2. Transfer systems and a generalized $\odot$ construction

We begin by recalling the definition of a transfer system on a poset from [Reference Balchin, MacBrough and Ormsby5, Reference Franchere, Ormsby, Osorno, Qin and Waugh9].

For a lattice $L$ , we write $\mathsf{Tr}(L)$ for the collection of all transfer systems of $L$ . In [Reference Balchin, Barnes and Roitzheim1], it was proved that $|\mathsf{Tr}([n])|$ coincides with the $(n+1)$ -th Catalan number. The proof for this proceeded in a recursive fashion which was governed by an operation denoted $\odot$ in loc. cit.

Let $\mathcal{R}$ be a transfer system on $[n]$ . Denote by $x$ the minimal element such that $x\, \mathcal{R}\, n$ , and consider the partition of $[n]$ as $[0,x-1] \amalg [x,n]$ . The key observation is that the restriction of $\mathcal{R}$ to $[0,x-1]$ and $[x,n]$ yields two transfer systems, and moreover, $\mathcal{R}$ can be recovered from the disjoint union of these two transfer systems. To retrieve the aforementioned $\odot$ operation from this, we note that the pivot element in [Reference Balchin, Barnes and Roitzheim1] is simply $x$ .

In this section, we will follow a suggestion of Mike Hill to extend this observation to transfer systems on an arbitrary finite lattice $L$ , which provides a method for producing recursive formulas for counting transfer systems. To this end, let $L$ be a finite lattice equipped with a transfer system $\mathcal{R}$ . Let $m$ be the maximal element of $L$ , we will say that $x \in L$ is fibrant if $x\, \mathcal{R}\, m$ . (This terminology is inspired by the corresponding statement under the equivalence of transfer systems to weak factorization systems.)

Proof. Let $\{x_i\}_i$ be the collection of fibrant elements of $\mathcal{R}$ . Then by the definition of transfer systems, it follows that $(\bigwedge _i x_i)\, \mathcal{R}\, m$ . In particular, $\bigwedge _i x_i$ is a fibrant element and $(\bigwedge _i x_i) \leqslant x_j$ for all $j$ as required.

The following is a standard result in the theory of finite lattices.

Via Lemma2.4, one sees that there is a well-defined notion of transfer system for $x_{\uparrow }$ and $x_{\uparrow }^{\mathsf{c}}$ . Indeed, the axioms for a categorical transfer system only depend on the meet operation, so we can use the same definition for meet-semilattices like $x_\uparrow ^{\mathsf{c}}$ . In particular, for $L$ a finite lattice and $\mathcal{R}$ a transfer system on $L$ , if $x$ is the unique minimal fibrant element, then $\mathcal{R}$ restricts to transfer systems $\mathcal{R}_2$ and $\mathcal{R}_1$ on $x_{\uparrow }$ and $x_{\uparrow }^{\mathsf{c}}$ , respectively.

Given $\mathcal{R}_1$ and $\mathcal{R}_2$ as above, we can construct a candidate transfer system $\mathcal{R}_1 \amalg \mathcal{R}_2$ by simply taking their disjoint union over $L$ . The following lemma proves that this retrieves the starting transfer system $\mathcal{R}$ .

Proof. Assume that there is some $y \in x_\uparrow ^{\mathsf{c}}$ and $z \in x_\uparrow$ such that $y\, \mathcal{R}\, z$ . By the property of a transfer system, we have $(x \wedge y )\, \mathcal{R}\, x$ , and by transitivity we have $(x \wedge y )\, \mathcal{R}\, x \mathcal{R}\, m$ , that is, $x \wedge y$ is fibrant. However, as $y \in x_\uparrow ^{\mathsf{c}}$ , $(x \wedge y ) \lt x$ , violating the assumed minimality of $x$ .

Example 2.6. Let $L = [1] \times [3]$ , and consider the following transfer system. Footnote ³

The minimal fibrant element $(0,1)$ is highlighted in red. One clearly sees pictorially that $\mathcal{R}$ splits as a transfer system $\mathcal{R}_2$ on $(0,1)_{\uparrow } \cong [1] \times [2]$ and a transfer system $\mathcal{R}_1$ on $(0,1)_{\uparrow }^{\mathsf{c}} \cong [1] \times [0] \cong [1]$ .

As such, every transfer system $\mathcal{R}$ on $L$ can be split into a pair of transfer systems $\mathcal{R}_1, \mathcal{R}_2$ on sub-(semi-)lattices $x_\uparrow ^{\mathsf{c}}$ and $x_\uparrow$ of $L$ where $\mathcal{R}_2$ has minimal fibrant element $x$ . In the converse direction, given two transfer systems $\mathcal{R}_1$ and $\mathcal{R}_2$ on $x_\uparrow ^{\mathsf{c}}$ and $x_\uparrow$ , respectively, we can ask when $\mathcal{R}_1 \amalg \mathcal{R}_2$ is a transfer system for $L$ . This happens if and only if the relations in $\mathcal{R}_2$ when restricted to $x_{\uparrow }^{\mathsf{c}}$ (i.e., those relations that are forced by the relations of $\mathcal{R}_2$ under the restriction property of a transfer system on the larger lattice $L$ ) are relations in $\mathcal{R}_1$ . We will say that such an $\mathcal{R}_1$ and $\mathcal{R}_2$ is a restriction closed pair over $L$ .

