Timeline for Non-existence of a prime generating polynomial recurrence relation
Current License: CC BY-SA 3.0
7 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Dec 11, 2015 at 0:22 | vote | accept | Gheehyun Nahm | ||
| Dec 10, 2015 at 17:13 | comment | added | Qiaochu Yuan | @Wojowu: that's a geometric series; you can sum it. | |
| Dec 10, 2015 at 14:47 | comment | added | David E Speyer | Regarding $a r^n + b$ -- Let $p=ar+b$, by hypothesis it is prime. Then $a r^p+b \equiv ar+b \equiv 0 \bmod p$. So $a r^p+b$ is not prime. | |
| Dec 10, 2015 at 13:49 | comment | added | Wojowu | @QiaochuYuan Repeatedly applying linear function will give you not $a+n=ar^n+b$, but rather $a_n=a^na_0+a^{n-1}b+...+ab+b$. | |
| Dec 10, 2015 at 7:52 | comment | added | Qiaochu Yuan | @Yaakov: even then I think it's hard. In that case $a_n = a r^n + b$ for integers $a, r, b$ and I'm not sure how to rule out the possibility that some such sequence has all prime values. For example, you might try finding some large prime $p$ with respect to which $r$ is a primitive root, and thereby hope to find a term divisible by $p$, but I don't know how to guarantee that such a $p$ exists (and as far as I know this sort of thing is also wide open). | |
| Dec 10, 2015 at 7:43 | comment | added | Yaakov Baruch | What about if $f$ is linear - is it a then a well known (trivial?) result? | |
| Dec 10, 2015 at 7:12 | history | answered | Qiaochu Yuan | CC BY-SA 3.0 |