I've come across a fun problem that I couldn't generalize.
Description
3 students arrive at a chairlift. They are free to use up to 3 consecutive chairlifts (no empty chairlifts between them). So they could all go on the same, split themselves and go on 2 or each one goes on its own. An empty chairlift in front of them does not count as separate possibility.
Each chairlift has 3 free seats (or $n$ seats for $n$ students). The students can arrange themselves on the seats freely.
So there are three ways how the students can arrange themselves on the lifts (slashes separate chairlifts): $3, 2/1, 1/2$ or $1/1/1$.
Questions
How many ways are there that the students ride the chairlift? Denote that number as $D_3$ and for $n$ students $D_n$.
What is $D_4$?
Is there a closed-form expression for $D_n$?
What I've tried
For $D_3$, I just listed the possibilities and got $276$, which seems to be correct. Also, $D_1=1, D_2=10$. Here is a picture that lists all possibilities for $D_2$:
For $D_4$ I've got $14\,712$.
The number of ways that $n$ students can distribute themselves on $n$ chairlifts with $n$ seats can be calculated with the partition function p, I think. So for 3 students, the possibilities are $\{3\}, \{2,1\}, \{1,1,1\}$. It is relatively easy to continue to perform the combinatorics in each partition. For the first, I got $3!=6$, for the second $3\cdot 2\cdot3\cdot2\cdot3=108$ and for the third $3^3\cdot3!=162$.
But I have trouble to generalize the combinatorical calculations.
Working for $D_3$
I have rechecked answers, the discrepancy is coming when only one lift is being used. There are three ways to choose the lift used, and $3!$ ways to place the persons in that lift $\to 18\;\;$ ways
$18+108+162=288$
However, I am not immediately able to see a simple way to generalize for $n$
Added
Here's a much simpler way to compute $D_3$
Imagine $3$ boxes with $3$ slots each arranged in a circle
If only one box is to be filled, first person can go to any of $9$ slots, but the next two are now restricted to the remaining two slots in that box, so $9\cdot2\cdot1 = 18$
If two boxes are to be filled, put the first person anywhere, and the other two in a neighbouring box, so $9\cdot6\cdot2 = 108$
If three boxes are to be filled, $9\cdot6\cdot3 = 162$
This is a much better formulation, and you could try out $D_4$ in this manner, and see if it leads anywhere !