From this discussion, we can form the basis of a recursive formula for computing $\mathsf{Tr}(L)$ based on the geometry of $L$ . We summarize this in the following theorem.

Construction 2.10. To rephrase this strategy in the form of an $\odot$ operation, we use the observation from the beginning of this section that the pivot element of the $\odot$ operation for $[n]$ was given by the minimal fibrant element of $\mathcal{R}$ . If $L$ is now an arbitrary finite lattice with $\mathcal{R}$ , $\mathcal{R}_1$ , and $\mathcal{R}_2$ as above, we let $\mathcal{R}_2 \smallsetminus \{x\}$ be the collection of relations where we have removed the initial element (i.e., we remove the fibrant element $x$ ). Then we could sensibly define $\mathcal{R} = \mathcal{R}_1 \odot (\mathcal{R}_2 \smallsetminus \{x\}$ ), where the operation $\odot$ inserts a new element $x$ and adds in the relation $x\, \mathcal{R}\, m$ and all relations induced by this.

In this notation, one would decompose the $\mathcal{R}$ of Example 2.6 as

3. Enumerating transfer systems for $D_{p^n}$

The eventual goal of this paper is to provide a recursive formula for transfer systems on $[1] \times [n]$ which we realize as the subgroup lattice of $G = C_{qp^n}$ for $p,q$ distinct primes. We will begin with a slightly more tame count, that of liftable transfer systems of $[1] \times [n]$ as we now define.

Definition 3.1. A transfer system $\mathcal{R}$ on $[1] \times [n]$ is liftable if $\mathcal{R}$ additionally satisfies

(L) \begin{equation} \text{If $(1,i)\, \mathcal{R}\, (1, j)$ for $i \lt j$ then $(0,i)\, \mathcal{R}\, (1,i)$.} \end{equation}

We write $L(n)$ to denote the collection of liftable transfer systems on $[1] \times [n]$ .

The relevance of liftable transfer systems in the general machinery of $N_\infty$ operads is provided by the following key result of [Reference Balchin, MacBrough and Ormsby5].

The link between liftable transfer systems and dihedral transfer systems arises via the lattice isomorphism $\operatorname{Sub}(D_{p^n})/D_{p^n} \cong [1] \times [n]$ . (Here the quotient is with respect to the conjugation action of $D_{p^n}$ on its subgroups.) Once we have given a recursive formula for $L(n)$ , we will, in the following section, exploit the involution on transfer systems as uncovered in [Reference Franchere, Ormsby, Osorno, Qin and Waugh9] to prove a recursion for $\mathsf{Tr}([1] \times [n])$ .

As a warmup, let us see how the liftable condition $(L)$ allows us to enumerate the number of saturated transfer systems in $L(n)$ . Recall that a transfer system is said to be saturated if it satisfies 2-out-of-3.

Proof. In [Reference Hafeez, Marcus, Ormsby and Osorno10], it was proved that there are $2^n$ saturated transfer systems on $[n]$ . Consider the bottom row of $[1]\times [n]$ given by the coordinates $(0,i)$ , then we can freely pick any saturated transfer system to fill this in. Let $j$ be the maximal number such that $(0,j) \, \mathcal{R} \, (1,j)$ . Clearly there are $n+2$ such choices (including the possibility of no such $j$ existing). We then claim that this data uniquely determines a saturated transfer system on $[1] \times [n]$ which moreover satisfies $(L)$ .

We begin by observing that if $j^{\prime},j^{\prime\prime} \gt j$ , we cannot have $(1, j^{\prime}) \, \mathcal{R} \, (1, j^{\prime\prime})$ . If this were the case then by $(L)$ , we would be forced to have $(0,j^{\prime}) \, \mathcal{R} \, (1,j^{\prime})$ which contradicts the maximality of $j$ . Instead consider the existence of $(1,j) \, \mathcal{R} \, (1, j+1)$ . By restriction, we must have $(0,j) \, \mathcal{R} \, (1, j+1)$ as well as $(0,j) \, \mathcal{R} \, (0, j+1)$ . As saturated transfer systems satisfy two-out-of-three, it would imply that we moreover have $(0,j+1) \, \mathcal{R} \, (1,j+1)$ , once again contradicting the maximality of $j$ . As such, everything to the right of $j$ on the top row of $[1] \times [n]$ must be empty.

As we have $(0,j) \, \mathcal{R} \, (1,j)$ , by restriction we also have $(0, \ell ) \, \mathcal{R} \, (1, \ell )$ for all $0 \leqslant \ell \leqslant j$ . By the saturation condition, this forces the top row to the left of $j$ to be identical to the bottom row.

