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Re: [PrimeNumbers] Euler Totient Result Corrected

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  • 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 1 of 4 , Jul 8 11:50 AM
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      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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