I'm learning math so this may seem obvious but its not to me.
In our other post titled
"Is this iteration involving primes known?"
An iteration is defined that "stepped on primes" along the natural number line with steps of length some prime plus one.
The full details are given in the post above.
So considering the minimal step length the primes stepped upon (for a minimal step length) are thus
3, 11, 29, 61, 151, 331, 691, 1453, 2953, 5923, 11863, 23761, 47563, 95203, 190471, 381001, 762049, 1524277, 3048679, 6097417, 12194857, 24389767 and so on
Now consider the difference in consecutive terms in this sequence minus one. This is the "sequence of differences"
7,17,31,89,179,359,761 and so on up to 12194909 and the terms to infinity (but we must firstly prove that there are an infinite number of terms here) can be computed by iterating as described in the post referred to above.
So the above sequence is the minimal step lengths from prime to prime which by definition is prime.
Then the question is: can we prove this "sequence of differences" is infinite and prove why from 89 onward to infinity are the primes of the form 3a+2 (for some integer a) ?
It is implied from the above question that a proof is required that proves this sequences of differences is infinite. It was asked in the post
if the sequence there was infinite and it obvious that it would follow then that the "sequences of differences" is infinite. So to proves these sequences are infinite requires proving an unproved statement about primes (this is given any prime $U$ find a prime $V$ such that $U+V+1$ is prime, $V$ can be minimal or not).
So we have not proved that this sequence is actually infinite (but suspect it is) see the post referred to above for more details.
So sadly this means the proof in the answer below is incorrect because it uses the property $x_n$ is infinite. It would be pleasantly surprising if the incorrect proof below was as short and simple as that, I'd love to know a correct answer.
The details why this proof is wrong is it assumes $x_n$ are infinite which we dont know to be true.The reason is because we do not have a proof of the following,if we are given any prime R, we must be able to prove that there exist another prime S, such that S is minimal, such that R+S+1 is prime. (Theres another unproved case when S isnt minimal) As far as I am aware there is no proof for this result. The question asks to prove if $x_n$ is infinite (not assume it is) and prove the form is 3a+2 of those integers in the infinite tail of the sequence
In addition to the above question the proof will also prove when the step length is not minimal. The non-minimal means the step length is now not the smallest possible prime. For example stepping on the prime 11 then a stepping by a non minimal length 41+1 to get to the prime 53. (The minimal step length is 17 this is why the term 29=11+17+1 appears above). This also means the proof below is incorrect as pointed out below
I don't understand why this proof got positive votes-maybe I'm missing something-so if you do vote the question up please leave the reason why.