On Fri, 2 Dec 2005, Bob Gilson wrote:

>

>

> "Werner D. Sand" <Theo.3.1415@...> wrote: Hi grostoon,

>

> my beginning of the product 2/3 * 7/5 * ... is quite correct.

> I thought I could avoid the ugly formula:

> product(p(n)/p(n+1)) * (p(n+3)/p(n+2)).

> It's an `alternating' product. Otherwise the limit would be simply = 0

> (I suspect).

>

> Werner

>

Hi,

I've calculated very rough estimation of products p1=2/3*11/13*...

and p2=7/5*19/17*... up to LIM primes (it is very rough since I used

Nextprime from GMP lib):

LIM= 1000 p1=0.131402 p2= 6.980669 p1p2=0.917275

LIM= 2000 p1=0.108394 p2= 8.584098 p1p2=0.930461

LIM= 4000 p1=0.088371 p2=10.297879 p1p2=0.910038

LIM= 8000 p1=0.073380 p2=12.474514 p1p2=0.915374

LIM= 16000 p1=0.060168 p2=15.118286 p1p2=0.909637

LIM= 32000 p1=0.049730 p2=18.303402 p1p2=0.910235

LIM= 64000 p1=0.041075 p2=22.084929 p1p2=0.907139

LIM= 128000 p1=0.033984 p2=26.647697 p1p2=0.905591

I have no time to analyze these series (e.g. quotients

18.303/15.118, etc), but I have another idea:

You investigate

Prod_{n=0} p(n)/p(n+1) * p(n+3)/p(n+2)

(I assume p(0)=2, for convenience). Another formulation is

Prod_{n=0} a_n/a_{n+1}, where

a_n=p(2n)/p(2n+1).

What about log of this product? I'm not good in logs for infinite products

but for finite one we have

log(Prod_{n=0} a_n/a_{n+1}) = Sum_{n=0} log a_n - log a_{n+1},

so it is an alternating series. Is it enough to apply Leibniz criterion?

log a_n= log [ p(2n)/p(2n+1)] = log p(2n) - log p(2n+1).

It is possible to say something about this difference? It is, of course,

negative, but what more? |log (a_n)| > |log(a_{n+1}| ????

There are large gaps in primes, so sometimes a_n is quite small, but for

twin primes it tends to 1.

Wojtek

===============================================

Wojciech Florek (WsF)

Adam Mickiewicz University, Faculty of Physics

ul. Umultowska 85, 61-614 Poznan, Poland

Phone: (++48-61) 8295033 fax: (++48-61) 8257758

email:

Wojciech.Florek@...