Assembling everything, it follows that there are $(n+2)2^n$ possibilities as claimed.

We will use the strategy outlined in Section 2 to give a recursion for $L(n)$ . In particular, we begin by considering the partition of $L(n)$ as:

\begin{equation*} L(n) = \coprod _{(a,b) \in [1] \times [n]} L(n,(a,b)) \end{equation*}

where $L(n,(a,b))$ is the set of liftable transfer systems such that $(a,b)$ is the minimal fibrant element (i.e., $(a,b)$ is minimal such that $(a,b)\, \mathcal{R}\, (1,n)$ ).

We will find it advantageous to further split up the elements $L(n,(a,b))$ based on further partitioning properties of transfer systems.

It is clear that any transfer system possesses at least one stationary and at least one extendable element, namely $(1,n)$ and $(0,n)$ respectively. We now have a different partition of $L(n)$ as:

\begin{equation*} L(n) = \coprod _{1 \leqslant k \leqslant n+1} L(n,k) \end{equation*}

where $L(n,k)$ is the collection of transfer systems such that exactly $k$ elements are stationary. We then further refine each $L(n,k)$ as:

\begin{equation*} L(n,k) = \coprod _{1 \leqslant \ell \leqslant n+1} L(n,k,\ell ) \end{equation*}

where $L(n,k,\ell ) \subseteq L(n,k)$ is the subset of those transfer system where exactly $\ell$ elements are extendable.

Finally, we can define $L(n,k,\ell, (a,b))$ to be $L(n,k,\ell ) \cap L(n,(a,b))$ . We provide a full description of these transfer systems in the following definition.

In the next collection of propositions, we will provide recursions for $L(n,k,\ell, (a,b))$ for varying families of $(a,b) \in [1] \times [n]$ . We require one more definition before continuing to the first case.

We will provide an in-depth exploration of $\mathrm{Tam}(n,k)$ in Section 5. In particular, we will explain their namesake (Tamari intervals) and provide explicit formulæ for $|\mathrm{Tam}(n)|$ and $|\mathrm{Tam}(n,k)|$ ; see Proposition5.2.

Proposition 3.8. Let $b \gt 0$ . Then

\begin{equation*} L(n,k,\ell, (0,b)) = \coprod _{0 \leqslant i \leqslant k} \mathrm {Tam}(b-1,i) \times L(n-b,k-i,\ell, (0,0)). \end{equation*}

Proof. Let $\mathcal{R} \in L(n, (0,b))$ . Then by restriction to $(0,b)_{\uparrow }^{\mathrm{c}} \cong [1] \times [b-1]$ , we obtain a transfer system $\mathcal{R}_1 \in L(b-1)$ , and by restricting to $(0,b)_\uparrow \cong [1] \times [n-b]$ , we obtain a transfer system $\mathcal{R}_2 \in L(n-b)$ .

Since $(0,b)$ is fibrant in $\mathcal{R}$ and the minimal element of $(0,b)_\uparrow$ , we see that $\mathcal{R}_2 \in L(n-b,(0,0))$ . Similarly, since $(0,b)\, \mathcal{R}\, (1,b)$ by assumption, we also have $(0,b-1)\, \mathcal{R}\, (1,b-1)$ . As such, $(1,b-1)$ is fibrant in $\mathcal{R}_1$ . In particular, we conclude that

\begin{equation*} \mathcal {R}_1 \in \mathrm {Tam}(b-1) \end{equation*}

and

\begin{equation*} \mathcal {R}_2 \in L(n-b,(0,0)). \end{equation*}

Conversely, given an arbitrary $\mathcal{R}_1 \in \mathrm{Tam}(b-1)$ and $\mathcal{R}_2 \in L(n-b,(0,0))$ , we can construct a relation $\mathcal{R}$ on $[1] \times [n]$ by placing $\mathcal{R}_2$ to the right of $\mathcal{R}_1$ (i.e., we reindex $\mathcal{R}_2$ via a horizontal shift of $b$ ). We will show that $\mathcal{R}_1$ and $\mathcal{R}_2$ form a restriction closed pair.

Clearly such an $\mathcal{R}$ is transitive and satisfies the lifting condition. Suppose that $(c,d)\, \mathcal{R}\, (e,f)$ and $(g,h) \leqslant (e,f)$ . If $(e,f) \leqslant (1,b-1)$ , then we know that $(c,d)\wedge (g,h)\, \mathcal{R}\, (g,h)$ by the virtue of $\mathcal{R}_1$ being a transfer system.

Else, if $(e,f) \not \leqslant (1,b-1)$ , then we must have $(e,f) \geqslant (0,b)$ , and since $(c,d)\, \mathcal{R}\, (e,f)$ and we have no relations from $(0,b)_{\uparrow }^{\mathsf{c}}$ to $(0,b)_\uparrow$ , we must also have $(c,d) \geqslant (0,b)$ . If $(g,h) \geqslant (0,b)$ as well, then $(c,d) \wedge (g,h)\, \mathcal{R}\, (g,h)$ from that fact that $\mathcal{R}_2$ is a transfer system.

