Timeline for A formula for this generating function that is similar to the $qt$-Catalan numbers
Current License: CC BY-SA 4.0
6 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Nov 12, 2020 at 23:00 | vote | accept | Drew | ||
| Nov 12, 2020 at 20:13 | answer | added | Gjergji Zaimi | timeline score: 7 | |
| Nov 12, 2020 at 19:11 | comment | added | Martin Rubey | I tried a bit to interpret the right hand side as a suitable specialization of the cycle index series for the species of permutations (i.e., the Frobenius character for the conjugation action of the symmetric group) but I didn't succeed. I would have thought that writing $\frac{1}{n (1-q^n) (1-t^n)}$ as $\frac{1}{(1-q)(1-t)}(n-1)! [n-1]_q! [n-1]_t! \frac{1}{n! [n]_q! [n]_t!}$ would be a good idea, but it seems that it isn't. | |
| Nov 12, 2020 at 18:37 | comment | added | Drew | You mean replace the z^n in the rhs with p_n, and then put some symmetric functions in the left hand side somehow, so that where you expand the symmetric functions in single variable z you recover the original? | |
| Nov 12, 2020 at 0:01 | comment | added | user35313 | Not an answer, but I am curious if there is a (conjectural?) symmetric function version of your claimed equality, so that we obtain the right hand side by a plethystic substitution into power sums, and the left hand side reduces to what you have there. | |
| Nov 11, 2020 at 23:07 | history | asked | Drew | CC BY-SA 4.0 |