# A $1,000,000 Question * 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
Homework Helper
I've just skimmed, so only a couple things for now.

Canute said:
* 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.

Canute said:
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.

shmoe said:
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 Science Advisor Homework Helper 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" [Broken] 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

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shmoe said:
They've gone much furthur than 1,500,000,000 zeros.
Over 1,500,000,000 Solutions not zeros

shmoe
Homework Helper
RandallB said:
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?

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shmoe said:
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
Homework Helper
RandallB said:
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
Staff Emeritus
Gold Member
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.

shmoe said:
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
Homework Helper
RandallB said:
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").

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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-
Canute said:
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
Homework Helper
Canute said:
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).

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

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Hurkyl
Staff Emeritus
Gold Member
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.

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shmoe
Homework Helper
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
Staff Emeritus
Gold Member
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.

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
Staff Emeritus
Gold Member
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.

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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).

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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.

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

Canute said:
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.

Canute said:
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.

Canute said:
[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.

Canute said:
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.

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shmoe said:
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.

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shmoe
Homework Helper
Just a comment this bit for now:

Canute said:
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

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

The product is over primes p, and $$\gamma$$ is euler's gamma~0.577.... So the probability N+/-1 is prime will be roughly $$e^{\gamma}\log(P)/P$$ 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.

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