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Re: [PrimeNumbers] Euler Totient Result Corrected

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  • Sebastian Martin
    Forget it, I go away to the beach, I need a few vacations. P.D. There are any bound? ... De: Sebastian Martin Para: Lista de Matemáticas
    Message 1 of 4 , Jul 8, 2008
      Forget it, I go away to the beach, I need a few vacations.
      P.D. There are any bound?


      ----- Mensaje original ----
      De: Sebastian Martin <sebi_sebi@...>
      Para: Lista de Matemáticas <matracas@...>; lista de primos <primenumbers@yahoogroups.com>
      Enviado: martes, 8 de julio, 2008 17:25:44
      Asunto: [PrimeNumbers] Euler Totient Result Corrected




      I have tested for n=1 to 200000 and I have new  possible bounds:

       
      Let  p(n-1), p(n), p(n+1) three consecutive prime numbers.
       
      We have:
       
      1/E <  Phi(p(n+1)+pn) /Phi(pn+p( n-1)) < E
       
       
      E=2.71828...
       
      Phi is the Euler Totient function.
       
      Additional Questions :
       
      1)      Can you caracterize the primes that take the integers values 1 and 2?

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    • Jens Kruse Andersen
      ... A good strategy can find better extremes than brute force. The prime sums p(n+1)+p(n) and p(n)+p(n-1) are both even (assuming n 2). Considering the
      Message 2 of 4 , Jul 8, 2008
        Sebastian Martin wrote:
        > I have tested for n=1 to 200000 and I have new possible bounds:
        > 1/E < Phi(p(n+1)+pn) /Phi(pn+p( n-1)) < E
        > E=2.71828...

        and later added:
        > Forget it, I go away to the beach, I need a few vacations.
        > P.D. There are any bound?

        A good strategy can find better extremes than brute force.
        The prime sums p(n+1)+p(n) and p(n)+p(n-1) are both even (assuming n>2).
        Considering the behaviour of Phi, we want both sums to have known
        factorization, and for an extreme we want one of them to have lots of
        small odd prime factors, and the other to have none.

        Let p = 3970433188*2411#+1, one of millions of prp's found by Paul Underwood
        and Markus Frind in 2003 during an AP8 search.
        p-2, p, p+2466 are consecutive primes.

        p+2466 + p = 2*3970433188*2411#+2468 = 2^2*617*117541*125887*q,
        where q is a 1023-digit prime.
        p + p-2 = 2^3*992608297*2411#.

        Phi(p+2466 + p)/Phi(p + p-2) = 6.94684...
        Proof:

        ? p=3970433188*prod(i=1,primepi(2411),prime(i))+1;
        ? p1=nextprime(p-2); p2=nextprime(p1+1); p3=nextprime(p2+1);
        ? \\ p1, p2, p3 are now consecutive prp's
        ? print("p+"p1-p", p+"p2-p", p+"p3-p": ",\
        eulerphi(p3+p2)/eulerphi(p2+p1)*1.0)
        p+-2, p+0, p+2466: 6.94684253993807043

        Computing p3 may take some time. The impatient can replace by p3=p+2466.
        The factorizations of the prime sums are easy to handle for PARI/GP's
        eulerphi.

        PrimeForm found all prp's and proved p and p-2.
        Marcel Martin's Primo proved p+2466 and q.

        --
        Jens Kruse Andersen
      • Jens Kruse Andersen
        ... Below is a similar case with the low value 0.14394..., close to 1/6.94684. ? p=5280767204*prod(i=1,primepi(2411),prime(i))-1; ? p1=nextprime(p-2466);
        Message 3 of 4 , Jul 8, 2008
          I wrote:
          > Let p = 3970433188*2411#+1
          ...
          > Phi(p+2466 + p)/Phi(p + p-2) = 6.94684...

          Below is a similar case with the low value 0.14394..., close to 1/6.94684.

          ? p=5280767204*prod(i=1,primepi(2411),prime(i))-1;
          ? p1=nextprime(p-2466); p2=nextprime(p1+1); p3=nextprime(p2+1);
          ? \\ p1, p2, p3 are now consecutive prp's
          ? print("p+"p1-p", p+"p2-p", p+"p3-p": ",\
          eulerphi(p3+p2)/eulerphi(p2+p1)*1.0)
          p+-2466, p+0, p+2: 0.143947882881142336

          Here p = 5280767204*2411#-1.
          p-2466, p, p+2 are consecutive primes with p+2 found by
          Paul Underwood and Markus Frind in 2003.

          p+2 + p = 2^3*19^2*3657041*2411#.
          p + p-2466 = 2^3*167*q, where q is a 1034-digit prime.
          PrimeForm proved p and p+2. Primo proved p-2466 and q.

          --
          Jens Kruse Andersen
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