- Bill Sindelar wrote:
>> Bill Sindelar wrote:

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

>>> For any integer k=3 or greater, and any integer n=0 or greater,

>>> one can find an arithmetic progression of prime numbers (PAP)

>>> of length k, such that the number n of consecutive primes

>>> between adjacent terms of the PAP is constant.

>

> Jens K Andersen wrote:

>> This follows from the widely believed conjecture that all

>> admissible prime constellations have infinitely many occurrences.

>> If the conjecture is true then one can first choose specific

>> admissible positions for all primes, and then choose additional

>> prime positions outside the whole PAP interval in a way which

>> fixes chosen primes modulo the interval start such that every

>> other number in the interval gets a known prime factor.

>

> I'm lost here. Seems like a convoluted approach.

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

previosly studied things. If you want something unconvoluted

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

right.

> A few weeks ago I realized that, in all the

There is plenty of literature about consecutive primes in arithmetic

> literature I read on PAP's, including Green and Tao, no reference to the

> numbers enclosed between adjacent primes of the PAP was made. The sole

> requirement was a constant difference between adjacent primes. I decided

> to add the requirement that the number of consecutive primes between

> adjacent primes also be constant

progression which is what you call (PAP-k, 0).

I have not seen mention of your (PAP-k, n>0) before.

> Sindelar wrote:

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

>>> Consecutive primes in arithmetic progression can be considered as

>>> a (PAP-k, n) since there are n=0 primes between adjacent terms. I

>>> would interpret Green and Tao as covering this type.

>

> Andersen wrote:

>> No. Green and Tao does not say that the primes in AP are

>> consecutive or have a constant number of primes between them.

>

> As I said in my first post, I do not have the ability to understand the

> Green-Tao proof to see if specific exceptions are included. Tao in a

> presentation slide showed the theorem worded as "The primes contain

> arbitrarily long arithmetic progressions". In another slide he says "In

> particular, for any given k, the primes contain infinitely many

> arithmetic progressions of length k". To me, this is a very broad claim

> covering any type of (PAP-k, n). Jens can you explain your answer a bit

> more?

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

A PAP-k (often just called AP-k) is k primes in arithmetic progression.

A PAP-k has no condition on how many other primes are between them.

Tao's two formulations are trivially equivalent.

1) "The primes contain arbitrarily long arithmetic progressions".

By definition of "arbitrarily long" this means that:

For any given natural number n, there is a PAP-k with k>=n.

2) "In particular, for any given k, the primes contain infinitely many

arithmetic progressions of length k".

By definition of "infinitely many" this means that:

For any given m and k, there are more than m PAP-k.

2) obviously implies 1) by setting m=1.

1) implies 2). Proof: If 1) is true then for any m and k, there exists

a PAP-(m+k), and that contains m+1 overlapping PAP-k.

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.

--

Jens Kruse Andersen - 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