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

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  • jbrennen
    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,
    Message 1 of 15 , Nov 19, 2002
      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!
    • 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 2 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 3 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 4 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 5 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
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