Quote from: bsunau7 on September 11, 2014, 04:46:54 AM

If you take the first p7 variant and subtract 2 you get the pattern:

Code:

-2 0 4 6 10 16 18

Aside from "12" is a very good match for the 6-tuplet pattern (for the second p7 variant you ignore the "4"). In both cases you just need to test p-2 and p+18 for a valid p7 chain. In effect a valid 6-tuplet means you know you have 5 out of 7 valid primes for the 7-tuplet.

A quick look at the others shows similar tricks to "re-use" valid 6-tuplets probably also exist.

Once again check my assumptions....

I'm afraid that doesn't work. Imagine that you did find p-2 and p+18 prime. So you have a septuplet. But, you started from a sextuple so p+12 is also prime. This means that your septuple is also an octuple: it has 8 primes in the range p-2 to p+18. A difference of 20. But Athony Forbes tells us that the minimum distance possible for 8-tuples is 26, so the 8-tuple with distance 20 cannot exist. But we said it did... this absurd comes from assuming you could find p-2 and p+18 both prime.

The thing is that ignoring the 12 is cheating, you won't have this:

Code:

-2 0 4 6 10 16 18

but actually this:

Code:

-2 0 4 6 10 12 16 18

and that is not possible because there is a prime q, with q<20 where any of the p(i) will be a multiple of q.
edit: let me see which one it is.....
ok, it's 5. Let's add 2 again for simplicity, so we have:

Code:

0 2 6 8 12 14 18 20

now:
if p+0 has the form 5x then it's not prime
if p+0 has the form 5x+1 then p+14 is actually 5x+1+14 = 5x+15 = 5(x+3) then it's not prime
if p+0 has the form 5x+2 then p+8 is actually 5x+2+8= 5x+10 = 5(x+2) then it's not prime
if p+0 has the form 5x+3 then p+2 is actually 5x+5 = 5(x+1) then it's not prime
if p+0 has the form 5x+4 then p+6 is actually 5x+10 = 5(x+2) then it's not prime

so it's not possible to find primes with that pattern...