Timeline for A generalization of partition function to the sums of squares
Current License: CC BY-SA 4.0
7 events
| when toggle format | what | by | license | comment | |
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| Nov 18, 2020 at 18:04 | comment | added | Brian Hopkins | $j$ appears in the left-hand side of the second generating function; it is the desired exponent of $z$, which tells the number of parts. Neither $n$ nor $j$ appear in the right-hand side of the generating function; $k$ does only because it is fixed and dictates the allowed power of summands. A good source for this material is Herb Wilf's book generatingfunctionality which is freely available online. | |
| Nov 18, 2020 at 8:58 | comment | added | user167505 | Why did the second generating function's value not include $j$? | |
| Nov 18, 2020 at 4:13 | comment | added | Brian Hopkins | I was following the usage of $k$ in the articles, which is different than in your question. If you just care about squares, set $k=2$ in my answer. The first generating function gives the number of partitions of $n$ into any number of squares, which is what I think is what you mean by $\sum_{k=0} r_k(n) x^n$. | |
| Nov 18, 2020 at 3:59 | comment | added | user167505 | Can you evaluate $\sum_{k=0}r_{k}(n)x^n$ rather than $\sum_{n=0}r_{k}(n)x^n$? It would be more helpful. Sorry, I am not good at evaluating generating functions. | |
| Nov 18, 2020 at 3:52 | vote | accept | CommunityBot | ||
| Nov 18, 2020 at 1:57 | vote | accept | CommunityBot | ||
| Nov 18, 2020 at 3:51 | |||||
| Nov 17, 2020 at 18:14 | history | answered | Brian Hopkins | CC BY-SA 4.0 |