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Re: [PrimeNumbers] Prime GAP of 364,188

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  • Kermit Rose
    From: Jens Kruse Andersen Date: 12/31/05 12:02:34 To: Prime Numbers Subject: Re: [PrimeNumbers] Prime GAP of 364,188 ... AFAIK, all efficient searches for
    Message 1 of 8 , Jan 2, 2006
      From: Jens Kruse Andersen
      Date: 12/31/05 12:02:34
      To: Prime Numbers
      Subject: Re: [PrimeNumbers] Prime GAP of 364,188

      Jose Ramón Brox wrote:

      > > There are better methods to choose which numbers have small factors.
      > > Such methods were used for the 2 known Megagaps:
      > > http://hjem.get2net.dk/jka/math/primegaps/megagap.htm
      >
      > Indeed a very intelligent approach: I was wondering how all that records
      > could be arrived at, and now I know. Are there any other approaches,
      > or yours is the main used one?

      AFAIK, all efficient searches for large gaps between large primes/prp's (e.g

      above 50 digits) have included some variant of these points:

      ********

      From Kermit

      kermit@...




      Has anyone tried this approach?


      To product a prime gap

      choose prime divisors,

      p1, p2, p3, .... pN.


      Lets pick p1 = 2, p2 = 3, p3 = 5, etc.



      Choose k = -1 mod 2. This makes k odd.

      choose k = -2 mod 3.

      Now we don't need to worry about -3, -5, -7, etc because these are all taken
      care of by
      choosing k = -1 mod 2.

      choose k = - 4 mod 5.

      not we don't need to worry about -5, - 8, - 11, - 14, etc, because these are
      already taken care of by k = -2 mod 3.

      choose k = -6 mod 7.

      Now we don't need to worry about -9, -14, -19, - 24, etc because these are
      taken care of
      by k = -4 mod 5.

      Choose k = - 10 mod 11

      Now we don't need to worry about -6, -13, -20, etc because these are taken
      care of by
      k = -6 mod 7.


      etc.



      So with the 5 primes, 2,3,5,7,11 we can assure a prime gape of length 11.

      Each prime we add to the list lengthen's the prime gap by more than 1.


      In fact, note that we just defined k = 1 mod 2, k=1 mod 3, k = 1 mod 5, k=1
      mod 7,

      So the minimum k satisfying this is

      k = 2 * 3 * 5 * 7 * 11 + 1 = 2311.


      In fact, I just realized why the pattern I chose will always yield p#+1 as
      the value of
      k. Thus I just proved that the prime gap of p# is at least p.




      Perhaps some other way of assigning the negative values will produce a sieve
      of higher
      merit.

      [Non-text portions of this message have been removed]
    • Jens Kruse Andersen
      ... ..... Yes, that is the primorial approach mentioned several times in this thread. ... Yes, that is exactly what the choices in 2) is about. Maybe you
      Message 2 of 8 , Jan 3, 2006
        I wrote:
        > 2) For each small prime, choose which numbers in the interval that
        > prime should divide, trying to minimize the "overlap" with other primes.

        Kermit Rose wrote:
        > Has anyone tried this approach?
        .....

        Yes, that is the primorial approach mentioned several times in this thread.

        > Perhaps some other way of assigning the negative values will produce
        > a sieve of higher merit.

        Yes, that is exactly what the choices in 2) is about.

        Maybe you didn't have time to read when you posted 6 mails in 2h 12min...

        --
        Jens Kruse Andersen
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