can be multiples of a prime < 2000 are at 6nN+123 and 6nN+431. So for each value of n we have two chances of finding a prime, and we know the two places to look.]

As P increases there is a constantly shifting relationship between N and the position of primes between P and sqrtN. That is to say, there is no reason for primes to occur between P and sqrtN in such as way as they will or will not be divisors of the numbers of N+/-1. Rather, the pattern of primes between P and sqrtN is determined to be ever-changing, in that the number and position of primes within this range, and thus their relationship to N+/-1, will be different for every value of P. As P increases the pattern of primes between P and sqrtN can never settle down in such a way as to ensure that one of N+/-1 must always have a divisor in this range. Equivalently, as P increases there is no reason that N+/-1 may not be a twin prime.

To say this differently, from the predictable way in which the combination wave of multiples of primes evolves as P increases, specifically the fact that this pattern continually changes its wavelength and the relative positions of its null points as each prime is added, then it seems inevitable that the pattern of primes between P and sqrtN must be such as to leave N+/-1 a twin prime infinitely many times.

This is a strange kind of attempted proof, but considering N +/-1 may be a twin prime even if all the numbers at 6n+/-1 between P and sqrtN are prime, then to me it seems fairly sound, if still very informal. Perhaps it would only become a proof there are infinitely many twin primes if we could show that N+/-1 must be a twin prime infinitely many times, while all I’ve shown is that there is in principle no reason why it shouldn’t be one. If there are infinitely many values of N and infinitely many values of 6nN, then it seems reasonable to suppose that there must be infinitely many twin primes. However, can this be an informal proof while I have still not shown it is impossible for there to be some value of P so large, and the sequence P to sqrtN so enormously long, that from then onwards there is always a divisor of either N+/-1 somewhere between P and sqrtN, even if only by coincidence? It seems to me that this objection doesn’t hold, but I’ll wait and see.

The infinitude of twin primes is made more likely by the fact that as P increases the ratio of primes to non-primes between P and sqrtN decreases. This may mean, and it should be possible to calculate this, that the probability of N+/-1 being a twin prime actually goes up rather than down as P increases. Also, the probability of an individual prime between P and sqrtN being a divisor of a number at N+/-1 falls as P increases.

If p = 101 then there will be 2 products of 101 at 6n+/-1 in every 606 numbers. We are not interested in any multiples except those at 6n+/-1, so we can discard the rest and say that there will be 2 products of 101 in every 202 numbers at 6n+/-1. In this way, the probability of a number above 606 at 6n+/-1 being a product of 101 is the reciprocal of 101. If P = 1,000,001 (let’s assume this is a prime, I don’t know) then the probability of N + 1 being a multiple of 103 are about a million to one. For a prime nearer to sqrtN where P = 1,000,001 the probability becomes vanishingly small. So not only does the ratio of primes to non-primes between P and sqrtN decrease as P increases, but so also does the probability that a particular one of them is a divisor of either N+/-1.

I have not tried to work out whether the probability of N+/-1 being a twin prime increases or decreases as P increases. I suspect that beyond a certain value of P it slowly increases. I can see how the calculation could be done but actually doing the sums would be beyond me, unless I could find a shortcut. But perhaps there is no need to show this much. If we could just show just that the probability of N+/-1 being a twin prime never falls to zero as P increases, then this would entail that there are infinitely many twin primes. Or so it seems so to me. To show this means examining the distribution of the primes.

Before going on to look at their distribution, which diverts us into some thoughts on RH, here is a summarised version of the proof I’m attempting. Depending on your comments I’ll either scrap it or try to complete it later.

1. For any value of P (P is defined as prime) there are an infinite quantity of pairs of numbers at 6n+/-1 that do not have divisors >1 in the set of numbers less than or equal to P.

After a bit of tutoring on another forum I think this can be stated as either:

a) For any finite collection F of primes there exist infinitely many pairs of the form {x, x+2} such that both x and x+2 are at 6n+/-1 and both are relatively prime to each f in F.

b) For any finite collection F of numbers, there exists an infinite set S = {(x,x+2) | such that for all f in F, gcd(x,f) = gcd(x+2,f) = 1}

(These three statements are supposed to be equivalent, but I’m not completely sure they are).

2. For every finite sequence of consecutive primes starting at zero there is an infinite quantity of pairs of numbers (x, x+2) at 6n+/-1 that have no divisors >1 in the sequence, and there are an infinite number of such sequences.

3. For the infinity of pairs of numbers at 6nN+/-1 that have no divisors >1 less than or equal to P, the rules governing the occurrence of primes do not ensure that one or the other of these numbers must have a prime divisor between P and sqrtN .

4. Therefore, twin primes must occur infinitely many times.

It is 3. that causes me problems. I think this can be shown, but I’m not sure if my way of doing it would count as a proof. Is it allowable to say that if something may happen then given infinitely many opportunities for it to happen it will happen infinitely many times? This seems to be the argument we use when saying that a coin will fall heads up infinitely many times given infinite tosses.

