5
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A $n$-LNakayama algebra (for $n \geq 2$) is a list $[c_0,c_1,...,c_{n-1}]$ with $c_{n-1}=1$, $c_i \geq 2$ for $i \neq n-1$ and $c_i -1 \leq c_{i+1}$ for all $i$. They are in bijection with Dyck paths, where one associates to such a list simply the Dyck path with area sequence $[c_0,c_1,...,c_{n-1}]$ , see page 6 of https://arxiv.org/pdf/1811.05846.pdf. Thus $n$-LNakayama algebras are enumerated by the Catalan numbers $C_{n-1}$. Call an $n$-LNakayama algebra strong in case for all $i \neq n-1$ we have that $c_{i+1}=c_i -1$ or ($c_{i+1}>c_i -1$ and $c_{i+c_i}=c_{i+1}-c_i +1$) (Those are exactly the $n$-LNakayama algebras with global dimension at most 2). Strong $n$-LNakayama algebras are enumerated by $2^{n-2}$ and thus in bijection with integer compositions. (although Im not sure what the best bijection might be)

The coarea sequence $[d_0,d_1,...,d_{n-1}]$ is the area sequence of the "opposite" Dyck path. Formally : $d_i = min \{ k | k \geq c_{i-k} \}$.

Now define a statistic $f(A)$ on the strong $n$-LNakayama algebras $A$: $f(A):= \{ i \in \{0,...,n-1 \} | c_i < d_{i+c_i -1}$ and $d_{i+c_i-1}-c_i=d_{i-1} \}$ (those are the number of indecomposable projective modules with injective dimension one. One can show that this is also the number of indecomposable injective modules with projective dimension one, since Nakayama algebras are QF-3 algebras).

I can prove that the strong $n$-LNakayama algebras with $f(A)=0$ are enumerated by the Fibonacci numbers (note that Fibonacci numbers are enumerated by integer compositions without parts equal to one, see https://oeis.org/A000045). Experimenting with the computer and findstat suggests that much more is true.

Guess 1: Via the Billey-Jockusch-Stanley bijection $g$ from Dyck paths to 321-avoiding permutations we have that $f(A)= p(g(A))$, where for a permutation $\pi$ we have that $p(\pi)$ is the number of fixed points of $\pi$.

See http://www.findstat.org/StatisticsDatabase/St001008 .

Guess 2: There is a bijection $h$ from strong $n$-LNakayama algebras to integer compositions such that $f(A)= t(h(A))$, where $t(U)$ for an integer composition $U$ counts the number of parts equal to one.

I arrived at guess 2 since the strong $n$-LNakayama algebras with $f(A)=1$ seem to be counted by https://oeis.org/A105423 (integer compositions with exactly one part equal to 1) and those with $f(A)=2$ are counted by https://oeis.org/A105423 (integer compositions with exactly two parts equal to one) (and of course the $f(A)=0$ case counted by the Fibonacci numbers as mentioned before).

For guess 2 there is a cyclic analogue for CNakayama algebras (Nakayama algebras with a cyclic quiver). But for simplicity I do not state this for now (since they do not have such a nice combinatorial model as Dyck paths). For the cyclic case the algebras with $f(A)=0$ where counted by the cyclic analogue of the Fibonacci numbers : https://oeis.org/A032190. And those with $f(A)=1$ were counted surprisingly again by the usual Fibonacci numbers.

Guess 1 is better tested I guess since it comes from findstat, while guess 2 is based just on small values of $f(A)$ (namely $f(A) \leq 2$) and oeis. Here is the statistic f for the 4-LNakayama algebras:

[ [ [ 2, 2, 2, 1 ], 0 ]

[ [ 3, 2, 2, 1 ], 1 ]

[ [ 2, 3, 2, 1 ], 1 ]

[ [ 3, 3, 2, 1 ], 0 ]

[ [ 4, 3, 2, 1 ], 3 ] ]

Here is the statistic f for the strong 5-LNakayama algebras:

[ [ 2, 3, 2, 2, 1 ], 0 ]

[ [ 4, 3, 2, 2, 1 ], 2 ]

[ [ 3, 2, 3, 2, 1 ], 2 ]

[ [ 4, 3, 3, 2, 1 ], 1 ]

[ [ 2, 4, 3, 2, 1 ], 2 ]

[ [ 3, 4, 3, 2, 1 ], 1 ]

