- 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 - Bill Sindelar wrote:

> (CP-4) means a set of 4 CONSECUTIVE PRIMES NOT IN ARITHMETICAL

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

> 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).

different values.

> Question (1); I discovered that the actual number of (CP-4)s that exist

Let a, b, c be random consecutive primes around N. They start a (CP-4) iff x =

> in a set of N CONSECUTIVE primes, counting 2 as the first, can be closely

> approximated by (N/logN).

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

One way to get a potential (CP-N) is to remove every 3rd number in an AP.

> 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.

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:

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.

I wrote:> The first (CP-10) has gaps 16 and 14 (sum 5#):

The second (CP-10) is 16 times larger with gaps 28 and 2:

> 5373097559,5373097573,5373097589,5373097603,5373097619,

> 5373097633,5373097649,5373097663,5373097679,5373097693

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