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Apr 9 at 14:27 comment added Somudro Gupto @MaxAlekseyev Thanks for this calrification. It is correct.
Apr 9 at 11:17 comment added Max Alekseyev For $k=(p_i-1)/2$ (and its odd multiples), we have $x_i=p_i-1$, which shows that no better upper bound is possible.
Apr 9 at 10:39 comment added Somudro Gupto @MaxAlekseyev Yes, $2<x_i< p_i$ trivially. An $\mathcal{O}(\log{p_i})$ bound would be more meaningful,
Apr 9 at 1:52 comment added Max Alekseyev Isn't $p_i$ a readily available upper bound for $x_i$?
Apr 8 at 19:56 comment added Somudro Gupto Yes, that is correct. Thank you again. It could be really helpful if this fraction could be expressed as a function of $N$. I think it needs a separate question.
Apr 8 at 19:22 comment added Will Sawin Not necessarily a non-negligible fraction in the sense of positive density, but I would expect this does happen for infinitely many primes.
Apr 8 at 18:47 comment added Somudro Gupto Thank you professor @WillSawin for this answer. If I understood correctly, in other words do you want say that $x_i=3$ for a non-negligible fraction of primes that have $2$ as one of their primitive roots?
Apr 8 at 17:28 comment added Will Sawin If I understand your question right, $x_i \geq 3$ should be the best possible lower bound (since $1$ and $2$ are impossible or obvious reasons). We can find examples where this bound is sharp by factoring $2^k-3$ - many prime factors should have $2$ as a primitive root. Some examples with $k/\log p$ small are $5,13,29, 61, 509$.
Apr 8 at 17:12 history edited Somudro Gupto CC BY-SA 4.0
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Apr 8 at 16:43 history edited Somudro Gupto CC BY-SA 4.0
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Apr 8 at 16:43 history edited Somudro Gupto CC BY-SA 4.0
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Apr 8 at 16:42 history edited Somudro Gupto CC BY-SA 4.0
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Apr 8 at 16:40 history edited Somudro Gupto CC BY-SA 4.0
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S Apr 8 at 16:40 review First questions
Apr 8 at 19:56
S Apr 8 at 16:40 history asked Somudro Gupto CC BY-SA 4.0