Timeline for Infinitely many primes of the form $2^n+c$ as $n$ varies?
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| Nov 13, 2009 at 10:43 | comment | added | Boris Bukh | I share your anxiety about these heuristics in general, and extreme naivete of mine in particular. For example, by these heuristics would give that if alpha is a real number, then floor(alpha^(3^n)) is prime finitely often. However, it is false for some alpha. | |
| Nov 13, 2009 at 10:17 | comment | added | Kevin Buzzard | Ok here is an explicit comment about your heuristic that I hope worries you. If 2 is not a primitive root mod a prime number q, then q might never divide 2^n+c (e.g. 7 never divides 2^n+9 and surely there will be other primes, possibly infinitely many more, with this property---certainly 2 never divides 2^n+9 either). Hence 2^n+9 is less likely to be composite than a random large number. For a proper heuristic you need to take this into account and recalculate. What I'm saying is that your heuristic might be too naive, and fixing it up to incorporate my comments might give a different answer | |
| Nov 13, 2009 at 10:13 | comment | added | Kevin Buzzard | Actually, suddenly I am nervous about this argument. If c=1 then one can say a lot about possible factors of 2^n+c, for example, and a competing heuristic on the Wikipedia page for Fermat primes seems to say that incorporating these facts screws up the heuristic that there are only finitely many Fermat primes. Does a general heuristic look like this: "here are some things I can think of, now let's assume everything else is random and sum 1/log"? So implicit in such a heuristic is the assertion that you've not missed anything? | |
| Nov 13, 2009 at 9:19 | comment | added | Kevin Buzzard | In general I am nervous about these sorts of heuristics, because applied naively they would predict infinitely many Fermat primes, whereas applied more sensibly (i.e. think about which n can occur from an elementary viewpoint first and then heuristicise) they predict finitely many. On the other hand I don't think this specific objection (my nervousness about heuristics) applies to your heuristics (because if we made intelligent elementary observations which ruled out certain n first then the sum would only get smaller). So I think you've probably done (2) and (3). | |
| Nov 13, 2009 at 9:13 | history | edited | Boris Bukh | CC BY-SA 2.5 |
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| Nov 13, 2009 at 9:08 | history | answered | Boris Bukh | CC BY-SA 2.5 |