Because 3. requires a look at the distribution of the primes I’ll come back to twin primes (if anyone’s still here) after a few comments on pi(x) and RH. The next bit is probably the make or break section. If by the end of that I haven’t said anything interesting then I’ll stop. I'll stop here for comments anyway.

 

[shmoe]

That's long. I don't have a lot of time to comment right now, so just a few of random things.

As there are infinitely primes then eventually P becomes infinitely large and the range N - P to N + P becomes infinitely large. That is, if we multiply together all the primes up to and including P = infinity then N must be divisible by every prime number, and the only prime numbers between N + infinity and N - infinity will be at N +/-1. These two numbers must be prime since they are one number away from a number divisible by every prime. Thus, there are two infinitely long sequences of consecutive non-primes on either side of N+/-1 where P = infinity.

This is the kind of rubbish that makes people stop reading. This isn't mathematics at all, P cannot be 'infinity', you cannot multiply an infinite quantity of numbers. There are reasons limits you see in calculus are defined the way they are.

We still do not know whether there are infinitely many occasions on which there are no primes between P and sqrtN, but it now seems more possible. As primes become more scarce as P increases perhaps they become so rare that there are almost always no primes in this range. On the other hand, as P increases, the range of non-primes N - 2 to N - P increases much more slowly than the range P to sqrtN. Perhaps, as a result, beyond a certain value of P there is always one or more primes between P and sqrtN.

There are always primes between P and sqrt(N) past a certain point. See Bertand's postulate which gives a prime between n and 2n for n large enough, sqrt(N) is much larger than 2P. N(P) is about e^P (this is the prime number theorem in disguise as it relates to the summatory function of the chebyshev function)). The prime number theorem then gives pi(sqrt(N))~(2/P)*e^(P/2) and pi(P)~P/log(P), and you actually have quite a few, really the number of primes up to P pales in an asymptotic comparison with the number of primes up to sqrt(N) (all these ~'s and 'abouts' can be made precise of course)

By the way, N(P) is often called the primorial function, and probably others.

[This allows another improvement to Aritosthenes’s sieve. Let P = 1999 and N = P! as usual. Say we sieve the first 500 numbers above N and we find primes at N + 123 and N + 431. We can now sieve an infinite number of similar ranges more easily. We know that for all value of n, between 6nN and 6nN+500 the only two numbers that can be multiples of a prime < 2000 are at 6nN+123 and 6nN+431. So for each value of n we have two chances of finding a prime, and we know the two places to look.]

The numbers in your example look bad, none of the 500 numbers after N=1999! can be prime except perhaps N+1. That aside, it still looks false as i understand what you're trying to do. Look up Dirichlet's theorem on primes in arithmetic progressions, it is a direct contradiction to what you're saying here as I understand it. Each of the residue classes mod 6nN that is relatively prime to 6nN will have infinitely many primes in it. It doesn't matter that in some smaller range a number in a residue class failed to be prime, so you cannot speed up a sieve this way.

Perhaps you could privde a more in depth example of what you mean here, maybe just with P=5, so I can be sure I get what you're saying.

The infinitude of twin primes is made more likely by the fact that as P increases the ratio of primes to non-primes between P and sqrtN decreases. This may mean, and it should be possible to calculate this, that the probability of N+/-1 being a twin prime actually goes up rather than down as P increases.

If we are to believe the more precise version of the twin prime conjecture given by Hardy and Littlewood (it provides an asymptotic) then we would believe the density of twin primes tails off just like the density of primes. Actually the fact that the sum of the recipricals of the twin primes converges (this has been proven by Brun) should let you prove for a fact that the density tails off, otherwise this sum would be similar in nature to the harmonic sum (I'll have to think about the details, though likely this has already been examined or follows from some simple fact about asymptotic densities I don't have at my fingertips).

I'll have a closer look as time permits.

 

[Canute]

That's long.

Yes, much too long. Sorry. I wanted to check my thought process, not just the end result. I'll try to keep everything brief from now.

This is the kind of rubbish that makes people stop reading. This isn't mathematics at all, P cannot be 'infinity', you cannot multiply an infinite quantity of numbers. There are reasons limits you see in calculus are defined the way they are.

I know I cannot actually multiply an infinity of numbers. I just find the relationship between zero and infinity interesting.

I forgot about Bertrand's postulate although, surprisingly, I did know about it. But what I've tried to do is re-invent the wheel, doing things a different way, to see how far I can get being as naive as possible (which I have to be).

The numbers in your example look bad, none of the 500 numbers after N=1999! can be prime except perhaps N+1.

Oh damn. Why can't I get this right? This is not what I meant to say at all. However, I now see that even what I meant to say is incorrect. Please forget this point entirely.

