Timeline for answer to Examples of integer sequences coincidences by Per Alexandersson
Current License: CC BY-SA 3.0
Post Revisions
7 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jan 8, 2018 at 3:50 | history | made wiki | Post Made Community Wiki by Todd Trimble | ||
| Jan 5, 2018 at 15:22 | comment | added | Max Alekseyev | I've edited the sequence and removed "conjectured" from the formula. | |
| Jan 5, 2018 at 14:15 | comment | added | Per Alexandersson | @MaxAlekseyev: Yes, indeed. I like the proof given in the comment - I have another family of combinatorial objects that seem to be in bijection with these, so hence, I was interested in the proof :) | |
| Jan 5, 2018 at 13:35 | comment | added | Max Alekseyev | "No proof" often means that the sequence did not attract enough attention. | |
| Jan 5, 2018 at 13:34 | comment | added | Max Alekseyev | Similar to oeis.org/A216239 | |
| Jan 5, 2018 at 13:09 | comment | added | user133281 | I understand that your point is more general, but the formula for $a_n$ you mention can -- if I'm not mistaken -- be proven by counting the number of cyclic permutations $\sigma$ for which $\{a,b\}$ is an inversion. Indeed, if $a<b$ we must have $\sigma(a)>\sigma(b)$ which (as $\sigma(a)=a$, $\sigma(b)=b$, and $\{a,b\}=\{\sigma(a),\sigma(b)\}$ are disallowed) gives $\binom{n}{2} - n + b -a$ options for $(\sigma(a),\sigma(b))$. Each of these options can be extended to $(n-3)!$ cyclic permutations. So $a_n = (n-3)! \sum_{a < b} (\binom{n}{2} - n + b - a) = n!(3n-1)/12$. | |
| Jan 5, 2018 at 11:51 | history | answered | Per Alexandersson | CC BY-SA 3.0 |