Timeline for Combinatorial identity involving number of cycles (of any length) in a permutation
Current License: CC BY-SA 3.0
13 events
| when toggle format | what | by | license | comment | |
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| Jul 24, 2016 at 20:40 | answer | added | Brian Hopkins | timeline score: 3 | |
| Jul 24, 2016 at 20:11 | comment | added | Brian Hopkins | I wonder if Hanlon is missing a $\text{sgn}(\sigma)$ term in the summation. Working out the $S_3$ example he has running through the article gives $2\alpha^2 - 3 \alpha + 1$ for the product but $2\alpha^2 + 3\alpha + 1$ for the sum (which is $\beta^3-3\beta^2+2\beta$ vs. $\beta^3 + 3\beta^2 + 2\beta$ in your formulation). | |
| Jul 24, 2016 at 20:08 | answer | added | Linus Hamilton | timeline score: 7 | |
| Jul 24, 2016 at 19:57 | history | edited | Ben McKay | CC BY-SA 3.0 |
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| Jul 24, 2016 at 19:53 | comment | added | I.K | Yes. I rewrote it. I used $ \beta = \alpha^{-1} $ Then it's the same equality, isn't it? | |
| Jul 24, 2016 at 19:51 | comment | added | Brian Hopkins | Looking at the paper, I think it's $\sum_{i=0}^{n-1} (1-i\alpha) = \prod_{\sigma \in S_n} \alpha^{n-c(\sigma)}$. | |
| Jul 24, 2016 at 19:51 | comment | added | I.K | I edited the question, I'm pretty sure that from the context, beta is positive strictly less than 1. | |
| Jul 24, 2016 at 19:48 | history | edited | I.K | CC BY-SA 3.0 |
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| Jul 24, 2016 at 19:46 | comment | added | Linus Hamilton | Are you missing a sign somewhere? If $\beta=1$ then the LHS is zero but the RHS is $n!$. EDIT: for a fixed $n$, both sides are polynomials in $\beta$, so if they are equal for $0<\beta<1$ then they are equal for $\beta=1$ too. | |
| Jul 24, 2016 at 19:44 | history | edited | I.K | CC BY-SA 3.0 |
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| Jul 24, 2016 at 19:39 | history | edited | Michael Albanese | CC BY-SA 3.0 |
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| Jul 24, 2016 at 19:37 | review | First posts | |||
| Jul 24, 2016 at 19:39 | |||||
| Jul 24, 2016 at 19:37 | history | asked | I.K | CC BY-SA 3.0 |