If we are to believe the more precise version of the twin prime conjecture given by Hardy and Littlewood (it provides an asymptotic) then we would believe the density of twin primes tails off just like the density of primes. Actually the fact that the sum of the recipricals of the twin primes converges (this has been proven by Brun) should let you prove for a fact that the density tails off, otherwise this sum would be similar in nature to the harmonic sum (I'll have to think about the details, though likely this has already been examined or follows from some simple fact about asymptotic densities I don't have at my fingertips).

Yes, but this is not quite relevant. While the density of twin primes tails off the probability of P!+/-1 being prime may nevertheless increase as P increases. Is this not so? Would it be of any interest to anyone if I could put a figure on the probability of P!+/-1 being prime as P increases?

I'll have a closer look as time permits.

I appreciate your efforts, but no need to take a closer look. I'm learning at high speed from your comments and see that much of what I'm saying is trivial, wrong or incompetently expressed. Could you just have a quick look at the first statement in my outline of a proof that there are infinite twin primes, and tell me whether the statements a and b are correct, comprehensible and/or contentious?

I will not inflict any more long posts on you, and sorry for the last one. I've not been quite sure what needs saying and what does not, since I've come at all this with no previous knowledge. The less the better seems to be the answer.

 

[shmoe]

Just a comment this bit for now:

Yes, but this is not quite relevant. While the density of twin primes tails off the probability of P!+/-1 being prime may nevertheless increase as P increases. Is this not so? Would it be of any interest to anyone if I could put a figure on the probability of P!+/-1 being prime as P increases?

It turns out to be quite relevant. The thing that's increasing your chances of N+/-1 being prime (N the product of primes less than or equal to P) is you know it's not divisible by the primes less than or equal to P. However there are essentially none of these compared to the primes between P and sqrt(N). The probability a number is not divisible by p is roughly 1-1/p. The probability that a number near N is prime is about 1/log(N)=1/P by the prime number theorem. However, we already know that N+/-1 is not divisible by 2, 3, ..., P, so we divide this by

[tex]\prod_{p\leq P}\left(1-\frac{1}{p}\right)\sim e^{-\gamma}/\log(P)[/tex]

The product is over primes p, and [tex]\gamma[/tex] is euler's gamma~0.577... So the probability N+/-1 is prime will be roughly [tex]e^{\gamma}\log(P)/P[/tex] so even the probability either one is prime is approaching zero as P tends to infinity and the probability they are a twin prime pair is necessarily going to zero as well.

Of course this kind of probabalistic argument isn't a proof, but it can work pretty darn well. Have a look at hardy and littlewoods argument for their twin prime conjecture, numerical evidence has shown it to be pretty accurate (their argument is in Hardy and Wright's intro number theory book and other places). The important idea here is that the primes up to P mean very little compared to the primes between P and sqrt(N) in the grand scheme of things.

I'll take a look at your numbered points later.

edit-sorry for the major edit, fixed one of my 'roughly's' to be more accurate. It's still a heuristic argument though.

 

[Canute]

Amazing! I actually understood much of that, I think. You must be a genius. I have a few questions.

What is "euler's gamma~.577..."?

If the probability of N+/-1 being a twin prime necessarily goes to zero as P increases then surely this is a proof that there are not infinitely many twin primes? If, on the other hand, it goes towards zero but never reaches it, then there are infinitely many twin primes. Or have I misunderstood something?

Am I correct in thinking that the probability of N+/-1 being prime is 1 minus the sum of the reciprocals of the primes up to and including sqrt p, and that the sum of the reciprocals can be calculated by means of the prime number theorem, euler's gamma and so forth? Is that anything like correct? If it is then I've understood something I've never understood before. I could see what had to be calculated but hadn't realized that this is what these functions did.

This would explain something else. My crude function for pi(x) counts the multiples of primes in the relevant range and deducts these from the quantity of 6n+/-1 numbers. It has a correction term in order to allow for the fact that many of the numbers at 6n+/-1 are the product of more than one prime. This correction term involves summing the reciprocals of the primes in the relevant range. So is this, or something like this, the reason they are summed within R's zeta function? I have assumed it is, and that this is the point of connection between my naive arithmetic and RH.

A Note

I've not yet fulfilled my absurd marketing hype claim about a proof of RH, but I haven't quite given up yet. It strikes me also that someone is missing out on great publishing opportunity. I’ve struggled through a couple of popular books on the primes, barely understanding any of the detailed mathematics, and have also wandered around the Internet looking for information, but never have I come across a straightforward explanation of why the primes behave as do.

Mathematician’s naturally focus on the mathematics when they write popular books, assuming that everybody knows why primes occur where they do. But most people cannot do mathematics and have no idea. There are millions of people out there who are fascinated by the primes and who believe it is a complete mystery why they occur where they do. I know a few of them. People can have the weirdest ideas about the primes, not realising that they are mechanistically caused and, in principle at least, perfectly predictable. Even after reading ‘Music of the Primes’ twice I didn’t know this, and if had known it would have saved me a lot of trouble.

