- Andrey,

--- Andrey Kulsha <Andrey_601@...> wrote:> Hello Bill!

IT WON'T BE THAT KIND OF SIEVE BECAUSE THAT SIEVE MUST

>

> > Audrey,

>

> Well, I'm not offended :-)

>

> > the sieve as it stands cannot avoid division,

> unless i

> > figure out something else.

>

> I think it will be Eratosthen's sieve :-)

START AT ONE AND CONTINUOUSLY REFERENCE ONE, BUT MINE

CAN START ANYWHERE. ALSO I PLAN ON HAVING AN ALGORITHM

THAT DOESN'T REQUIRE DIVISIONS TO SIMPLIFY THE

SIEVING. AFTER THAT I PLAN TO HAVE A REVERSE ALGORITHM

TO FIND THE PRIMES DIRECTLY.>

YES, THAT'S RIGHT.

> > So the sieve can be started anywhere on the

> fragment,

> > but I thought the square of a prime would be as

> good a

> > place as any.

> >

> > Some definitions:

> >

> > N# = a number to be factored

> > R# = the rank of a number to be factored

> > F# = the factor of a composite

> > ArmA = the left half of the fragment

> > ArmB = the right half of a fragment

>

> Well, if M is a center of the fragment, then

> R=|N-M|.

>

TRUE.

> > For example, on the frag 2^6 to 2^7, we start at

> 121

> > on Arm B. Its factors are F11 and F11. its rank is

> > R25. look for all ranks that occupies a gcd=1

> position

>

> All N of a kind 6k+-1, and only them, occupies a

> gcd=1

> position. Hence, theirs (and only theirs) ranks have

> this

> form too.

>

IT MUST BE DIVISIBLE BY F11, NOT R11.

> > on arm A and which, when added to R25, will be

> > divisible by F11.

>

> Let this rank will be Rx, and number of this rank

> will be

> Nx=M-Rx.

> M+R25 is divisible by 11. R25+Rx must be divisible

> by R11.

> Hence, (M+R25)-(R25+Rx)=Nx will be too.

RIGHT.

>

RIGHT. I BOO-BOO'ED HERE. I WILL ENDEAVOR TO HAVE THE

> > That rank is R19 and the number is N77.

>

> All that you are doing there, - a trial sieving the

> numbers

> 6k+-1 divisible by 11 from the ArmA. There are

> around M/33

> such numbers, so if you take M, for example, 3*2^50,

> you

> will have more than 10^14 such numbers!

SIEVE SIEVE FOR COMPOSITES DIRECTLY WITHOUT SIEVING

FOR RANKS, AND EVENTUALLY, AS I'VE MENTIONED ABOVE, NO

SIEVING. BUT EVEN AS IT STANDS, WE DON'T NEED TO

REFERENCE "1" ALL THE TIME AND CAN SIEVE ANY FRAGMENT

NEEDED, UNLIKE ERATOSTHENE'S SIEVE.>

TRUE, BUT EVEN IF I SIEVE IN AN ERATOSTHENIAN WAY, I

> > N77/F11=F7, so

> > now look for a rank on arm B where gcd=1 which,

> when

> > added to R19 is divisible by F7 or F11.

>

> And now you have sieved:

>

> a) all the numbers 6k+-1 divisible by 11 from the

> ArmB (and

> now you have sieved them from both fragments);

>

> b) all the numbers 6k+-1 divisible by 7 from the

> ArmB (at

> the next step you will sieve such numbers from

> another

> fragment too).

>

> > There are none for F11 so stop this thread. But

> there

> > is one for F7. N119 at R23: (N119/F7=F17). So look

> for

> > a number on arm A that occupies gcd=1, whose rank

> when

> > added to R23 is divisible by F17 or F7.

> >

> > There are 2 new numbers: N85 at R11,

> (R23+R11)/F17=2

> > and N91 at R5 (R23+R5/F7=4). N91 has no number or

> rank

> > that fits the criteria on arm B, so stop this

> thread.

> >

> > N85/F17=5, so continue as above. There are 2

> numbers

> > on arm B derived from N85: N115 and N125 at ranks

> R19

> > and R29 respectively.

> >

> > There are no ranks on arm B that when added to R29

> = a

> > number divisible by 25, so stop this thread. The

> other

> > number, 115 when divided by 5=23. Look for a

> number on

> > arm B at gcd=1 whose rank when added to 19 is

> > divisible by 5. 65 and 95 fulfil this criteria at

> > ranks 31 and 1 respectively. 95's rank (rank=1)

> has no

> > number whose rank will fit the criteria, so stop

> this

> > thread. Continue with 65. There should be no

> > composites left at gcd's=1.

>

> Well, now you have sieved all the numbers 6k+-1

> divisible by

> 11,7 and 5 from both fragments. Hence, all left

> numbers

> 6k+-1 must be prime.

>

> But Eratosthen's sieve is much faster, because:

>

> a) generating a list of primes less or equal to

> sqrt(N) is

> much faster than irregular receiving them when

> dividing

> ranks (lots of them appears for large N, about

> 3*2^50);

CAN SIEVE WITHOUT REFERENCING "1" AND THEREFORE CAN

SIEVE OTHER FRAGMENTS.>

NO, YOU MUST ALWAYS START AT 1 AND SIEVE UP TO THE

> b) we can work with any fragments, not with two

> given;

DESIRED POINT.>

THE SQUARES OR POWERS IS JUST A STARTING POINT BECAUSE

> c) there are no squares or any another powers of

> primes in

> Eratosthen's sieve.

