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Re: [PrimeNumbers] Equivalence for twin primes

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  • Maximilian Hasler
    { for(n=1,9,Pn=prime(n);for(k=1,n-1,Pk=prime(k);Pnk=prime(n-k); (Pn-Pnk==(n-k)*Pk) == (istwin(Pn)&istwin(Pk)) | print( Counter-example: (n,k)= ,[n,k] ,
    Message 1 of 5 , Mar 5, 2011
      { for(n=1,9,Pn=prime(n);for(k=1,n-1,Pk=prime(k);Pnk=prime(n-k);
      (Pn-Pnk==(n-k)*Pk) == (istwin(Pn)&istwin(Pk)) |
      print("Counter-example: (n,k)=",[n,k]",
      P(n)-P(n-k)-(n-k)*P(k)=",Pn"-"Pnk"-",n-k,"*"Pk))) }

      Counter-example: (n,k)=[4, 2], P(n)-P(n-k)-(n-k)*P(k)=7-3-2*3
      Counter-example: (n,k)=[5, 2], P(n)-P(n-k)-(n-k)*P(k)=11-5-3*3
      Counter-example: (n,k)=[5, 3], P(n)-P(n-k)-(n-k)*P(k)=11-3-2*5
      Counter-example: (n,k)=[5, 4], P(n)-P(n-k)-(n-k)*P(k)=11-2-1*7
      Counter-example: (n,k)=[6, 2], P(n)-P(n-k)-(n-k)*P(k)=13-7-4*3
      Counter-example: (n,k)=[6, 3], P(n)-P(n-k)-(n-k)*P(k)=13-5-3*5
      Counter-example: (n,k)=[6, 4], P(n)-P(n-k)-(n-k)*P(k)=13-3-2*7
      Counter-example: (n,k)=[7, 2], P(n)-P(n-k)-(n-k)*P(k)=17-11-5*3
      Counter-example: (n,k)=[7, 3], P(n)-P(n-k)-(n-k)*P(k)=17-7-4*5
      Counter-example: (n,k)=[7, 4], P(n)-P(n-k)-(n-k)*P(k)=17-5-3*7
      Counter-example: (n,k)=[7, 5], P(n)-P(n-k)-(n-k)*P(k)=17-3-2*11
      Counter-example: (n,k)=[7, 6], P(n)-P(n-k)-(n-k)*P(k)=17-2-1*13
      Counter-example: (n,k)=[8, 2], P(n)-P(n-k)-(n-k)*P(k)=19-13-6*3
      Counter-example: (n,k)=[8, 3], P(n)-P(n-k)-(n-k)*P(k)=19-11-5*5
      Counter-example: (n,k)=[8, 4], P(n)-P(n-k)-(n-k)*P(k)=19-7-4*7
      Counter-example: (n,k)=[8, 5], P(n)-P(n-k)-(n-k)*P(k)=19-5-3*11
      Counter-example: (n,k)=[8, 6], P(n)-P(n-k)-(n-k)*P(k)=19-3-2*13



      On Sat, Mar 5, 2011 at 6:13 AM, Sebastian Martin Ruiz <s_m_ruiz@...> wrote:
      > Hello all:
      >
      > Let n and k positive integers k<n.
      >
      > Let P(i) the ith-prime number
      >
      > We have:
      >
      > P(n)-P(n-k)-(n-k)P(k)=0   IF AND ONLY IF    P(n) and P(k) are Twin Primes
      >
      >
      > Sincerely
      >
      > Sebastián Martín Ruiz
      >
      >
      >
      >
      > [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 Links
      >
      >
      >
      >
    • Peter Kosinar
      ... I believe the Sebastian meant ... IFF P(n) and P(k) form a Twin pair , i.e. P(n) - P(k) = 2. The IF direction is trivial, of course. Peter [Non-text
      Message 2 of 5 , Mar 5, 2011
        > { for(n=1,9,Pn=prime(n);for(k=1,n-1,Pk=prime(k);Pnk=prime(n-k);
        > (Pn-Pnk==(n-k)*Pk) == (istwin(Pn)&istwin(Pk)) |
        > print("Counter-example: (n,k)=",[n,k]",
        > P(n)-P(n-k)-(n-k)*P(k)=",Pn"-"Pnk"-",n-k,"*"Pk))) }
        >
        > [lots of counterexamples]

        > > P(n)-P(n-k)-(n-k)P(k)=0   IF AND ONLY IF    P(n) and P(k) are Twin Primes

        I believe the Sebastian meant "... IFF P(n) and P(k) form a Twin pair",
        i.e. P(n) - P(k) = 2. The "IF" direction is trivial, of course.