I suspect that most non-mathematicians, the few who have ever thought about the primes, assume that even mathematicians are mystified as to why they occur where they do. If someone were to write a book explaining simply and in more or less natural language why they behave the way they do, establishing an understanding of the basic mechanisms and principles before going on to explore how these relate to the twin primes conjecture, the prime number theorem, RH and so on, then it would have sizeable niche market. Is there anything like this already on the market?

 

[shmoe]

What is "euler's gamma~.577..."?

gamma is the constant defined as:

[tex]\gamma=\lim_{n\rightarrow\infty}\sum_{k=1}^{n}\frac{1}{k}-\int_{1}^{n}\frac{dt}{t}=\lim_{n\rightarrow\infty}\sum_{k=1}^{n}\frac{1}{k}-\log n[/tex]

It's about .57721566... and turns up all over number theory. You can think of it as the difference between the area under the graph of 1/t from 1 to infinity and the rectangles of width 1 above it that are trying to approximate this area (both areas are infinite, but their difference is finite)

If the probability of N+/-1 being a twin prime necessarily goes to zero as P increases then surely this is a proof that there are not infinitely many twin primes? If, on the other hand, it goes towards zero but never reaches it, then there are infinitely many twin primes. Or have I misunderstood something?

This is not a complete argument. the probability that a number is divisiple by p is (1-1/p) is really only true for ranges of numbers that are divisible by p. For example, the probability a number in {1,2,3,4,5,6} is divisble by 2 is 1/2, but in {1,2,3,4,5,6,7} it's not. However the larger your range of numbers, the more accurate this 1-1/p is, and since P is so much smaller than N, it's reasonable to expect the analysis in my last post to hold. Look up hardy and littlewoods version for twin primes to see their justifications for more details on where the holes in this kind of argument lie.

Another thing, the probability a number near x is prime is 1/log(t), this goes to zero yet there are still infinitely many primes. It just means they become 'sparse'.

It's also worth noting that 'probability' is really referring to some kind of asymptotic density statements. The primes of course aren't randomly distributed.

Another way of measuring 'how many are tehre' is by looking at the sum of the reciprocals. The sum of the reciprocals of the primes diverges so there must be infinitely many of them. The sum of the reciprocals of the squares converges, so even though there are infinitely many squares, they are in a sense less numerous than the primes. The sum of the reciprocals of the twin primes converges, so they are decidedly less numerous than the primes, but this gives no info on whther there are infinitely many or not.

Am I correct in thinking that the probability of N+/-1 being prime is 1 minus the sum of the reciprocals of the primes up to and including sqrt p, and that the sum of the reciprocals can be calculated by means of the prime number theorem, euler's gamma and so forth? Is that anything like correct? If it is then I've understood something I've never understood before. I could see what needed to be calculated but hadn't realized that this is what these functions do.

Pretty much, but the (1-1/p) probability deal get's less and less accurate as p get's close to your number. This is why I used the prime number theorems prediction of 1/log(N) for the bulk of it and kept this product over 1-1/p to include terms less than P, which is extremely small compared to N.

Merten's proved [tex]\prod_{p\leq x}\left(1-\frac{1}{p}\right)\sim\frac{e^{-\gamma}}{\log{x}}[/tex]

This would explain something else. My crude function for pi(x) counts the multiples of primes in the relevant range and deducts these from the quantity of 6n+/-1 numbers. It has a correction term in order to allow for the fact that many of the numbers at 6n+/-1 are the product of more than one prime. This correction term involves summing the reciprocals of the primes in the relevant range. So is this, or something like this, the reason they are summed within R's zeta function?

You 'correction term' sounds like it might be the principle of inclusion/exclusion you find in combinatorics classes, and is what Legendre used as the first way to compute pi(x) without needing to find all primes less than x. If you have two primes p and q, and the set M={1,2,...,m} there are [m/p] multiples of p and [m/q] multiples of q in M ([] is the floor function). However, there are [m/pq] multiples of pq in M which are counted twice. Therefore there are [m/p]+[m/q]-[m/pq] multiples of p or q in M. It get's much more complicated with more primes to sieve out.

The 1-1/p terms turn up in the product version of Zeta beacuse of the geometric series [tex](1-1/p)^{-1}=1+1/p+1/p^2+\ldots[/tex]. the product of these series over all the primes gives the term 1/n exactly once for each n by unique factorization. This is why you can think of the equivalence of the product and the sum versions as an analytic version of the fundamental theorem of arithmetic.

