Attempted reductio ad absurdum assault on the objection about twin candidates not being the same as twin primes…


*****EDIT*****

The guts of the following post should be ignored. It contains some “stream of consciousness” thinking on different ideas at different times, and isn’t well written, well connected or well edited. There are a number of logical flaws that I can’t be bothered to try to fix, as I’ve ended up on a different track. The main message I would like to leave is:

“… if we can guarantee for all n that there must be at least one candidate twin prime pair less than P(n+1)^2 remaining after sieving to the n-th prime then the twins are infinite.”

I am working on the distribution of candidate twin primes to see if I can get anywhere.

************

This post is offered in rebuttal to some thoughtful comment from Hugh. In my last post I offered what I thought might possibly be a nice, elegant “proof” of the twin prime conjecture and prime k-tuples conjecture.

This is a little bit convoluted, which in my mind means the chances of there being an error is fairly high. It is also not a complete argument, just takes me part way.

The basic argument offered previously is that the step-wise process of stripping out all composite numbers from the set of natural numbers must by necessity leave a rational fraction of the infinite set of natural numbers as primes, prime twins and prime k-tuples. I’ll not repeat the entire logic here, it’s in the previous post, and presented in a substantially more convoluted form a couple of posts prior.

Imagine that all primes to Pn have been repeated to infinity, thus removing all composites that are multiples of all primes to Pn. We’ve already shown that at this point, there remain infinite numbers of candidate primes, candidate twins and candidate k-tuples, and we have developed formulae for the density of candidate primes, twins or k-tuples after n stages of sieving (i.e. stripping all multiples of all primes to Pn).

We also know that in any single stage of sieving, we remove 1/Pn, 2/Pn or k/Pn of the total remaining candidate primes, twin pairs or k-tuples.

We know from Euclid’s simple proof that the primes are infinite. We have not (yet) proven the same for twins or k-tuples.

Let us assume the set of twin primes is finite, therefore there exists a finite P_l which is the lower member of the highest twin prime pair, and P_u which is the upper member.

Since there are no twin primes beyond P_u, all candidate twins higher than P_u must eventually be eliminated as multiples of other primes. The primes up to P_u could not eliminate all possible candidate twin pairs to leave only the finite set of twin pairs, since they each only eliminate a rational fraction of the initial infinite set of candidate twins.

Of the remaining candidate twin pairs, since none are actual twin pairs, then in every case either the upper or lower member, or both, is/are a composite number. Therefore, the next lowest candidate twin pair {TC_l , TC_u} meets this criteria, and its members must be greater than P_l^2, otherwise at least one member would either already have been removed as multiples of primes up to P_l, or both must be prime. In other words, if the next candidate twin prime pair were less than P_u^2, it would have to be an actual twin prime pair. Since it cannot be a twin prime pair, as {P_l, P_u} constitute the highest twin pair, it must be greater than or equal to P_u^2.

Therefore, after any subsequent stage of sieving to a new prime, Pm, all candidate twins must be removed to at least Pm^2, and the next lowest candidate twin pair is greater than Pm^2.

There cannot be two or more candidate twin pairs remaining between Pm^2 and P(m+1)^2, since the next round of sieving by P(m+1) cannot remove more than one prime (and therefore one twin candidate) between Pm^2 and P(m+1)^2. Therefore, after any given m-th round of sieving, there can never be two or more candidate pairs remaining between Pm^2 and P(M+1)^2.

Therefore, in order that the set of twin primes to be finite and confined to Pk, then after k rounds of sieving to Pk there cannot exist any candidate twin pairs between Pk and Pk^2, and there cannot exist more than one candidate twin pair between Pk^2 and P(k+1)^2.

We can extend this reasoning to show that after k rounds of sieving, if a single candidate twin pair exists at less than P(k+1)^2, it must be an actual twin prime pair. Therefore, if we can show that it is not possible for sieving up to Pk to remove all possible twin candidates less than P(K+1)^2, there must be infinitely many twin primes.

About petequinn

I'm a Canadian geotechnical engineer specializing in the study of landslides. I started this page to discuss some mathematical topics that interest me, initially this involved mostly prime numbers, but more recently I've diverted focus back to a number of topics of interest in geotechnique, geographic information systems and risk. I completed undergraduate training in engineering physics at Royal Military College (Kingston, Ontario), did a masters degree in civil (geotechnical) engineering at University of British Columbia (Vancouver), and doctorate in geological engineering at Queen's University (Kingston). I was a military engineer for several years at the beginning of my career, and did design and construction work across Canada and abroad. I've worked a few years for the federal government managing large environmental clean up projects in Canada's arctic, and I've worked across Canada, on both coasts and in the middle, as a consulting geotechnical engineer. My work has taken me everywhere in Canada's north, to most major Canadian cities and many small Canadian towns, and to Alaska, Chile, Bermuda, the Caribbean, Germany, Norway, Sweden, Bosnia, and Croatia. My main "hobby" is competitive distance running, which I may write about in future.
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52 Responses to Attempted reductio ad absurdum assault on the objection about twin candidates not being the same as twin primes…

  1. Surely there’s a confusion in this one about where we have sieved to when we reach {P_l, P_u} – we can only assume we sieved the primes up to the square root of P_u.

    “Of the remaining candidate twin pairs, since none are actual twin pairs, then in every case either the upper or lower member, or both, is/are a composite number. Therefore, the next lowest candidate twin pair {TC_l , TC_u} meets this criteria, and its members must be greater than P_l^2, otherwise at least one member would either already have been removed as multiples of primes up to P_l, or both must be prime. “

    No – the members of {TC_l , TC_u} only need be greater than P_u as they can’t have been sieved out by any of the primes up to the *the square root of P_u.* There are a lot of primes to choose from between the square root of P_u and P_u, and what needs to be proved is that these primes can’t sieve out the candidates between P_u and P_u^2.

