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Primes and Squares1- clarifications

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  • billoscarson
    Hi Richard and all: Thanks for your question. What I mean by prime divisors is definitely any and all divisors
    Message 1 of 1 , Dec 31, 2003
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      Hi Richard and all:
      Thanks for your question.
      What I mean by prime divisors is definitely any and all divisors
      < sqrt x, so the 15 and 1 would qualify. I didn't mention them
      because I wanted to give an example, not necessarily a comprehensive
      example, but I should have included all of them. If you look at the
      definition of LE you will see that 1 is eliminated from the list of
      primary divisors used in the analysis.
      Let me walk you through the part of the solution that has to do
      with primary divisors. Most of this information is in the first
      example (N=12), but perhaps a bit criptically.

      Given N =12, how many primes, if any, are between 12^2 and 13^2?

      First we calculate the number of possible primes (KE) between 144
      and 169. We know that the only possible primes are the odd numbers,
      being 145,147,149,151,153,155,157,159,161,163,165,and 167. These
      numbers define the KE set. Note that the set has 12 values, so KE
      =12 (KE itself is not a set, just the number of values in the KE
      set). We could use some intelligence and say that we know some of
      thes numbers are not prime (like 155 and 165, each divisible by 5)
      but it will be difficult to keep track of all that, so we will just
      leave them in as possible candidates for this analysis.
      Next we must calculate the number of primary divisors that may be
      legitimately used to disprove that a number in the KE set is not
      prime, i.e. not divisible by anything other than itself or 1. We
      first calculate the primary divisors,i.e. odd numbers less than N
      (for N even), and in this example they are 1,3,5,7,9,11. Then we
      calculate how many of thes values we can use as divisors to negate
      primality of numbers in the KE set. The result I defined as LE, and
      in this example are 3,5,7,9, and 11, all the primary divisor values
      except 1.
      Therefore, LE = 5 (3,5,7,9,11 constitute 5 values). Note that the
      definition of LE in my post does exclude 1 so everything is OK so
      far.
      This takes us to a point where we have the number of possible
      primes defined, as well as the values that can be used as their
      divisors. Rather than explaining further, I'll wait for questions so
      I can address specifically what is not clear. Don't forget to look
      at the examples (especially N=12) to aid in following my thoughts.

      PS I don't think I have yet mastered responding to posts. Do you
      just type in the original title with a RE: at the start?

      Regards, Bill
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