Timeline for Fibonacci-Motzkin paths and J-type continued fractions
Current License: CC BY-SA 4.0
12 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| May 5, 2021 at 23:24 | comment | added | Peter Taylor | @InesInstitoris, I don't know. The closest I've been able to find in about twenty minutes' research is the Hankel continued fractions of this paper, but they should have numerators which are cubic monomials rather than cubic polynomials. | |
| May 5, 2021 at 17:06 | comment | added | Jeanne Scott | Thanks Peter. Does this kind of continued fraction (at the very end of your post) have name (i.e. it's like a $J$-type except for the extra quadratic and cubic terms which occur in the expansion)? | |
| May 5, 2021 at 16:19 | vote | accept | Jeanne Scott | ||
| May 5, 2021 at 7:52 | history | edited | Peter Taylor | CC BY-SA 4.0 |
Horizontals come before descents
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| May 5, 2021 at 7:46 | comment | added | Peter Taylor | Ah, I got the definition of the Fibonacci-Motzkin paths slightly wrong. | |
| May 5, 2021 at 7:07 | comment | added | Peter Taylor | @InesInstitoris, the square consists of one term for the level-0 $\rightarrow$ before moving to level 1 and one term for the level-0 $\rightarrow$ after coming back from level 1. | |
| May 5, 2021 at 2:12 | comment | added | Jeanne Scott | To put it differently, shouldn't $a_k = z^2\beta_k (1 -z\gamma_k )^{-1}$ for $k \geq 1$ ? | |
| May 5, 2021 at 0:42 | comment | added | Jeanne Scott | Otherwise we don't recover the Fibonacci generating function $(1 - z -z^2)^{-1}$ upon specializing all the $\gamma$'s and $\beta$'s to 1. | |
| May 4, 2021 at 23:29 | comment | added | Jeanne Scott | In fact, shouldn't all the denominators be multiplicity free --- i.e. the height $k$-terms contributing $z^{2k}\beta_1 \cdots \beta_k (1 - z\gamma_0)^{-1} \cdots (1 - z\gamma_k)^{-1}$ ? | |
| May 4, 2021 at 23:23 | comment | added | Jeanne Scott | Why do you have a square $(1 -z\gamma_0)^2$ in the denominator for the $\beta_1$-terms? It seems to me that the height one terms contribute $z^2 \beta_1 \sum_{n \geq 0} {\gamma_0^{n+1} -\gamma_1^{n+1} \over {\gamma_0 - \gamma_1}} z^n = {z^2\beta_1 \over {(1-z\gamma_0)(1-z\gamma_1) } }$. | |
| May 4, 2021 at 22:49 | history | edited | Peter Taylor | CC BY-SA 4.0 |
added 21 characters in body
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| May 4, 2021 at 22:20 | history | answered | Peter Taylor | CC BY-SA 4.0 |