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

=====

Refining my sieve into a simple, non-stochastic algorithm

Bill Krys

Email: billkrys@...

Toronto, Canada (currently: Beijing, China)

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