As given above, the product will diverge, you really have terms [tex](1-1/p^{s})^{-1}[/tex] for suitable choices of s, so you never really have a 1-1/p term. You can still think of these things as probabilities though, 1/Zeta(2) gives the probability two integers are relatively prime for example (formulated suitable as an asymptotic density statement of course).

Is there anything like this already on the market?

I really have no idea. The popsci books seem pretty empty of mathematics from my view. I would think that some kind of exposition on a basic sieve would cover much of what you are hoping for, but I don't really know. Besides du Satouy's, have you read any of the other books?

 

[shmoe]

to reply to your earlier post-

1. For any value of P (P is defined as prime) there are an infinite quantity of pairs of numbers at 6n+/-1 that do not have divisors >1 in the set of numbers less than or equal to P.

After a bit of tutoring on another forum I think this can be stated as either:

a) For any finite collection F of primes there exist infinitely many pairs of the form {x, x+2} such that both x and x+2 are at 6n+/-1 and both are relatively prime to each f in F.

b) For any finite collection F of numbers, there exists an infinite set S = {(x,x+2) | such that for all f in F, gcd(x,f) = gcd(x+2,f) = 1}

(These three statements are supposed to be equivalent, but I’m not completely sure they are).

2. For every finite sequence of consecutive primes starting at zero there is an infinite quantity of pairs of numbers (x, x+2) at 6n+/-1 that have no divisors >1 in the sequence, and there are an infinite number of such sequences.

These all seem to be saying the same thing, and are certainly true. Take N to be the product of primes less than P and consider mN+/-1 for any natural number m. mN+/-1 has no prime factor less than P.

Though I have no idea why you're obsessed with 6n+/-1. There's really nothing special about 6. You can consider mod 30, then the prime pairs are 30n+1 and 30n-1, 30n+11 and 30n+13, or 30n+17 and 30n+19. The sequence of pairs {30n+11, 30n+13} gives a sequence that fits your criterea for the set of primes 2, 3 and 5 and says something even stronger, homing in on where it lands mod 30, not just mod 6 (of course you can say the same thing about 30n+/-1 and {30n+17,30n+19}). Similar deal for any modulus.

3. For the infinity of pairs of numbers at 6nN+/-1 that have no divisors >1 less than or equal to P, the rules governing the occurrence of primes do not ensure that one or the other of these numbers must have a prime divisor between P and sqrtN .

True statement- you haven't been able to find a reason why one of these numbers must have a prime divisor between P and sqrt(N)

False conclusion- the converse must be true infinitely often.

There's a lot of things we can't prove, this doesn't mean they aren't true.

Is it allowable to say that if something may happen then given infinitely many opportunities for it to happen it will happen infinitely many times? This seems to be the argument we use when saying that a coin will fall heads up infinitely many times given infinite tosses.

No, that is not at all a valid proof, especially when you're defining 'opportunity' as 'something you were unable to rule out'.

Saying that the limit of the ratio of heads to tosses approaches 1/2 as the number of tosses approaches infinity is just a restatement of how we defined the probability as a limit of the number of trials. You are never guaranteed any outcome with a real coin and a necessarily finite number of trials.

 

[Canute]

Sorry to be slow - I haven;t been able to log in for a couple of days. Has there been a fault at this end? I'll respond to your second post first.

Though I have no idea why you're obsessed with 6n+/-1. There's really nothing special about 6... Similar deal for any modulus.

Ok, I see that. There is something special about 6 to me, but that's just because of the way I've approached all this. As long as the modulus is a multiple of 6 the deal is the same.

True statement- you haven't been able to find a reason why one of these numbers must have a prime divisor between P and sqrt(N).

False conclusion- the converse must be true infinitely often.

There's a lot of things we can't prove, this doesn't mean they aren't true.

I'm not sure about this. I think it can be shown that the converse must be true infinitely often - although I cannot actually show it yet. At least, I'm, not simply suggesting that I cannot find a reason that these numbers must have such a divisor. Rather, I'm suggesting there are reasons why infinitely often they will not have such divisors.

Would it be true to say that given an infinite number of spins a roulette wheel must output zero infinitely many times? Or, can an infinite number of spins produce a finite number of zeros? The latter seems counterintuitive to me.

Saying that the limit of the ratio of heads to tosses approaches 1/2 as the number of tosses approaches infinity is just a restatement of how we defined the probability as a limit of the number of trials. You are never guaranteed any outcome with a real coin and a necessarily finite number of trials.

I'm ok with that. But what about where there are an infinite number of trials? Do we just take the limit as the outcome?

 

[Canute]

This is not a complete argument. the probability that a number is divisiple by p is (1-1/p) is really only true for ranges of numbers that are divisible by p. ... Look up hardy and littlewoods version for twin primes to see their justifications for more details on where the holes in this kind of argument lie.