The last remaining case is $(0,b) \leqslant (c,d)$ and $(g,h) \leqslant (1,b-1)$ , then $(c,d) \wedge (g,h) = (c \wedge g, h)$ . If $c \wedge g = g$ , then we have $(c \wedge g, h ) \mathcal{R}(g,h)$ . Otherwise, $c=0$ and $g=1$ , so that we just need to ensure that $(0,h)\, \mathcal{R}\, (1,h)$ fo all $h \leqslant b$ , but this follows from that fact that $(0,b-1)$ is fibrant in $\mathcal{R}_1$ .

As such, we have shown that

\begin{equation*} L(n,(0,b)) = \mathrm {Tam}(b-1) \times L(n-b,(0,0)). \end{equation*}

Now we note that $(1,d)$ is stationary in $\mathcal{R}$ if and only if $d \leqslant b-1$ , and $(1,d)$ is stationary in $\mathcal{R}_1$ , or $d \geqslant b$ and $(1,d-b)$ is stationary in $\mathcal{R}_2$ . Furthermore, $(0,d)$ is extendable in $\mathcal{R}$ if and only if $d \geqslant b$ and $(0,d-b)$ is extendable in $\mathcal{R}_2$ . As such, we can refine the above formula to the desired result.

Proposition3.8 is valid whenever $b \gt 0$ . However, we see that the formula is trivial if $b=0$ , even though there are liftable transfer systems on $[1] \times [n]$ where the minimal fibrant element is $(0,0)$ (in particular the maximal transfer system is such). The next proposition resolves that case of $b=0$ .

Proposition 3.9.

\begin{equation*} L(n,k,\ell, (0,0)) = \coprod _{k-1 \leqslant k^{\prime} \leqslant n} L(n-1, k^{\prime}, \ell - 1). \end{equation*}

Proof. Let $\mathcal{R} \in L(n,k,(0,0))$ , and $\mathcal{R}^{\prime} \in L(n-1)$ be the restriction to $(0,1)_\uparrow \cong [1] \times [n-1]$ . Let $b \leqslant n$ be the maximal such that $(1,0)\, \mathcal{R}\, (1,b)$ . Then $\mathcal{R}$ is determined by $\mathcal{R}^{\prime}$ and $b$ . Further, the only constraint that we have on $\mathcal{R}^{\prime}$ and $b$ is that if $b \neq 0$ then $(1,b-1)$ must be stationary in $\mathcal{R}^{\prime}$ .

We then note that an element $(1,d)$ is stationary in $\mathcal{R}$ if and only if $d \geqslant b$ , and either $d=b=0$ or $(1,d-1)$ is stationary in $\mathcal{R}^{\prime}$ . Conversely, an element $(0,d)$ is extendable if and only if $d=0$ or $(0,d-1)$ is extendable in $\mathcal{R}^{\prime}$ . The result follows.

Via Propositions3.8 and 3.9, we have now computed $|L(n,k,\ell, (0,b))|$ for all possible values of $b$ . As such we now turn our attention to computing $|L(n,k,\ell, (1,b))|$ . The case where $b \lt n$ is trivial as we now prove.

Proposition 3.10. Let $b \lt n$ , then

\begin{equation*} L(n,k,\ell, (1,b)) = \varnothing . \end{equation*}

Proof. Recall that $L(n)$ consists of the liftable transfer systems. Assume that $\mathcal{R} \in L(n,k,\ell, (1,b))$ then by assumption we have $(1,b)\, \mathcal{R}\, (1,n)$ . However, by the lifting condition, this implies that $(0,b)\, \mathcal{R}\, (1,b)\, \mathcal{R}\, (1,n)$ , violating the minimality of $(1,b)$ . Therefore, no such $\mathcal{R}$ exists.

The simplicity of the case of $(1,b)$ where $b \lt n$ is one of the reasons why working with liftable transfer systems is simpler than the general case. Given Proposition3.10, it now suffices to compute $L(n,k,\ell, (1,n))$ :

Proposition 3.11. We have

\begin{equation*} L(n,k,\ell, (1,n)) = \coprod _{\ell - 1 \leqslant \ell ^{\prime} \leqslant n} L(n-1,k-1, \ell ^{\prime}). \end{equation*}

Proof. Let $\mathcal{R} \in L(n,k,(1,n))$ , and $\mathcal{R}^{\prime} \in L(n,-1)$ be the restriction to $[1] \times [n-1]$ . Clearly there are no constraints on $\mathcal{R}^{\prime}$ except from that it must have $k-1$ stationary elements.

