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Re: All composites?

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  • 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 1 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 2 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 3 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 4 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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