Solution:

Answer: $2+\sqrt{2}$

We first find $A_{n}$ in terms of $n$. Let $a_{n}$ be the number of ways to get to the point $(n, n+3)$, and let $b_{n}$ be the number of ways to get to the point $(n+1, n+2)$. By symmetry, $a_{n}$ is also the number of ways to get to $(n+3, n)$ and $b_{n}$ is also the number of ways to get to the point $(n+2, n+1)$.

We can easily see that $a_{0}=1$ and $b_{0}=3$. This also means that $A_{n}=a_{n-3}+3 b_{n-3}+3 b_{n-3}+a_{n-3}=2 a_{n-3}+6 b_{n-3}$.

We also get the recurrence:
$$\begin{aligned} a_{i+1} & = a_{i} + b_{i} \\ b_{i+1} & = a_{i} + 3 b_{i} \end{aligned}$$
We have both $3 a_{i+1} = 3 a_{i} + 3 b_{i}$ and $a_{i+2} = a_{i+1} + b_{i+1}$. Subtracting these gives
$$\begin{aligned} a_{i+2} - 3 a_{i+1} & = a_{i+1} - 3 a_{i} + b_{i+1} - 3 b_{i} \\ a_{i+2} - 3 a_{i+1} & = a_{i+1} - 3 a_{i} + a_{i} \\ a_{i+2} & = 4 a_{i+1} - 2 a_{i} \end{aligned}$$
Now we can solve this recurrence using its characteristic polynomial $x^{2} - 4x + 2$, which has roots of $2+\sqrt{2}$ and $2-\sqrt{2}$. We can then write $a_{i} = A(2+\sqrt{2})^{i} + B(2-\sqrt{2})^{i}$ for some constants $A$ and $B$. Now, $a_{0}=1$ and $a_{1}=a_{0}+b_{0}=4$. Using this, we solve for $A$ and $B$ to get
$$a_{i} = \left(\frac{1+\sqrt{2}}{2}\right)(2+\sqrt{2})^{i} + \left(\frac{1-\sqrt{2}}{2}\right)(2-\sqrt{2})^{i}$$
Then,
$$\begin{aligned} b_{i} & = a_{i+1} - a_{i} \\ & = \left(\frac{1+\sqrt{2}}{2}\right)\left((2+\sqrt{2})^{i+1} - (2+\sqrt{2})^{i}\right) + \left(\frac{1-\sqrt{2}}{2}\right)\left((2-\sqrt{2})^{i+1} - (2-\sqrt{2})^{i}\right) \\ & = \left(\frac{1+\sqrt{2}}{2}\right)(1+\sqrt{2})(2+\sqrt{2})^{i} + \left(\frac{1-\sqrt{2}}{2}\right)(1-\sqrt{2})(2-\sqrt{2})^{i} \\ & = \left(\frac{3+2\sqrt{2}}{2}\right)(2+\sqrt{2})^{i} + \left(\frac{3-2\sqrt{2}}{2}\right)(2-\sqrt{2})^{i} \end{aligned}$$
Therefore,
$$A_{n} = 2 a_{n-3} + 6 b_{n-3} = (10+7\sqrt{2})(2+\sqrt{2})^{n-3} + (10-7\sqrt{2})(2-\sqrt{2})^{n-3}$$
We can then easily see that $A_{n} < (2+\sqrt{2}) A_{n-1}$. Also, since $2-\sqrt{2} < 1$, as $n$ approaches infinity, the ratio $\frac{A_{n}}{A_{n-1}}$ approaches $2+\sqrt{2}$. Hence the least upper bound of $\frac{A_{n}}{A_{n-1}}$ is $2+\sqrt{2}$.