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 $$(p, q, p + h, p + 2h)$$

  • $\begingroup$ Nope, this (counting primes in a linear relation) is an entirely different business. I don't see how a proof of the infinitude of primes helps here. $\endgroup$ – dohmatob Sep 15 '15 at 8:10
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    $\begingroup$ Sounds like an unsolved problem to me. $\endgroup$ – Ivan Neretin Sep 15 '15 at 8:34
  • $\begingroup$ Mathematicians generally don't make conjectures unless they have a strong reason for believing a statement to be true. Why do you think there are infinitely many such triples? $\endgroup$ – Gerry Myerson Sep 16 '15 at 9:06
  • $\begingroup$ @GerryMyerson: Well, my intuition tells me so :). OK, jokes aside. We know there are arbitrarily long polynomial progressions of primes (Tao-Ziegler arxiv.org/pdf/math/0610050v2.pdf). I think this should be true too with the added restriction that the primes be consecutive (But I'm unable to prove it). This would prove the conjecture I stated. Is this "reason" strong enough ? :) $\endgroup$ – dohmatob Sep 16 '15 at 9:27
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    $\begingroup$ No.${}{}{}{}{}$ $\endgroup$ – Gerry Myerson Sep 16 '15 at 13:27

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.]


Your equation is equivalent to


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,"  "
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
  • $\begingroup$ Note that not every constellation leads to a solution, you must find an $x$, such that $x,x+2,x+6,x+12$ are prime, but not $x+8$, for instance. I am not sure, it is also conjectured that such constellations appear infinite many times, but probably this is the case. $\endgroup$ – Peter Sep 16 '15 at 18:09
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    $\begingroup$ @dohmatob I just noticed that you had this idea. A proof of such conjectures seems to be out of reach. $\endgroup$ – Peter Sep 16 '15 at 18:23
  • $\begingroup$ I also had the idea of considering constellations of the form $(0,2g,6g,12g)$ and then abandoned it, since I'm actually interested in consecutive primes, whilst the conclusions of the $k$-tuples conjecture, etc. are quite modest, in this regards. $\endgroup$ – dohmatob Sep 17 '15 at 3:43
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    $\begingroup$ You can add the assumption that $x\equiv6\pmod7$, still get (conjecturally) infinitely many primes, and then they'll be consecutive, since $7\mid x+8$. $\endgroup$ – Gerry Myerson Sep 17 '15 at 5:22
  • $\begingroup$ @GerryMyerson: Thanks for the great insight. Finally, something constructive :) I've commented your mathoverflow comment. $\endgroup$ – dohmatob Sep 17 '15 at 7:08

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