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Twin primes

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  • Guy
    Hello all, I ve been working on the twin prime conjecture for about five years and would like to share some of my ideas (no proofs as yet though, sorry!) The
    Message 1 of 9 , Mar 18 5:24 PM
      Hello all,

      I've been working on the twin prime conjecture for about five years
      and would like to share some of my ideas (no proofs as yet though,
      sorry!)

      The moebius function: M(n)

      M(1)=1
      M(n)=0 if n is divisible by m^2 for some m>1
      M(n)=(-1)r if n is the product of r distinct primes,

      The Mangoldt function: A(n)

      A(n)= log p if n=p^k, k>0
      0 otherwise

      a function that defines the primes: F(n)
      F(n)= M(n)*A(n) gives 0 if n is square-divisible or divides a product
      of primes, or log(1/n)=-log n if and only if n is prime

      therefore, F(n)*F(n+2) is equal to log n * log (n+2) if and only if n
      is the first prime in a twin pair. (The second prime n+2 would not
      produce a nonzero result since n+4 is not prime, and n>3) So then,

      sum n<=x(F(n) * F(n+2))/(n * log n * log (n+2))

      would sum the reciprocals of the first prime in a prime pair (by
      using the error term (2*c)/log n (c=.6601618158 the twin prime
      constant) I've supposed this limits to about 1.05893... (*)

      I'm using (2*c)/log n instead of Prof Brent's term (4*c)/ log n
      because I've summed this using one term in the pair instead of the
      usual two.

      The constant I marked (*) appears to be constructed out of a ratio of
      prime power products:

      product n=1, infinity, 1-M(n)*n^-2 divided by
      product odd primes 1-(p-1)^-2

      (The denominator being the constant c=.66016...) This appears to be
      the same as the sum of reciprocals of the twins defined above, in
      terms of a limit (I dont think this assumes infinitely many twins
      since Brun's constant is the limit of 1/p+1/(p+2) regardless of the
      number of terms in the sum). Any one like to prove this? Or, if I am
      wrong please show me where. (you might complain about using the
      moebius function in a product...!) If this is proven, what will it
      say about the total number of twin pairs?

      Thanks,
      from Guy H Bearman
    • d.broadhurst@open.ac.uk
      ... Has this been checked numerically to Nicely-level (Intel destruction level:-) accuracy? As a sum, it looks like a subset of the Brun reciprocals, which
      Message 2 of 9 , Mar 18 7:38 PM
        "Guy" <anastasis_1999@y...> wrote:

        > appears to be constructed out of a ratio

        Has this been checked numerically to Nicely-level
        (Intel destruction level:-) accuracy?

        As a sum, it looks like a subset of the Brun reciprocals,
        which Thomas Nicely had to about 1 part in a billion,
        as I recall.

        As a product it looks like the sort of thing
        that Pieter Moree or Gerhard Niklasch might eat for breakfast.

        So, if you use best practice for each form,
        how good is the match, please?

        If it's a few parts per billion,
        then it's a capital empirical Ansatz, and makes
        me wonder whether you can guess a product form for Brun?
        If not, why for one and not for the other?

        Intrigued..

        David
      • d.broadhurst@open.ac.uk
        Prod_n (1-mu(n)/n^2) looks easy to get to (say) 18 sf. 1) Take product up to N=10^6. 2) For the rest take logs and retain only sum_{n N} mu(n)/n^2 = 6/pi^2 -
        Message 3 of 9 , Mar 18 9:28 PM
          Prod_n (1-mu(n)/n^2) looks easy to get to (say) 18 sf.
          1) Take product up to N=10^6.
          2) For the rest take logs and retain only
          sum_{n>N} mu(n)/n^2 = 6/pi^2 - sum_{n<=N} mu(n)/n^2.
          3) Then combine with twin-prime constant.
          4) Mail 18-digit prediction for "semi-Brun" sum
          to Thomas Nicely, and pray hard
          that the first 9 digits hold up.
          David
        • d.broadhurst@open.ac.uk
          Finally: you do not actually need Nicely help, since 2*semi-Brun - Brun is much better convergent and you can test it deep yourself, when you have fixed the
          Message 4 of 9 , Mar 18 11:18 PM
            Finally: you do not actually need Nicely help, since
            2*semi-Brun - Brun is much better convergent
            and you can test it deep yourself, when you have fixed
            the glitch that prod_{n>=1} (1-mu(n)/n^2) is zero!
          • d.broadhurst@open.ac.uk
            Pm=prod_{n 1} (1-mu(n)/n^2) [cannot include n=1, else vanishes!] C2=twin-prime, as per usual SB=semi-Brun, only take first reciprocal for each twin At speed, I
            Message 5 of 9 , Mar 19 10:42 AM
              Pm=prod_{n>1} (1-mu(n)/n^2) [cannot include n=1, else vanishes!]
              C2=twin-prime, as per usual
              SB=semi-Brun, only take first reciprocal for each twin
              At speed, I make those
              Pm=1.430380788
              C2=0.660161816
              SB=1.059064265
              You said SB*C2=Pm, which is clearly no good.
              SB*C2*Pm=1 is better, but not good enough.
              I make it 1.0000561 [needs checking]
              Back to the drawing board?
            • d.broadhurst@open.ac.uk
              ... Guy Bearman has lost his Yahoo connectivity. Offlist, he confirmed that I had guessed what he meant and said ... It would be in the spirit of this list if
              Message 6 of 9 , Mar 19 3:30 PM
                I wrote:

