Timeline for unique integer partitions
Current License: CC BY-SA 2.5
3 events
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| Jul 10, 2010 at 1:58 | comment | added | Qiaochu Yuan | It's always a good idea to give your actual numerical constraints. It is actually quite feasible to read off the leading terms of p_k(n) in terms of n from the generating function; the leading term is something like 1/k! {n+k-1 choose k} and this should give a pretty reasonable approximation. | |
| Jul 10, 2010 at 1:45 | comment | added | B Rivera | +1 thanks for the answer. perhaps i should have given my actual numerical constraints. i was looking for some type of method that would allow me to solve for cases of $n > 10^{6}$ and $k > 1000$ in a reasonable amount of time. perhaps it is not possible? | |
| Jul 9, 2010 at 2:29 | history | answered | Qiaochu Yuan | CC BY-SA 2.5 |