- The only pertinent web reference I could find is Green and Tao's theorem

which says that for every integer k=1 or greater, one can find an

arithmetic progression of prime numbers of length k. The proof is very

complicated. I do not have the ability to see if it would cover the

following statement.

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.

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

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

I discovered that there are such PAPs. For convenience, let's call that

type (PAP-k, n). Here are random examples.

(PAP-3, 4) is (1549, 1579, 1609). (PAP-4, 4) is (6047, 6089, 6131, 6173).

(PAP-5, 4) is (28381081, 28381141, 28381201, 28381261, 28381321). Each

one of these has four consecutive primes between adjacent terms.

(PAP-5, 7) is (4550617, 4550737, 4550857, 4550977, 4551097). Constant

difference is 120. There are seven consecutive primes between adjacent

terms. Its associated (AP) of ordinal numbers is (319206, 319214, 319222,

319230, 319238). Constant difference is 8.

(PAP-3, 1000) is (363589, 376609, 389629). Constant difference is 13020.

There are 1000 consecutive primes between adjacent terms. Its associated

(AP) of ordinal numbers is (31033, 32034, 33035). Constant difference is

1001.

To my considerable frustration after many calculations, I was not able to

find a (PAP-k, n) of length k greater than 5. This prompted the obvious

question as to whether no prime arithmetic progression of primes of

length greater than five exists, such that the number of consecutive

primes between adjacent terms of the progression is equal, or that I

simply did not go far enough.

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.

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

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

Also their thoughts on the above statement. Is there a limit on k? Thanks

folks for your time.

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