Timeline for Sum of products of binomials
Current License: CC BY-SA 3.0
14 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Feb 23, 2018 at 17:29 | answer | added | Zach Teitler | timeline score: 2 | |
| Feb 20, 2018 at 13:46 | comment | added | Simone Melchiorre Chiarello | @ZachTeitler you are right, the question as I wrote it suggests your answer. In my mind I was hoping some interpretations with lattice paths or graphs, but I guess I can figure it out using the obvious combinatorial interpretation. I upvote you because it actually helped! IraGessel that's amazing! Abdelmalek I don't bother you with my actual matrix, but it is not the product of two matrices with easier determinants. Thank you all anyway! | |
| Feb 20, 2018 at 2:06 | comment | added | Ira Gessel | You can use Petkovšek's algorithm (en.wikipedia.org/wiki/Petkovšek%27s_algorithm) to prove that there is no closed form for the sum. | |
| Feb 19, 2018 at 18:10 | comment | added | Abdelmalek Abdesselam | why does Binet not work? which determinant are you trying to compute? | |
| Feb 19, 2018 at 16:57 | comment | added | Zach Teitler | You ask for a combinatorial interpretation. Have you already considered the obvious one? It's the number of ways to choose $k$ objects out of $a+b+1$, where the first $a$ objects are chosen without repetition, the last $b+1$ with repetition. This is exactly the same interpretation that arises from $(1+q)^a(1-q)^{-1-b}$, so I don't know if it helps, but since you asked for a combinatorial interpretation... | |
| Feb 19, 2018 at 16:17 | answer | added | user64494 | timeline score: 1 | |
| Feb 19, 2018 at 16:13 | answer | added | Andrey Rukhin | timeline score: 4 | |
| Feb 19, 2018 at 15:16 | history | edited | Simone Melchiorre Chiarello | CC BY-SA 3.0 |
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| Feb 19, 2018 at 14:00 | comment | added | Simone Melchiorre Chiarello | Unfortunately Binet's formula does not seem to work in my case. I think a combinatorial interpretation would deal with it better. | |
| Feb 19, 2018 at 13:44 | comment | added | Abdelmalek Abdesselam | The sum is good for computing determinants because det(AB)=(det A)(det B). | |
| Feb 19, 2018 at 12:07 | comment | added | Henri Cohen | Using Leibnitz' formula one has $$S(a,b)=\sum_{j=0}^{\min(a,b)}(-1)^j\binom{k+b-j}{k}\binom{a}{j}2^{a-j}$$ but I don't know if that helps. | |
| Feb 19, 2018 at 11:38 | comment | added | Simone Melchiorre Chiarello | Thank you for the trial, yes I confirm your calculation. Anyway, for future people reading this, even some combinatorial explanation could be useful. | |
| Feb 19, 2018 at 11:33 | comment | added | Christian Stump | If I computed it correctly (please recheck), then $S(12,13) = 3 \cdot 101 \cdot 1370899$ for $k = \min\{a,b\} = 12$. It thus looks impossible to me to provide any product formula not involving any sum. (Looking for big prime factors is my first test to check to hope for product formulas.) | |
| Feb 19, 2018 at 10:23 | history | asked | Simone Melchiorre Chiarello | CC BY-SA 3.0 |