Timeline for When are there infinitely many primes in a sequence generated by a simple recurrence relation?
Current License: CC BY-SA 4.0
8 events
| when toggle format | what | by | license | comment | |
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| May 29, 2022 at 2:22 | history | edited | Timothy Chow | CC BY-SA 4.0 |
Added a link and corrected the non-primality claim
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| Dec 29, 2016 at 18:20 | history | edited | Kevin Buzzard | CC BY-SA 3.0 |
added 9262111 comment; added 2017 question.
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| Dec 29, 2016 at 18:05 | comment | added | Kevin Buzzard | Thanks user102986. I had completely forgotten about my previous question; it absolutely qualifies as a covering congruence argument. I am not sure I can make my question more precise; I was just idly observing that I didn't understand the dichotomy on a conceptual level. | |
| S Dec 29, 2016 at 16:38 | comment | added | user102986 | (and maybe others I don't know about)." | |
| S Dec 29, 2016 at 16:38 | comment | added | user102986 | You might like to look at some of the answers to Buzzard's question here: mathoverflow.net/questions/5323/… (Why does $2^n+9262111$ not qualify as a covering congruence argument which applies in your case?) More seriously, in order to avoid a repeat of the low quality answers to that question, perhaps you can make your question more precise and ask what you really want to know, which seems to be something like "is there a known and accepted way to make conjectures about these questions which avoids all the pitfalls I just mentioned ( | |
| Dec 29, 2016 at 16:35 | comment | added | Aaron Meyerowitz | You can also interlace two or more sequences so have $3^k-1$ for $n=2k$ and $(4^k-1)/3$ for $n=2k-1.$ | |
| Dec 29, 2016 at 16:19 | answer | added | user40023 | timeline score: 17 | |
| Dec 29, 2016 at 15:55 | history | asked | Kevin Buzzard | CC BY-SA 3.0 |