Timeline for Combinatorics problem related to Motzkin numbers with prize money I
Current License: CC BY-SA 3.0
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48 events
| when toggle format | what | by | license | comment | |
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| S Aug 18, 2017 at 14:07 | history | bounty ended | Mare | ||
| S Aug 18, 2017 at 14:07 | history | notice removed | Mare | ||
| Aug 17, 2017 at 2:50 | history | edited | Mare | CC BY-SA 3.0 |
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| Aug 17, 2017 at 2:28 | history | edited | Mare | CC BY-SA 3.0 |
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| Aug 17, 2017 at 2:25 | vote | accept | Mare | ||
| S Aug 17, 2017 at 2:25 | history | bounty started | Mare | ||
| S Aug 17, 2017 at 2:25 | history | notice added | Mare | Draw attention | |
| S Aug 16, 2017 at 19:54 | history | bounty ended | Mare | ||
| S Aug 16, 2017 at 19:54 | history | notice removed | Mare | ||
| Aug 16, 2017 at 19:53 | history | edited | Mare | CC BY-SA 3.0 |
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| Aug 11, 2017 at 22:35 | comment | added | Gerhard Paseman | After the last of the prize money is awarded, I recommend editing the title to remove the phrase "prize money". In general, there may be issues with advertising problems this way; for future, I suggest bounties to start and mention in the beginning of the post (and not in the title) any status regarding bounty or prize money, updated responsibly. Gerhard "Doesn't Do Raffles Here Either" Paseman, 2017.08.11. | |
| Aug 11, 2017 at 22:01 | history | edited | Mare | CC BY-SA 3.0 |
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| Aug 11, 2017 at 15:20 | history | edited | Mare | CC BY-SA 3.0 |
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| Aug 11, 2017 at 14:32 | answer | added | Anton Mellit | timeline score: 3 | |
| S Aug 10, 2017 at 22:20 | history | bounty started | Mare | ||
| S Aug 10, 2017 at 22:20 | history | notice added | Mare | Draw attention | |
| Aug 9, 2017 at 19:54 | answer | added | FindStat | timeline score: 8 | |
| Aug 9, 2017 at 17:00 | history | edited | Mare | CC BY-SA 3.0 |
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| S Aug 9, 2017 at 12:55 | history | suggested | Christian Stump | CC BY-SA 3.0 |
(hopefully) fixed definition of d_i
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| Aug 9, 2017 at 12:23 | review | Suggested edits | |||
| S Aug 9, 2017 at 12:55 | |||||
| Aug 9, 2017 at 8:38 | history | edited | Mare | CC BY-SA 3.0 |
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| Aug 9, 2017 at 8:12 | comment | added | Mare | @GerryMyerson It is Herbert Kupisch. Sorry for the typos. | |
| Aug 9, 2017 at 8:09 | history | edited | Mare | CC BY-SA 3.0 |
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| Aug 9, 2017 at 8:05 | comment | added | Mare | @MartinRubey cant thank you enough. Im very sorry, it is $d_{a+b-1} = d_{a+b + c_{a+b}-1} - c_{a+b}$ instead of $d_{a+b} = d_{a+b + c_{a+b}-1} - c_{a+b}$. Was a little late when I posted the question... Now in this example one has $c_2=2<c_1=3$ but $d_2=2$ and $d_{2+c_2-1}=d_3=3=d_1+c_2=1+2$. For $a=1$ and $b=2$ one has $c_3 =2 \geq 2=c_2$. | |
| Aug 9, 2017 at 7:58 | history | edited | Mare | CC BY-SA 3.0 |
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| Aug 9, 2017 at 0:23 | comment | added | Martin Rubey | My next problem is the 2-Gorenstein condition. For example, if $c=[3,2,2,2,1]$, I get $d=[1,2,3,2,2]$, only descent $a=1$, $X_1=\{1,2\}$. Now for $a=1$ and $b=1$ I have neither $c_2 \geq c_1$ nor $d_2 = d_{2+c_2-1} -c_2$. | |
| Aug 9, 2017 at 0:00 | comment | added | Martin Rubey | The position of a descent above is (essentially) the number of down steps before an occurrence of a factor DUU in the Dyck path. | |
| Aug 8, 2017 at 23:43 | comment | added | Gerry Myerson | Is it Kupish, or Kupisch? Both appear more than once in the body of the question. Perhaps someone who knows could edit. | |
