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From: Pubkeybreaker on 2 Dec 2009 10:59
On Dec 2, 8:33 am, eestath <stathopoulo...(a)gmail.com> wrote:
> How can anyone can model Erratosthenes sieve??

It's been done.

Look up the following paper by DeBruijn:

On The Number of Uncancelled Elements in the Sieve
of Eratosthenes.

It uses the Buchstab function.

From: jbriggs444 on 2 Dec 2009 15:13
On Dec 2, 9:20 am, eestath <stathopoulo...(a)gmail.com> wrote:
> > then you exclude one third of them but half of them have factor 2 so
> > you actually exclude 1/3 - 1/(2*3)
> > then you exclude one fith of them but 1 in 2 has as its factor 2 and 1
> > in 3 have factor 3 so you exlude
>
> 1/5 - 1/(2*5)-1/(3*5)
>
> Is the correct answer ...and many simply failed to understand!!!!!!!

No. That's not correct. Please check your work. How many integers
from zero to 29 are multiples of 5 but not multiples of 2 or 3.

Your prediction is 1.
I can find 2.

Is it just barely possible that you have over-excluded multiples of 30?
From: eestath on 2 Dec 2009 22:56
Aruro has the correcet formula:

On Dec 1, 11:07 pm, eestath <stathopoulo...(a)gmail.com> wrote:

> try to understand what i' am saying not what you are taught!

> If you model Eratosthenes sieve:

> first you exclude 1/2 of the numbers: 2,4,6,8,10,...then
> you exclude 1/3 of them but 1/2 of them are allready excluded, so you
> actually exclude 1/(2*3) then
> you exclude 1/5 of them but 1/(2*3) of them are allready excluded so
> you actually exclude 1/(2*3*5)
> ....

Your first two counts are correct; the third and all subsequent counts
are not correct, because you are failing the inclusion-exclusion
principle.

Try it with the numbers from 1 through 60.

Crossing out all the multiples of 2 does indeed get rid of (1/2) of
them (namely, thirty).

Then crossing out the multiples of 3 accounts for (60/3) = 20;
however, those that were multiples of 2 as well as of 3 were already
crossed; so (60/6)=10 have been counted twice. So we are only crossing
(1/3)-(1/2)(1/3) = (1/3)-(1/6) = (1/6) of the numbers, that is, 10;
they are 3, 9, 15, 21, 27, 33, 39, 45, 51, and 57. This still agrees
with what you have.

Then you say: let's cross out the multiples of 5. Fine: there are
(60/5) = 12 such multiples.

You claim the number we will cross but haven't crossed yet is
(60/2*3*5)=2. What you are doing is saying: we already crossed out
the multiples of 2 among them (that's (1/5)(1/2) = (1/10), so 6 of
them), and we already crossed out the multiples of 3 (that's (1/5)(/
13) = (1/15), so 4 of them). That means 12-(6+4)=2 to cross.

However, we actually cross 4: 5, 25, and 35.

Where did you go wrong? Well, when you counted the multiples of 2, you
were also counting multiples of 6; and when you counted multiples of
3, you *also* counted multiples of 6. So you took out the multiples of
6 twice. So you need to add them back again. That means adding back
(1/5)(1/6) = (1/30) of the numbers; since (60/30)=2, that means that
we really have 12 - (6+4) + 2 = 4 numbers to cross. And indeed, that
is what happens.

So the correct fraction is not (1/2*3*5); the correct fraction is
(1/5)-(1/2*5)-(1/3*5)+(1/6*5) = 1/15, not 1/30.

When you get to 7, you will have (1/7) - (1/2*7) - (1/3*7) - (1/5*7) +
(1/2*3*7) + (1/2*5*7) + (1/3*5*7) - (1/2*3*5*7). Or, easier,

(1/7)*(1-(1/2))*(1-(1/3))*(1-(1/5)).

Etc.

PS: Do not send me private e-mails in response.

--
Arturo Magidin


From: eestath on 2 Dec 2009 23:02
WHO SAID THAT I WAS CORRECT I MADE A SUGESTION,
AND I WAITE FOR YOUR RESPONSE:)

From: eestath on 2 Dec 2009 23:35
WHO SAID THAT I WAS CORRECT I MADE A SUGESTION,
AND I WAIT FOR YOUR RESPONSE:)
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