Timeline for Restricted integer partitions modulo k
Current License: CC BY-SA 3.0
3 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jan 2, 2016 at 13:21 | comment | added | Fedor Petrov | Yes, something like this, but is it minimal? | |
| Jan 2, 2016 at 12:23 | comment | added | Ofir Gorodetsky | At least when $k$ is squarefree, it seems we can take the period as $m! k^m$. Explanation: Let $f(k,m)$ be the period. Use the recurrence $p(n,m) = p(n-m, m) + p(n-1,m-1)$ to deduce that $a_n \mod k$ satisfies the recurrence corresponding to the polynomial $(1-x^m)(1-x^{f(k,m-1)})$. We can find a multiple of $f(k,m)$ by finding $t$ satisfying $(1-x^m)(1-x^{f(k,m-1)}) \mid 1-x^t \mod k$. In characteristic 0 this is unsolvable ($1-x^n$ has no repeated roots), but for instance, if $k$ is a prime, we can take $t = k \cdot m \cdot f(k,m-1)$, hence $f(k,m) \mid km \cdot f(k,m-1)$. | |
| Jan 2, 2016 at 11:56 | history | answered | Fedor Petrov | CC BY-SA 3.0 |