I've probably missed your point, but as far I can tell this doesn't seem to address my question. If, as I think you said, the probability of N+/-1 being a twin prime necessarily goes to zero as P increases, then doesn't this mean that there are not infinitely many twin primes? (Although, I suppose there could be an infinity of twin primes not at N+/-1. Is this what you meant by saying it was not a complete argument?) On the other hand, if this probability does not go to zero (or never reaches its limit) then wouldn't this mean there is no highest twin prime?

It's also worth noting that 'probability' is really referring to some kind of asymptotic density statements. The primes of course aren't randomly distributed.

Ok. I realize that in reality the probability of a number being prime is 1 or 0.

Pretty much ...

I thought so but expected to be wrong. This is the sort of thing a popmaths book ought to mention IMHO. It came as a revelation to me when I first figured it out.

You 'correction term' sounds like it might be the principle of inclusion/exclusion you find in combinatorics classes, and is what Legendre used as the first way to compute pi(x) without needing to find all primes less than x.

Good grief. I can do that as well. It seems that although I've been very naive I have at least arrived at the right sort of conclusions. Mind you, I expect Legendre worked this out when he was at primary school.

I really have no idea. The popsci books seem pretty empty of mathematics from my view. I would think that some kind of exposition on a basic sieve would cover much of what you are hoping for, but I don't really know. Besides du Satouy's, have you read any of the other books?

They seem chock a block full of mathematics from my view. M. du Sautoy's was bad enough but John Derbyshire's 'Prime Obsession' was worse. These are the only two books I've read on the topic. I'd like to read Hardy and Littlewood, as you've suggested, but I've assumed I wouldn't be capable of understanding it. If their book is any more mathematical than Derbyshire's then I haven't got a chance of following it.

By the way - tell me when you've had enough. At the moment I'm finding your comments amazingly helpful so would like to keep going, but I know the discussion must be rather tedious from your point of view.

 

[shmoe]

Ok, I see that. There is something special about 6 to me, but that's just because of the way I've approached all this. As long as the modulus is a multiple of 6 the deal is the same.

The modulus doesn't need to be a multiple of 6 either!

I'm not sure about this. I think it can be shown that the converse must be true infinitely often - although I cannot actually show it yet. At least, I'm, not simply suggesting that I cannot find a reason that these numbers must have such a divisor. Rather, I'm suggesting there are reasons why infinitely often they will not have such divisors.

That was just how your statements (3) and (4) looked to me.

It may very well be that N+/-1 are primes infinitely often. It may also be the case that the N+1's are prime infinitely often and the N-1's are prime infinitely often yet there are only finitely many twin primes to be had this way (like what we can say about 30n+11 and 30n+13 for example).

Would it be true to say that given an infinite number of spins a roulette wheel must output zero infinitely many times? Or, can an infinite number of spins produce a finite number of zeros? The latter seems counterintuitive to me.

I'm ok with that. But what about where there are an infinite number of trials? Do we just take the limit as the outcome?

You can't spin a roulette wheel infinitely many times . Any (reasonable) model for the space of infinitely many roulette spins will have the probability of the set of events with only a finite number of 0's as being zero. This doesn't mean these events don't exist (at least in an abstract sense, none of the events exist in the real world). We can get into a more detailed discussion of what we mean when we talk about probability, but this is getting a little off topic. You of course know the primes aren't really random, this is really an imprecise way of thinking about densities of sequences which I'll talk about more below.

I've probably missed your point, but as far I can tell this doesn't seem to address my question. If, as I think you said, the probability of N+/-1 being a twin prime necessarily goes to zero as P increases, then doesn't this mean that there are not infinitely many twin primes? (Although, I suppose there could be an infinity of twin primes not at N+/-1. Is this what you meant by saying it was not a complete argument?) On the other hand, if this probability does not go to zero (or never reaches its limit) then wouldn't this mean there is no highest twin prime?

By 'not a complete argument', I mean it makes some leaps that haven't been justified, and is possibly not true. These kinds of probabilistic arguments, like Hardy&Littlewoods for twin primes, are not proofs. However, they can give results that look suprisingly accurate. The conjectured density of twin primes has held up to the actual data quite remarkably. In the case of N+/-1, there is much less data to go on, but tests up to something like P<120,000 do support the suggested result (N(P) gets unworkably huge fast, so the data is more limited), and the assumptions that can't be proven rigorously are still very 'reasonable'.

Let me put what the 'probability goes to zero' means in more precise terms. Let's just consider the sequence of perfect squares, 1, 4, 9, 16, 25, ... Let S(x) be the number of perfect squares less than or equal to x, so S(10)=3, S(19)=4, and so on. We actually have S(x)=[sqrt(x)], where [] is the floor function, i.e. S(19)=[sqrt(19)]=[4.358...]=4.

