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15Bill Krys' sieve: response to Andrey and others

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  • Bill Krys
    Dec 31, 2000
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      Andrey and all,

      I concede that Eratosthene's sieve can be used for any
      fragment, but I still assert that we must indirectly
      reference "1" to generate the primes for sieving,
      however, mine (or should I say my version of
      Eratosthene�s?) would require the same.

      However, doesn't Eratosthene's sieve require a program
      to at least look at numbers that have already been
      sieved? My version, I hope will not have to look to
      see if a number has been tossed out as composite. I�m
      looking for a way to go directly from number to number
      and toss them out only once. Would this make sieving
      significantly quicker?

      You are telling me that after looking at the
      fragments, or based on reason, that there is no
      symmetry for numbers other than gcd=2 or 3 and so the
      gcd=1 has no symmetry. I hope you are wrong. I will
      work on a proof that each composite (less an
      identifiable group of numbers, say, squares or other
      powers of primes) can be matched in a programmable way
      to one and only one composite on the other fragment
      which shares a factor.

      The symmetry to which I refer is not obvious. The gcd
      between composites for numbers divisible by at least 2
      or 3 is, yes, obvious. That gets rid of them for the
      purposes of developing my sieve (and as asserted
      below, a proof of Riemann). The other composites (less
      the identifiable group) are displaced along the same
      fragment but still share at least one common factor on
      the opposite arm. If this isn�t true, while I�m
      working on a proof that it is true, show me the
      smallest fragment where this isn�t true or show me a
      proof it isn�t true.

      Since I�ll try to prove that each gcd not = 1 (less
      the identifiable group, say squares or even powers of
      primes) is symmetrical for any and all fragments, then
      the only other numbers left on any given fragment are
      3*2^(n-1), the primes and the identifiable group. The
      numbers 3*2^(n-1) are not symmetrical (or obviously
      symmetrical, depending on how one defines symmetry),
      but they make up the line about which there is
      symmetry, so I needn�t worry about them. Finally, I
      must show that the identifiable group is symmetrical
      over all fragments, and I�ll have proved primes must
      be symmetrical about the same line by default.

      Finally, for now, you said that such elementary math
      wouldn�t prove such a complicated theory. Maybe you
      are right, but I�m working on the assumption that
      complicated math is just quicker simple math.


      Refining my sieve into a simple, non-stochastic algorithm

      Bill Krys
      Email: billkrys@...
      Toronto, Canada (currently: Beijing, China)

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