15Bill Krys' sieve: response to Andrey and others

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• 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.

Bill

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Refining my sieve into a simple, non-stochastic algorithm

Bill Krys
Email: billkrys@...