The ratio of squares to the total number of integers less than x is S(x)/x, considering x an integer for simplicity. If we randomly select an integer from 1, 2, 3, ..., x, the probability we get a perfect square is simply S(x)/x. Now S(x)/x=[sqrt(x)]/x~1/sqrt(x) (when y is very large, [y] is pretty close to y), so we can in a sense say the probability an integer less than x is a perfect square is 1/sqrt(x). Since 1/sqrt(x)->0 as x->infinity, S(x)/x->0 as x-> infinity as well. Hence we are justified in saying "the probability an integer less than x is a perfect square goes to zero as x goes to infinity", the true meaning behind this probability statement is in behavior of S(x)/x.

Now, even though the density of the squares goes to zero (or their probability if you prefer), there are still infinitely many of them. So even if we could prove the density of the twin primes goes to zero, it would not tell us that there are finitely many of them- you can have an infinite sequence whose density tails off and approaches zero.Derbyshire's book was trying to explain some very complicated maths without using much maths. A healthy knowledge of complex analysis is really needed to appreciate and understand the Zeta function. Something like Hardy and Wright's Number theory text covers elementary number theory. 'elementary' in this sense means no use of anything outside a calculus of one variables course, and in reality you don't need a background beyond high school math to understand most of it. Same goes for most number theory texts with 'elementary' in the title. If you haven't already, you should give some a shot.

I'll talk about number theory all day, I'm not bored yet.

 

[Canute]

You can't spin a roulette wheel infinitely many times .

Quite right. Just as well given that I once was half-addicted to playing it. But could we not say that for a non-halting roulette wheel the quantity of zeros will be greater than any finite quantity of zeros?

Let me put what the 'probability goes to zero' means in more precise terms. Let's just consider the sequence of perfect squares, 1, 4, 9, 16, 25, ... Let S(x) be the number of perfect squares less than or equal to x, so S(10)=3, S(19)=4, and so on. We actually have S(x)=[sqrt(x)], where [] is the floor function, i.e. S(19)=[sqrt(19)]=[4.358...]=4.

The ratio of squares to the total number of integers less than x is S(x)/x, considering x an integer for simplicity. If we randomly select an integer from 1, 2, 3, ..., x, the probability we get a perfect square is simply S(x)/x.

I'm fine up to here.

Now S(x)/x=[sqrt(x)]/x~1/sqrt(x) (when y is very large, [y] is pretty close to y), so we can in a sense say the probability an integer less than x is a perfect square is 1/sqrt(x).

I see that S(x)/x=sqrt(x)/x, but do not understand ~1/sqrt(x). Also, where did y suddenly appear from, and what do the square brackets around [y] signify? I'd like to fully understand the point you're making here.

Since 1/sqrt(x)->0 as x->infinity, S(x)/x->0 as x-> infinity as well. Hence we are justified in saying "the probability an integer less than x is a perfect square goes to zero as x goes to infinity", the true meaning behind this probability statement is in behavior of S(x)/x.

Lets' put this on hold until I've conquered the previous para.

Now, even though the density of the squares goes to zero (or their probability if you prefer), there are still infinitely many of them. So even if we could prove the density of the twin primes goes to zero, it would not tell us that there are finitely many of them- you can have an infinite sequence whose density tails off and approaches zero.

Yes, I see that. My question was badly phrased. If, as I think you said, the probability of N+/-1 being a twin prime necessarily goes to zero as P increases, then there seem to be two possibilities. First, the probability actually falls to zero, in which case there are not infinitely many twin primes at N+/-1, or, second, the probability does not actually fall to zero, in which case there are infinitely many of them. Is this not so?

Something like Hardy and Wright's Number theory text covers elementary number theory. 'elementary' in this sense means no use of anything outside a calculus of one variables course, and in reality you don't need a background beyond high school math to understand most of it. Same goes for most number theory texts with 'elementary' in the title. If you haven't already, you should give some a shot.

Are you seriously suggesting that I would understand any of H & L's book? I find that hard to believe. Even your definition of 'elementary' is beyond me. I have no idea what is either inside or outside a calculus of one variables course, whatever that is. It seem a little unlikely I'll get anywhere with H & L.

 

[shmoe]

Quite right. Just as well given that I once was half-addicted to playing it. But could we not say that for a non-halting roulette wheel the quantity of zeros will be greater than any finite quantity of zeros?

If you had some roulette wheel that will spin forever, even thousands of times a second, it's still possible that if you check in on it in 20 years that you haven't spun a zero yet. You might start to wonder if your wheel isn't rigged, but there's no way to be absolutely sure.

We can make a decent analogy with coin flipping. An infinite sequence of flips can be modeled by a sequence of 0's and 1's. If you think about the binary expansion of real numbers on the interval [0,1] these are exactly the same as the coin flips. The set of numbers in [0,1] that have only finitely many zeros is a countable set, the rest are uncountable. This is the sense that getting only a finite number of heads in an infinite sequence of coin flips is extremely rare, it's like a countable set vs an uncountable set.

