Loading ...
Sorry, an error occurred while loading the content.
 

Re: k*2^n-1 and k*2^n+1 are twins

Expand Messages
  • Robert
    ... Jack This goes to show it sometimes takes a while to reply to posts, in this case 2 1/2 years! In any case the following k has 13 twins to n=10000
    Message 1 of 15 , May 21, 2005
      --- In primenumbers@yahoogroups.com, "jbrennen" <jack@b...> wrote:
      > Can anybody find a value of k which yields more twin primes than
      > k=202507305 (3*5*7*11*13*13487) ?
      >
      > When k=202507305, k*2^n+/-1 are twin primes for n:
      >
      > 2, 12, 17, 28, 31, 33, 42, 55, 62, 86, 89, 91
      >
      > (and most likely for no other values of n)
      >
      > At the time that I found this one, I remember searching far and
      > wide for a "better" k with more than 12 twin primes, with no luck.
      >
      > If you find a value of k with more than 12 twin primes, please
      > let me know!

      Jack

      This goes to show it sometimes takes a while to reply to posts, in
      this case 2 1/2 years!

      In any case the following k has 13 twins to n=10000

      7985650262654529465

      And the twins are n=1,3,17,37,38,39,70,97,485,556,561,1082,1086

      So there is a nice bitwin length 3 in there are well.

      This resulted in a new search I have started to determine highest
      scoring k.2^n+ & - 1 series, where 1 point is awarded for each twin
      or cunningham chain length 2. (a CC length 3 gets two points, etc).

      The k quoted has 13 twins, and 15 points for cunningham chains (up
      to n=10000), so 28 is a nice easy target to beat.

      Regards

      Robert Smith
    • Jack Brennen
      ... That s a pretty good one, and the last 5 of those exponents... wow! I would have expected my record to be beaten by a k value with an abundance of small
      Message 2 of 15 , May 22, 2005
        Robert wrote:
        > This goes to show it sometimes takes a while to reply to posts, in
        > this case 2 1/2 years!
        >
        > In any case the following k has 13 twins to n=10000
        >
        > 7985650262654529465
        >
        > And the twins are n=1,3,17,37,38,39,70,97,485,556,561,1082,1086

        That's a pretty good one, and the last 5 of those exponents... wow!
        I would have expected my "record" to be beaten by a k value with an
        abundance of small twin primes (n <= 100) -- just because it's so
        much easier to selectively prune the search tree. I'm sure that my
        hypothetical search algorithm would have given up on this k value
        once it found only 8 twins with n <= 100.

        The factorization of this k:

        3 5 7 11 13 19 29 37 251 1009 103007

        Any significance to that? I'm not surprised to see the abundance
        of small primes, but the product of the "larger" primes in the list
        (251 * 1009 * 103007 == 26087449813) seems pretty large for you to
        have done a brute force search for a large number of twins...
        Maybe you were sieving for bi-twin chains and came across this
        lucky coefficient???

        Good find, regardless of how or why you found it. :)
      • Robert
        ... in ... wow! I know, extraordinary to get 5 after n=484 ... I agree, but they are not so easy to find - 10 twins to n=100 is not too hard, but getting those
        Message 3 of 15 , May 22, 2005
          --- In primenumbers@yahoogroups.com, Jack Brennen <jb@b...> wrote:
          > Robert wrote:
          > > This goes to show it sometimes takes a while to reply to posts,
          in
          > > this case 2 1/2 years!
          > >
          > > In any case the following k has 13 twins to n=10000
          > >
          > > 7985650262654529465
          > >
          > > And the twins are n=1,3,17,37,38,39,70,97,485,556,561,1082,1086
          >
          > That's a pretty good one, and the last 5 of those exponents...
          wow!

          I know, extraordinary to get 5 after n=484

          > I would have expected my "record" to be beaten by a k value with an
          > abundance of small twin primes (n <= 100) -- just because it's so
          > much easier to selectively prune the search tree. I'm sure that my
          > hypothetical search algorithm would have given up on this k value
          > once it found only 8 twins with n <= 100.
          >
          I agree, but they are not so easy to find - 10 twins to n=100 is not
          too hard, but getting those to be twins after 100 is a pig.

          > The factorization of this k:
          >
          > 3 5 7 11 13 19 29 37 251 1009 103007
          >
          > Any significance to that?

          Yes, this is a Payam M(37-) E(51-) number, meaning it is
          3*5*11*13*19*29*37 = M(37), and in fact has no factors with order
          base 2 of less than 52, for the k.2^n-1 series. In this case it is
          also not that divisible on the + side either.

          But there are zillions of M(37-) numbers, they will always be really
          prime, but sometimes they are also prime on the + side, as in this
          case, because no small factors appear there either. When this is the
          case, a happy coincidence will occasionally occur, and the twins and
          cunninghams line up.

          My score of 28 (twins +cunninghams) is really a good target to beat.
          Been going all day and only got one 24.

          Regards

          Robert Smith
        • thefatphil
          ... See also my announcement on NMBRTHRY a couple of years back. I believe I injected the 15-twin k into a usenet post at about the same time, if you really
          Message 4 of 15 , May 23, 2005
            --- In primenumbers@yahoogroups.com, Jack Brennen <jb@b...> wrote:
            > Robert wrote:
            > > This goes to show it sometimes takes a while to reply to posts, in
            > > this case 2 1/2 years!
            > >
            > > In any case the following k has 13 twins to n=10000
            > >
            > > 7985650262654529465
            > >
            > > And the twins are n=1,3,17,37,38,39,70,97,485,556,561,1082,1086
            >
            > That's a pretty good one, and the last 5 of those exponents... wow!
            > I would have expected my "record" to be beaten by a k value with an
            > abundance of small twin primes (n <= 100) -- just because it's so
            > much easier to selectively prune the search tree. I'm sure that my
            > hypothetical search algorithm would have given up on this k value
            > once it found only 8 twins with n <= 100.
            >
            > The factorization of this k:
            >
            > 3 5 7 11 13 19 29 37 251 1009 103007
            >
            > Any significance to that? I'm not surprised to see the abundance
            > of small primes, but the product of the "larger" primes in the list
            > (251 * 1009 * 103007 == 26087449813) seems pretty large for you to
            > have done a brute force search for a large number of twins...
            > Maybe you were sieving for bi-twin chains and came across this
            > lucky coefficient???
            >
            > Good find, regardless of how or why you found it. :)

            See also my announcement on NMBRTHRY a couple of years back.
            I believe I injected the 15-twin k into a usenet post at about the
            same time, if you really want its numerical value.

            Phil
          Your message has been successfully submitted and would be delivered to recipients shortly.