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From: master1729 on 12 May 2010 07:10
i wrote :

> anonymous.rubbertube wrote :
>
> > On May 6, 10:14 am, master1729
> <tommy1...(a)gmail.com>
> > wrote:
> > > no response.
> > >
> > > surely you must understand parts of it ??
> >
> >
> >
> > I just looked over your "proof." It looks as
> though
> > you have
> > overlooked that any number which is the product of
> > more than one prime
> > will be sieved more than once. This means your
> > formula for the maximum
> > number of primes in the interval (a,b), which you
> > obtain by taking b-a
> > and subtracting 1 for each number sieved from the
> > interval (a,b) for
> > each prime less than sqrt(b-a), this formula is
> > wrong--there can be
> > more than that many primes in the interval (a,b).
>
> my proof gave C or C - 1.
>
> C does not mean floor(x/p).
>
> C and C - 1 are to illustrate that a prime sieves an
> interval of certain length a potential amount of
> times which differs by 1 , C or C - 1.
>
> we then sum the worst case difference of 1 and get
> pi(sqrt(c))
>
> if you think this affects the equation strongly you
> are wrong.
>
> we get another correction which cancels itself out.
>
> let me clarify with an example.
>
> interval length 100.
>
> 3 sieves it 33 or 34 times. ( "or 1)" )
> 5 sieves it 20 times.
>
> multiples of 15 are sieved by 3 and 5.
>
> but the number of times a multiple of 15 or
> equivalently a sieve of both 3 and 5 occurs is not
> independant or random !.
>
> 15 sieves it 6 or 7 times. ( "or 2)" )
>
> note that "or 2" and "or 1" depend on eachother
> rather than being independant or random.
>
> and therefore it is already accounted for by "or 1" ,
> which i used in my proof.
>
>
> regards
>
> tommy1729

ok.

this explains why an interval of size n sieved by numbers smaller than n never violates H-L 2nd.

and this has been quite well explained.

but lets continue with the example , interval of lenght 100.

what if it contained a multiple of 3*5*7 = 105 ?
( thus 3,5,7 sieve less -> more primes ? )

then note that the matter depends on the density of primes.

that might be counterintuitive , since h-l seems to hold better when there are less primes.

however since 105 + 100/2 = 155 , we need to sieve all primes up to sqrt(155).

if you wonder how i got at "100/2" , this simply places 105 roughly in the middle of the interval , which is its best possible position in trying to disprove H-L 2nd.

but sqrt(155) = 12.44 and 155 > 120(11^2 - 1) so we need to sieve multiples of 11.

this (sieve p=11) again (at worst case) cancels with the multiples 3*5*7 (105) and 2*3*5*7 (210).

we sieve above 120 ( and below 155) ;

121 (11*11) , 143 (11*13).

thats the general procedure.

as for the density of primes (mentioned above), we need to show that for large integer m , there is a prime between m and sqrt(3/2) m.

in fact by the prime number theorem we can even estimate the amount of primes within ;

1 - sqrt(3/2) is about 1/5.

so between m and sqrt(3/2) m there are about

1/5 m / (log ( 1/5 m ) - 1) =

m / (5 log m - 5 log(5) - 5).

setting that equal to 2 we get :

m / (5 log m - 5 log(5) - 5) = 2.

m = 37. that might look weird.

because m must be integer and the equation needs to be reasonable (stable), we take the minimum value right before strictly increasing occurs ; at m = 37.

of course the interval is now of length in the order of m^2 + 1 = 37^2 + 1 = 1370.


thus for intervals of length > 1370 :

H-L 2nd holds.

note that "Bau's counterexample" is of lenght shorter than 1370 , thereby arriving at an important testing boundary constant of 1370 !!

( and "statisticly" , interval of size 1370^2 = 1876900 or smaller should contain the first counterexample )


thus as i mentioned before H-L needs only be proven for large intervals of size n.

*assuming* all small ones have been checked.

however *small* turns out to be 1370.

so i proved a slighty weaker statement :

If there is no counterexample with interval lenght (about) n =< 1370 , then hardy-littlewood 2nd conjecture holds for all interval lenghts m >= 3.

or even slightly stronger :

if there is such a counterexample then if we replace in H-L 2nd the function 'pi' with 'pic' defined as max[ pi or the largest amount in a counterexample of length =<1370 ] then the modified conjecture holds.


regards

tommy1729

the master
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