Timeline for OEIS Sequence A002846 and properties of matrix inverses
Current License: CC BY-SA 3.0
21 events
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| Mar 14, 2018 at 9:39 | history | edited | domotorp |
added oeis tag
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| Aug 3, 2017 at 9:38 | vote | accept | Helmut | ||
| Aug 3, 2017 at 9:37 | answer | added | Helmut | timeline score: 4 | |
| Jul 20, 2017 at 4:54 | comment | added | მამუკა ჯიბლაძე | @Helmut Actually exactly one of them has all entries nonzero along the diagonal (and determinant is the product of these). So a positive linear combination is nonsingular if this one enters with nonzero coefficient in the combination (but also this is not necessary) | |
| Jul 20, 2017 at 2:55 | comment | added | Gerry Myerson | Sorry, Helmut, I overlooked the "linear combinations" part of the question. | |
| Jul 19, 2017 at 23:35 | comment | added | Helmut | @Gerry Myerson. True. But linear combinations of the matrices are generally non-singular (in fact with probability 1 if the coefficients are selected randomly). | |
| Jul 19, 2017 at 22:32 | comment | added | Gerry Myerson | Of the 15 matrices given, 13 have an all-zero column (and thus are singular). | |
| Jul 19, 2017 at 21:50 | comment | added | მამუკა ჯიბლაძე | @AlexM. Do you mean it is easy? Or known? | |
| Jul 19, 2017 at 12:41 | comment | added | Alex M. | @Helmut: This question might be better suited for Math.SE. While it is a research problem for you, it might not necessarily be considered so by the standards of the mathematics community. | |
| Jul 19, 2017 at 12:20 | comment | added | Helmut | @HughThomas I have corrected the errors (matrices should be nxn although top row and rightmost column are uninformative). Matrices are triangular above antidiagonal as you say. I have also added examples as suggested (the row entries are separated by dots for some reason). I hope this helps and thanks for your interest. | |
| Jul 19, 2017 at 12:15 | history | edited | Helmut | CC BY-SA 3.0 |
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| Jul 19, 2017 at 12:04 | comment | added | მამუკა ჯიბლაძე | Thus e. g. the $n$th (last) row is always $000\cdots01$, the first row is always $n00\cdots00$, and the second row is always $(n-2)10\cdots00$. | |
| Jul 19, 2017 at 12:01 | history | edited | Helmut | CC BY-SA 3.0 |
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| Jul 19, 2017 at 12:00 | comment | added | მამუკა ჯიბლაძე | @HughThomas I figured it out actually. Matrices are $n\times n$ lower triangular if by "$i$th row" one understands the $(n+1-i)$th row, i. e. refinements go from down to up. Example: $5\to41\to311\to2111\to11111$; \begin{align*}&\text{the $1$st row corresponding to $11111$ is $50000$,}\\&\text{the $2$nd row corresponding to $2111$ is $31000$;}\\&\text{the $3$rd row corresponding to $311$ is $20100$;}\\&\text{the $4$th row corresponding to $41$ is $10010$;}\\&\text{the $5$th (last) row corresponding to $5$ is $00001$.} \end{align*} | |
| Jul 19, 2017 at 12:00 | history | edited | Helmut | CC BY-SA 3.0 |
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| Jul 19, 2017 at 11:54 | history | edited | Helmut | CC BY-SA 3.0 |
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| Jul 19, 2017 at 11:24 | comment | added | Hugh Thomas | There seem to be some convention issues here which would be alleviated by including the relevant matrices for n=3 and n=4 in the question. (I did try to download the suggested file, but for whatever reason was not able to gunzip it.) If the 1st row is really supposed to represent the situation after 1-1=0 partitionings, then it would be a vector of length $n$, with a 1 in the final position, contrary to the assertion that the matrix is $n-1\times n-1$. Also, it seems to me that the resulting matrices are not upper triangular since they are above the antidiagonal not the main diagonal. | |
| Jul 19, 2017 at 10:38 | comment | added | მამუკა ჯიბლაძე | At the Wikipedia page it is explained that $a(n)$ is the number of chains of length $n-1$ in the poset of partitions with refinement order. Could you restate definition of your matrices in these terms? What I understand is that each of your matrices is determined by one such chain. Can you tell how to obtain it from that chain? | |
| Jul 19, 2017 at 10:26 | comment | added | Helmut | You will also see some R code at the site which you can run to generate the trees yourself. It takes a while to run for n=8 or 9. | |
| Jul 19, 2017 at 10:19 | comment | added | მამუკა ჯიბლაძე | What if I am not a python expert? | |
| Jul 19, 2017 at 9:50 | history | asked | Helmut | CC BY-SA 3.0 |