We say a sequence $a_0,a_1,...,a_{2n}$ that it's a good sequence if for every $1\le i\le 2n$, $a_i$ is an integer, $a_0 = a_{2n} = 1$, and $$\frac{a_i}{a_{i-1}} \in \frac{1}{6}, \frac{2}{3}, \frac{3}{2}, 6 $$.
How many good series are? The answer has to do with the number of 2s and 3s in every $a_i$ and somehow use the Catalan numbers to solve it.
You may notice that every "multiplier" $\frac{a_i}{a_{i-1}}$ is a number of the form $2^a 3^b$ with $(a,b)\in\{(-1,-1),(1,-1),(-1,1),(1,1)\}=E$. Since $$ a_{m} = \frac{a_{m}}{a_{m-1}}\cdot \frac{a_{m-1}}{a_{m-2}}\cdots\frac{a_2}{a_1}\cdot\frac{a_1}{a_0}\cdot a_0 $$ every $a_m$ a number of the form $2^a 3^b$ with $(a,b)\in\mathbb{Z}^2$. The number of good sequences is given by the number of "double strings" in $\{-1,1\}^{2n} \times \{-1,1\}^{2n}$ such that any component has only prefixes with a non-negative sum and a total sum equal to zero. There is a bijection between the set of strings having such property and the paths in a $n\times n$ grid made only by unit steps towards east or north, starting at $(0,0)$ and ending at $(n,n)$, such that they never go below the diagonal.
The relation with Catalan numbers should now be evident: the answer is given by $\left[\frac{1}{n+1}\binom{2n}{n}\right]^2$.