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Re: [PrimeNumbers] another attempt at Twin prime conjecture...

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  • Décio Luiz Gazzoni Filho
    ... Forget about using density arguments for your `proof . The bounds on prime density, even assuming the Riemann Hypothesis, are much too loose for this.
    Message 1 of 7 , Nov 5, 2004
      On Friday 05 November 2004 14:27, you wrote:
      > <snip>
      >
      > Proof:
      > ------
      > Let p and p+2 be a twin pair. Then, p^2 and (p+2)^2 will be the their
      > square range.
      > If this twin exists in the primes less than p+3, then in the range of
      > numbers between p^2 and (p+2)^2, the prime density should be less
      > than the previous range for us to not have a twin prime in this
      > square range. This would invalidate our assumption.
      > The contrary will prove the conjecture.
      >
      > We compare these prime densities using Prime Number Theorum.

      Forget about using density arguments for your `proof'. The bounds on prime
      density, even assuming the Riemann Hypothesis, are much too loose for this.
      Although you seem to be foregoing any kind of bounds and assuming the PNT
      gives an exact answer. Sorry, but that's wrong.

      > For primes including p+2, we take p+3.
      >
      > Primes less than p+3 can be given by p + 3 / log (p+3)
      > This can be written as p + 3 / log p + log 3
      > Ignoring log 3 we get, p + 3 / log p ---- 1

      Nitpick: log(p+3) != log(p) + log(3). You've repeated that mistake somewhere
      in the remainder of your `proof'. However this isn't what invalidates the
      `proof', as you can show that log(p+k) -> log(p) as p -> oo and k is fixed;
      my arguments above are the real reason why your argument fails.

      Décio


      [Non-text portions of this message have been removed]
    • Décio Luiz Gazzoni Filho
      ... No, I mean that you can t use the PNT to count primes in an interval in an exact fashion; there ll always be an error bound. According to the book by
      Message 2 of 7 , Nov 5, 2004
        On Friday 05 November 2004 15:42, you wrote:
        > Do you mean that one cannot compare 1,p+3 and
        > p^2,(p+2)^2 bounds or ranges?

        No, I mean that you can't use the PNT to count primes in an interval in an
        exact fashion; there'll always be an error bound. According to the book by
        Crandall & Pomerance, the PNT as currently proved could be stated as

        pi(x) = li(x) + O(x e^(-C sqrt(log x))),

        where pi(x) is the prime counting function and li(x) is the logarithmic
        integral int(dt/log t) from 2 to x, and is asymptotic to x/log x. The problem
        is that you're neglecting the error term O(...). Even if one were to assume
        the Riemann Hypothesis, through which the error term would be decreased to
        O(x^(1/2 + eps)) for eps -> 0, the bounds are not tight enough for your
        method of proof to work.

        While you replied to me in private, I'm going to get this back into the list,
        as it might help others pursuing similar lines of proof, whether for the twin
        prime conjecture or something else.

        Décio


        [Non-text portions of this message have been removed]
      • Décio Luiz Gazzoni Filho
        Also, here s another argument regarding your method of proof, which I ll recall involves showing that if p, p + 2 is a twin prime pair, then there exists
        Message 3 of 7 , Nov 5, 2004
          Also, here's another argument regarding your method of proof, which I'll
          recall involves showing that if p, p + 2 is a twin prime pair, then there
          exists another twin prime pair between p^2 and (p + 2)^2.

          The book by Hardy & Wright claims that it is an unsolved problem whether there
          exists a prime between n^2 and (n+1)^2. In a sense this is a weaker form of
          your statement, as it requires the existence of a single prime, not a prime
          pair. On the other hand it is a little stronger, as your interval is (p +
          2)^2 - p^2 = (p + 2 - p)(p + 2 + p) = 2(2p + 2) = 4p + 4, while (n+1)^2 - n^2
          = (n + 1 - n)(n + 1 + n) = 2n + 1. Also, your argument is only required to
          hold for primes p, p + 2, whereas no restrictions are placed on n. Still, I
          don't see why your argument would be easier to prove.

          Décio


          [Non-text portions of this message have been removed]
        • Suresh Batta
          OK. If the range isn t tight, why not consider the bounds p^2 and p*(p+2)? The range would be just 2p. Then, the difference would be.. p^2 + 2p / log (p)*(p+2)
          Message 4 of 7 , Nov 5, 2004
            OK. If the range isn't tight, why not consider the bounds p^2 and
            p*(p+2)? The range would be just 2p.

            Then, the difference would be..

            p^2 + 2p / log (p)*(p+2) - p^2 / log p^2

            p^2 + 2p / log p * log (p+2) - p^2 / 2 log p

            Reducing log (p+2) to log p when p--> Infinity, we get,

            p^2 + 2p / 2 log p - p^2 / 2 log p

            2p / 2 log p

            p / log p ----------- 1 (PNT)

            Compare this to the original range, p+3 / log p.
            This reduces to p / log p for large p.

            So, both of the ranges show same order.

            Can we now not say that, since both the ranges have comparable prime
            quantities, if a P+3 has a twin prime, then squared range
            ((p+2)^2, p^2) should have too have a twin prime?

            My effort stems from the encouragement of a recent observation i
            made. I counted how may twin primes are there between the squares of
            a twin prime.
            It's something like:
            5-7 2
            11-13 2
            17-19 2
            29-31 2
            41-43 3
            59-61 5
            71-73 3
            101-103 7
            107-109 6
            137-139 6
            149-151 10
            179-181 13
            191-193 7
            197-199 8

            When I graphed this and looked at the regression equation, it looked
            like the same rate of growth as p/log p. This made me believe that
            may be Squaring a number or infact any exponential order does not
            change any thing. pi(x) stays the same. Hence this attempt.

