Timeline for The number of partitions of a positive integer allowing at most r repetitions of any part
Current License: CC BY-SA 4.0
8 events
| when toggle format | what | by | license | comment | |
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| Apr 22, 2023 at 3:18 | comment | added | Gerry Myerson | As OP asks whether there is a recursive formula, I imagine OP would settle for something that does for $q_2(n)$ what Euler's Pentagonal Number Theorem does for $r=\infty$. I would recommend calculating a few values of $q_2(n)$ and then consulting the Online Encyclopedia of Integer Sequences, oeis.org – OK, I took my own advice, it's here: oeis.org/A000726 with many references and links. | |
| Apr 21, 2023 at 23:21 | comment | added | sqd | Thanks for the reference to Glaisher's theorem. | |
| Apr 21, 2023 at 23:19 | history | undeleted | sqd | ||
| Apr 21, 2023 at 23:18 | history | deleted | sqd | via Vote | |
| Apr 21, 2023 at 23:12 | comment | added | Sam Hopkins | See also: en.wikipedia.org/wiki/Glaisher%27s_theorem. | |
| Apr 21, 2023 at 23:06 | comment | added | Sam Hopkins | But it's unclear what you mean by a "formula." For example, what "formula" did you have in mind in the case $r=1$ (or $r=\infty$, for that matter)? | |
| Apr 21, 2023 at 22:53 | comment | added | Sam Hopkins | Stating the obvious: the generating function of these numbers is $\sum_{n\geq 0} q_r(n) x^n = \prod_{i\geq 1} \frac{1-x^{i(r+1)}}{1-x^i}$. | |
| Apr 21, 2023 at 22:36 | history | asked | sqd | CC BY-SA 4.0 |