Timeline for If $n=2m$, what is the order of the permutation $\sigma(k)=2k , \quad \sigma(m+k)=2k-1$
Current License: CC BY-SA 4.0
8 events
| when toggle format | what | by | license | comment | |
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| Feb 24, 2023 at 6:20 | comment | added | მამუკა ჯიბლაძე | Namely, orders of $68$ modulo $113$, $113^2$ and $113^3$ are all equal to $112$. | |
| Feb 24, 2023 at 6:19 | comment | added | მამუკა ჯიბლაძე | Sorry, this is not actually true. The smallest counterexample I found is this: order of $2$ modulo $1093$ is the same as order of $2$ modulo $1093^2$, namely $364$. For a proof that once the order of some $1<a<p$ modulo $p^k$ is divisible by $p$ then order of $a$ modulo $p^{k+1}$ is $p$ times order of $a$ modulo $p^k$, see math.stackexchange.com/a/2217138/214353 It seems also true that these orders remain the same before becoming divisible by $p$. The smallest example when order modulo $p^3$ is not divisible by $p$ seems to be $p=113$, $a=68$. | |
| Feb 23, 2023 at 17:11 | comment | added | Christophe Leuridan | @მამუკაჯიბლაძე Is it easy to prove that the order of 2 modulo $p^\nu$ is $p^{\nu-1}$ times the order of 2 modulo $p$? I am a bit surprised by this statement. | |
| Feb 23, 2023 at 5:56 | comment | added | მამუკა ჯიბლაძე | In more detail, this reduces to the case when $2m+1$ is a prime $p$, since the order of $2$ modulo $p^\nu$ is $p^{\nu-1}$ times the order of $2$ modulo $p$, while the order of $2$ modulo $mn$ for $m$, $n$ coprime is the lcm of orders modulo $m$ and $n$. | |
| Feb 18, 2023 at 9:01 | vote | accept | ABB | ||
| Feb 18, 2023 at 9:01 | comment | added | ABB | @Per Alexandersson, Yes, you are right. If this problem could be figured out, then probably one may find an approach to compute the eigenvalues/eigenvectors of some non-symmetric discrete trigonometric transforms which are coming from signal processing in electrical engineering! | |
| Feb 18, 2023 at 7:18 | comment | added | Per Alexandersson | I wonder if this was what OP was actually interested in, but ABB reformulated it to a permutation type question.. | |
| Feb 17, 2023 at 20:25 | history | answered | Christophe Leuridan | CC BY-SA 4.0 |