- 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