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Twin Prime Conjecture

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• Hi everybody What about this little recipe using the Erathostenes sieve to proof the conjecture. 1.) Take all integers 2.) Take first prime P(1)=2, sieve with
Message 1 of 3 , Sep 5 8:23 PM
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Hi everybody

What about this little recipe using the Erathostenes sieve to proof
the conjecture.

1.) Take all integers

2.) Take first prime P(1)=2, sieve with multiples of 2.

3.) Take second Prime P(2)=3, sieve with multiples of 3
This creates an ifinite number N of twin pairs with a spacing of 6.
The position of the twins is n*(2*3) or n*(P1(1)*P(2))

4.) Take P(3)=5, sieve with multiples of 5.
This will eliminate some twin primes but never those with the
position n*(2*3*5)=n*30=n*(P(1)*P(2)*P(3)), because at these
numbers all prime sieves are 'synchronized'.

2 comes from 28 hops onto 30 and continues at 32
3 comes from 27 hops onto 30 and continues at 33
5 comes from 25 hops onto 30 and continues at 35

There is an infinite number of twin pairs with the spacing 30 left.

5.) Take P(4)=7, sieve with multiples of 7.
This will eliminate some twin primes but never those with the
position n*(2*3*5*7)=n*210, because at these numbers all prime
sieves are 'synchronized'. There is an infinite number of twin pairs
with the spacing 210 left.

6.) Continue sieving with primes into infinity. Any sieve P(m) will
have an infinite number of twin primes at the positions
n*(P(1)*P(2)*P(3)....*P(m), therefore the number of twin primes is
infinite.

Regards

Uli
• Hi Uli, The problem with this is that the infinite pairs are not necessarally prime. For instance the wel know Phil Carmody pseudoprime 91= 7*13 which belongs
Message 2 of 3 , Sep 5 8:40 PM
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Hi Uli,
The problem with this is that the infinite pairs are not necessarally
prime. For instance the wel know Phil Carmody pseudoprime 91= 7*13
which belongs to (89,91) one of the infinite pairs left after sieving
ou 2,3 and 5.
So I'm afraid all you can say about your pairs is they are coprime to
the primes you have already sieved out. You can make no conclusion as
to their divisibility by larger primes.
Cheers
Ken
--- In primenumbers@yahoogroups.com, uli@i... wrote:
> Hi everybody
>
> What about this little recipe using the Erathostenes sieve to proof
> the conjecture.
>
> 1.) Take all integers
>
> 2.) Take first prime P(1)=2, sieve with multiples of 2.
>
> 3.) Take second Prime P(2)=3, sieve with multiples of 3
> This creates an ifinite number N of twin pairs with a spacing of 6.
> The position of the twins is n*(2*3) or n*(P1(1)*P(2))
>
> 4.) Take P(3)=5, sieve with multiples of 5.
> This will eliminate some twin primes but never those with the
> position n*(2*3*5)=n*30=n*(P(1)*P(2)*P(3)), because at these
> numbers all prime sieves are 'synchronized'.
>
> 2 comes from 28 hops onto 30 and continues at 32
> 3 comes from 27 hops onto 30 and continues at 33
> 5 comes from 25 hops onto 30 and continues at 35
>
> There is an infinite number of twin pairs with the spacing 30 left.
>
> 5.) Take P(4)=7, sieve with multiples of 7.
> This will eliminate some twin primes but never those with the
> position n*(2*3*5*7)=n*210, because at these numbers all prime
> sieves are 'synchronized'. There is an infinite number of twin
pairs
> with the spacing 210 left.
>
> 6.) Continue sieving with primes into infinity. Any sieve P(m) will
> have an infinite number of twin primes at the positions
> n*(P(1)*P(2)*P(3)....*P(m), therefore the number of twin primes is
> infinite.
>
> Regards
>
> Uli
• I have been working on twin prime conjecture with this perspective -- a theorum that proves that one can find atleast one twin prime in the range of numbers
Message 3 of 3 , Dec 3, 2003
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I have been working on twin prime conjecture with this perspective --
a theorum that proves that one can find atleast one twin prime in
the range of numbers between the squares of a twin prime.
I am excited with this since a proof would not only prove the twin
prime conjecture but establish a more easy way for us to sieve a
range of numbers for higher primes/twin primes.

Batta
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