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Density of primes between square numbers

  1. May 14, 2009 #1
    Is the density of primes considerably greater nearer the geometric average of two consecutive square numbers?

    [Think of deconstructing a square of integral area n2 into composite rectangles of diverging (n-1)(n+1), (n-2)(n+2), (n-3)(n+3)... .]

    This reasoning may work to a lesser yet significant degree with powers greater than two.
     
  2. jcsd
  3. May 14, 2009 #2

    Hurkyl

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    I'm boggled as to why you would say "the geometric average of two consecutive square numbers" rather than "the product of two consecutive numbers".



    Unless you stick to very small numbers, I can see no reason why the answer to your question would be "yes". What would suggest it?



    I can't make any sense out of the rest of your post. I mean that literally: not a judgement of right or wrong, but as a judgement of whether or not I can extract meaning from that sequence of words and symbols.
     
  4. May 14, 2009 #3
    Thank you for being understanding, Hurkyl. Mine is a half-baked idea.
     
  5. May 15, 2009 #4

    CRGreathouse

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    How near?

    I'm almost sure this isn't true unless taken to the trivial extreme that n(n+1) is itself composite for n > 1.
     
  6. May 15, 2009 #5
    CR,

    I appreciate your input. I should look before I lemma.
     
  7. May 15, 2009 #6

    CRGreathouse

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    Loren: I didn't see your other post (#3) before I posted -- sorry. I wouldn't have posted otherwise.
     
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