- Hi All,

Over the last 6 months I've been searching for a 7716 digit triple.

Using a method first used by David Broadhurst I have been prping

numbers of the form

P=((N*35+x)*328963*6011#*(328963*6011#+1)+210)*(328963*6011#-1)/35+c

where x = 11 or 31

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?

cheers

Ken - Ken Davis wrote:
> These are contructed such that they are BLS provable for c =

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

> 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?

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%.

--

Jens Kruse Andersen - --- In primenumbers@yahoogroups.com, "Jens Kruse Andersen"

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

in 90

> 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

> > (360/4) to be a prp

extremely

> > 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

> > unlucky?

unlucky.

>

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

>

numbers in

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

> 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.

taken

>

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

> 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.

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

> 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%.

of 1 in 120.

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

> Jens Kruse Andersen

>