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Equivalence for twin primes

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  • 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 1 of 5 , Mar 5, 2011
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      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 2 of 5 , Mar 5, 2011
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         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]
        >
        >
        >
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