    The logic you are using here would mean that when we reach 197 and 199, all twin prime candidates up to 199 squared have already been sieved out. But we need only sieve up to 13 to establish that 197 and 199 are twin primes, while future candidates such as 361 and 359 remain, to be eliminated in this case by 19*19.

    “In other words, if the next candidate twin prime pair were less than P_u^2, it would have to be an actual twin prime pair. Since it cannot be a twin prime pair, as {P_l, P_u} constitute the highest twin pair, it must be greater than or equal to P_u^2.

    Therefore, after any subsequent stage of sieving to a new prime, Pm, all candidate twins must be removed to at least Pm^2, and the next lowest candidate twin pair is greater than Pm^2.”

    So this bit doesn’t work.

  2. Or a more succinct way ot putting this, it doesn’t take k rounds of sieving to eliminate the primes up to Pk, as you suggest in your last para. k rounds of sieving would sieve up to Pk^2, whereas only (sqr rt Pk) rounds of sieving are needed to sieve up to Pk.

    • petequinn says:

      I haven’t made my point very clearly at all, I can see. I started writing that post several days ago, and wound up following a few different lines of reasoning to get to the point I thought I made but evidently didn’t.

      What I’m claiming is that after sieving by primes to an arbitrary Pn, any remaining candidate twin pair less than P(n+1)^2 must necessarily be an actual twin prime pair.

      If it can be shown that there must be such a twin prime pair remaining regardless of how big n becomes, then there must be an infinite number of twins.

      I’ll try to present the logic more cleanly later, meanwhile I’m investigating the necessity of a candidate twin prime pair remaining lower than P(n+1)^2 after sieving to Pn.

      • HI Pete,

        “What I’m claiming is that after sieving by primes to an arbitrary Pn, any remaining candidate twin pair less than P(n+1)^2 must necessarily be an actual twin prime pair.”

        I’ve already given you a clear counterexample above. To sieve for candidate twin primes up to 289 (17^2) you only need to use the primes up to 13. So 197/199 is an actual twin when you have sieved to 13 and 359/361 is a candidate twin pair (but not an actual twin). Your assumptions in the argument above are that you need to sieve using *all the primes up to 199* to establish that 197/199 is an actual twin prime.

        If you had sieved to 199 then sure, you would know all the actual twins up to 199^2.

        You’re mixing up “sieving to Pn” with “sieving to the number which guarantees that Pn will be a prime” (which is the last prime you reach before the sq rt of Pn).

      • petequinn says:

        No, sorry Hugh, I see you’re not following where I’m going. You’re saying precisely the same thing I’ve just said here: “To sieve for candidate twin primes up to 289 (17^2) you only need to use the primes up to 13.”

        The important part of my comment is the second half, where I’ve claimed that if we can guarantee for all n that there must be at least one candidate pair (which must be an actual pair as we’ve both just argued) less than P(n+1)^2 then the twins are infinite. I don’t think you’ve challenged that part of the argument, and perhaps I haven’t made the case very clearly (or at all, I guess, although for the moment I’m focussing on seeing how twin prime candidates evolve to Pn^2).

  3. OK, let me try reading it again from scratch…

  4. No, sorry but I’ve looked again and it really honestly doesn’t work.

    The pattern of how twin primes develop between Pm^2 and P(m+1)^2 is indeed important, I agree on that and I’m also looking at that.

    But the details of your specific argument here don’t work. I’ve broken down the points into detail, in the hope that you’ll see what I mean.

    “Since there are no twin primes beyond P_u, all candidate twins higher than P_u must eventually be eliminated as multiples of other primes.”

    *Yes, but only if you take this to mean that all candidate twins higher than P_u must be eliminated as multiples of primes other than those that eliminated the primes up to P_u (in other words the primes up to sq rt of P_u).

    “The primes up to P_u could not eliminate all possible candidate twin pairs to leave only the finite set of twin pairs, since they each only eliminate a rational fraction of the initial infinite set of candidate twins.”

    *Indeed. But they must eliminate all candidate twin pairs that aren’t actual twin primes up to P_u^2, right?

    “Of the remaining candidate twin pairs, since none are actual twin pairs, then in every case either the upper or lower member, or both, is/are a composite number. Therefore, the next lowest candidate twin pair {TC_l , TC_u} meets this criteria, and its members must be greater than P_l^2, otherwise at least one member would either already have been removed as multiples of primes up to P_l…”

    *Not true. Its members only need have prime factors higher than the sq rt of P_u. 359/361 is eliminated by 19^2. If we were checking whether 197/199 was the highest twin prime pair, we wouldn’t neglect to sieve using the primes from 17 up to 181, would we?

    “In other words, if the next candidate twin prime pair were less than P_u^2, it would have to be an actual twin prime pair. Since it cannot be a twin prime pair, as {P_l, P_u} constitute the highest twin pair, it must be greater than or equal to P_u^2.

    *Not true. It could be a candidate twin prime pair that is eliminated by any of the primes between P_u and P_u^2. Just as 19 eliminates the pair 359/361.

    “Therefore, after any subsequent stage of sieving to a new prime, Pm, all candidate twins must be removed to at least Pm^2, and the next lowest candidate twin pair is greater than Pm^2.”

    *This is correct, but it is just what we knew from basic maths, which is that a composite number K can’t have factors that are all higher than the square root of K. If we have gone on and sieved from 17 up to 199, we know that there will be no candidate twin primes that aren’t actual twin primes under 199^2.

    “There cannot be two or more candidate twin pairs remaining between Pm^2 and P(m+1)^2, since the next round of sieving by P(m+1) cannot remove more than one prime (and therefore one twin candidate) between Pm^2 and P(m+1)^2.”

