Timeline for A simple looking problem in partitions that became increasingly complex
Current License: CC BY-SA 3.0
12 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jan 22, 2013 at 11:46 | vote | accept | Nilotpal Kanti Sinha | ||
| May 9, 2012 at 10:25 | answer | added | Aaron Meyerowitz | timeline score: 2 | |
| May 9, 2012 at 6:11 | comment | added | Daniel Parry | This might help as well. We can develop a generating function. If for (h,k)=1 and k≤n! we define s(h/k) as the number of solutions to the equation $$k_1+ \frac{k_2}{2} + \frac{k_3}{3} + ... \frac{k_n}{n} = \frac{h}{k}$$ then $$\sum_{(h,k)=1,\ k\le n!}s(h/k)e(\frac{h\tau}{k})= \prod_{j=1}^n \frac{1}{1-e(\tau/j) }.$$ | |
| May 8, 2012 at 12:56 | comment | added | Max Alekseyev | I asked a question about computing terms of related sequence oeis.org/A020473 at mathoverflow.net/questions/96334/… It may happen that the same approach can be used for efficient counting number of solutions to the discussed equation. | |
| May 8, 2012 at 5:48 | comment | added | Daniel Parry | It appears that the $k_1$ term might be unnecessary. If we define $s(n,m)$ to be the number of solutions to $$\frac{k_2}{2}+\frac{k_3}{3}+...\frac{k_n}{n}=m $$ then $s(n)=\sum_{i=0}^n s(n,i).$ | |
| May 8, 2012 at 4:36 | comment | added | Nilotpal Kanti Sinha | @Dougals, yes now actually, I am finding the second problem of restricted partitions more interesting than the one I started with. | |
| May 7, 2012 at 22:31 | answer | added | William J. Keith | timeline score: 3 | |
| May 7, 2012 at 15:22 | answer | added | Max Alekseyev | timeline score: 8 | |
| May 7, 2012 at 14:47 | comment | added | Douglas Zare | By weakening your original condition, you seem to be adding a huge number of spurious solutions. This seems to make the problem more difficult, and even a complete solution to the new problem wouldn't say much about the original. If you are interested in the original problem, I think you should turn around. Of course, you may find the second problem of restricted partitions more interesting. By the way, the partitions of $n$ into $k$ parts which are at least $m$ correspond with the partitions of $n−km$ into at most $k$ parts. | |
| May 7, 2012 at 14:46 | comment | added | Eric Naslund | It seems more natural if we define $\tilde{s}(n)$ to be the number of solutions to $$1=k_1 +\frac{k_2}{2}+\cdots+\frac{k_n}{n}.$$ This more closely mimics the partition problem which is the number of solutions to $$1=k_n+\frac{n-1}{n}k_{n-1}+\dots + \frac{2}{n}k_2+\frac{1}{n}k_1.$$ If $n=p$ is prime, then $\tilde{s}(n)=1+\tilde{s}(n-1)$ | |
| May 7, 2012 at 12:08 | comment | added | Brendan McKay | Do you agree with 1,3,10,55,196,2730,10032 ? Not in OEIS. | |
| May 7, 2012 at 11:27 | history | asked | Nilotpal Kanti Sinha | CC BY-SA 3.0 |