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Re: All composites?

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  • Maximilian Hasler
    ... For which n 1 is 4*n!^8 + 1 prime ?? Maximilian
    Message 1 of 6 , Jul 1, 2009
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      --- In primenumbers@yahoogroups.com, Sebastian Martin Ruiz <s_m_ruiz@...> wrote:
      >
      > Hello all:
      > �
      > 32(n!)**8+1 is composite for all n�positive integer�??

      For which n>1 is 4*n!^8 + 1 prime ??

      Maximilian
    • David Broadhurst
      ... Sebastian Martin Ruiz wrote:   ... On the contrary, the PNT makes it probable (but not yet provable) that there is an *infinite* number of
      Message 2 of 6 , Jul 1, 2009
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        --- In primenumbers@yahoogroups.com,
        Sebastian Martin Ruiz <s_m_ruiz@...> wrote:
         
        > 32(n!)**8+1 is composite for all n positive integer ??

        On the contrary, the PNT makes it probable (but not yet provable)
        that there is an *infinite* number of primes of this form.

        The probability that (2*(n!))^8+1 is prime
        is (heuristically) asymptotic to exp(Euler)/(8*n).
        [If I got the constant wrong, Chris will correct me.]
        The integral of 1/n diverges.

        Moral: just because you didn't find a small one,
        doesn't mean there isn't an infinity of larger ones.

        David
      • David Broadhurst
        correcting a typo, which does not seem to me to affect the argument: The probability that ((2*(n!))^8)/2+1 is prime is (heuristically) asymptotic to
        Message 3 of 6 , Jul 1, 2009
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          correcting a typo, which does not seem to me to
          affect the argument:

          The probability that ((2*(n!))^8)/2+1 is prime
          is (heuristically) asymptotic to exp(Euler)/(8*n).
          [If I got the constant wrong, Chris will correct me.]
          The integral of 1/n diverges.

          David
        • David Broadhurst
          ... For none. Proof: Set x = (n!)^2 in the identity 4*x^4 + 1 = (2*x^2 + 1)^2 - (2*x)^2 David (per proxy Léon François Antoine)
          Message 4 of 6 , Jul 1, 2009
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            --- In primenumbers@yahoogroups.com,
            "Maximilian Hasler" <maximilian.hasler@...> wrote:

            > For which n>1 is 4*n!^8 + 1 prime ??

            For none. Proof: Set x = (n!)^2 in the identity
            4*x^4 + 1 = (2*x^2 + 1)^2 - (2*x)^2

            David (per proxy Léon François Antoine)
          • Maximilian Hasler
            ... Right... of course! Bad example - that s not what I meant to say. Maybe, Is 8*n!^8+1 composite for all n 4 ? Probably not. Actually I don t mind at all.
            Message 5 of 6 , Jul 1, 2009
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              --- In primenumbers@yahoogroups.com, "David Broadhurst" <d.broadhurst@...> wrote:
              >
              > > For which n>1 is 4*n!^8 + 1 prime ??
              >
              > For none. Proof: Set x = (n!)^2 in the identity
              > 4*x^4 + 1 = (2*x^2 + 1)^2 - (2*x)^2

              Right... of course! Bad example - that's not what I meant to say.
              Maybe,

              Is 8*n!^8+1 composite for all n>4 ?

              Probably not. Actually I don't mind at all. It was just to express the same "Moral" given elsewhere more explicitely. But I should have chosen my example more carefully...
              At least this shows that the "32" is in some sense simply the 3rd possibiliy, after 2 and 8. For 2^7 there happen to be 3 small primes; for 2^9 and 2^11 only two. Thus, whenever the first prime is found for 2^5, then the same post can be made replacing 32 with 2^13.)

              Maximilian
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