- Thank you once again, Jens. I value your responses. I would like to add a

few comments in reply.

> Bill Sindelar wrote:

Jens K Andersen wrote:

> > 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.

> This follows from the widely believed conjecture that all

I'm lost here. Seems like a convoluted approach. Here's how I came to

> 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.

making the above statement. A few weeks ago I realized that, in all the

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, and see if I could find any PAPs of

this type.

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.

Sindelar wrote:> Consecutive primes in arithmetic progression can be considered as

Andersen wrote:

> a

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

> would

> interpret Green and Tao as covering this type.

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

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

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

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?

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