We begin by assuming that there are no relations $(0,d)\, \mathcal{R}\, (0,n)$ for $d \lt n$ , that is, $\mathcal{R}$ has a single extendable element. Thus, all relations in $\mathcal{R}$ are contained in $(0,n)^{\mathsf{c}}_{\uparrow }$ and hence $\mathcal{R}$ is determined solely by $\mathcal{R}^{\prime}$ . In particular, we have

\begin{equation*} L(n,k,1,(1,n)) = L(n-1,k-1) = \coprod _{0 \leqslant \ell ^{\prime} \leqslant n} L(n-1,k-1,\ell ^{\prime}). \end{equation*}

We now move to the case where $\mathcal{R}$ has more than a single extendable element. That is, we assume that there exists some $d \lt n$ with $(0,d)\, \mathcal{R}\, (0,n)$ for $d \lt n$ . The set of all $d \lt n$ admitting arrows $(0,d)\, \mathcal{R}\, (0,n)$ is determined by the maximal such $d$ , which we denote $b$ . Indeed, if $(0,d)\, \mathcal{R}\, (0,n)$ then by restriction we also have $(0,d)\, \mathcal{R}\, (0,b)$ , and hence by definition $(0,d) \mathcal{R}^{\prime} (0,b)$ . Conversely if $(0,d) \mathcal{R}^{\prime} (0,b)$ , then by transitivity we have $(0,d)\, \mathcal{R}\, (0,b)\, \mathcal{R}\, (0,n)$ . The only constraint that we have on $b$ is that $(0,b)$ must be extendable in $\mathcal{R}^{\prime}$ . Note that if $b$ is the $i$ -th smallest extendable element in $\mathcal{R}^{\prime}$ , then $\mathcal{R}$ has $i+1$ extendable elements. If $\mathcal{R}^{\prime}$ has $k$ stationary elements, then $\mathcal{R}$ has $k+1$ stationary elements. Thus we have shown that

\begin{equation*} L(n,k,\ell, (1,n)) = \coprod _{\ell - 1 \leqslant \ell ^{\prime} \leqslant n} L(n-1,k-1,\ell ^{\prime}). \end{equation*}

as required. We note that this retrieves the above formula when $\ell = 1$ .

We are now in a position to combine the above results into the main theorem of this section which provides an explicit recursive algorithm for computing $|L(n)|$ .

Theorem 3.12. Let $n \geqslant 0$ . Then for $1 \leqslant \ell, k \leqslant n+1$ and $(a,b) \in [1] \times [n]$ we have

\begin{align*} |L(n,k,\ell, (0,0))| = & \sum _{k-1 \leqslant k^{\prime} \leqslant n} |L(n-1,k^{\prime},\ell -1)|&\\ |L(n,k,\ell, (1,n))| = & \sum _{\ell -1 \leqslant \ell ^{\prime} \leqslant n} |L(n-1,k-1,\ell ^{\prime})|&\\ |L(n,k,\ell, (0,b))| = & \sum _{0 \leqslant i \leqslant k} |\mathrm{Tam}(b-1,i)| \cdot |L(n-b,k-i,\ell, (0,0))|& (b\gt 0)\\ |L(n,k,\ell, (1,b))| = &{\hspace{1ex}} 0 & (b\lt n). \end{align*}

By convention, set $|L(n,k, \ell, (a,b))| = 0$ when $k, \ell$ are out of range. Then

\begin{equation*} |L(n)| = \sum _{\substack {1 \leqslant k, \ell \leqslant n+1 \\ (a,b) \in [1] \times [n]}} |L(n,k,\ell, (a,b))|. \end{equation*}

Computation 3.15. By starting at $m=0$ , we inductively store the values of $|L (m, k, \ell, (a,b))|$ for all $m \leqslant n$ , $1 \leqslant k, \ell \leqslant m+1$ and $(a,b) \in [1] \times [m]$ . The entries for all $m \lt m^{\prime}$ allow us to compute the values for $m=m^{\prime}$ via Theorem 3.12 and Proposition 5.2 ; some small values are recorded in Table 1, and large-scale structure appears at the end of the paper in Figure 1 .

Table 1. Values of $|L(n)|$ , the number of transfer systems for $D_{p^n}$ ( $p\ne 2$ ), computed via Theorem3.12.

Figure 1. A (nonlogarithmic) plot of $|L(n)|/|T(n)|$ for $0\leqslant n\leqslant 80$ . The semilogarithmic plot appears approximately linear.

4. Enumerating transfer systems for $C_{qp^n}$

In the previous section, we enumerated the liftable transfer systems on $[1] \times [n]$ . We will now move to the general case. For clarity, we will write $T(n)$ for the collection of all transfer systems $[1] \times [n]$ (i.e., $L(n) \subseteq T(n)$ ). We will employ the same strategy as before, and $T(n,k, \ell, (a,b))$ will be as before (but without the lifting condition).

