- Sindelar wrote:
> > I'm lost here. Seems like a convoluted approach.

Andersen wrote:

> You asked for comments on your statement (which is an unproven

Jens, I think I may have offended you by writing "I'm lost here. Seems

> guess). I briefly showed that it would follow from a well-known

> and trusted conjecture, sometimes called the prime k-tuple

> conjecture

> (that name is sometimes restricted to special cases). Of course

> it's

> less "convoluted" to not relate your guess to anything else like

> previously studied things. If you want something unconvoluted

> (but not very useful by itself) then here it is: I guess your guess

> is

> right.

like a convoluted approach." Looking back at this, I can see that it can

be taken as arrogant criticism. Gad, that is not what I meant it to be. I

should have written "I am unable to follow your explanation. It seems

complicated to me because I know nothing about admissible prime

constellations, but I accept your opinion". I regret my choice of words

and hope you accept my sincere apology.

Sindelar wrote in regard to Green and Tao:>>>To me, this is a very broad claim covering any type of (PAP-k, n).

Jens can you explain your answer a bit more?

Andersen wrote:> Just to be clear: My "No" was only to your second sentence:

(PAP-k, n=0))"

> "I would interpret Green and Tao as covering this type. (meaning

>

Jens thank you. Nothing more to explain. You made it clear that the Green

> You defined (PAP-k, n) as a PAP-k with n primes between each of

> the k-1 pairs of successive primes in the AP. Tao and Green don't

> mention this concept of equal prime counts and their theorem says

> nothing about your (PAP-k, n) for n=0 or any other n value.

> I don't know what else you want me to explain.

> All I can say is that the theorem simply doesn't say it.

and Tao theorem does not apply to type (PAP-k, n=0 or greater). And thank

you for an example of a (PAP-8, 5). Don't know how you calculated that so

quickly. I was beginning to think there might be a limit on k.

Bill Sindelar

[Non-text portions of this message have been removed] - On Thu, 27 Sep 2007 01:55:00 -0000 "elevensmooth"

<elevensmooth@...> writes:> --- In primenumbers@yahoogroups.com, w_sindelar@... wrote:

William Lipp wrote:

>

> > Jens can you explain your answer a bit more?

> I'll try.

I'm glad you did, and I thank you. You must be a mind reader. You somehow

sensed why I got lost trying to follow Jens reasoning. Right off the bat

I'm confronted with "admissible prime constellations" and right there I'm

lost.

Your neat little introductory on this greatly helped me understand what

Jens meant. I'm going to study this concept in more detail. Regards with

appreciation.

Bill Sindelar - Bill Sindelar wrote:
> Jens, I think I may have offended you by writing "I'm lost here. Seems

No problem. You can search more information about admissible constellations

> like a convoluted approach."

with a search engine.

If a prime p <= k does not divide the common difference in an AP-k then

p will divide at least one of the terms in the AP. In order to be

admissible, a PAP-k must therefore have a common difference which is

a multiple of k# (k primorial).

I guess a PAP-k with small difference (and therefore relatively few primes

between the terms) will have a better chance of being a (PAP-k, n),

because the number of primes can vary between fewer values.

A PAP-11 has minimal difference 11# = 2310, so 10 intervals of 2309

numbers must have the same prime count to produce a (PAP-11, n)

with minimal difference. That appears computationally too hard for me.

PAP-7 to PAP-10 all have minimal difference 10# = 7# = 210.

I used my old tuplet finder to systematically search a lot of PAP-10 with

difference 210 and count whether there happened to be an equal number

of primes between the terms. There were other things to use my only

computer for so the search stopped when only (PAP-8, n) had been found.

Hans Rosenthal is more patient and has found many (PAP-9, 0), also called

CPAP-9, with a version of the same program. (PAP-9, n>3) looks easier.

In 2004 he found the smallest known CPAP-8 = (PAP-8, 0) with

another version. I just tested the other PAP-8 from the search and found

a (PAP-8, 1) with difference 210:

64881326075217862991473794035228920286672784697 +

0,36,210,264,420,564,630,784,840,942,1050,1086,1260,1360,1470

--

Jens Kruse Andersen - Jens K. Anderson wrote:
> No problem. You can...

I'm relieved. I was trying to get up some nerve to ask you what sort of

>>

approach you used on (PAP-8, 5) when your mail arrived with the answer:

> I used my old tuplet finder to systematically search a lot of PAP-10

I just tested the other PAP-8 from the search and

> with

> difference 210 and count whether there happened to be an equal

> number

> of primes between the terms. There were other things to use my only

> computer for so the search stopped when only (PAP-8, n) had been

> found.

> found

Sindelar wrote (Yahoo #19096):

> a (PAP-8, 1) with difference 210:

> 64881326075217862991473794035228920286672784697 +

> 0,36,210,264,420,564,630,784,840,942,1050,1086,1260,1360,1470

>>The approach I used required making the following assumption, which is

equivalent to the above statement; Let S(p, n) represent an infinite

subset of the universal set of all consecutive odd primes, where p is the

first prime of the subset, and n (including 0) represents the number of

consecutive primes from the universal set that have been omitted between

adjacent primes of the subset.Then any S(p, n) contains a set of any

number k of primes in arithmetic progression. The program I wrote is

based on this.>

I used Pari-gp for this. For every set of k consecutive primes, which has

n skipped consecutive primes between its adjacent terms, after an

inputted integer, it checks if the terms of that set are in arithmetic

progression. Jens, is this slower than your approach with your tuplet

finder? If one could prove the above assumption, would that also prove

that all admissible prime constellations have infinitely many occurrences

as you put it, or only those that have a (PAP-k, n) subset?

Sindelar wrote (Yahoo #19093):

>>Obviously, the ordinal numbers of the primes in such a PAP are also in

arithmetic progression (AP) with a constant difference of (n+1).>

This suggested trying this assumption which is just a fancy way of

defining a (PAP-k, n): In any infinite arithmetic progression of positive

integers with a common difference d, there exists a subset of k

consecutive integers, so that if each integer in that subset is

considered to represent the ordinal number of a prime, the associated

primes will be in arithmetic progression of length k with (d-1)

consecutive primes between adjacent terms of that arithmetic progression.

(Ordinal number of a prime means its position in the numerically ordered

set of all primes, with prime 2 being number 1). It works but is more

computationally complicated. What do you think?

Bill Sindelar

Bill Sindelar - Bill Sindelar wrote:
> I used Pari-gp for this. For every set of k consecutive primes, which has

I would expect your method to be much slower based on how

> n skipped consecutive primes between its adjacent terms, after an

> inputted integer, it checks if the terms of that set are in arithmetic

> progression. Jens, is this slower than your approach with your tuplet

> finder?

"randomly" consecutive prime gaps appear to be distributed.

> If one could prove the above assumption, would that also prove

No, and also no to the only-part. Your assumption says nothing

> that all admissible prime constellations have infinitely many occurrences

> as you put it, or only those that have a (PAP-k, n) subset?

about the existence of specific differences between primes,

so it says nothing about any admissible constellation.

> This suggested trying this assumption which is just a fancy way of

.....

> defining a (PAP-k, n):

> It works but is more

"computationally complicated" refers to something computational,

> computationally complicated. What do you think?

for example the time to compute something with a given algorithm.

You have made another formulation of your conjecture but not

described an algorithm so "computationally complicated" is a

concept which does not apply.

I don't have time to discuss more.

--

Jens Kruse Andersen