In how many ways can someone choose a permutation $w \in S_n$ and color each one of the integers $1,2,\ldots,n$ white, yellow or blue so that the minimum element of every cycle of $w$ is white?
Comments: I know that the number of permutations of $[n]$ is $n!$. I know that there are $${n \choose a_1,a_2,...,a_r}$$ colorings s.t. exactly $a_i$ elements of $[n]$ are colored with the colour $i$, for $i \in [r]$. For those colours there are $a_i! $ ways to choose a cycle with the color i from w. (I don't know if any of these is useful)
The unsigned Stirling numbers of first kind ${n\brack k}$ count the number of permutations of $n$ with $k$ disjoint cycles.
Then, the answer is $$ A(n)=\sum_{k=1}^n {n\brack k}3^{n-k}. $$
To further simplify it, note that the unsigned Stirling numbers of first kind satisfy $$ \sum_{k=0}^n {n\brack k} x^k = x(x+1)\cdots (x+n-1). $$
With this in mind, it should be easy to check that $$ A(n)=(3n-2)!!!=\prod_{k=1}^n(3k-2). $$