THEY ARE EASY TO FIND.>

ARE YOU REFERRING TO ERATOSTHENE'S? BECAUSE I DON'T

> > I am working on finding a pattern to the factors

> > without relying on division. Some interesting

> teasers

> > appear, but I need more numbers to look at.

>

> I've just showed you a method.

>

SEE ANY OTHER METHOD YOU'VE SHOWN.

> > Could you

BUT ANDREY, THE COMPOSITES NOT DIVISIBLE BY 2 OR 3

> > generate, in an excel-readable file, fragments

> that

> > include the gcd's and the greatest factor (not the

> > gcd), for those numbers that are at gcd=1, but are

> > composite.

>

> There will be 1/3 of all composites, namely of form

> 6k+-1,

> i.e. non-divisible by 2 or 3.

ALSO SHOW A SYMMETRY. I REALLY THINK THIS IS YOUR AND

OTHER CRITICS' MAIN AREA OF CONFUSION. I AM TRYING TO

FIND A PATTERN FOR THIS SYMMETRY WITHOUT SIEVING FOR

THEM AND THERE ARE PATTERNS, I JUST NEED TO SEE MORE

NUMBERS.

>

KIND OF RIGHT. THE FACT THAT THERE IS SYMMETRY OF

> > I've included the excel files for reference.

>

> I'm surprized that you missed that your "symmetry"

> exists

> only for divisibility by 2 and 3. All of exceptions

> from

> your symmetry are divisible by prime greater than 3,

> and you

> are "restoring" this symmetry by trial sieving of

> these

> numbers.

THESE OTHER COMPOSITES IS A PATTERN WORTHY OF

INVESTIGATION, NOT DEBUNKING BEFORE IT'S UNDERSTOOD

WHAT YOU'RE DEBUNKING.>

WHAT DOES THIS MEAN? I COULD GO ON AND ON ABOUT

> > Reimann

>

> Well, I'm not offended on "Audrey", but please don't

> distort

> a surname of great scientist!

UNQUESTIONED AUTHORITY IS DANGEROUS, ETC, ETC. I'M NOT

CLAIMING TO BE IN HIS LEAGUE OR ATTEMPTING TO BELITTLE

HIM OR HIS LEGACY.>

I DON'T KNOW WHAT THIS MEANS. BUT ALL I UNDERSTAND

> > will wait.

>

> I think he never will wait. :-)

>

ABOUT REIMANN IS THAT THERE IS A LINE WHICH ON EITHER

SIDE HAS AN EQUAL NUMBER OF PRIMES. IF THIS

UNDERSTANDING IS WRONG THEN I'M SORRY. HOWEVER, IF

EACH FRAGMENT CAN BE SHOWN TO HAVE AN EQUAL NUMBER OF

PRIMES AND POWERS OF PRIMES (PERHAPS ONLY SQUARES OF

PRIMES), THAT GOES A LONG WAY TO PROVING PRIMES ALONE

ARE BALANCED IF YOU SHOW POWERS OF PRIMES ARE

BALANCED.

PLEASE ATTACK WHEN YOU ARE READY.

BILL

=====

Refining my sieve into a simple, non-stochastic algorithm

Bill Krys

Email: billkrys@...

Toronto, Canada (currently: Beijing, China)

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http://photos.yahoo.com/ - Bill Krys wrote:

> > > Reimann

What he means is that you are misspelling the name -- it's spelled:

> >

> > Well, I'm not offended on "Audrey", but please don't

> > distort

> > a surname of great scientist!

>

> WHAT DOES THIS MEAN? I COULD GO ON AND ON ABOUT

> UNQUESTIONED AUTHORITY IS DANGEROUS, ETC, ETC. I'M NOT

> CLAIMING TO BE IN HIS LEAGUE OR ATTEMPTING TO BELITTLE

> HIM OR HIS LEGACY.

Riemann - Jack and Andrey,

Thanks for the correction.

I guess I've been in my little "mirror" world so long

that "ie" and "ei" are the same to me. Just joking.

I'm sorry to have been so dense.

Bill

--- Jack Brennen <jack@...> wrote:> Bill Krys wrote:

=====

>

> > > > Reimann

> > >

> > > Well, I'm not offended on "Audrey", but please

> don't

> > > distort

> > > a surname of great scientist!

> >

> > WHAT DOES THIS MEAN? I COULD GO ON AND ON ABOUT

> > UNQUESTIONED AUTHORITY IS DANGEROUS, ETC, ETC. I'M

> NOT

> > CLAIMING TO BE IN HIS LEAGUE OR ATTEMPTING TO

> BELITTLE

> > HIM OR HIS LEGACY.

>

> What he means is that you are misspelling the name

> -- it's spelled:

>

> Riemann

>

Refining my sieve into a simple, non-stochastic algorithm

Bill Krys

Email: billkrys@...

Toronto, Canada (currently: Beijing, China)

__________________________________________________

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