--- In

primenumbers@yahoogroups.com, "Jens Kruse Andersen"

<jens.k.a@...> wrote:

>

> Ken Davis wrote:

> > These are contructed such that they are BLS provable for c =

> > 1,5,7,11,13.

> > I sieved to a level where I expected around 1 in 360 to be PRPs

> > So far I have tested 17,221,365 numbers (of the c=7 type)

> > Finding 49949 prps (1 in 344 tests. This is good)

> > For each of these 49949 I prp'ed the other 4 types, expecting 1

in 90

> > (360/4) to be a prp

> > This has yielded 557 pairs (1 in 89.6 stil going to plan).

> > Based on the same logic as above I would have expected 1 in 120

> > (360/3)

> > of these pairs to have yielded a triple of some sort

> > However as yet I have had no luck.

> > Is there something wrong with my logic, or have I just been

extremely

> > unlucky?

>

> There is something wrong *and* you are unlucky, but not extremely

unlucky.

>

> Suppose that after getting a prp for c=7, you test the four other

numbers in

> some fixed order. If a second prime is found, there can be 0, 1, 2

or 3

> candidates left for a third prime, depending on where the second

was.

> Your calculation corresponds to saying it's always 3.

>

> The problem is that "the second prime" and "the third prime" are

taken

> from the same pool of 4 candidates. Here is another way of looking

at it:

> For each of the 49949 initial prp's, there are 6 ways to get a

triple of

> some sort, because there are 6 ways to choose 2 out of 4 numbers.

> Each way has probability 1 in 360^2, so the expected number of

> triples of some sort is 49949*6 / 360^2 = 2.31.

> The probability of having no triples of any sort is around

> 1/exp(2.31) = 10%.

So the probability of a pair yielding a triplet is 1 in 240 instead

of 1 in 120.

I still think poisson has it in for me :-(

> --

> Jens Kruse Andersen

>