Sigh. Nevermind... taking the link down.

The other expression is also flawed... it fails badly

when i exceeds 35 (that is, (p(i))^2=22201).

Back to the drawing board...

--- Jeremy Wood <

mickleness@...> wrote:

> After work I'll revisit everything again. In the

> meantime I kept the file online, but put red notes

> around the incorrect section.

>

> But the paper still points out -- although it

> doesn't

> formally prove -- that:

> pi(p(i+1)^2)>=r(i)*(p(i+1)^2)+i-1

> where r(i) =

> (p(1)-1)/p(1)*(p(2)-1)/p(2)*...(p(i)-1)/p(i)

>

> If anyone has any thoughts as to how one could apply

> this towards Brocard's Conjecture, please let me

> know.

> Or if this is also flawed, please let me know.

>

> Cheers

> - Jeremy Wood

>

> --- Patrick Capelle <patrick.capelle@...>

> wrote:

>

> > --- In primenumbers@yahoogroups.com, Jeremy

> > <mickleness@y...> wrote:

> > >

> > > Hi everyone... I just joined the list.

> > >

> > > I wrote a little paper on primes recently,

> > offering an informal proof

> > > of Brocard's Conjecture. a few notes on twin

> > primes. and other

> > > observations.

> > >

> > > I was wondering if people on this list could

> look

> > it over and let me

> > > know... well... if it has any merit. I'm

> > competent at math, but

> > > proofs and high level math are a little foreign

> to

> > me...

> > >

> > > http://homepage.mac.com/bricolage1/essays/

> > >

> >

> >

> > Hello Jeremy,

> >

> > At the beginning of your proof of Brocard's

> > conjecture,you wrote :

> > "Well if d-b >= k, and a >= b and c >= d, then

> > surely c-a >= k ".

> > Surely not.There are cases where c-a < k.

> > Take for instance a = 5, b = 2, c = 7, d = 6 and k

> =

> > 3.

> >

> > Regards,

> > Patrick Capelle.

> >

> >

> >

> >

>

>