| Aug 8, 2017 at 23:31 | comment | added | Martin Rubey | So, essentially, the Kupisch and the co-Kupisch sequence are the sequences of heights of the down and up steps, respectively, after prepending an up-step and appending a down-step to your Dyck path. | |
| Aug 8, 2017 at 21:01 | comment | added | Mare | In fact I used not my usual convention that one start with $c_0$ and goes to $c_{n-1}$ for Nakayama algebras (so calculating mod n is more natural, as is needed when the quiver is a circle), I hope this leads not to more such things. | |
| Aug 8, 2017 at 20:59 | history | edited | Mare | CC BY-SA 3.0 |
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| Aug 8, 2017 at 20:58 | comment | added | Mare | @MartinRubey Thanks for testing. I think the problem is resolved when looking at the $c_i$ mod $n$ so they are defined for every $i \in \mathbb{Z}$. Then $2 \geq c_{2-2}=c_0=c_n=1$ and thus $d_2=2$ always as it should be by using quiver interpretation. | |
| Aug 8, 2017 at 20:50 | comment | added | Martin Rubey | I still do not understand the definition of co-Kupisch: for $c=[3,2,1]$, we want $d_2$ to be the smallest $k$ which is at least $c_{2-k}$. For $k=0$, we have $0 \not\geq c_2$, and for $k=1$ we have $1\not\geq c_1$. | |
| Aug 8, 2017 at 20:16 | comment | added | Mare | I added the representation theoretic background. $d_1$ is always 1, I added that. | |
| Aug 8, 2017 at 20:15 | history | edited | Mare | CC BY-SA 3.0 |
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| Aug 8, 2017 at 20:10 | comment | added | Sylvain JULIEN | Is it me or does the sequence of Motzkin numbers grow like the sequence of Fibonacci numbers of even index ? | |
| Aug 8, 2017 at 19:46 | comment | added | Martin Rubey | might there be a typo in the definition of the co-Kupisch sequence? For example, let $c=[3,2,1]$. Then $d_1$ is supposed to be the smallest number $k$ which is at least $c_{1-k}$, but there is no such number, is there? | |
| Aug 8, 2017 at 19:46 | comment | added | Gerhard Paseman | If the sets X are intervals of integers and are only used in one place, then I see them as obscuring the definition. It would be more clear to me to just mention descents and say "b with $c_{a-1} \leq b \lt c_a$". Just having the current nomenclature for the objects is a mental challenge, as it defies not suggest a picture or framework for the named objects. Gerhard "Use Surnames For Holiday Cards" Paseman, 2017.08.08. | |
| S Aug 8, 2017 at 19:42 | history | suggested | Alex Meiburg | CC BY-SA 3.0 |
missing the $\ge 2$
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| Aug 8, 2017 at 19:30 | review | Suggested edits | |||
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| Aug 8, 2017 at 19:18 | comment | added | Mare | @AlexMeiburg sorry, I forgot the condition that $c_i \geq 2$ for all $ i \neq n$. | |
| Aug 8, 2017 at 19:15 | history | edited | Mare | CC BY-SA 3.0 |
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| Aug 8, 2017 at 19:11 | comment | added | Alex Meiburg | Is there a constraint that that in an $n$-Kupish series, only the last term may be $1$? Otherwise I don't see why (for instance) [1,1,1,1,1] is not 5-Kupish as well. | |
| Aug 8, 2017 at 18:10 | history | edited | Mare | CC BY-SA 3.0 |
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| Aug 8, 2017 at 17:55 | history | edited | Mare | CC BY-SA 3.0 |
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| S Aug 8, 2017 at 17:52 | history | suggested | Jacob White | CC BY-SA 3.0 |
edited for grammar, and replacing the word 'special' with the word 'descent', which is often used in combinatorics.
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| Aug 8, 2017 at 17:51 | review | Suggested edits | |||
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| Aug 8, 2017 at 17:41 | history | asked | Mare | CC BY-SA 3.0 |