- Could somebody please give me some online references to information

about the benefits of sieving.

E.g. Suppose a number K is a random integer then it has a 1 in 1/ln K

chance of being prime but if K is not divisible by 2 or 3 then the

chance of it being prime is 2/(3lnK)

etc

That type of thing.

Thanks in advance

Gary - As usual a typo on my formula!

I know it was just an example but if K is not divisible by 2 or 3 then the

chance of it being prime is 3/(lnK)

Gary

garychaffey2 <garychaffey2@...> wrote:

Could somebody please give me some online references to information

about the benefits of sieving.

E.g. Suppose a number K is a random integer then it has a 1 in 1/ln K

chance of being prime but if K is not divisible by 2 or 3 then the

chance of it being prime is 2/(3lnK)

etc

That type of thing.

Thanks in advance

Gary

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[Non-text portions of this message have been removed] - hi,

The following observation might help.

From 1-10, the numbers not divisible by 2,3 are 1,5,7.

I.e 3 in number.

Between 11-20,the numbers not divisible by 2,3 are

11,13,17,19.

I.e 4 in number.

Between 21-30,the numbers not divisible by 2,3 are

23,25,29.

I.e 3 in number.

So, between any 10 consecutive numbers of the form xx1

and xx0.

Either of the 3 cases occur.

3 | xx1 and so 3| xx1,xx4,xx7.(Case 1)

3 | xx1+1 and so 3| xx2,xx5,xx8.(Case 2)

3 | xx1+2 and so 3| xx3,xx6,xx9.(Case 3)

The favourable condition in Case 1 is the occurance of

xx1 and xx7 because 2|xx4 anyway. So, the number of

favourable case is 2.

The favourable condition in Case 2 is the occurance of

xx5 because 2|xx2 and 2|xx8 anyway. So, the number of

favourable case is 1.

The favourable condition in Case 3 is the occurance of

xx3 and xx9 because 2|xx6 anyway. So, the number of

favourable case is 2.

Between any 10 consecutive numbers xx1 and xx0, all

the time 3 divides (2 of the numbers) or (one of the

numbers) in accordance to the favourable cases in Case

1,2 and 3.

Since 5 of the numbers between 10 consecutive numbers

xx1 and xx0 is divisibe by 2, the total of numbers

divisible by 2 and 3 is

Case 1: 5+2=7

Case 2: 5+1=6

Case 3: 5+2=7.

Therefore 6 out of every 10 or 7 out of every 10

consecutive numbers is divisible by 2 and 3, which are

respectively 70% and 60% respectively. So, the total

of numbers divisible by 2 and 3 between 1 and 100 is

7+6+7+7+6+7+7+6+7+7=67 out of 100, which is 67%.

Between 100 and 200.

6+7+7+6+7+7+6+7+7+6= 66%.

Trying out a few times and finding the average we see

that the percentage of numbers divisible by 2 and 3 is

about 66.67%.

According to Dr.Schneier's book,testing an odd number

is not divisible by 3,5,7 eliminates 54% of odd

numbers under test. It is found in general that the

fraction of odd candidates which is not a multiple of

any prime less than n is 1.12/ln n.

Hope this helps.

Sarad.

--- Gary Chaffey <garychaffey2@...> wrote:

> As usual a typo on my formula!

____________________________________________________

> I know it was just an example but if K is not

> divisible by 2 or 3 then the

> chance of it being prime is 3/(lnK)

> Gary

> garychaffey2 <garychaffey2@...> wrote:

> Could somebody please give me some online references

> to information

> about the benefits of sieving.

> E.g. Suppose a number K is a random integer then it

> has a 1 in 1/ln K

> chance of being prime but if K is not divisible by 2

> or 3 then the

> chance of it being prime is 2/(3lnK)

> etc

> That type of thing.

> Thanks in advance

> Gary

Sell on Yahoo! Auctions no fees. Bid on great items.

http://auctions.yahoo.com/ - From: "garychaffey2" <garychaffey2@...>
> Subject: Benefits of sieving. References please

Mertens' Theorem.

>

> Could somebody please give me some online references to information

> about the benefits of sieving.

Phil

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