Here is an heuristic argument in behalf of the conjecture:

The twin primes results from the pairing of the following A.P.

6n - 1 = 5 11 17 23
29 35 41 47 53 59 65 .....

6n + 1 = 7 13 19 25 31 37 43 49 55 61 67 .....

Mean = 6 12 18 30
42 60 ...

Also with the mean (3 + 5) / 2 = 4 , I have the available differences:

4 , 6 , 12 , 18 , 30 , 42 , 60 , 72 ... (4 & Multiples of 6)

The prime numbers >= 5 are contained in the sequence:

5 7 11 13 17 19 23 25 29 31 35 37 41 43 47 49 ...

(With differences: 2 , 4 , 2
, 4 , 2 , 4 ....)

That is: With this differences I can compose the available differences and summing it, from each prime , I can reach another prime. So:

5 + 6 = 11

7 + 4 = 11

11 + 6 = 17

13 + 4 = 17

17 + 6 = 23

19 +
4 = 23

23 + 6 = 29

29 +12 = 41

31 + 6 = 37

. . . . . . . . .

113 + 18 = 131

. . . . . . . . .

El Lunes 25 de agosto de 2014 8:36, Luis Rodriguez <luiroto@...> escribió:

Sure. There are sufficient differences between primes to be pairing with the even numbers resulting from the mean of two twin primes.

The list of means of two twin primes are:

4,6,12,18,30,42,50,72,102,....

Its impossible that never we will find a difference between two primes that cannot be one of that list.

Ludovicus