Timeline for Growth of $r_{2}(n)$
Current License: CC BY-SA 3.0
7 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Aug 19, 2018 at 23:23 | comment | added | GH from MO | If you like my answer, please accept it officially (so that it turns green). Thanks in advance! | |
| Dec 23, 2014 at 17:00 | comment | added | Noam D. Elkies | Sorry, if I had references to hand I'd have given an answer rather than a comment. I see that meanwhile GH from MO supplied a more thorough answer with sources, so I refer you to his answer for further information. | |
| Dec 23, 2014 at 12:13 | answer | added | GH from MO | timeline score: 18 | |
| Dec 23, 2014 at 6:55 | answer | added | Charles | timeline score: 3 | |
| Dec 23, 2014 at 6:50 | comment | added | M.Souf | Thank you Noam D. Elkies, could you give me the references for these statements. | |
| Dec 23, 2014 at 6:37 | comment | added | Noam D. Elkies | It is known that $r_2(n)$ is usually zero: the number of $n<x$ such that $r_2(n) > 0$ is asymptotic to a multiple of $x\,\big/\sqrt{\log x} = o(x)$. But if $n$ is a product of $k$ distinct primes each congruent to $1 \bmod 4$ then $r_2(n) = 2^{k+2}$, which is $n^{o(1)}$ but grows much faster than polynomial in $\log n$ if we use the first $k$ primes in that congruence class. | |
| Dec 23, 2014 at 6:00 | history | asked | M.Souf | CC BY-SA 3.0 |