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All composites?

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  • Sebastian Martin Ruiz
    Hello all:   32(n!)**8+1 is composite for all n positive integer ??   Sincerely   Sebastian [Non-text portions of this message have been removed]
    Message 1 of 6 , Jul 1, 2009
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      Hello all:
       
      32(n!)**8+1 is composite for all n positive integer ??
       
      Sincerely
       
      Sebastian




      [Non-text portions of this message have been removed]
    • Maximilian Hasler
      ... For which n 1 is 4*n!^8 + 1 prime ?? Maximilian
      Message 2 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 3 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 4 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 5 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 6 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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