    *And this is wrong, since all the primes in between Pm and Pm^2 will be eliminating pairs, not just P (m+1). There is no reason why P(m+2), P(m+3) etc can’t remove additional candidates between Pm^2 and P(m+1)^2. Again, 19 eliminates 359/361 even though it is lower than 197.

    “Therefore, after any given m-th round of sieving, there can never be two or more candidate pairs remaining between Pm^2 and P(M+1)^2.”

    * And this doesn’t hold for the same reason. (It also wouldn’t hold anyway, since large prime gaps with a large gap between Pm^2 and P(m+1)^2 would allow for more candidate pairs than this to be eliminated by P(m+1), but that is less important.)

    “Therefore, in order that the set of twin primes to be finite and confined to Pk, then after k rounds of sieving to Pk there cannot exist any candidate twin pairs between Pk and Pk^2, and there cannot exist more than one candidate twin pair between Pk^2 and P(k+1)^2.

    * And this doesn’t hold either.

    • petequinn says:

      I think you should just just ignore that post. It wasn’t very well written, and obviously not very well edited. Luckily it desn’t worry me greatly to be wrong from time to time.🙂

      The main point I intended as the primary takeaway message remains: “… if we can guarantee for all n that there must be at least one candidate twin prime pair less than P(n+1)^2 (remaining after sieving to the n-th prime) then the twins are infinite.”

  5. Yes, I agree with that. And also if you can guarantee that above P^2 the primes above P can’t “conspire” together to eliminate all twin candidates between subsequent squares of primes then the twins are infinite, which is a slightly weaker statement of the same idea because it allows the possibility that there isn’t always a twin pair between every two consecutive squared primes.

    I’ll put something up on my blog in a bit about this stuff. I’ve got an attempted proof that feels like it almost works.

  6. Hugh says:

    OK, I’ve put up a new attempted proof here:

    http://barkerhugh.blogspot.com/2011/11/new-attempted-twin-prime-proof-probably.html

    It follows on from our conversation, but I’m trying to show that the number of actual twin primes below P^2 always keeps growing rather than relying on the infinitude of twin prime candidates above P^2. I’d appreciate it if you could take a look. If it works we can call it the Barker-Quinn twin prime proof, but if it fails I’ll take the blame…

    • petequinn says:

      Hey Hugh, I’ve grabbed a copy to have a look. Won’t promise to comment today, as this is pretty long and I have a short attention span. 🙂

      I tried to place this comment on your blog but it wouldn’t let me. Not sure if it has something to do with the blog settings or my browser settings.

    • petequinn says:

      Hugh, your blog still won’t let me post a comment. Here is an initial comment I just tried to insert over there:

      Hugh I’ve been starting to work my way through this. I’m not sure I follow, here are some thoughts (in ALL CAPS) on the first part, about K numbers. Some are notes to myself to help me follow along.

      K numbers

      1) Let P be any prime number and Pn the next prime number after P.

      2) Let a Kp number be a number that is either prime or has no prime factors of P or lower.

      3) We can use Kp numbers to test for primality in the region 1 … P^2:

      4) Let Z# be the primorial of a prime ≥ P, for instance Pn#.

      Z# is a multiple of all primes up to P, by definition.

      5) Let A be a Kp number that can be expressed as 6N+1 or 6N-1
      (where N is a positive integer). OK THE SET OF KP VALUES ARE THE SET OF POSSIBLE PRIMES (OR PRIMES) AFTER SIEVING BY 2 AND 3

      (So A = 5, 7, 11, 13 etc – these are the only kinds of numbers which can be primes).

      6) Let Z# – A = R NOT OBVIOUS WHY THIS IS NECESSARY YET. A AND R ARE MIRROR SYMMETRIC WITHIN Z#

      If R

      > P^2:

      R can’t be divisible by a prime of P or lower (since Z# is a multiple of these numbers) and R can’t be divisible by a prime higher than P since R is smaller than P^2.

      Therefore R cannot be a composite number. WHY NOT? WITH WHAT YOU’VE DEFINED OF R, IT COULD BE ANY “POSSIBLE PRIME” LESS THAN P^2, AND DOES NOT HAVE TO BE A PRIME. IT COULD BE A COMPOSITE NUMBER. DOESN’T 25, FOR EXAMPLE, MEET YOUR DEFINITION FOR R?

      7) It is too complex to work out the ratio of Kp numbers in any given region of numbers so we need to simplify this strategy when it comes to searching for twin primes:

      • Hi Pete,
        OK some replies below (the stuff about K numbers isn’t actually crucial to the attempted proof, it just sets up a way of looking at things, but…)

        1) Let P be any prime number and Pn the next prime number after P.
        2) Let a Kp number be a number that is either prime or has no prime factors of P or lower.
        3) We can use Kp numbers to test for primality in the region 1 … P^2:
        4) Let Z# be the primorial of a prime ≥ P, for instance Pn#.
        Z# is a multiple of all primes up to P, by definition.
        5) Let A be a Kp number that can be expressed as 6N+1 or 6N-1
        (where N is a positive integer). OK THE SET OF KP VALUES ARE THE SET OF POSSIBLE PRIMES (OR PRIMES) AFTER SIEVING BY 2 AND 3

        * No, first you limit the set of numbers to possible primes above 2 and 3, then you further restrict it to K numbers, which are numbers that are either prime or have no prime factors of P or lower. For P=5 any multiple of 5 is not a K number. For P = 7, any multiple of 5 or 7 isn’t a K number etc.

        (So A = 5, 7, 11, 13 etc – these are the only kinds of numbers which can be primes).