            Suresh
          • Décio Luiz Gazzoni Filho
            ... This is a gain of a constant factor. I believe you need asymptotically tighter bounds to achieve a proof. That is, instead of p times a constant, you would
            Message 5 of 7 , Nov 5, 2004
              On Saturday 06 November 2004 01:17, you wrote:
              > OK. If the range isn't tight, why not consider the bounds p^2 and
              > p*(p+2)? The range would be just 2p.

              This is a gain of a constant factor. I believe you need asymptotically tighter
              bounds to achieve a proof. That is, instead of p times a constant, you would
              need sqrt(p) or log(p) or whatever. I'm not an expert in analytical number
              theory, I'm just sure the bounds are not tight enough or this proof would
              have surfaced much earlier.

              > <snip>
              >
              > My effort stems from the encouragement of a recent observation i
              > made. I counted how may twin primes are there between the squares of
              > a twin prime.
              > It's something like:
              > 5-7 2
              > 11-13 2
              > 17-19 2
              > 29-31 2
              > 41-43 3
              > 59-61 5
              > 71-73 3
              > 101-103 7
              > 107-109 6
              > 137-139 6
              > 149-151 10
              > 179-181 13
              > 191-193 7
              > 197-199 8
              >
              > When I graphed this and looked at the regression equation, it looked
              > like the same rate of growth as p/log p. This made me believe that
              > may be Squaring a number or infact any exponential order does not
              > change any thing. pi(x) stays the same. Hence this attempt.

              In fact there exists a conjecture for the number of twin primes up to a
              certain bound

              Here's an heuristic for the number of twin primes between p^2 and (p+2)^2. We
              start with the approximation pi2(x) ~ 2 C2 x/log^2(x), where pi2(x) is the
              twin-prime counting funcion and C2 is the twin-prime constant 0.660161858...
              This formula was taken from the trusty book of Crandall & Pomerance. Now

              pi2((p+2)^2) - pi2(p^2) ~ 2 C2 ( (p+2)^2/log^2((p+2)^2) - p^2/log^2(p^2) )
              ~ 2 C2 ( (p+2)^2/log^2(p^2) - p^2/log^2(p^2) )
              = 2 C2/log^2(p^2) ( (p+2)^2 - p^2 )
              = 2 C2/(2^2 log^2(p)) ( 4p + 4 )
              = 2 C2 (4p + 4)/(2^2 log^2(p))
              ~ 2 C2 4p/(2^2 log^2(p))
              = 2 C2 p/log^2(p)

              Notice that the first approximation is very good -- for p ~ 1e3 the error is
              on the fourth decimal place. The second approximation is good also: the term
              I discarded is 2 C2/log^2(p), which is already 0.5 for p as small as 5.

              Now for the interpretation of this result: in the interval (p^2,(p+2)^2),
              there are as many twin primes as in the interval (0,p). This makes sense:
              while there are ~4p primes in the former interval, they are larger than the
              primes in the latter interval, and this difference is accounted for by the
              factor 1/4 -- which is precisely the square of the ratio between the number
              of digits of a typical number in each interval, not coincidently the
              dependence embodied by the log^2(p) term in the approximation to pi2(x).
              Thus your regression equation is off by a factor 1/log(p).

              Please, do not interpret these heuristics as validating your proof: they are
              just *unproven heuristics*. Actually if this formula were exact, we wouldn't
              need to employ your proof at all; the infinity of twin primes would follow
              directly from the formula.

              Décio


              [Non-text portions of this message have been removed]
            • Jens Kruse Andersen
              ... Definitely no. The only comparable prime quantities are about the _expected_ number of primes based on unproven heuristics. Even if this expectation
              Message 6 of 7 , Nov 6, 2004
                Suresh Batta wrote:

                > Can we now not say that, since both the ranges have comparable prime
                > quantities, if a P+3 has a twin prime, then squared range
                > ((p+2)^2, p^2) should have too have a twin prime?

                Definitely no. The only "comparable prime quantities" are about the _expected_
                number of primes based on unproven heuristics. Even if this expectation could be
                proven reasonably accurate, it would not say anything about the number of twin
                primes.

                p/log p is the approximate number of primes below p.
                The prime number theorem says it is asymptotically right. This means the
                relative error (NOT the absolute error) tends to 0 when p tends to infinite.

                However, as Decio notes, this and better known approximations are too poor to
                say anything about the number of primes (let alone twin primes) from p to p+x
                when x is much smaller than p.

                It can only be used to say things like:
                The _average_ number of primes in an interval of x numbers _around_ the size of
                p is approximately x/log p.

                Little is known about how large the deviation from such averages can be.
                To illustrate possible deviations, here are the most extreme values known around
                100 digits:

                There are 11 primes (0.16 expected) among the 104-digit numbers p to p+36 for
                p = 24698258*239# + 28606476153371, found by Norman Luhn and I.

                There are 0 primes (30 expected) among the 93-digit numbers c to c+6378 for
                c = 5629854038470321802219554908853741163682800524507382035301697914566243\
                83980052820124370178769, found by Torbjörn Alm and I.

                There probably exists far more extreme cases.

                As far as twin primes go, there is not even anything known about averages.

                --
                Jens Kruse Andersen
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