From Remark3.14, we see that we need only consider how to resolve the case of $T(n,k,\ell, (1,b))$ for $b \lt n$ , and in the other cases the recursion for $T(n,k, \ell, (a,b))$ is the same as the recursion for $L(n,k, \ell, (a,b))$ . To resolve this remaining case, we will use a powerful observation regarding the duality of transfer systems as described in [Reference Franchere, Ormsby, Osorno, Qin and Waugh9] which we now recall. For a transfer system $\mathcal{R}$ on a finite lattice $L$ , we write

\begin{equation*} \mathcal {E}(\mathcal {R}) = \{ (z,y) \mid \text { there exists } x \in L \text { such that } z \leqslant x \lt y \text { and } x\, \mathcal {R}\, y \} \end{equation*}

for the downward closure of $\mathcal{R}$ .

We recall that a lattice $L$ admits a self-duality if there exists a bijection $\nabla : L \to L$ such that $ x \leqslant y$ if and only if $y^\nabla \leqslant x^\nabla$ for all $x,y \in L$ .

Definition 4.1. Let $L$ be a finite lattice admitting a self-duality $\nabla$ and let $\mathcal{R}$ be a transfer system on $L$ . The dual of $\mathcal{R}$ is the transfer system $\mathcal{R}^\ast$ defined as:

\begin{equation*} \mathcal {R}^\ast = ((\mathcal {E}(\mathcal {R})^{op})^\nabla )^{\mathsf {c}}. \end{equation*}

From [Reference Franchere, Ormsby, Osorno, Qin and Waugh9, Theorem 4.21], we have that $(\mathcal{R}^\ast )^\ast = \mathcal{R}$ so this provides an involution on the set $\mathsf{Tr}(L)$ . In the case that $L = [1] \times [n]$ , we use the canonical duality given by . In particular, we have $(a,b) \, \mathcal{R} \, (c,d)$ if and only if .

Proposition 4.2. The duality on $\mathsf{Tr}([1] \times [n])$ restricts to a duality:

\begin{equation*} T(n,k,\ell, (a,b)) \longleftrightarrow T(n,\ell, k,(1-a,n-b)). \end{equation*}

This duality does not preserve the property of being liftable.

Proof. Let $\mathcal{R} \in T(n,k,\ell, (a,b))$ . We first observe that $(1-a,n-b)$ is the minimal fibrant element in $\mathcal{R}^\ast$ . Indeed, we have $(a,b)$ as the minimal element such that $(a,b)\, \mathcal{R}\, (1,n)$ . From the definition, we see that $(1-a,n-b)\, \mathcal{R}^\ast \, (1,n)$ if and only if . Assume that we have $(0,0)\, \mathcal{E}(\mathcal{R}) \, (a,b)$ , then this implies the existence of some $(i,j) \lt (a,b)$ such that $(i,j)\, \mathcal{R}\, (a,b)$ , but by transitivity we would get $(i,j)\, \mathcal{R}\, (a,b)\, \mathcal{R}\, (1,n)$ , contradicting the minimality of $(a,b)$ . The minimality of $(1-a,n-b)$ as a fibrant element of $\mathcal{R}^\ast$ is afforded by the minimality of $(a,b)$ in $\mathcal{R}$ .

We will now explore the duality between the extendable and stationary elements. Let $y \geqslant 1$ , we will show that $(1,y-1)$ is stationary in $\mathcal{R}$ if and only if $(0,n-y)$ is extendable in $\mathcal{R}^\ast$ . Again by definition, this happens if and only if . Assume that $(1,0)\, \mathcal{E}(\mathcal{R})\, (1,y)$ , then this implies the existence of some $0 \leqslant z \lt y$ with $(1,z)\, \mathcal{R}\, (1,y)$ . However, we have assumed that $y-1$ is stationary, so no such $z$ can exist.

The remaining case of $y=0$ is covered by the fact that $(1,n)$ is always stationary and $(0,n)$ is always extendable.

Combining Proposition4.2, Theorem3.12, and Remark3.14, we arrive at our desired result.

Theorem 4.3. Let $n \geqslant 0$ . Then for $1 \leqslant \ell, k \leqslant n+1$ and $(a,b) \in [1] \times [n]$ , we have

\begin{align*} |T(n,k,\ell, (0,0))| = & \sum _{k-1 \leqslant k^{\prime} \leqslant n} |T(n-1,k^{\prime},\ell -1)|&\\ |T(n,k,\ell, (0,b))| = & \sum _{0 \leqslant i \leqslant k} |\mathrm{Tam}(b-1,i)| \cdot |T(n-b,k-i,\ell, (0,0))|& (b\gt 0)\\ |T(n,k,\ell, (1,b))| = &{\hspace{1ex}} |T(n,\ell, k,(0,n-b))|. & \end{align*}

By convention, set $|T(n,k, \ell, (a,b))| = 0$ when $k, \ell$ are out of range. Then

\begin{equation*} |T(n)| = \sum _{\substack {1 \leqslant k, \ell \leqslant n+1 \\ (a,b) \in [1] \times [n]}} |T(n,k,\ell, (a,b))|. \end{equation*}

Computation 4.5. Following the methods of Computation 3.15, we are once again able to efficiently compute values of $|T(n)|$ ; for small values of $n$ , these are presented in Table 2, and larger-scale behavior appears at the end of the paper in Figure 1 .

