Let $A=A_n$ be the algebra of upper triangular matrices over a field $K$ with $n$ simple modules. It is a nice result that there are $C_{n+1}=1,2,5,14,...$ (Catalan numbers for $n \geq 1$) tilting $A_n$-modules, where a tilting module $T$ is a module with $n$ indecomposable summands (we assume all modules are basic) and projective dimension 1 and $Ext^{1}(T,T)=0$. Let $J$ be the Jacobson radical of $A_n$ and $B_{n,l}:=A_n /J^l$ for some $n-1 \geq l \geq 2$.

Computer experiments with small n and l suggest the following generalisation:

The number of tilting $B_{n,l}$ modules equals $C_l$.

Is this true? If yes, there should be a simple reason, which I do not see at the moment.

If no, what is the correct number of $B_{n,l}-$tilting modules?

My guess goes as follows: Let $e$ be the idempotent of $B=B_{n,l}$ such that $eB$ is minimal faithful projective-injective. Then $eB$ is a summand of any $B$-tilting module. Thus any tilting module is of the form $T=eB \oplus X$ and $X$ corresponds to a tilting-module of $B/BeB$ (why?) which can be identified with $A_{l-1}$. Thus there are as many tilting $B$-modules as $A_{l-1}$ tilting modules which is $C_l$.