- Bill Sindelar wrote:
> For any integer k=3 or greater, and any integer n=0 or greater, one can

This follows from the widely believed conjecture that all

> 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.

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.

> Consecutive primes in arithmetic progression can be considered as a

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

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

> interpret Green and Tao as covering this type.

consecutive or have a constant number of primes between them.

> I would very much appreciate the advice of those that have worked on

(PAP-8, 5): 1376308269430349 + 210*i is prime for i = 0 to 7,

> (PAP-k, 0) as to the computational complexity of finding (PAP-k, n>3).

and there are 5 primes between in each case.

It's probably not the smallest (PAP-8, n).

(PAP-9, n>3) and (PAP-10, n>3) look easier than n=0 which has known

cases (a single case for PAP-10) at http://hjem.get2net.dk/jka/math/cpap.htm

(PAP-11, n>0) looks very hard, but not as hard as for n=0 which

could take trillions of GHz years.

A PAP-11 has a common difference which is a multiple of 11# = 2310,

and 10 intervals of that size would very rarely have the same number

of primes - unless the numbers are so large that the intervals have few

or no primes, but then the 11 primes in the PAP-11 would be large

and extremely hard to find.

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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