        Peter

        [Non-text portions of this message have been removed]
      • Kermit Rose
        On 3/5/2011 8:04 AM, Sebastian Martin Ruiz s_m_ruiz@yahoo.es s_m_ruiz ... It is trivial that if p(n) , p(n-k), and p(k) are distinct primes, and n k, then
        Message 3 of 5 , Mar 5, 2011
          On 3/5/2011 8:04 AM, "Sebastian Martin Ruiz" s_m_ruiz@... s_m_ruiz
          wrote:
          > ________________________________________________________________________
          > 1. Equivalence for twin primes
          > Posted by: "Sebastian Martin Ruiz" s_m_ruiz@... s_m_ruiz
          > Date: Sat Mar 5, 2011 2:13 am ((PST))
          >
          > Hello all:
          >
          > Let n and k positive integers k<n.
          >
          > Let P(i) the ith-prime number
          >
          > We have:
          >
          > P(n)-P(n-k)-(n-k)P(k)=0 IF AND ONLY IF P(n) and P(k) are Twin Primes
          >
          >
          > Sincerely
          >
          > Sebastián Martín Ruiz
          >
          >
          >


          It is trivial that

          if p(n) , p(n-k), and p(k) are distinct primes,
          and n>k,
          then
          P(n)-p(n-k) -(n-k)P(k)=0

          if and only if p(n-k) = 2, and n-k = 1.




          In the case where p(n) and p(k) belong to the same set of twin primes,
          we would have

          example:
          p(n) = 7, p(k) = 5

          n = 4, k = 3

          p(n-k) = p(1) = 2

          P(n)-2 -(n-k)P(k)=0

          7 - 2 - 1*5 = 5 - 5 = 0

          In general , if p(n) and p(k) belong to the same set of twin primes,
          and n > k,

          it is trivial that

          P(n)-P(n-k)-(n-k)P(k)
          = p(n) - p( n - [n-1]) - ([n-(n-1)] )p(n-1)
          = p(n) - p(1) - 1 * (p(n-1))
          = p(n) - 2 - p(n-1)
          = (p(n) - p(n-1)) - 2
          = 2 - 2 = 0


          If
          P(n)-P(n-k)-(n-k)P(k) = 0
          p(n) - p(n-k) = (n-k) p(k)

          n = 2
          what are permitted values of k?
          k = 1?
          3 - p(1) = (2-1)*p(1) ?
          3 - 2 = 1 * 2 ?
          1 = 2 ?
          no

          n = 3
          p(3) - p(3-k) - (n-k) p(k) = 0
          p(3) - p(3-k) = (n-k) p(k)
          5 - p(3-k) = (n-k)

          5 = p(3-k) + (3-k)
          2 = p(3-k) - k

          2 + k = p(3-k)

          k is odd



          In the case where p(n) and p(k) belong to different
          sets of twin primes,

          example:

          p(n) = 13
          p(k) = 7

          n = 6
          k = 4
          n-k = 2

          P(n)-P(n-k)-(n-k)P(k)
          =13 - 3 - 2*7
          = -4
          is not zero.
        • Sebastian Martin Ruiz
           P(n)-P(n-k)-(n-k)P(k)=0   IIF    P(n) and P(k) are a Twin Primes pair IF P(n) and P(k) are a Twin Primes pair Then k=n-1 P(n-k)=P(1)=2 
          Message 4 of 5 , Mar 5, 2011
             P(n)-P(n-k)-(n-k)P(k)=0   IIF    P(n) and P(k) are a Twin Primes pair

            IF P(n) and P(k) are a Twin Primes pair

            Then k=n-1 P(n-k)=P(1)=2  P(n)-2-(n-(n-1))P(n-1)=0

            Then P(n)-P(n-1)=2

            On the other hand If

            P(n)-P(n-k)-(n-k)P(k)=0  then

             P(n) and P(k) are a Twin Primes pair

             is more dificult but i think it is also true.