        6) Let Z# – A = R NOT OBVIOUS WHY THIS IS NECESSARY YET. A AND R ARE MIRROR SYMMETRIC WITHIN Z#

        *Yes, that is all I am setting up here, two sets of numbers A and R which are mirror symmetric within Z#. Then we consider the region up to Z^2 where this allows us to identify primes.

        If R > P^2:

        *Not sure what happened here – the proper version is *If R < P^2* !! Maybe an earlier typo surviving…***

        R can’t be divisible by a prime of P or lower (since Z# is a multiple of these numbers) and R can’t be divisible by a prime higher than P since R is smaller than P^2.

        Therefore R cannot be a composite number. WHY NOT? WITH WHAT YOU’VE DEFINED OF R, IT COULD BE ANY “POSSIBLE PRIME” LESS THAN P^2, AND DOES NOT HAVE TO BE A PRIME. IT COULD BE A COMPOSITE NUMBER. DOESN’T 25, FOR EXAMPLE, MEET YOUR DEFINITION FOR R?

        * No, because R = 210 – 185, 185 is a multiple of 5 and therefore for A = 185, A is not a K5 number. This is obviously true of any number from 0-210. It's only when you subtract primes or composites which only have factors higher than 7 from 210 that anything interesting emerges.*

        In the sum 7# – A = R, if A is a multiple of 5 or 7 then R is also a multiple of 5 or 7. However if A is a prime of a composite which doesn't have any prime factors of 7 or under *and* R < 7^2 then R is a prime. Examples – 210 – 169 = 41. 210 – 187 = 23.

        When you get up to higher numbers, Z# – A can be a different K number. For instance

        A = 29969 (= 1303*23)
        30030 – 29969 = 361
        So R = 361. We know that this can't have 5, 7, 11 or 13 as a prime factor (because if it did, so would 29969) so it must either be prime or a composite with higher factors. In fact it is 19^2.
        However look at this sum:
        A = 29989 (=421 * 71)
        30030 – 29989 = 131
        So, again, we know this can't have 5, 7, 11 or 13 as a factor.
        However 131 < 169 so we also know it can't have a factor of 13 or under.
        So 131 has to be prime.
        (We could actually test up to 289 this way, couldn't we, so I probably need to tighten this up.. But hopefully this at least makes it clear what the point of K numbers is. And the bit from J numbers onward is more crucial).

      • petequinn says:

        OK, I’m still not sure I follow 100%, but will read these comments carefully before moving on to the J numbers.

  7. Hugh says:

    Yes, it is pretty long as it needs to set up a few things along the way. No rush, but will be interested to have someone try to see if it works. I think I’ve fixed the comment thing now.

  8. Hugh says:

    I’ve also made a few minor edits to typos, nothing major but I had a few fractions the wrong way round. There’s probably still typos in there…

  9. Hugh says:

    Hi Pete,

    I think I’ve spotted a fairly basic problem – if you haven’t read it yet, don’t bother till I try to fix the glitch.

  10. Hugh says:

    OK I fixed the glitch and it wasn’t fatal. If you printed the other out, best to chuck it out though as it needed a fairly big re-edit.

  11. I think actually maybe the whole thing can be boiled down without all the J number stuff and K number stuff – that’s more a record of my train of thought than the core of the proof. Also I wanted to use J twins rather than twin prime candidates because I think it is cleaner since “candidates” is a problematic word (since any 6n+/-1 pair either is or isn’t a twin prime, regardless of whether we have sieved to it or not). But all it is really doing is getting us by a longer route to the point where we know that the pairs which don’t have a factor of P or below must be prime if they are below P^2.

    I think I’ll put up a boiled down version of the key bit that relies on actual twin primes. Maybe wait for that instead of worrying about the J numbers right now.

  12. OK, I’ve put down a simpler version which doesn’t rely on J and K numbers, which hopefully makes it relate more easily to the way you have been looking at the patterns.

  13. I’ll be astonished if there aren’t still some typos etc in there, since I did it quite quickly, but I’m trying to shift your attention to the fundamental part of the train of thought, which is about comparing the (P-2)/P * (Pn-2)/Pn etc product to how rapidly the ratio of twin primes would necessarily fall in the set of numbers above a “highest twin prime pair”.

    • petequinn says:

      Hey Hugh,

      I’m struggling to make my way through it so far. See some preliminary comments in ALL CAPS:

      Here is a version that leaves out some of the waffle and handwaving in the previous version.

      What I want to do is to show that the set of actual twin primes always keeps growing between Pa^2 and a much higher prime Pz^2, where Pa and Pz are two prime numbers.

      1) Let the highest prime we have sieved for be P and the next prime be Pn. OKAY, AT THIS POINT ALL “POSSIBLE PRIMES” TO P^2 ARE ACTUAL PRIMES

      2) The change (MEANS WHAT, “CHANGE?”) in the number of all 6n+/-1 pairs (YOU MEAN PAIRS THAT REMAIN UNDER CONSIDERATION AS POSSIBLE TWINS, NOT YET SIEVED OUT BY PRIMES TO Pn?) as we go from P^2 to Pn^2 (where Pn is the next prime) can be calculated thus:

      The number of pairs under consideration increases from (P^2– 1)/6 to (Pn^2-1)/6 (SORRY BUT WHERE DOES THIS COME FROM? DO YOU MEAN ALL 6n+/-1 PAIRS, NOTHING IS YET SIEVED OUT? SINCE YOU’VE SIEVED TO P^2 SHOULDN’T THE INCREASE BE SOMETHING DIFFERENT?)

