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

Twin prime curio

Expand Messages
  • David Litchfield
    Let n be (p(p+2)+1)/2 where p and p + 2 are potential twin primes. If p and p+2 are twin primes then the sum of the products of (n+x)(n-x) that divide
    Message 1 of 2 , Oct 3, 2004
    • 0 Attachment
      Let n be (p(p+2)+1)/2 where p and p + 2 are potential twin primes.

      If p and p+2 are twin primes then the sum of the products of (n+x)(n-x) that divide integrally by p or p+2, with x starting at 0 to x = n-2, will equal p(p+2)(a) ((6a(6a-1) + (6a-1)(6a+1))) - (3(a(a-1)+1)) where a = (p+1) / 6. If either of p or p+2 are composite then the sum will be greater than p(p+2)(a)((6a(6a-1) + (6a-1)(6a+1))) - (3(a(a-1)+1)).



      As I'm not that good at expressing myself mathematically as an example

      p = 5, p+2 = 7



      18 * 18 = 324

      17 * 19 = 323

      16 * 20 = 320

      15 * 21 = 315

      ..

      2 * 34 = 68

      Note that we only have x to n-2 = 16 i.e. we don't include 1 and 35.

      Of all the generated products 5 will divide 155, 180, 260, 275, 315 and 320. 7 will divide 203, 224, 308 and 315. As 315 divides by both 5 and 7 it is added only once. The sum of all these products is 2240.

      So with p = 5 then (p+ 1) / 6 = 1 in this example so



      p(p+2)(a)((6a(6a-1) + (6a-1)(6a+1))) - (3(a(a-1)+1))

      (35)(1) ((6(6-1) + (6-1)(6+1))) - (3(1(1-1)+1))

      35(30 + 35 - 1) = 2240

      The reason this works out this way is due to the fact if p or p+2 is composite then more of the products are divisible than with two primes when one expects only 2p of the products to be divisible.

      Anybody fancy putting this into more formal mathematical language?

      Cheers,

      David


      ---------------------------------
      ALL-NEW Yahoo! Messenger - all new features - even more fun!

      [Non-text portions of this message have been removed]
    • David Litchfield
      Hi all, (Hope this formatting comes out) n-2 ____ / (x+n)(x-n) ... k=0 I think (?) this is the sum of all (x+n)(x-n) for all n starting from 0 to n-2. But
      Message 2 of 2 , Oct 4, 2004
      • 0 Attachment
        Hi all,
        (Hope this formatting comes out)

        n-2
        ____
        \
        / (x+n)(x-n)
        -------
        k=0

        I think (?) this is the sum of all (x+n)(x-n) for all n starting from 0 to n-2. But how would I modify it to show that only those (x+n)(x-n) that divide by p or p+2 should be summed?
        Cheers,
        David


        David Litchfield <dwlitchfield@...> wrote:


        Let n be (p(p+2)+1)/2 where p and p + 2 are potential twin primes.

        If p and p+2 are twin primes then the sum of the products of (n+x)(n-x) that divide integrally by p or p+2, with x starting at 0 to x = n-2, will equal p(p+2)(a) ((6a(6a-1) + (6a-1)(6a+1))) - (3(a(a-1)+1)) where a = (p+1) / 6. If either of p or p+2 are composite then the sum will be greater than p(p+2)(a)((6a(6a-1) + (6a-1)(6a+1))) - (3(a(a-1)+1)).



        As I'm not that good at expressing myself mathematically as an example

        p = 5, p+2 = 7



        18 * 18 = 324

        17 * 19 = 323

        16 * 20 = 320

        15 * 21 = 315

        ..

        2 * 34 = 68

        Note that we only have x to n-2 = 16 i.e. we don't include 1 and 35.

        Of all the generated products 5 will divide 155, 180, 260, 275, 315 and 320. 7 will divide 203, 224, 308 and 315. As 315 divides by both 5 and 7 it is added only once. The sum of all these products is 2240.

        So with p = 5 then (p+ 1) / 6 = 1 in this example so



        p(p+2)(a)((6a(6a-1) + (6a-1)(6a+1))) - (3(a(a-1)+1))

        (35)(1) ((6(6-1) + (6-1)(6+1))) - (3(1(1-1)+1))

        35(30 + 35 - 1) = 2240

        The reason this works out this way is due to the fact if p or p+2 is composite then more of the products are divisible than with two primes when one expects only 2p of the products to be divisible.

        Anybody fancy putting this into more formal mathematical language?

        Cheers,

        David


        ---------------------------------
        ALL-NEW Yahoo! Messenger - all new features - even more fun!

        [Non-text portions of this message have been removed]



        Unsubscribe by an email to: primenumbers-unsubscribe@yahoogroups.com
        The Prime Pages : http://www.primepages.org/




        Yahoo! Groups SponsorADVERTISEMENT


        ---------------------------------
        Yahoo! Groups Links

        To visit your group on the web, go to:
        http://groups.yahoo.com/group/primenumbers/

        To unsubscribe from this group, send an email to:
        primenumbers-unsubscribe@yahoogroups.com

        Your use of Yahoo! Groups is subject to the Yahoo! Terms of Service.



        ---------------------------------
        ALL-NEW Yahoo! Messenger - all new features - even more fun!

        [Non-text portions of this message have been removed]
      Your message has been successfully submitted and would be delivered to recipients shortly.