Timeline for Permanent of a matrix
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
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| Feb 9, 2018 at 20:52 | comment | added | Mare | Thanks to the help/suggestions of Martin Rubey a proof has now been found. I might post it soon. | |
| Aug 31, 2017 at 10:46 | history | edited | Fedor Petrov | CC BY-SA 3.0 |
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| Aug 29, 2017 at 21:25 | comment | added | Martin Rubey | The formulas should now follow from en.wikipedia.org/wiki/Eulerian_number#Identities, very likely the first one. Note the exceedances and ascents have the same distribution via the so-called first fundamental transformation, findstat.org/MapsDatabase/Mp00087. | |
| Aug 29, 2017 at 21:18 | comment | added | Mare | @MartinRubey thanks, does this explain the infinite sum identities? | |
| Aug 29, 2017 at 21:14 | comment | added | Martin Rubey | More precisely: take out a factor $a$ from each row, move your first row last, then compute the permanent from its definition. | |
| Aug 29, 2017 at 21:13 | comment | added | Martin Rubey | $M(a, b, n)$ is $a^n$ times the generating polynomial for the number of exceedances findstat.org/StatisticsDatabase/St000155 (in $b$). | |
| Aug 29, 2017 at 21:09 | comment | added | Ofir Gorodetsky | A possible approach: There is an identity that relates the permanent of $A+B$ to the permanents of submatrices of $A$ and $B$ (see, for instance, this paper tandfonline.com/doi/abs/10.1080/03081088708817770 ). By writing your matrix as $aJ+B$ where $J$ is the all-ones matrix, this identity leaves you the following problem: calculate the permanent of your matrix, and its submatrices, in the case $a=0$. | |
| Aug 29, 2017 at 20:56 | history | edited | Mare | CC BY-SA 3.0 |
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| Aug 29, 2017 at 20:47 | history | asked | Mare | CC BY-SA 3.0 |