Does Poisson hate me?

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• 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
Message 1 of 3 , Jul 9, 2008
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
• ... 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
Message 2 of 3 , Jul 9, 2008
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%.

--
Jens Kruse Andersen
• ... in 90 ... extremely ... unlucky. ... numbers in ... or 3 ... was. ... taken ... triple of ... So the probability of a pair yielding a triplet is 1 in 240
Message 3 of 3 , Jul 9, 2008
--- 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
>
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