      Cancelling the 6, the number of pairs in the range increases by (Pn^2– 1)/(P^2– 1) (BUT THIS ISN’T NECESSARILY A WHOLE NUMBER? I GUESS YOU MEAN INCREASES IN A MULTIPLICATIVE SENSE, NOT ADDITIVE)

      So for instance, when we go from 121 to 169, the number of (CANDIDATE? THERE ARE 10 AND 12 ACTUAL TWIN PRIME PAIRS LOWER THAN 121 AND 169) pairs increases from 20 to 28, and the ratio between the number of pairs up to 121 and the number of pairs up to 169 is 20:28.

      (Note – this is only exactly right if we accept -1 and 1 as the first 6n+/1 pair. This doesn’t affect this attempted proof as nothing hinges on whether or not you count -1 and 1 as prime).

      3) Assume there is a highest twin prime pair.

      Consider the primes P^2 and Pn^2 (P^2 AND Pn^2 ARE BY DEFINITION NOT PRIME) which are higher than the largest twin prime pair.

      4) Since the number of twin primes doesn’t grow in this region (WHAT DOES THIS MEAN, AND ON WHAT BASIS DO YOU CLAIM THIS?), the ratio of actual twin primes in the region 0… Pn^2 will be the actual twin prime ratio of the set up to P^2 multiplied by (P^2-1)/(Pn^2-1).( Because this is the inverse of the fraction above – THE FRACTION ABOVE HAD TO DO WITH POSSIBLE TWINS, SO FAR AS I COULD TELL, NOT ACTUAL TWIN PRIME PAIRS – IN ANY EVENT I’M NOT SURE THE BASIS FOR THIS CLAIM…?)

  14. HI Pete,

    Off on a business trip today. I’ll get back to you properly later. Two quick things though:

    YOU MEAN PAIRS THAT REMAIN UNDER CONSIDERATION AS POSSIBLE TWINS, NOT YET SIEVED OUT BY PRIMES TO Pn

    No, I mean all pairs of 6n+/1 pairs, including those we already know aren’t twins and those we know are. There’s 20 of these (inc 1 and -1) to 121 and 28 to 169. We expect the ratio of twin primes in this set to fall steadily, but how fast it falls is the question.

    OKAY, AT THIS POINT ALL “POSSIBLE PRIMES” TO P^2 ARE ACTUAL PRIMES

    Yes, and at step 3 I assumed this:

    “3) Assume there is a highest twin prime pair.

    Consider the primes P^2 and Pn^2 which are higher than the largest twin prime pair”

    So we are talking about a region where there are *no new twin primes* between P^2 and Pn^2. So the ratio of twin primes to ^N+/-1 pairs must fall by the ratio (P^2-1)/(Pn^2-1). It’s intended to be a reductio ad absurdum proof in which it is shown that the ratio can’t fall this fast.

    P^2 AND Pn^2 ARE BY DEFINITION NOT PRIME

    Yes, sloppy of me, I meant “the squares of primes P and P^2, which are higher than the largest twin prime pair.” These squares are higher than the last twin prime, so in this region, by definition, the ratio of twin primes to ^N+/-1 pairs must fall by the ratio (P^2-1)/(Pn^2-1). If it doesn’t, it can *only* be because there is a twin prime between P^2 and Pn^2.

    • petequinn says:

      OK thanks, I’m not being purposefully obtuse, just want to make sure I understand

      Cheers,

      Pete

    • petequinn says:

      I’m not sure how you get this:

      “…the ratio of twin primes to ^N+/-1 pairs must fall by the ratio (P^2-1)/(Pn^2-1).”

      You’ve claimed that the ratio of 6N+/1 pairs falls by this ratio (without, I think, proving it to be the case?) but so far as I can tell said nothing in particular about the pattern/distribution of the actual twins themselves?

  15. Firstly there is a typo there: “^N+/-1” should be “6N+/1”

    Secondly, remember we’ve assumed that we are in the rgion above the highest twin prime pair, so this follows very simply. Think about the ratio of “the pair of numbers 5/7” within the set of numbers. We know this never happens after the number 7 so it is easy to see what we should expect about the ratio of “5/7 numbers in the set of all numbers up to N.

    When we have sieved to 121 we know that the ratio of 6N+/1 pairs that are “5/7” is 1/20 [(121-1)/6]. When we sieve up to 169 we know that it is 1/28 [(169-1)/6]. If the ratio didn’t fall from 1/20 to 1.28 in this region it could *only* be because there was another 5/7 pair between 121 and 169. So to get from the ratio at 121 to the ratio at 169 we have to multiply by 20/28.

    It’s a fairly simple idea and on its own says nothing whatsoever about the distribution of twin primes – it is only important because it establishes a benchmark for how fast the ratio of twin primes would have to fall between P^2 and Pn^2 if we had gone past the last twin prime, And I think that is interesting when you realise that (p-2)/p is never going to fall continually as fast as this.

  16. And just to clarify the maths. There is one 6N+/1 pair every 6 integers, at 5/7, 11/13, 17/19 etc. P^2 will always be a 6N+1 number, whether P is a 6N+1 number or a 6N-1 number. So if we subtract 1 and divide by 6 we get the number of pairs up to P^2.

    • petequinn says:

      P^2 will always be a 6N+1 number

      Why is that? I just checked and it seems to be true for primes > 3, up to the limit I can check in Excel. It would seem there must be an obvious explanation, but I’ll admit it escapes me. Is this a consequence of Fermat’s Little Theorem?

  17. No, it’s just basic algebra – a prime P >3 is always equal to 6N+1 or 6N-1.

    (6N+1)^2 is (36N^2 + 12N + 1) = 6(6N^2 + 2N)+1, which has to be a 6N+1 number.

    (6N-1)^2 is (36N^2 – 12N +1) = 6(6N^2 – 2N)+1 which also has to be 6N+1 number.

    So any P^2 – 1 = 6N and the number of pairs up to this point (inc 1 and -1) is N.

