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11
  • 1
    $\begingroup$ Interesting conjecture. I am sure it is open. $\endgroup$ Commented Jan 28, 2017 at 23:05
  • 5
    $\begingroup$ I am not aware of anything in that direction. Even $P(p)\geq 3$ sounds difficult to me. $\endgroup$ Commented Jan 28, 2017 at 23:12
  • 7
    $\begingroup$ Regarding $p=11$: the smallest sequence is $11+n\times 210\times 7315048$ for $0\le n \le 10$. $\endgroup$ Commented Jan 29, 2017 at 13:47
  • 4
    $\begingroup$ A good question. If true, the proof would have to be very delicate (not at all like the regularity-lemma-like proofs of Green-Tao). If false, I think a proof would be extremely strange. Perhaps current techniques can give some lower bound on $P(p)$, but this too sounds tricky to me. See the following (especially the last page or so) for some discussion of other related results. people.maths.ox.ac.uk/~conlond/green-tao-expo.pdf $\endgroup$ Commented Jan 31, 2017 at 0:24
  • 2
    $\begingroup$ I am not sure about the complete history of the problem $\ P(p)=p?\ $ I know that Siemion Fajtlowicz proposed this conjecture in 1991/2 or earlier. At that time I've got an algorithm and coded a program which gave me $\ P(13)=13.\ $ Once again, I am not a specialist, I don't know the full history here. My feeling was that $\ P(17) < 17\ ($ perhaps $\ \le 15).\ $ I feel strongly that $\ P(p) < p\ $ for every prime $\ p>13;\ $ I'd even conjecture that $\ p-P(p)\rightarrow \infty\ $ for $\ p\rightarrow\infty$. $\endgroup$ Commented Jan 31, 2017 at 5:37