Cartan determinants of subsets Let $n \geq 3$ be fixed.
We associate to every subset $S \subseteq \{1,...,n-1 \}$ a number, which we call Cartan determinant of $S$ (see the end of this post for a representation theoretic background).
Define for $k,t \in \{1,...,n \}: m_{k,t}:= t-max(0,t-k)=min(k,t)$ (so $m_{k,t}=t$ for $k>t$ and $m_{k,t}=k$ for $t \geq k$).
Write $S= \{s_1,...,s_{r-1} \}$ with the $s_i$ in increasing order and $r-1$ the cardinality $|S|$ of $S$ and set $s_r:=n$.
Let $a_{i,j}:=m_{s_i,s_j}=min(s_i,s_j)$ for $i,j \in \{1,...,r\}$.
Let $A_S$ be the matrix with entries $a_{i,j}$, called the Cartan matrix of $S$ and let $d_S$ be defined as the determinant of $A_S$, called the Cartan matrix of $S$.
We set $A_{\emptyset}=n$ and thus $d_{\emptyset}=n$.

Question 1: Is there an explicit formula for $d_S$?
Question 2: I can prove that $d_S$ is non-zero. Is it more general true that $d_S >0$?

Now call $S$ great in case the following holds.
-In case $n$ is even, $|S|$ is odd,$min(S)$ is odd and $max(S)$ is odd and all differences $s_i-s_{i-1}$ are odd for $i=2,...,r-1$.
-In case $n$ is odd, $|S|$ is even, $min(S)$ is odd, $max(S)$ is even and all differences $s_i-s_{i-1}$ are odd for $i=2,...,r-1$.
-$\emptyset$ is great if and only if $n$ is odd.

Question 3: Is it true that $d_S$ is odd if and only if $S$ is great? Are the great subsets $S \in \{1,...,n-1  \}$ enumerated by the Fibonacci numbers $F_n$?

Question 2 and 3 have a positive answer for $n \leq 14$ and thus there is good evidence that it is true and something nice is going on. I can prove (looking at entries mod 2) that if $S$ is great, then $d_S$ is odd, but Im not sure about the converse. The result being so nice, I would think that there is a short proof that avoids heavy computations.
Now here come some more weird observations that might be much harder to show.

Question 4:
Call a subset $S$ prime in case $d_S$ is prime. When is $S$ prime?
For $n \leq 20$ it was true that the number of prime $S$ is given by  $\sum\limits_{k=1}^{m}{\pi(k)}$, where $\pi(k)$ is the prime counting function. See https://oeis.org/A046992 . Is this true in general?

Some other findings that occured in the oeis:

The number of subsets $S$ with $d_s=1 mod 3$ seems to be given by the seqeunce https://oeis.org/A075111 , related to the tribonacci numbers. Is this true?
The number of subsets $S$ with $d_S=1 mod 4$: https://oeis.org/A005252 .
With $d_S=3 mod 4$: https://oeis.org/A024490.

Note that all the 3 previous sequences are related to the Fibonacci sequence, which might indicate that this all has a secret connection to Fibonacci combinatorics (which I can not see at the moment).

The number of $S$ with $d_S$ such that $Gcd(d_S,n)=1$ seem to be given by https://oeis.org/A100347.

Representation theoretic background.
Let $A=A_n=K[x]/(x^n)$. Then there is a unique indecomposable $A$-module $M_i$ of dimension $i$ for $i=1,...,n$.
The subsets $S \subseteq \{1,...,n-1 \}$ are in bijection with the (faithful) module $M_S := A \oplus \bigoplus\limits_{i \in S}^{}{M_i}$. Then $d_S$ is the Cartan determinant of the algebra $End_A(M_S)$.
 A: If $s_1<s_2<\ldots<s_{r-1}<s_r=n$, the matrix $A_S$ is the matrix of the quadratic form
$$
s_1(x_1+\ldots+x_r)^2+(s_2-s_1)(x_2+\ldots+x_r)^2+\\(s_3-s_2)(x_3+\dots+x_r)^2+\ldots+(s_r-s_{r-1}) x_r^2.
$$
This quadratic form is obviously positive-definite which already implies $d_S>0$. Next, we get $$(A_Sx,x)=(D_STx,Tx)\Rightarrow A_S=T^tD_ST,$$
where $D_S$ is diagonal with elements $s_1,s_2-s_1,\ldots,s_r-s_{r-1}$ and $T$ is the triangular map $x\to (x_1+\ldots+x_r,x_2+\ldots+x_r,\ldots,x_r)$. Thus
$$
\det A_S=\det D_s=s_1(s_2-s_1)(s_3-s_2)\ldots (s_r-s_{r-1})
$$
which answers other questions. 
A: For any set $S=\{s_1,\dots,s_k\}\subset[0,\infty)$ with $s_1<\dots<s_k$, the matrix $A_S=(\min(s_i,s_j))_{i,j=1}^k$ is the covariance matrix of the random vector $(B(s_1),\dots,B(s_k))$, where $B(\cdot)$ is a standard Brownian motion. So, $A_S$ is positive semidefinite. Moreover, since the random variables $B(s_1),\dots,B(s_k)$ are clearly linearly independent, $A_S$ is positive definite, and hence $\det A_S>0$.
