Timeline for Number of hereditary modules of a hereditary algebra
Current License: CC BY-SA 4.0
8 events
| when toggle format | what | by | license | comment | |
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| Oct 15, 2018 at 14:23 | comment | added | Julian Kuelshammer | Thanks. It seems then that $b_Q$ is not in OEIS although there are many sequence starting $2,6,23$ of which some seemed like they could be related. | |
| Oct 15, 2018 at 13:12 | comment | added | Mare | @JulianKuelshammer The next term is 114. I added it. | |
| Oct 15, 2018 at 13:12 | history | edited | Mare | CC BY-SA 4.0 |
added 3 characters in body
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| Oct 15, 2018 at 12:49 | comment | added | Mare | @JulianKuelshammer Yes the $n=2$ case should not be hard since there can occur no relations. Ill calculate now one more term for $b_Q$ but it takes forever and the next term will probably the last one my computer can do in a reasonable time. | |
| Oct 15, 2018 at 12:06 | comment | added | Julian Kuelshammer | Probably a better way to see this is to note that $a_{Q,2}(m)=\sum_{k=1}^{m}\frac{k(k-1)(k+4)}{6}$. What is summed over is the number of 2-element subsets which contain an injective module (on the $k$-th layer). There are $\frac{m(m+1)}{2}$ which contain two injectives and $\sum_{j=1}^{k-1}k(n-k)=\frac{k(k-1)(k+1)}{6}$ many which contain one injective and one other module. | |
| Oct 15, 2018 at 11:31 | comment | added | Julian Kuelshammer | For $a_{Q,2}$ I think you should be able to use the recursion $a_{Q,2}(m)=(m+1)m+a_{Q,2}(m-1)+(m-1)m(m+1)/6$ where $(m+1)m$ are all the pairs which contain a simple module, $a_{Q,2}(m-1)$ are all the sequences which come from the "upper subtriangle" in the Auslander-Reiten quiver and $(m-1)m(m+1)/6$ are the remaining ones that appear because in the "upper subtriangle" there is some interchange between zero-relations and commutativity relations. | |
| Oct 15, 2018 at 9:24 | comment | added | Julian Kuelshammer | Just out of curiosity: How does $b_Q$ continue, i.e. what are the next two numbers after 23? | |
| Oct 14, 2018 at 22:52 | history | asked | Mare | CC BY-SA 4.0 |