            P(n)-P(n-k)-(n-k)*P(k)=7-3-2*3=/=0
            Counter-example: (n,k)=[4, 2], P(n)-P(n-k)-(n-k)*P(k)=7-3-2*3
            then your contraexample is not true.





             



            ________________________________
            De: Maximilian Hasler <maximilian.hasler@...>
            Para: Sebastian Martin Ruiz <s_m_ruiz@...>
            CC: primenumbers@yahoogroups.com
            Enviado: sáb,5 marzo, 2011 14:12
            Asunto: Re: [PrimeNumbers] Equivalence for twin primes

             
            { for(n=1,9,Pn=prime(n);for(k=1,n-1,Pk=prime(k);Pnk=prime(n-k);
            (Pn-Pnk==(n-k)*Pk) == (istwin(Pn)&istwin(Pk)) |
            print("Counter-example: (n,k)=",[n,k]",
            P(n)-P(n-k)-(n-k)*P(k)=",Pn"-"Pnk"-",n-k,"*"Pk))) }

            Counter-example: (n,k)=[4, 2], P(n)-P(n-k)-(n-k)*P(k)=7-3-2*3
            Counter-example: (n,k)=[5, 2], P(n)-P(n-k)-(n-k)*P(k)=11-5-3*3
            Counter-example: (n,k)=[5, 3], P(n)-P(n-k)-(n-k)*P(k)=11-3-2*5
            Counter-example: (n,k)=[5, 4], P(n)-P(n-k)-(n-k)*P(k)=11-2-1*7
            Counter-example: (n,k)=[6, 2], P(n)-P(n-k)-(n-k)*P(k)=13-7-4*3
            Counter-example: (n,k)=[6, 3], P(n)-P(n-k)-(n-k)*P(k)=13-5-3*5
            Counter-example: (n,k)=[6, 4], P(n)-P(n-k)-(n-k)*P(k)=13-3-2*7
            Counter-example: (n,k)=[7, 2], P(n)-P(n-k)-(n-k)*P(k)=17-11-5*3
            Counter-example: (n,k)=[7, 3], P(n)-P(n-k)-(n-k)*P(k)=17-7-4*5
            Counter-example: (n,k)=[7, 4], P(n)-P(n-k)-(n-k)*P(k)=17-5-3*7
            Counter-example: (n,k)=[7, 5], P(n)-P(n-k)-(n-k)*P(k)=17-3-2*11
            Counter-example: (n,k)=[7, 6], P(n)-P(n-k)-(n-k)*P(k)=17-2-1*13
            Counter-example: (n,k)=[8, 2], P(n)-P(n-k)-(n-k)*P(k)=19-13-6*3
            Counter-example: (n,k)=[8, 3], P(n)-P(n-k)-(n-k)*P(k)=19-11-5*5
            Counter-example: (n,k)=[8, 4], P(n)-P(n-k)-(n-k)*P(k)=19-7-4*7
            Counter-example: (n,k)=[8, 5], P(n)-P(n-k)-(n-k)*P(k)=19-5-3*11
            Counter-example: (n,k)=[8, 6], P(n)-P(n-k)-(n-k)*P(k)=19-3-2*13

            On Sat, Mar 5, 2011 at 6:13 AM, Sebastian Martin Ruiz <s_m_ruiz@...> wrote:
            > Hello all:
            >
            > Let n and k positive integers k<n.
            >
            > Let P(i) the ith-prime number
            >
            > We have:
            >
            > P(n)-P(n-k)-(n-k)P(k)=0   IF AND ONLY IF    P(n) and P(k) are Twin Primes
            >
            >
            > Sincerely
            >
            > Sebastián Martín Ruiz
            >
            >
            >
            >
            > [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 Links
            >
            >
            >
            >






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