[ [ 4, 4, 3, 2, 1 ], 0 ]

[ [ 5, 4, 3, 2, 1 ], 4 ]

edit: Here is the statistic on strong $n$-LNakayama algebras for $n \leq 6$ where they are displayed as permutations (via the Billey-Jockusch-Stanley bijection). Has anyone seens those $2^{n-2}$ or can recognize them?

n=2: [1] => 1

n=3: [2, 1] => 0

[1, 2] => 2

n=4: [1, 3, 2] => 1

[2, 1, 3] => 1

[3, 1, 2] => 0

[1, 2, 3] => 3

n=5: [2, 1, 4, 3] => 0

[1, 2, 4, 3] => 2

[1, 3, 2, 4] => 2

[1, 4, 2, 3] => 1

[2, 1, 3, 4] => 2

[3, 1, 2, 4] => 1

[4, 1, 2, 3] => 0

[1, 2, 3, 4] => 4

n=6: [2, 1, 4, 3, 5] => 1

[2, 1, 3, 5, 4] => 1

[2, 1, 5, 3, 4] => 0

[2, 1, 3, 4, 5] => 3

[1, 3, 2, 5, 4] => 1

[1, 3, 2, 4, 5] => 3

[3, 1, 2, 5, 4] => 0

[3, 1, 2, 4, 5] => 2

[1, 2, 4, 3, 5] => 3

[1, 4, 2, 3, 5] => 2

[4, 1, 2, 3, 5] => 1

[1, 2, 3, 5, 4] => 3

[1, 2, 5, 3, 4] => 2

[1, 5, 2, 3, 4] => 1

[5, 1, 2, 3, 4] => 0

[1, 2, 3, 4, 5] => 5

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  • 1
    $\begingroup$ The layered permutations (en.wikipedia.org/wiki/Layered_permutation) are in obvious bijection with integer compositions. They are not a subset of the $321$-avoiding permutations, but they are the intersection of the $231$- and $312$-avoiding permutations. Recall that for any permutation $\sigma$ of length 3, the permutations avoiding $\sigma$ are counted by the Catalan numbers. A classical bijection due to Simion and Schmidt, in terms of left-to-right minima, bijects the $321$- and $231$- avoiding permutations, and might be what you're looking for. $\endgroup$ – Sam Hopkins Dec 16 '19 at 22:05
  • $\begingroup$ Although another obvious subset of Catalan objects in bijection with integer compositions are Dyck paths which are composed of "mountains" that go up A steps and then down A steps. $\endgroup$ – Sam Hopkins Dec 16 '19 at 22:16
  • $\begingroup$ @SamHopkins This subset of Dyck paths is not equal to the strong LNakayama algebras (see for example the n=5). But of course maybe there is a nice bijection. $\endgroup$ – Mare Dec 16 '19 at 22:21
  • 1
    $\begingroup$ In arxiv.org/abs/math/0203033 it is shown (bijectively) that the distribution of fixed points for $321$- and $132$-avoiding permutations is the same, relevant to the above discussion. $\endgroup$ – Sam Hopkins Dec 16 '19 at 22:29
4
+50
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It looks like the permutations are of the form $\pi_1 \oplus \pi_2 \oplus \cdots \oplus \pi_k$ (see https://en.wikipedia.org/wiki/Skew_and_direct_sums_of_permutations) where the $\pi_i$ are long-cycles of the form $(1,\ell,\ell-1,\ldots,2)$. These evidently correspond to compositions, and in the correspondence the number of fixed points is evidently the number of parts of size one (as you desired). They seem like a kind of variant of layered permutations.

$\endgroup$
  • 1
    $\begingroup$ The long-cycle $w=(1,\ell,\ell-1,\ldots,2)$ (or its inverse) is sometimes called the "standard Coxeter element"; it has a reduced decomposition $w=s_1 s_2 \cdots s_{n-1}$ (or $s_{n-1}\cdots s_2 s_1$ depending on your convention for multiplying permutations). These permutations might be called "parabolic standard Coxeter elements"; they are what you get from the product $s_1 s_2 \cdots s_{n-1}$ by deleting some subset of factors. $\endgroup$ – Sam Hopkins Dec 18 '19 at 23:12
  • $\begingroup$ Thanks, this looks very helpful. I will try to find a proof of my guesses in the next days. It seems to be not so hard in case all guesses are correct. $\endgroup$ – Mare Dec 20 '19 at 12:48

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