  18. Oh, I see a little confusion from step 7 over whether we are concerned with (P-2)/P or (P-4)/P – it doesn’t really matter but most of the actual working is based on the former, not the latter.

    I could fix this bit up if it causes confusion.

  19. Actually ignore the previous comment, I was managing to confuse myself…

    • petequinn says:

      I’ve carried on a little further but managed to get stuck again:

      Let (P^2-1)/(Pn^2-1) be the zero twin prime rate. OK, THIS IS THE RATE THE RATIO OF TWIN PRIMES HAS TO FALL FROM P^2 TO Pn^2 IF NO TWIN PRIMES BETWEEN THEM

      This is the rate at which the twin prime ratio would fall between P^2 and Pn ^2 if we had sieved past the “highest twin prime pair”

      5) If the ratio doesn’t fall at the zero twin prime rate between P^2 and Pn^2, there must be an additional twin prime in this region. GOTCHA

      6) To get the approximate twin prime ratio between zero and Pn^2, multiply the number of actual twin primes under P^2 by (P-2)/P – I DON’T “GET” WHAT YOU MEAN BY RATIO HERE??? SHOULDN’T THE RATIO BE THE NUMBER OF ACTUAL TWINS DIVIDED BY P^2?

      • petequinn says:

        6) To get the approximate twin prime ratio between zero and Pn^2, multiply the number of actual twin primes under P^2 by (P-2)/P – I DON’T “GET” WHAT YOU MEAN BY RATIO HERE??? SHOULDN’T THE RATIO BE THE NUMBER OF ACTUAL TWINS DIVIDED BY P^2?

        So I threw some numbers into a spreadsheet to check what you really meant here. I think what you meant to write was:

        To get the approximate twin prime ratio between zero and Pn^2, multiply the twin ratio between zero and P^2 by (P-2)/P (or did you mean multiply by (Pn – 2)/Pn? This gives a slightly closer approximation, although the difference is negligible as n grows)

        I’ll work with that and start to read further.

  20. Sorry I explain things so badly – when you’re done, if it is still standing, I’ll try rewriting it to make it clearer.

    Anyhow, by the “approximate twin prime ratio” I am simply talking about the approximation we have both already been using to look at the likely decline of twin primes in a given area. We know that across a primorial length pattern, the prime P will remove 2/P of the remaining twin prime candidates, leaving (P-2)/P still standing – so the product of this fraction seems a reasonable way of *estimating* how fast we would expect remaining twin prime candidates to decline in a smaller area, for instance between P^2 and Pn^2 (you’ve done the work there showing that it isn’t a terrible estimate, although it slightly overestimates in that region)..

    We know that above the highest twin prime there will still be an infinite repeating pattern of twin prime candidates, so the question is whether they can possibly be eliminated fast enough between P^2 and Pn^2 to meet the zero twin prime rate.

    So there are two stages. First we use the approximate twin prime ratio as a rough estimate of how fast we would expect remaining twin prime candidates to be eliminated between consecutive P^2 and Pn^2 etc. Then we show that this approximation will always be insufficient to satisfy the zero twin prime rate. So if this estimate were a rigidly accurate one, we would know right away that there isn’t a highest twin prime.

    But we know it is possible for there to be variations in how accurate this estimate is, which leaves the possibility of some kind of “conspiracy” where the primes between P^2 and Pn^2 do always get eliminated.

    So the final stage of the argument is to suggest (rather than rigorously prove) that the approximate twin prime ratio does indeed give us an accurate estimate of how fast twin prime candidates will be eliminated between Pa^2 and Pz^2, the squares of two primes (higher than the highest twin pair) that are separated by a large region.

    If this could last step be rigorously shown I believe it would prove there couldn’t be a highest twin prime pair – because it would be impossible for the remaining twin prime candidates to be eliminated at the rate required by the zero twin prime rate. And this would mean there must always be actual twin primes left (not just that there is always an infinity of candidates remaining to be eliminated).

    • petequinn says:

      I’ve carried on a little further. I’m not all the way through yet. No further hangups following the writing/logic yet, although a little skeptical that things will end well. 🙂 I’ll work my way through the rest as time is available:

      6) To get the approximate twin prime ratio between zero and Pn^2, multiply the number of actual twin primes under P^2 by (P-2)/P – I DON’T “GET” WHAT YOU MEAN BY RATIO HERE??? OK, I’VE CORRECTED THIS IN MY HEAD TO: “To get the approximate twin prime ratio between zero and Pn^2, multiply the twin ratio between zero and P^2 by (P-2)/P”

      (We know there is an infinite repetitive pattern of numbers that for which the primes up to the last prime we sieved to are not factors).

      Note that it doesn’t matter at this stage that we know this may be an inaccurate measure. However we do know that it is a good estimate of the number of remaining “twin prime candidates” that will be removed by P over a large region of numbers.

      Let (P-2)/P be the approximate twin prime rate. OK, BUT I EXPECT AT THIS POINT TO FIND IT NOT VERY USEFUL, SINCE THE APPROXIMATE TWIN RATE CHANGES VERY LITTLE AS n GROWS, AND WILL NOT BE VERY SENSITIVE TO PRESENCE/ABSENCE OF TWIN PRIMES BETWEEN P^2 AND Pn^2 FOR LARGE n

      7) If P and Pn are higher than the highest twin prime pair then:
      P must be ≤ Pn-4 OK

      8) The most important equation in this attempted proof is this:

      P-2/P > (P^2-1)/(Pn^2-1) for any value of P and Pn.

