It is well known that for the algebra of $n \times n$-upper triangular matrices over a field the number of tilting modules is equal to the Catalan number $C_n$. This is just the (hereditary) Nakayama algebra with Kupisch series $[n,n-1,...,2,1]$ and can be viewed as the "mother" of all Nakayama algebras with a linear quiver because it has each such algebra as a quotient. The cyclic analogue of this algebra is the Nakayama algebra with Kupisch series $[n,2n-1,2n-2,....,n+2,n+1]$ for $n \geq 2$, which can be viewed as the "mother" of all Nakayama algebras with a cyclic quiver of finite global dimension because it has each such algebra as a quotient. Now let $A$ be this Nakayama algebra with Kupisch series $[n,2n-1,2n-2,....,n+2,n+1]$. This is an algebra with global dimension 2 (while the algebra of upper triangular matrices has global dimension 1).

I wondered what the tilting modules over this algebra are. The problem seems to contain the problem of classifying the tilting modules over the algebra of upper triangular matrices as a special case because the indecomposable modules with projective dimension one in $A$ just behave like modules over the algebra of upper triangular matrices and thus the number of 1-tilting modules of $A$ should be also equal to the Catalan numbers.

The number of tilting modules of $A$ starts with 1,3,10,35,126 and this suggests that the number of tilting modules equals $\binom{2n-1}{n}$ which are the monotone maps $\{1,...,n \} \rightarrow \{1,...,n \}$, see https://oeis.org/A001700.

This leads to the following guess:

There is a natural bijection from the set of monotone maps $\{1,...,n \} \rightarrow \{1,...,n \}$ to the set of tilting modules of $A$.

Note that the monotone maps $f$ with $f(i) \leq i$ are counted by the Catalan numbers (see for example exercise 78. in the book "Catalan numbers" by Richard Stanley) and thus the above bijection (if it exists) should restrict to a bijection between monotone maps $f$ with $f(i) \leq i$ and the 1-tilting modules of $A$ (at least if it is a nice bijection).

I wanted to ask whether there is a quick proof of this guess in case it is true using some advanced tools. I am able to translate the problem into a purely combinatorial problem but it looks very complicated at the moment and maybe there is an easy trick to obtain such a bijection or maybe this is even known.

The combinatorial translation gives the problem where $n$ points (corresponding to the indecomposable summands of the basic tilting module) are drawn into two triangles (whose points correspond to the 2-rigid indecomposable modules in the Auslander-Reiten quiver of the algebra). There is one bigger triangle with $\frac{n(n+1)}{2}$ points and one smaller triangle with $\frac{n(n-1)}{2}$ points so that both triangles have together $n^2$ points. Here the tilting modules for $n=3$ (maybe someone can see how they correspond to monotone sequences?): https://www.docdroid.net/YwBhi0k/monotonetilting.pdf

Here the configurations where the red market points only occur in the smaller triangle or on the leftmost boundary of the bigger triangle count the 1-tilting modules so that for n=3 we get 5 1-tilting modules.