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In order to understand Theorem 9 from https://arxiv.org/pdf/1106.0952 I like to check with SageMath that

$$ x_{n} = \frac{1+x_{n-1}^r}{x_{n-2}}, \ \ \text{for any}\ n\geq 4, \ r\geq 2\ \text{ with initial conditions }\ x_1=x, \ x_2 = y \quad (x \text{ and } y\text{ are indeterminates})$$ agrees with $$x_{n}=\displaystyle\frac{\displaystyle\sum_{\beta\in\mathcal{F}(\mathcal{D}_n)}x_{1}^{r|\beta|_1}x_{2}^{r(c_{n-1}-|\beta|_2)}}{x_{1}^{c_{n-1}}x_{2}^{c_{n-2}}}$$ where the sequence $c_{n}$ is defined by the linear recurrence

$$c_{n} = r c_{n-1}-c_{n-2}$$

where $c_1=0$, $c_2=1$ (see page 3) and $\mathcal{F}(\mathcal{D}_{n})$ is the family of subpaths of the maximal Dyck path $\mathcal{D}_{n}$ from $(0,0)$ to $(c_{n-1}-c_{n-2},c_{n-2})$ with steps NORTH or EAST.

My question(s): for the following cases

r=2 for n = 4, 5, 6,...,12

r=3, n = 4, 5

r=4, n = 4, 5

r=5, n = 4, 5

r=6, n = 4, 5

up to

r=14, n = 4, 5

everything works fine.

But when $n\geq 6$ and $r\geq 3$ the two expressions for $x_n$ don't agree apparently. I am afraid I simply don't understand correctly the rule about adding preceding edges to another subpath if a green subpath is present in a given subset of subpaths of $\mathcal{F}(\mathcal{D}_n)$. For instance, how many preceding edges we should add to another subpath when $r=3$ and $n=6$ and $\alpha(i,k)$ is present ?

I think that when $r=3$ and $n=6$ all green subpaths are these:

$$\alpha(1, 3)$$ $$\alpha(1, 4)$$ $$\alpha(1, 5)$$ $$\alpha(1, 6)$$ $$\alpha(1, 7)$$ $$\alpha(1, 8)$$ $$\alpha(4, 6)$$ $$\alpha(4, 7)$$ $$\alpha(4, 8)$$ $$\alpha(6, 8)$$

If I am not mistaken, all paths above are $(3,1)$-green.

Is this true ?

Are they actually green ?

I am missing any subpath ?

Are all subpaths above actually $(3,1)$-green ?

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    Cross-posted: https://mathoverflow.net/questions/489634/number-of-preceding-edges-for-green-subpaths-of-a-maximal-dyck-path-for-r-geq-3 – RobPratt Mar 21 '25 at 17:26

1 Answers1

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My $\alpha(3,8)$ subpath color is wrong (Thanks to A. Burcroff and K. Lee.). Instead of red, the subpath $\alpha(3,8)$ is actually $(4,2)$-green (pointed by A. Burcroff). Then the algorithm works at least for the cases below (now with $r=3$ and $n=6$ working)

r=2 for n = 4, 5, 6,...,12

r=3, n = 4, 5, 6

r=4, n = 4, 5

r=5, n = 4, 5

r=6, n = 4, 5

up to

r=14, n = 4, 5

Finally, we should add 2 preceding edges to the path $\alpha(3,8)$. K. Lee pointed that when $r=3$ and $n=6$ we should add 2 preceding edges to some subpath (this subpath is $\alpha(3,8)$).