      This can be established in two steps. First:

      (P^2– 1)/(Pn^2– 1) X
      (X-1)/X < (Y-1)/Y
      To get from X/Y to (X-1)/(Y-1) we multiply by [(X-1)/X] * [Y/(Y-1)] which must be less than 1.
      Therefore (X-1)/(Y-1) (P^2)/(Pn^2)

      To show this, assume Pn = P + 4
      (P^2)/(Pn^2) = (P^2)/[(P+4)^2] = (P^2)/(P^2 +8P +16)
      Multiply this by (P^2 + 8P +16) and you get (P^2)
      If you multiply (P-2)/P by (P^2 + 8P +16) you get
      (P^3 + 6P^2 -32)/P = (P^2 +6P – 32/P)
      (P^2 +6P – 32/P) > P^2 for any positive integer larger than 2

      Therefore P-2/P > (P^2)/(Pn^2) > (P^2– 1)/(Pn^2– 1) where Pn is P +4

      It’s also easy to show that where the gap between P and Pn is more than 4, it is still true that Pn-2/Pn > (P^2– 1)/(Pn^2– 1) since P^2/(P+X)^2 falls as X rises.

      9) Therefore P-2/P > (P^2– 1)/(Pn^2– 1) for all values of P and Pn OK NO PROBLEM WITH THIS.

      10) Therefore the rate of change of the approximate twin prime ratio > the zero twin prime rate OK, AGREE THAT THIS APPLIES FOR WHAT YOU’VE DEFINED AS THESE THINGS. NOT SURE AT THIS POINT HOW USEFUL THE “APPROXIMATE TWIN PRIME RATIO” COULD TURN OUT TO BE, SINCE IT IS ONLY AN ESTIMATE THAT IS UNPROVEN

      This means that if the actual twin prime ratio between P^2 and Pn^2 changes by Pn-2/Pn (INSTEAD OF P-2/P I GUESS?) the twin prime ratio doesn’t fall as fast as it would if there were no more actual twins and there must be an additional twin prime in this region. AT THIS POINT I CAN’T SEE HOW THIS WILL HELP YOU, SINCE FOR LARGE n THERE’S NEGLIGIBLE DIFFERENCE IN THIS “APPROXIMATE” RATION FOR P VERSUS Pn

      11) However we can’t rely on the approximate twin prime ratio being correct over short regions. AGREED

      So the last step in this attempted proof is to show that we can assume that the rate of change in the approximate twin prime ratio does give us an accurate estimate of the actual change in the twin prime ratio in the space between Pa^2 and Pz^2 (where Pa and Pz are primes and Pz is much larger than Pa, with many primes in between Pa and Pz). THIS SEEMS TO BE A STRONG CLAIM THAT WOULD NEED TO BE DEFENDED

  21. And to boil that down a bit:

    (P-2)/P is a rough estimate of how the ratio of twin candidates change when P eliminates candidates in a region above P^2 (in particular we have been looking at the region from P^2 to Pn^2 for which it seems to be an OK but not perfect estimate).

    (P^2-1)/(Pn^2-1) is the exact way the ratio of twin primes must change in between P^2 and Pn^2 if there are no actual twins in this region.

    I show (I think) that (P-2)/P > (P^2-1)/(Pn^2-1) for any P and Pn where P and Pn are separated by 4 (eg they are in the region above the last twin prime pair).

    So then the question is whether the actual rate at which twin prime candidates are eliminated is always higher than the rate (P-2)/P between successive squared primes. My conjecture/suggestion at the end is that over large regions, (P-2)/P * Pn-2)/Pn etc will converge on being an accurate estimate.

    • petequinn says:

      Ahhhh, OK, I follow now. In a nutshell, I think we’re working toward exactly the same thing, but from slightly different angles.

      This is what I see as the weakest point:

      “15) So I would conjecture that when we take Pz higher and higher, and start approaching infinity, it is safe to assume that the change in the actual twin prime ratio between Pa^2 and Pz^2 will get closer and closer to (Pa-2)/P * (Pb-2)/P * …. (Pz-2)/Pz”

      My plots have shown that the approximation in fact gets a little worse as P increases, unfortunately. There must be an easy way to explain this away mathematically, but for the moment it eludes me. I’ve been focussing on trying to understand the PATTERN of removed candidate primes at each sieving stage, rather than the NUMBER, which we’ve been looking at so far. Hence my foray into this strange numbering system, and illustration of eliminated candidates in matrix form. I see obvious patterns emerging, but at the moment am unable to describe them clearly.

  22. Hi Pete,

    Yes, I think we are looking at some of the same things in slightly different ways.

    And I agree that clearly is the potential weak point. It’s worth bearing in mind that you were looking at the number of twins from 0 to P^2 whereas I am looking at the change in the ratio of twins between P^2 and Pn^2 or Pa^2 and Pz^2. I need to think a little about what difference this makes but one relevant thing is that for this calculation, it doesn’t matter if the estimate for twins under P^2 or Pa^2 is inaccurate – the question is how the actual count of twins will change in higher regions.

    I like the primorial counting system a lot, but not convinced it will reveal a great deal. It’s obvious that the gaps in P# will form a repeating pattern of gaps in Pn# – but the way that twin candidates are eliminated within Pn# depends on higher primes.

    The point of my K numbers thing was to look at the way that higher primes impact on the area under P^2. There is a simple “formula” for identifying primes there using K numbers subtracted from Pn#, but I don’t think it is much help in identifying patterns either unfortunately.

  23. “My plots have shown that the approximation in fact gets a little worse as P increases, unfortunately.”

    A possible simple explanation for this might come from the fact that P^2 is a smaller and smaller proportion of P# as we go up through the numbers? I’ll think about that.

  24. Thinking aloud:

    There is no particular reason to expect P to remove 2/P of the twin prime candidates between P^2 and Pn^2 – we would only expect this to be exactly correct between P^2 and (P#+P^2) (a full primorial).

