Express $F(n,k)$ through Stirling Numbers if
$1)\ F(n,k)$ represents the number of sequences $a_1,..., a_n$ of natural numbers not surpassing the number $k$.
$2)$ The first occurrence of the number $i$ happens before the first occurrence of the number $i +1$.
$3)$ Each of the numbers $1,..,k$ appears at least one time.
I would like some help with this. For sure the numbers $1, ...,k$ are our elements and point $3)$ seems to suggest the application of Stirling Number of the 2nd Kind, but I don't understand what $2)$ means and don't know how to put the $3$ requirements together.
For instance if I put $F(2,1)$ we would have $2$ sequences of natural numbers not surpassing $1$. My guess is that the sequences would be
$(1),\ (1,1)$
Now for $F(2,2)$ we have $2$ sequences of natural numbers not surpassing $2$:
$(1), (1,2)$
For $F(3,2)$
$(1),\ (1,1),\ (1,2)$
I'm assuming the requirement $2)$ is for each sequence, that is for $F(3,2)$ the sequence $(2,1)$ and $2$ are not allowed (the first $2$ in a sequence has to appear after the first $1$).
In any case, the approach I followed doesn't seem to give any clue about how to express this function through Stirling.
HINT: The third condition says that you’re partitioning $[n]$ into exactly $k$ parts, one for each the numbers $1,\ldots,k$: the $i$-th part is the set of positions in which $i$ appears. The second condition says in effect that if you have a partition of $[n]$ into $k$ parts, and the minimum elements of the parts are $m_1<m_2<\ldots<m_k$, then the part with minimum $m_i$ must be the part that contains the element $i$ of $[k]$. Thus, each partition of $[n]$ into $k$ parts determines a unique sequence.