See <https://en.wikipedia.org/wiki/Nakayama_algebra> for the definition of Nakayama algebras and define the permanent of such an algebra to be the permanent of its Cartan matrix.
(all algebras are assumed to be connected and finite dimensional)
Conjecture:
Let $X$ be the set of Nakayama algebras of finite global dimension with $n$ simple modules. 
Then the maximum of the permanents of algebras in X is given by $\sum\limits\_{k=0}^{\infty}{\frac{k^n}{2^{k+1}}}$ and it is uniquely attained.
See <https://oeis.org/A000670> for the conjectured sequence.


The algebras with a line as a quiver should always have permanent equal to one. Here the algebras with permanent higher than 1 and finite global dimension with 3 simple modules (given by their Kupisch series as first entry and the permanent as second entry):


[ [ [ 2, 2, 3 ] 3 ], [ [ 2, 4, 3 ], 5 ], [ [ 3, 4, 4 ], 11 ], [ [ 3, 5, 4 ], 13 ] ]



Permanent of Nakayama algebras
Mare's answer reduces the problem to showing that the permanant of the $n \times n$ matrix
$$\begin{bmatrix} 
1 & 1 & 1 & \cdots & 1 \\
1 & 2 & 2 & \cdots & 2 \\
1 & 1 & 2 & \cdots & 2 \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
1 & 1 & 1 & \cdots & 2 \\
\end{bmatrix}$$
is the number of weak orders on $[n]:=\{ 1,2, \ldots, n \}$. This answer will prove that. 


First of all, permanent is unchanged by reordering the rows, so I will modify the matrix to 
$$\begin{bmatrix} 
1 & 2 & 2 & \cdots & 2 \\
1 & 1 & 2 & \cdots & 2 \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
1 & 1 & 1 & \cdots & 2 \\
1 & 1 & 1 & \cdots & 1 \\
\end{bmatrix}.$$
By the definition of the permanent, the permanent is
$$\sum\_{\sigma \in S\_n} 2^{\# \{ j \ :\ \sigma(j) > j \}}.$$


We can give a similar formula for the number of weak orders. Given a weak order $\succeq$ on $[n]$, define a total order on the same set by break all ties in according to the standard order on $[n]$. For example, $13 \succeq 5 \succeq 24$ turns into $3 > 1 > 5 > 4 > 2$. Total orders on $[n]$ correspond to permutations in $S\_n$; given a permutation $\sigma$, the number of weak orders which give rise to it is $2^{\# \{ k : \sigma(k) > \sigma(k+1) \}}$. For example, to choose a weak order giving rise to the total order $3 > 1 > 5 > 4 > 2$, we just have to decide whether or not to group $(3,1)$, $(5,4)$ and $(4,2)$ together, and these choices are independent. So A000670 is
$$\sum\_{\sigma \in S\_n} 2^{\# \{ k \ :\ \sigma(k) > \sigma(k+1) \}}.$$


For a permutation $\sigma \in S\_n$, the exponent $\# \{ k : \sigma(k) > k \}$ is called the number of exceedances of $\sigma$, and $\# \{ k \ :\ \sigma(k) > \sigma(k+1) \}$ is called the number of descents of $\sigma$. So the conjecture follows from the following stronger result:


**Theorem** The number of permutations of $n$ elements with $r$ descents is the same as the number of permutations of $n$ elements with $r$ exceedances.


This was originally proved by MacMahon in 1915 and has been reproved many times since. The number of permutations of $n$ elements with $r$ descents is called the [Eulerian number](https://en.wikipedia.org/wiki/Eulerian_number), denoted $\left\langle \begin{smallmatrix} n\\r \end{smallmatrix} \right\rangle$ and a statistic $f : S\_n \to \mathbb{Z}$ which 
takes the value $r$ a total of $\left\langle \begin{smallmatrix} n\\r \end{smallmatrix} \right\rangle$ times on $S\_n$ is called "Eulerian". 


A quick proof can be found as Proposition 3.14 in [these notes](https://www.math.ust.hk/~mabfchen/Math5380/Bijective-Counting.pdf). (No originality claimed for this source; just the first one I found without a lot of extraneous material.)