    However we should expect that the product X = 3/5 * 5/7 * …. (P-2)/P will be an accurate estimate for the ratio of twin primes *at some point between P^2 and P#*. This is because at P# we know it will be lower than X (because the primes up to P will have left a ratio of X candidates and higher primes will have removed more candidates) – while at P^2 we would expect that P won’t have had enough impact to reduce the ratio of actual twin primes by as much as it will eventually do.

    So X is likely to be an accurate estimate of the ratio of actual twin primes at a point a bit beyond P^2, and will almost certainly fall a bit short for the ratio in the region 0…P^2.

    I need to think more about how this relates to our arguments, just wanted to make a note while it occurred to me. What I would be very interested to see is a graph of whether X can be used to get an accurate estimate of the change in the ratio of twin primes between P^2 and a higher Px^2 (for instance between 19^2 and 101^2, starting from the actual number of twin primes up to 19^2 and then multiplying by 17/19 * 21/23 * …. 99/101).

    • petequinn says:

      What I would be very interested to see is a graph of whether X can be used to get an accurate estimate of the change in the ratio of twin primes between P^2 and a higher Px^2 (for instance between 19^2 and 101^2, starting from the actual number of twin primes up to 19^2 and then multiplying by 17/19 * 21/23 * …. 99/101).

      I looked at this when I was trying to understand your ratio definition. It does give a very good estimate, but so does the zero twin prime ratio, sinc they both approach 1 fairly quickly (but the latter less quickly).

  25. Thanks, that’s interesting.

    • petequinn says:

      I think you’re heading in a good direction, mind you. I think the answer lies in the fact that as we go from P^2 to Pn^2, sieving by Pn, we remove exactly one more possible prime that is an actual prime (Pn itself) and exactly one possible prime that is non-prime (Pn^2). I think a mathematical elaboration of this would be fruitful, and this is one thing I’ve been working on, indirectly, by exploring patterns.

      I’m going to throw some more illustrations up over the next day or two for contemplation and comment.

      **** minor correction – sieving with Pn also removes P*Pn, so it removes one prime and two possible primes up to Pn^2

      • I think one trouble with this theory is that it ignores larger prime gaps. Obviously it is harder to eliminate the primes in these, but on the other hand it allows for the possibility that P removes P*Pn, P*P(n+1), P*P(n+2), P*P(n+3) etc. There’s no good reason why you can’t have quite a lot of multiples of P lower than Pn^2.

      • petequinn says:

        Prior sieving by P has already removed all those. I’m referring only to what sieving by Pn removes. And to correct my earlier correction, P*Pn was also already removed by sieving by P, so again we’re left with only removing one actual prime (i.e. Pn) and one (non-prime) candidate prime up to Pn^2 (i.e. Pn^2) when sieving by Pn.

  26. Just following up on that, I think the most definite you can be is that for P to eliminate more than 2 twin prime candidate pairs between P^2 and Pn^2, you would need a gap =/> 6P between P^2 and Pn^2, for P to eliminate more than 4 candidate pairs you would need a gap =/> 12P and so on.

    This is simply because a region of 6P can’t contain more than 2 multiples of P that are a 6N+1 or 6N-1 number. For instance between 169 and [169 + (6*13)}] = 247 we get 13*13, 13*17, 13*19 (=169, 221, 247), and we know there will only be 2 further multiples of 13 up to (247+78).

    289 – 169 = 120 = approx 9.2 x 13 – so from 13^2 to 17^2 we know that 13 can only eliminate between 3 and 5 twin prime candidates.

    (I wonder if there is a way to relate that back to the zero twin prime rate stuff also?)

  27. Minor correction – we don’t know that 13 will eliminate any candidates in this region since it could fall on pairs that already contain a multiple of lower primes, so it’s not right to say it can only eliminate between 3 and 5, only that *it can’t eliminate more than 4 (since 120 < 2*6*13)*

  28. Hi Pete,

    I talked about P rather than Pn for a reason. You say:

    “Prior sieving by P has already removed all those. I’m referring only to what sieving by Pn removes. And to correct my earlier correction, P*Pn was also already removed by sieving by P, so again we’re left with only removing one actual prime (i.e. Pn) and one (non-prime) candidate prime up to Pn^2 (i.e. Pn^2) when sieving by Pn.”

    There’s a confusion here between which region we are sieving in. P is the prime that sieves in the region between P^2 and Pn^2, and it can eliminate numerous multiples of P in this region. Multiples of lower primes have already been eliminated in this region, and Pn can’t eliminate any non-primes before Pn^2 by definition.

    Pn is the prime that sieves in the region between Pn^2 and P(n+1)^2, and it can eliminate numerous multiples of Pn in this region.

    By definition, Pn can only eliminate one non-prime in the region up to Pn^2 because Pn^2 is the first composite with Pn as a factor and no lower primes as a factor. But this doesn’t tell us anything at all about how many non-primes will be eliminated between successive instances of P^2 and Pn^2. In this region the question is whether any remaining candidates can be eliminated by multiples of P.

    • petequinn says:

      There’s a confusion here between which region we are sieving in.

      Not sure what to tell you here Hugh. I meant what I wrote. You may have a different (possibly better) angle in mind, and this particular observation may not bear any fruit, but what I wrote is correct, and I don’t think I’m confused. At the moment anyway…🙂

  29. Fair enough, you’re probably not confused, I just don’t see that that particular observation is going to be of any use, because it’s a bit of a tautology. Pn^2 is the first non-prime that Pn can possibly sieve out, so if you consider the region up to Pn^2 of course Pn will only sieve out one non-prime. It’s no more useful than pointing out that in the region up to Pn, there aren’t any numbers divisible by Pn other than Pn.

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