Timeline for Partition of a number as the sum of k integers, with repetitions but without counting permutations.
Current License: CC BY-SA 4.0
4 events
| when toggle format | what | by | license | comment | |
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| Jun 5 at 0:31 | comment | added | TravorLZH | You can prove using Ferrer diagram that the number of ways to partition some number into $\le k$ components is the same as the number of ways to partition some number into arbitrarily number of components, each $\le k$. | |
| Jun 3 at 11:28 | comment | added | SomeCallMeTim | To me, it seems unlikely. You would need to find a subtitute for $f$ with singularities only at one of each permutation of each partitition - or perhaps quotienting the space in some way to remove the multiplicity coming from permutations. Distinguishing between such permutations in an analytic way seems difficult. "Fixing" this method for your needs would basically constitute a new method, would be my guess. | |
| S May 30 at 16:11 | review | First questions | |||
| May 30 at 16:36 | |||||
| S May 30 at 16:11 | history | asked | Lorenzo Alvarado | CC BY-SA 4.0 |