I see that S(x)/x=sqrt(x)/x, but do not understand ~1/sqrt(x). Also, where did y suddenly appear from, and what do the square brackets around [y] signify? I'd like to fully understand the point you're making here.

S(x) isn't exactly sqrt(x), sqrt(x) may not even be an integer, we actually have S(x)=[sqrt(x)]. This [] function is just telling you to round down to the nearest integer, so S(10)=[sqrt(10)]=[3.162...]=3 for example.

However when x is big, so is sqrt(x) and you will have [sqrt(x)] very close to sqrt(x) relative to the size of sqrt(x). For example, sqrt(12345)=111.10... but we actually have S(12345)=[sqrt(12345)]=111, the percent difference is very small and gets smaller as x grows.

If you're happy with saying S(x)~sqrt(x) when x is large, that's enough. So S(x)/x~sqrt(x)/x=1/sqrt(x).

The key point from this if you take x large enough, the proportion of squares in the set {1, 2, ..., x} is as small as we like.

Yes, I see that. My question was badly phrased. If, as I think you said, the probability of N+/-1 being a twin prime necessarily goes to zero as P increases, then there seem to be two possibilities. First, the probability actually falls to zero, in which case there are not infinitely many twin primes at N+/-1, or, second, the probability does not actually fall to zero, in which case there are infinitely many of them. Is this not so?

The probability can go to zero while you still have infinitely many of them. Think about the sequence of perfect squares for the moment.

Are you seriously suggesting that I would understand any of H & L's book? I find that hard to believe. Even your definition of 'elementary' is beyond me. I have no idea what is either inside or outside a calculus of one variables course, whatever that is. It seem a little unlikely I'll get anywhere with H & L.

You never know until you try. Most elementary number theory books have no calculus in them at all. Hardy and Wright do use the big O notation and the like, so you will have plenty of asymptotic statements, so some idea of limits will be desirable. Hardy's book admitedly isn't the easiest intro, but it has some great stuff in it. You might find a book like Silverman's "A Friendly Introduction to Number Theory" to be a good place to start, but there are lots of options- check nearby libraries.

 

[Canute]

If you had some roulette wheel that will spin forever, even thousands of times a second, it's still possible that if you check in on it in 20 years that you haven't spun a zero yet. You might start to wonder if your wheel isn't rigged, but there's no way to be absolutely sure.

I'm sure you're right about this, but I don't get it. 20 years is a finite time. If after 500 million years there are no zeros I'd be sure the wheel is rigged, and even that is the blink of an eye in eternity.

... but we actually have S(12345)=[sqrt(12345)]=111, the percent difference is very small and gets smaller as x grows.

Got that.

If you're happy with saying S(x)~sqrt(x) when x is large, that's enough. So S(x)/x~sqrt(x)/x=1/sqrt(x).

Ok. Does ~ mean equivalent to, proportional to, roughly the same as, or something like that?

The key point from this if you take x large enough, the proportion of squares in the set {1, 2, ..., x} is as small as we like.

Is it? I can't see how it could become as small as zero.

The probability can go to zero while you still have infinitely many of them. Think about the sequence of perfect squares for the moment.

I'm confused about this point. I see that the proportion of squares below x approaches zero as x increases, but I can also see that it never reaches zero. Likewise, the probability of some n being prime falls to zero as n increases, but we know it never actually reaches zero.

But whether it does or does not reach zero my question still seems relevant. If you are saying that N+/-1 eventually falls to zero as P increases then there are not infinitely many twin primes of the form N+/-1. If it does not then there are. Is there something wrong with this reasoning?

 

[shmoe]

I think I see the problem. Whenever I've said "goes to zero" it doesn't necessarily mean whatever I was talking about is ever actually equal to zero, rather it gets arbitrarily close as in the sense of a limit.

[tex]\lim_{x\rightarrow\infty}f(x)=0[/tex]

means that for any [tex]\epsilon>0[/tex] we can find a [tex]N[/tex] so that if [tex]x\geq N[/tex] then [tex]|f(x)-0|<\epsilon[/tex]. "as small as we like" is in the choice of [tex]\epsilon[/tex], "as x->infinity" is in the lower bound of big N making sure x is 'large enough'.

S(x)/x is always greater than zero, but gets as close to zero as we like. By taking large enough x values, we can make S(x)/x very close to zero:

S(10^10)/10^10=0.00001
S(10^20)/10^20=0.0000000001

But it will never reach zero. I think you understood this and it was just a terminology gap.

Ok. Does ~ mean equivalent to, proportional to, roughly the same as, or something like that?

Asymptotic. Writing f(x)~g(x) as x goes to infinity means that f(x)/g(x) approaches 1, as in a limit. This means the relative error between the two get's very small (the percent difference) but the absolute error may still be very 'large' (|f(x)-g(x)| is the absolute error). You can think of it as "close for large values of x".