Timeline for Rationality of the sum of the reciprocals of the values of a polynomial function at the positive integers
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Oct 3, 2016 at 21:09 | comment | added | Stefan Kohl♦ | @TimothyChow: This is quite possible. Do you know of any heuristics which suggest that $S(f)$ is irrational except in cases where it is obviously rational (I mean one which is better than "the rationals are countable, but the irrationals are not, so assuming some 'well-behaved' kind of random distribution, almost all $S(f)$ should be irrational")? | |
| Oct 3, 2016 at 18:04 | comment | added | Timothy Chow | Asking about algorithmic decidability is probably the wrong question. It could be that $S(f)$ is irrational except in certain cases where rationality is obvious. Then it would be decidable, but we wouldn't be able to prove that it is decidable. I think you're really just interested in which cases the irrationality is known. | |
| Oct 3, 2016 at 10:56 | answer | added | joro | timeline score: 1 | |
| Oct 3, 2016 at 1:12 | answer | added | Valborf | timeline score: 1 | |
| Oct 2, 2016 at 22:17 | comment | added | Stefan Kohl♦ | @SashaP: As far as I know, yes. -- But still it is conceivable that someone knows how to prove algorithmic undecidability, or can give criteria which apply to classes of polynomials which do not include $n^{2k+1}$. | |
| Oct 2, 2016 at 21:40 | comment | added | T. Amdeberhan | I'd be very surprised if there is any such. | |
| Oct 2, 2016 at 20:47 | comment | added | SashaP | Isn't it already open for arbitrary $n^{2k+1}$? | |
| Oct 2, 2016 at 20:46 | history | asked | Stefan Kohl♦ | CC BY-SA 3.0 |