## Re: [PrimeNumbers] New Type of Prime Arithmetic Progression?

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• ... You asked for comments on your statement (which is an unproven guess). I briefly showed that it would follow from a well-known and trusted conjecture,
Message 1 of 12 , Sep 26, 2007
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Bill Sindelar wrote:
>> Bill Sindelar 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.
>
> 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
>> 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
> 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

There is plenty of literature about consecutive primes in arithmetic
progression which is what you call (PAP-k, 0).
I have not seen mention of your (PAP-k, n>0) before.

> Sindelar wrote:
>>> 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?

Just to be clear: My "No" was only to your second sentence:
"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
• ... I ll try. Are you aware of the concept of admissible constellations? For example, twin primes, x and x+2, are believed to occur infinitely often. A
Message 2 of 12 , Sep 26, 2007
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I'll try.

Are you aware of the concept of "admissible constellations?" For
example, twin primes, x and x+2, are believed to occur infinitely
often. A triplet of the form x, x+2, x+4 is not admissible because
one of these numbers always divisible by 3. Hence (3, 5, 7) is the
only example - a finite number. However x, x+2, x+6 is admissible and
is believed to be prime infinitely often.

Jens point is that you can construct an admissible constellation that
has all kn-1 primes in fixed locations so the the arithmetic
progression and primes between are honored. This is more restrictive
than your rules require, but would be an example as long as none of
the the other values within the constellation's range are also prime.
Jen's further point is that you could further restrict the
constellation so that the intermediate values were divisible by
selected primes, and hence composite. Again more restrictive than you
rules, but it would qualify as an example. Finally, from the
constellation conjecture there would be an infinite number of these
constellation, each and every one an example.

For example, x, x+2, x+6, x+8, x+12 is an example of admissible
constellation that has 3 primes in arithmetic progression with(al
least) one prime between them. In this case there is exactly one
prime because x+4 and x+10 must be divisible by 3. In general, we
would need to jiggle the start point as x=ay+b for selected fixed
values if a and b to ensure the omitted points are composite.

Finally, every example that you find can could be reverse-engineered
to such an admissible constellation, so this search would be a subset
of a search for examples of admissible constellations.

William Lipp
Poohbah of OddPerfect.org
• ... Jens, I think I may have offended you by writing I m lost here. Seems like a convoluted approach. Looking back at this, I can see that it can be taken as
Message 3 of 12 , Sep 29, 2007
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Sindelar wrote:
> > I'm lost here. Seems like a convoluted approach.

Andersen wrote:
> 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.

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

Andersen wrote:
> Just to be clear: My "No" was only to your second sentence:
> "I would interpret Green and Tao as covering this type. (meaning
(PAP-k, n=0))"
>
> 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 thank you. Nothing more to explain. You made it clear that the Green
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]
• ... Jens, I think I may have offended you by writing I m lost here. Seems like a convoluted approach. Looking back at this, I can see that it can be taken as
Message 4 of 12 , Sep 29, 2007
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Sindelar wrote:
> > I'm lost here. Seems like a convoluted approach.

Andersen wrote:
> 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.

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

Andersen wrote:
> Just to be clear: My "No" was only to your second sentence:
> "I would interpret Green and Tao as covering this type. (meaning
(PAP-k, n=0))"
>
> 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 thank you. Nothing more to explain. You made it clear that the Green
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 ... I m glad you did, and I thank you. You must be a mind reader. You somehow sensed why I got lost trying to
Message 5 of 12 , Sep 29, 2007
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On Thu, 27 Sep 2007 01:55:00 -0000 "elevensmooth"
<elevensmooth@...> writes:
> --- In primenumbers@yahoogroups.com, w_sindelar@... wrote:
>

William Lipp wrote:
> 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
• On Thu, 27 Sep 2007 01:55:00 -0000 elevensmooth ... I m glad you did, and I thank you. You must be a mind reader. You somehow sensed why I got lost trying to
Message 6 of 12 , Sep 29, 2007
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On Thu, 27 Sep 2007 01:55:00 -0000 "elevensmooth"
<elevensmooth@...> writes:
> --- In primenumbers@yahoogroups.com, w_sindelar@... wrote:
>

William Lipp wrote:
> 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
Message 7 of 12 , Sep 29, 2007
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Bill Sindelar wrote:
> Jens, I think I may have offended you by writing "I'm lost here. Seems
> 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
• ... I m relieved. I was trying to get up some nerve to ask you what sort of ... I just tested the other PAP-8 from the search and ... equivalent to the above
Message 8 of 12 , Oct 2, 2007
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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
> 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.
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

Sindelar wrote (Yahoo #19096):

>>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
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
• ... I would expect your method to be much slower based on how randomly consecutive prime gaps appear to be distributed. ... No, and also no to the only-part.
Message 9 of 12 , Oct 2, 2007
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Bill Sindelar wrote:
> 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?

I would expect your method to be much slower based on how
"randomly" consecutive prime gaps appear to be distributed.

> 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?

No, and also no to the only-part. Your assumption says nothing
about the existence of specific differences between primes,

> This suggested trying this assumption which is just a fancy way of
> defining a (PAP-k, n):
.....
> It works but is more
> computationally complicated. What do you think?

"computationally complicated" refers to something computational,
for example the time to compute something with a given algorithm.