Timeline for Ternary cyclotomic polynomials with $n=15r$
Current License: CC BY-SA 3.0
12 events
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Jun 15, 2020 at 7:27 | history | edited | CommunityBot |
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Feb 10, 2016 at 14:37 | comment | added | Yusuf Gurtas | My conjecture about the gap is probably not wrong but it's far from explaining the big picture. $q \mod p \not = \pm 1$ is a much better conjecture (not mine of course). | |
Jan 24, 2016 at 13:38 | comment | added | user9072 | Interesting observation. Thanks for sharing them. | |
Jan 24, 2016 at 4:56 | comment | added | Yusuf Gurtas | For example, it looks like for $3<p<q<r$ where $q=p+2,r=mpq+3$ the coefficient $c_{mpq+q}$ of $\Phi_n(x)$ is $-2$. Here $n=pqr$ and $ (p,q,r)$ is a prime triple. | |
Jan 24, 2016 at 0:06 | comment | added | Yusuf Gurtas | If my calculations are correct then $(5,7)$ is another pair like that. I mean the flat ones occur only within the margin of $\pm 1$. My wild guess is the gap $q-p=2$ is responsible for that. | |
Jan 23, 2016 at 20:13 | comment | added | user9072 | The pair $(3,7)$ is different in that $7$ is $1$ modulo $3$ and thus $\pm 2$ is flat. This is true for any pair $p<q$. I have no explanation for $\pm 10$. I have no good intuition on this problem. One could do some numerical tests. | |
Jan 23, 2016 at 20:08 | comment | added | Yusuf Gurtas | Is $(3,5)$ the only pair where the flat ones are squeezed in the margin of $\pm 1$? For example the "next" pair $(3,7)$ has a "gap" for $w=10,11$. I mean the flat ones occur at margin of $\pm 1,\pm 2$ as well as $\pm 10$ using the theorem you mentioned. | |
Jan 23, 2016 at 19:45 | comment | added | Yusuf Gurtas | Not a problem. It answers my question. Thank you very much. | |
Jan 23, 2016 at 19:43 | vote | accept | Yusuf Gurtas | ||
Jan 23, 2016 at 18:51 | comment | added | user9072 | Come to think of it I could have used smaller primes. But I will leave it, at least for now. | |
Jan 23, 2016 at 18:48 | history | edited | user9072 | CC BY-SA 3.0 |
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Jan 23, 2016 at 18:23 | history | answered | user9072 | CC BY-SA 3.0 |