Partitions with pairwise small intersection Let $n$ and $k$ be positive integers, such that $k$ is a divisor of $n$. I am interested in creating a sequence of partitions of $\{1,\dots,n\}$, like the one below. The rules are this:

*

*Each row is a partition of $\{1,\dots,n\}$ into $n/k$ disjoint sets, each with size $k$.


*For any two subsets $S$ and $T$ appearing in the list of partitions, $S$ and $T$ have at most one element in common.
Example with $n=12$ and $k=3$. You can make a list of three partitions:
$$\{\{1,2,3\},\{4,5,6\},\{7,8,9\},\{10,11,12\}\},\\
\{\{1,4,7\},\{2,5,10\},\{3,8,11\},\{6,9,12\}\},\\
\{\{1,5,8\},\{2,4,12\},\{3,9,10\},\{6,7,11\}\}$$
This is valid because, for example, $\{1,2,3\}$ and $\{1,4,7\}$ have at most one element in common, and the same for $\{4,5,6\}$ and $\{1,5,8\}$, and for every other way to choose two subsets from this list of partitions.
The total number of possible partitions is equal to $\frac{n!}{(k!)^{n/k}\cdot (n/k)!}$, but the size of the largest list will be much smaller than this due to the at-most-one condition.

Question: What is known about the length largest possible list of partitions satisfying these conditions?

 A: Let $P(n,k)$ be number of interest, the size largest list of partitions for a given $n$ and $k$. Here is what I could figure out about $P(n,k)$.

*

*We have the upper bound
$
P(n,k)\le \frac{n-1}{k-1},
$
which follows because in each partition, $1$ needs to be grouped with $k-1$ new numbers, and there are only $n-1$ other numbers.


*If $k> \sqrt n$, then $P(n,k)=1$.  If $k$ numbers are together in one partition, then those $k$ numbers all need to be in different parts in the next partition. Since there are $n/k$ parts, this means we need $n/k\ge k$ in order to have two or more partitions.


*If $k^2<n$, and $n/k$ is a prime power, then $P(n,k)\ge \frac{n}k$. The strategy is to construct $k$ mutually orthogonal Lain squares using the finite field construction. Specifically, choose binary operations $\oplus$ and $\otimes$ on the set $F:=\{0,\dots,n/k-1\}$ such that $(F,\oplus,\otimes)$ is a finite field, where $0$ is the zero element. Then, construct $k$ Latin squares $L^1,\dots,L^k$, each of order $n/k$, via the formula $$L^m_{i,j}=m\otimes i\oplus j \qquad i,j\in F,m\in \{1,\dots,k\}$$Then, concatenate these together to get an $(n/k)\times n$ matrix $[L^1\mid L^2\mid \dots \mid L^k]$. Each row of this matrix represents a partition of $\{1,\dots,n\}$, where for each $i\in \{1,\dots,n/k\}$, the $i^\text{th}$ part consists of the column indices of the entries containing $i$. 
Example: Using this strategy, we can improve on your result for $n=12,k=3$. The finite field construction produces these three squares:
1 2 3 4  1 2 3 4  1 2 3 4
2 1 4 3  4 3 2 1  3 4 1 2
3 4 1 2  2 1 4 3  4 3 2 1 
4 3 2 1  3 4 1 2  2 1 4 3

Translating this into a list of four partitions, the result is
$$
\begin{array}{c|c|c|c}
1,5,9&2,6,10& 3,7,11&4,8,12\\
2,8,11&1,7,12& 4,6,9&3,5,10\\
3,6,12&4,5,11& 1,8,10&2,7,9\\
4,7,10&3,8,9& 2,5,12&1,6,11
\end{array}
$$


*If $k^2=n$, and there exist $\ell-2$ mutually orthogonal Latin squares (MOLS), then $P(n,k)\ge \ell$. In particular, if $k$ is a prime power, then $P(n,k)=k+1$.
The strategy is to convert the MOLS into an orthogonal array, $OA(\ell, k)$, following the instructions in the Wikipedia article. The result is a $k^2\times \ell$ matrix. Each column of that matrix can be turned into a partition, using the same strategy as in the last section, so we get $\ell$ partitions which satisfy the conditions.


*$P(2n,2)=2n-1$, via a Round robin tournament schedule.


*$P(6n+3,3)=3n+1$, via a Kirkman Triple System.


*$P(n,k)$ attains its optimal value of $\frac{n-1}{k-1}$ if and only if a resolvable Steiner system $S(2,k,n)$ exists.  A Steiner system, $S(t,k,n)$, is a collection of blocks, where each block a size-$k$ subset of $\{1,\dots,n\}$,  such that every size-$t$ subset of $\{1,\dots,n\}$ is contained in exactly one block. This is called resolvable if the blocks can be partitioned into parallel classes, where each parallel class is set of pairwise disjoint blocks whose union is $\{1,\dots,n\}$. This is essentially a tautology, but it gives a connection with existing research.