Table 2. Values of $|T(n)|$ , the number of transfer systems for $C_{qp^n}$ , computed via Theorem4.3.

5. Restricted Tamari intervals

In this section, we will study the term $\mathrm{Tam}(n,k)$ which appears in Proposition3.8, Theorems3.12, and 4.3.

We begin by recalling from Definition3.7 that we denote by $\mathrm{Tam}(n)$ the set of all transfer systems $\mathcal{R}$ with $(0,n)\, \mathcal{R}\, (1,n)$ and $\mathrm{Tam}(n,k) =\mathrm{Tam}(n) \cap L(n,k)$ .

Recall from [Reference Balchin, Ormsby, Osorno and Roitzheim6, Proposition 2.23, Proposition 4.17] that $\mathsf{Tr}([n])$ admits a lattice bijection to the Tamari lattice. As such for $\mathcal{R}$ and $\mathcal{R}^{\prime}$ transfer systems on $[n]$ , we say that $\mathcal{R} \leqslant \mathcal{R}^{\prime}$ is a Tamari interval. The next lemma justifies our choice of notation for $\mathrm{Tam}(n)$ . We note that if $\mathcal{R}$ is a transfer system on $[n]$ , then we—analogously to the $[1] \times [n]$ case—say that $b \in [n]$ is stationary for $\mathcal{R}$ if there exists no $d \gt b \geqslant c$ with $(1,c)\, \mathcal{R}\, (1,d)$ .

From work of Chapoton [Reference Chapoton8], we know that the cardinality of the set of Tamari intervals for $[n]$ has closed form given by:

\begin{equation*} |\mathrm {Tam}(n)| = \dfrac {2}{(n+1)(n+2)}\binom {4n+5}{n}. \end{equation*}

(We warn the reader that our indexing conventions differ from those of [Reference Chapoton8]). The following proposition provides a closed formula for $|\mathrm{Tam}(n,k)|$ .

Proposition 5.2. Let $n \geqslant 0$ and $1 \leqslant k \leqslant n+1$ . Then

\begin{equation*} |\mathrm {Tam}(n,k)| = \dfrac {2(2k + 1)!(4n - 2k + 3)!}{(k - 1)!(k + 1)!(n - k + 1)!(3n - k + 4)!} \end{equation*}

Proof. An explicit bijection between transfer systems on $[n]$ and rooted binary trees with $(n+1)$ internal nodes was constructed in [Reference Balchin, Barnes and Roitzheim1]. One can check that under this bijection that the number of stationary elements of a transfer system $\mathcal{R}$ on $[n]$ is equal to the number of elements on the corresponding binary tree whose path from the root is a straight line to the left.

Using the classical bijection between binary trees with $n+1$ nodes and Dyck paths of length $2n+2$ , we see that the number of stationary elements is also given by one less than the number of times the corresponding Dyck path touches the horizontal axis.

The result then follows from [Reference Bousquet-Mélou, Fusy and Préville-Ratelle4, Corollary 11] and the first remark following it.

Table 3. Values of $|\mathrm{Tam}(n,k)|$ for small values of $n$ and $k$ using Proposition5.2.

6. Maximally extendable $D_{p^n}$ transfer systems and Schröder numbers

In the remainder of this paper, we will explore a certain type of transfer system on $[1] \times [n]$ in the liftable and nonliftable cases, the enumeration of which retrieve interesting number sequences. We begin by introducing the class that we are interested in.

We will once again begin with the case of liftable transfer systems.

Proposition 6.4 ([Reference Qi and Guo15]). If $n = 0$ then $\mathfrak{S}_n = 1$ . For $n\ge 1$ , the large Schröder numbers satisfy the recurrence:

\begin{equation*} \mathfrak S_n = 3\mathfrak S_{n-1} + \sum _{k=1}^{n-2} \mathfrak S_k \mathfrak S_{n-k-1}. \end{equation*}

Additionally, for $n \geqslant 1$ we have

\begin{equation*} \mathfrak {S}_n = \sum _{1 \leqslant k \leqslant n} \mathrm {Nar}(n,k) \cdot 2^k = \sum _{1 \leqslant k \leqslant n} \dfrac {1}{n} \binom {n}{k} \binom {n}{k-1} \cdot 2^k \end{equation*}

where $\mathrm{Nar}(n,k)$ is the $(n,k)$ -th Narayana number.

Computation 6.5. Table 4 provides a calculation of $\mathfrak{S}_n$ for small values of $n$ .

Table 4. Values of $\mathfrak{S}_n$ computed via Proposition6.4.

We can produce refined statistics on royal paths by tracking the number of diagonal returns each path has, using the convention that a northeast step on the diagonal counts as a return. Let $\mathfrak S_n(k)$ denote the number of royal $n$ -paths with $k$ returns to the diagonal; call these the refined Schröder numbers.

