I conjecture that there are infinitely many linear relations $p_n + p_{n + 3} = 2 p_{n + 2}$ in the sequence of primes!

Question

Let $p_1 < p_2 < p_3 < \ldots < p_n < \ldots$ be the sequence of primes (with $p_1 := 2$ as usual).

Conjecture: There are infinitely many positive integers $n$ for which \begin{eqnarray} p_n + p_{n + 3} = 2 p_{n + 2}. \end{eqnarray}

Or equivalently,

There are infinitely many quadruplets of consecutive primes of the form $$(p, q, p + h, p + 2h)$$

Answer 1

Your equation is equivalent to

$$p_{n+3}-p_{n+2}=p_{n+2}-p_n$$

Consider the prime-constellation $0,2,6,12$ , which is a so-called admissible prime constellation (See https://en.wikipedia.org/wiki/Prime_k-tuple for more details).

It is conjectured that there infinite many prime constellations for every admissible vector. So, probably your claim is correct, but I think there is no proof.

The first few constellations solving your equation :

? x=1;while(x<1000,x=x+1;while(1-(isprime(x)==1)*(isprime(x+2)==1)*(isprime(x+6)
==1)*(isprime(x+8)==0)*(isprime(x+12)==1),x=x+1);forprime(j=x,x+12,print1(j,"  "
));print)
17  19  23  29
41  43  47  53
227  229  233  239
347  349  353  359
641  643  647  653
1091  1093  1097  1103

The first constellation, for which $p_n>10^7$ ,seems to be the $1485-th$

1485   10017011  10017013  10017017  10017023

Answer 2

On standard conjectures in Number Theory, there are infinitely many $p$ such that $p\equiv6\bmod7$, and $p$, $p+2$, $p+6$, and $p+12$ are all prime. The congruence guarantees that the four primes are consecutive primes. Then $p_n=p$ gives a solution to the equation in the question.

The standard conjectures are probably not going to be proved any time soon.

[Note: the idea of using $p$, $p+2$, $p+6$, and $p+12$ came from a poster on MO.]