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Re: All composites?

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  • 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 1 of 6 , Jul 1, 2009
      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 2 of 6 , Jul 1, 2009
        --- 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 3 of 6 , Jul 1, 2009
          --- 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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