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Union of 2 Sets of Primes in Arithmetical Progression

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• Two humble questions for AP gurus and theorists, but first 2 definitions. (CP-4) means a set of 4 CONSECUTIVE PRIMES NOT IN ARITHMETICAL PROGRESSION such that
Message 1 of 4 , Feb 4, 2005
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Two humble questions for AP gurus and theorists, but first 2 definitions.

(CP-4) means a set of 4 CONSECUTIVE PRIMES NOT IN ARITHMETICAL
PROGRESSION such that the smallest prime of the set is equal to the sum
of the two next smallest primes minus the largest prime of the subset.
For example the smallest possible (CP-4) is (5, 7, 11, 13).
(CP-N) means a set of N CONSECUTIVE PRIMES NOT IN ARITHMETICAL
PROGRESSION such that EACH POSSIBLE SUBSET of 4 CONSECUTIVE primes is a
(CP-4). For example (5, 7, 11, 13, 17, 19, 23) is a (CP-7) containing 4
(CP-4)s. Every (CP-N) is ALWAYS the UNION of 2 sets of primes in
arithmetical progression both with the same constant difference.
Question (1); I discovered that the actual number of (CP-4)s that exist
in a set of N CONSECUTIVE primes, counting 2 as the first, can be closely
approximated by (N/logN). For example, in a set of N=100,000 consecutive
primes there will be about (N/logN)=8685 (CP-4)s. The actual number is
8690. Seems to work pretty well, but is it true? If it is, then it would
strike me as a quite remarkable connection to the Prime Number Theorem. I
don't have the ability to understand the math underlying the proof of the
theorem to determine whether it anticipates a case like this.
Question (2); How large a (CP-N) can one expect to find? I found (CP-N)s
with up to N=7 consecutive primes, but so far no luck with N=8. I can't
go much further with my limited resources. They get increasingly harder
to find.
Here are 2 examples of the smallest and largest examples of (CP-7) that I
could find; (5, 7, 11, 13, 17, 19, 23) and (48887063, 48887071, 48887081,
48887089, 48887099, 48887107, 48887117).
Surely, considering all the work done on the PNT, and on PAPs and CPAPs,
someone must have looked into this and also its flip side where you look
for the largest (CP-N) where NO possible (CP-4) exists. I found nothing
on this subject. I admit it's sort of a dumb exercise but I'm curious and
would appreciate any comments. Thanks folks.
Bill Sindelar
• ... A (CP-N), N =4, is N consecutive primes with gaps alternating between two different values. ... Let a, b, c be random consecutive primes around N. They
Message 2 of 4 , Feb 4, 2005
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Bill Sindelar wrote:

> (CP-4) means a set of 4 CONSECUTIVE PRIMES NOT IN ARITHMETICAL
> PROGRESSION such that the smallest prime of the set is equal to the sum
> of the two next smallest primes minus the largest prime of the subset.

> (CP-N) means a set of N CONSECUTIVE PRIMES NOT IN ARITHMETICAL
> PROGRESSION such that EACH POSSIBLE SUBSET of 4 CONSECUTIVE primes is a
> (CP-4).

A (CP-N), N>=4, is N consecutive primes with gaps alternating between two
different values.

> Question (1); I discovered that the actual number of (CP-4)s that exist
> in a set of N CONSECUTIVE primes, counting 2 as the first, can be closely
> approximated by (N/logN).

Let a, b, c be random consecutive primes around N. They start a (CP-4) iff x =
c+b-a is prime and there are no primes between c and x.
x is odd when a>2. If we otherwise assume x is random (maybe not accurate),
then x has around 2/log N chance of being prime by PNT.
There are no primes between c and x iff the prime gap following c is not
smaller than the random gap from a to b. The probability is near 1/2.
The probability of x completing a (CP-4) should then be
1/2 * 2/log N = 1/log N.
It is not surprising if the number of (CP-4) among the first N primes is
around N/log N.

Most of the first N primes are above N so the prime chance of x should
actually be a little below 1/log N. I haven't analyzed the detailed
heuristics. Maybe they should take into account that x is not a completely
random odd number.

> Question (2); How large a (CP-N) can one expect to find? I found (CP-N)s
> with up to N=7 consecutive primes, but so far no luck with N=8. I can't
> go much further with my limited resources. They get increasingly harder
> to find.

One way to get a potential (CP-N) is to remove every 3rd number in an AP.
If the prime tuple conjecture (the variant with no other primes inside the
pattern) is true then (CP-N)s can be arbitrarily long.
I expect that to hold.

Sextuplets are (CP-6). The largest known is at
http://www.ltkz.demon.co.uk/ktuplets.htm#largest6 :
138765468778 * 850# + 2822707 + 0, 4, 6, 10, 12, 16
(362 digits, 2004, Norman Luhn)

--
Jens Kruse Andersen
• A (CP-N), N =4, is N consecutive primes with gaps alternating between two different values. The first (CP-8) has gaps 10 and 8:
Message 3 of 4 , Feb 4, 2005
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A (CP-N), N>=4, is N consecutive primes with gaps alternating between two
different values.

The first (CP-8) has gaps 10 and 8:
67944073,67944083,67944091,67944101,67944109,67944119,67944127,67944137

The first (CP-9) has gaps 2 and 28 (sum 5#):
1860017189,1860017191,1860017219,1860017221,1860017249,
1860017251,1860017279,1860017281,1860017309

The first (CP-10) has gaps 16 and 14 (sum 5#):
5373097559,5373097573,5373097589,5373097603,5373097619,
5373097633,5373097649,5373097663,5373097679,5373097693

--
Jens Kruse Andersen
• Sorry for not making a single post after completing computations. I have stopped now. ... The second (CP-10) is 16 times larger with gaps 28 and 2: 87873432313
Message 4 of 4 , Feb 4, 2005
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Sorry for not making a single post after completing computations.
I have stopped now.

I wrote:
> The first (CP-10) has gaps 16 and 14 (sum 5#):
> 5373097559,5373097573,5373097589,5373097603,5373097619,
> 5373097633,5373097649,5373097663,5373097679,5373097693

The second (CP-10) is 16 times larger with gaps 28 and 2:
87873432313 + 0,28,30,58,60,88,90,118,120,148

A probably non-minimal (CP-11) with gaps 16 and 14:
196723765163557 + 0,16,30,46,60,76,90,106,120,136,150

And a probably non-minimal (CP-12) with gaps 14 and 16:
438536033046239 + 0,14,30,44,60,74,90,104,120,134,150,164

A (CP-13) must contain an AP6 inside an AP7. The common difference in an AP7
not starting at 7 is always a multiple of 7# = 210.
Inside the AP7 there has to be at least (210-2)*6 = 1248 simultaneous
composites.
That makes a (CP-13) computionally infeasible.
If the numbers are small then 1248 composites is too hard.
If the numbers are large then 13 primes is too hard.

--
Jens Kruse Andersen
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