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Let $p_n$ be the $n$-th prime.

For each integer $k \ge 0$, do there exist an infinite number of $k+3$ consecutive primes $(p_n, p_{n+1}, \ldots, p_{n+2+k})$ so that

  • (1) The gap between the 1st and 2nd, and between the 2nd and last, are equal: $p_{n+1}-p_n = p_{n+2+k}-p_{n+1}$.

  • (2) There are $k$ primes between the 2nd and last, i.e., between $p_{n+1}$ and $p_{n+2+k}$.

For $k=0$, the answer is Yes by the recent breakthroughs on prime gaps. Here are some examples:
     PrimeGapsk
One could whimsically imagine "skipping" a flat stone on the primes, where the first bounce covers the gap between the 1st and 2nd primes, followed by $k+1$ smaller bounces that together cover the same gap before sinking on the last prime.

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Likely yes. No proof yet. Stay tuned. Also, this would follow from the k-primes conjecture, which seems more likely to be answered before your question. –  The Masked Avenger Feb 5 at 16:39
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@TheMaskedAvenger as far as I can see it does not follow from results on progressions in primes (since nothing guarantees the primes to be consecutive, which is also the point of my first comment) –  quid Feb 5 at 16:52
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@quid, its quite possible that I am confusing my idea with what is actually in the literature. My recall is that if an admissible pattern of prime gaps occur, then that pattern occurs infinitely often WITH no extra primes "inside" the pattern. What might be in the literature instead would have the constellation occur as a proper subconstellation, allowing extra primes. –  The Masked Avenger Feb 5 at 16:59
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@quid: I removed the claim that the $k=0$ pattern---three consecutive primes in arithmetic progression---is resolved by recent results. Thanks for the correction! –  Joseph O'Rourke Feb 5 at 17:26
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In the illustration for $k=1$, 47 should be 43 (I'd fix it myself, but pretty displays are beyond my TeXnical level). –  Gerry Myerson Feb 6 at 0:00

1 Answer 1

up vote 11 down vote accepted

This would follow from the $k$-tuple conjecture in the following way.

Choose an admissible tuple $d_1, \ldots, d_{k+2}$, such that $d_2-d_1=d_{k+2}-d_2$. If $n\in[d_1, d_{k+2}]$ is an integer, such that $\{d_1, \ldots, d_{k+2}\}\cup\{n\}$ is admissible, pick a prime number $p$ such that for all $i$ we have that $n-d_i$ is not divisible by $p$. Pick integers $e_i>d_{k+2}$, such that $\{d_1,\ldots, d_{k+2}, e_1, \ldots, e_\ell\}$ is admissible, and the set $\{d_1,\ldots, d_{k+2}, e_1, \ldots, e_\ell\}$ covers all residue classes modulo $p$ with the exception of $n\pmod{p}$. Repeat until you arrive at an admissible set $D$, such that $D\cap[d_1, \ldots, d_{k+2}]=\{d_1, \ldots, d_{k+2}\}$, and for all $n\in[d_1, d_{k+2}]\setminus\{d_1, \ldots, d_{k+2}\}$ we have that $D\cup\{n\}$ is not admissible. Apply the $k$-tuple conjecture to the set $D$. If $x+d$ is prime for all $d\in D$, and $x$ is sufficiently large, then $x+d_1, \ldots, x+d_{k+2}$ are consecutive prime numbers.

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Wow! That is an impressively nontrivial connection to the $k$-tuple conjecture! –  Joseph O'Rourke Feb 6 at 0:41

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