                > Pm=prod_{n>1} (1-mu(n)/n^2) [cannot include n=1, else vanishes!]
                > C2=twin-prime, as per usual
                > SB=semi-Brun, only take first reciprocal for each twin
                > SB*C2*Pm=1 is better, but not good enough.
                > I make it 1.0000561 [needs checking]

                Guy Bearman has lost his Yahoo connectivity.
                Offlist, he confirmed that I had guessed what he meant and said

                > Thank you for making the time to refute this idea

                I replied:

                > It was fun. When I worked out what you meant and got 1.00005,
                > at low accuracy, the adrenaline started to flow.
                > I hope that someone will check my numerics,
                > lest I have missed a gem.

                It would be in the spirit of this list if someone
                (Andrey Kulsha where are you?)
                checks that I am not mistaken.

                Had it worked, it would have been sensational!

                Goodness knows how Guy was led to guess something that got so close.
                It's a salutory reminder of how much dedicated private study
                surfaces in our amateur (fine word!) pages.

                David
              • Guy H Bearman
                Yes, it s me again :) (all logs are to base e.) Brent (1975) uses the formula, sum(k=1, +inf, (k+1)/zeta(k+1) - k/zeta(k+2))*(log x)^(k-1)/(k+1)! in his work
                Message 7 of 9 , Oct 5, 2001
                  Yes, it's me again :)

                  (all logs are to base e.)
                  Brent (1975) uses the formula,

                  sum(k=1, +inf, (k+1)/zeta(k+1) - k/zeta(k+2))*(log x)^(k-1)/(k+1)!

                  in his work on Brun's sum.

                  Just now I was thinking (again), how do I do a product equalling
                  Brun's sum? And came up with this:

                  4*c*prod(k=1, +inf, (k+1)/zeta(k+1) - k/zeta(k+2)), where c is the
                  twin prime constant, 0.6601618158...

                  this comes to the value (approx.) 0.40888 (each term in the product
                  tends to 1 as k->+inf) - yes I know this doesn't look like Brun's
                  constant!

                  call this constant M. Then calculate -1/(log(1-M)). With only twenty
                  terms of the product, and using 9 dp for each zeta function, I get
                  1.90204 already. Would anyone like to do this with say, 100 dp of
                  each zeta function, for several hundred terms, and see if it gets any
                  closer to 1.90216?

                  Thanks, from Guy
                • navid.altaf@gstt.sthames.nhs.uk
                  I ve got a bit of a fascination at the moment for composites (c) of the form c = p^2 One of the things I seem to be observing is that there seems to be twin
                  Message 8 of 9 , Sep 9, 2003
                    I've got a bit of a fascination at the moment for composites (c) of the form

                    c = p^2

                    One of the things I seem to be observing is that there seems to be twin
                    primes in close
                    proximity to composites of the above form.

                    eg.

                    5^2 = 25 17,19 and 29,31

                    7^2 = 49 41,43 and 59,61

                    317^2 = 100489 100391,100393 and 100517,100519


                    Is this significant?


                    Navid





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                  • Andy Swallow
                    ... Not particularly. The first two examples are far too small to draw any reasonable conclusions from. It seems strange that you suddenly have to jump from 7
                    Message 9 of 9 , Sep 9, 2003
                      > I've got a bit of a fascination at the moment for composites (c) of the form
                      >
                      > c = p^2
                      >
                      > One of the things I seem to be observing is that there seems to be twin
                      > primes in close
                      > proximity to composites of the above form.
                      >
                      > eg.
                      >
                      > 5^2 = 25 17,19 and 29,31
                      >
                      > 7^2 = 49 41,43 and 59,61
                      >
                      > 317^2 = 100489 100391,100393 and 100517,100519
                      >
                      >
                      > Is this significant?

                      Not particularly. The first two examples are far too small to draw any
                      reasonable conclusions from. It seems strange that you suddenly have to
                      jump from 7 to 317, but maybe there are other examples in between, I
                      can't be bothered to check.

                      According to the prime number theorem, the average distance from a prime
                      p to the next prime is about log p. Suppose now that p is part of a twin
                      prime pair, and that q is the first of the next twin prime pair. Then
                      the expected distance between p and q is roughly log^2 p instead. So the
                      twin primes are expected to be fairly common, and certainly a lot more
                      common than numbers of the form p^2. So it's not surprising that you may
                      well find twin primes "fairly close" to some prime squares.

                      Andy
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