Proposition 6.6 ([14]). The refined Schröder numbers take the values

\begin{equation*} \mathfrak S_n(k) = \begin {cases} 0 &\text {if }n\lt k,\\ 2^n &\text {if }n=k,\\ 2^k\frac {k}{n-k}\sum _{p=1}^{n-k}\binom {n-k}{p}\binom {n-1+p}{p-1}&\text {if }n\gt k. \end {cases} \end{equation*}

Furthermore,

\begin{equation*} \sum _{k=1}^n \mathfrak S_n(k) = \mathfrak S_n. \end{equation*}

Computation 6.7. Table 5 provides a calculation of $\mathfrak{S}_n(k)$ for small values of $n$ and $k$ .

Table 5. Values of $\mathfrak{S}_n(k)$ computed via Proposition6.6.

While the above result is standard, we were unable to find a proof of the following recurrence in the literature. It will be extremely useful in linking large Schröder numbers with transfer systems.

Lemma 6.8. The refined Schröder numbers satisfy the recurrence relation:

\begin{equation*} \mathfrak S_n(k) = 2\mathfrak S_n(k-1) + \sum _{p=k}^n \mathfrak S_{n-1}(p) \end{equation*}

for $1\le k\le n$ .

Proof. Write $E$ for east steps, $N$ for north steps, and $D$ for diagonal steps. Given a royal $n$ -path, consider the location of its first diagonal return. If this return is in position $(1,1)$ , then the path begins $EN$ or $D$ and the rest of the path may be filled in by any royal $(n-1)$ -path with $k-1$ returns. This accounts for the $2\mathfrak S_n(k-1)$ term. If the first return is in position $(r,r)$ with $r\gt 1$ , then we may write our path as $EPNQ$ where $P$ is a royal $(r-1)$ -path and $Q$ is a royal $(n-r)$ -path. The path $PQ$ is then a royal $(n-1)$ -path with at least $k$ diagonal returns. This decomposition is unique and each such royal $(n-1)$ -path arises in this fashion, so this accounts for the term $\sum _{p=k}^n \mathfrak S_{n-1}(p)$ .

We are now prepared to demonstrate the connection between (refined) large Schröder numbers and maximally extendable transfer systems on $D_{p^n}$ .

Proof. It suffices to prove the second statement since the refined Schröder numbers $\mathfrak S_{n+1}(k)$ add up to $\mathfrak S_{n+1}$ . Write $\mathfrak m(n,k) := |L(n,k,n+1)|$ for the number of maximally exetndable transfer systems on $D_{p^n}$ with exactly $k$ stationary nodes. Clearly $\mathfrak m(0,1) = 2 = \mathfrak S_1(1)$ , so it suffices to prove that $\mathfrak m(n,k)$ satisfies the recurrence of Lemma6.8, that is,

\begin{equation*} \mathfrak m(n,k) = 2\mathfrak m(n,k-1) + \sum _{p=k}^n \mathfrak m(n-1,p). \end{equation*}

This is a direct consequence of Theorem3.12 specialized to the case $\ell = n+1$ .

Proposition 6.11. Let $L^{\mathrm{max}}(n)$ denote the collection of maximally extendable liftable transfer systems on $[1]\times [n]$ . The asymptotics of $|L^{\mathrm{max}}(n)|$ satisfy

\begin{equation*} |L^{\mathrm {max}}(n)|\sim C\frac {(3+\sqrt {8})^n}{n^{3/2}} \end{equation*}

where

\begin{equation*} C = \frac {\sqrt {2}(3+2\sqrt {2})}{2\sqrt {\pi ({-}4+3\sqrt {2})}}\approx 4.720408926. \end{equation*}

This provides a lower bound on the asymptotic behavior of $|L(n)|$ .

Proof. This follows directly from Theorem6.9 and the asymptotics of the large Schröder numbers (see for instance Exercise 12 on p. 539 of [Reference Knuth12]).

We now shift our attention back to all transfer systems on $[1] \times [n]$ , and consider the maximally extendable transfer systems here. Recall from Proposition4.2 that there is a duality:

\begin{equation*} T(n,k,\ell, (a,b)) \longleftrightarrow T(n,\ell, k,(1-a,n-b)). \end{equation*}

In particular, by summing up over all elements $(a,b) \in [1] \times [n]$ we obtain a duality:

\begin{equation*} T(n,k,\ell ) \longleftrightarrow T(n,\ell, k). \end{equation*}

Proposition 6.17 ([Reference Ruskey17, Reference Klazar11]]). Let $n \geqslant 1$ . Then

\begin{equation*} \mathfrak {A}_n = \dfrac {\displaystyle {\sum _{0 \leqslant i \lt n}} \binom {2i+1}{i} \binom {2n-1}{n-i-1}}{2n-1}. \end{equation*}

Moreover, $\mathfrak{A}_n$ satisfies the recurrence $\mathfrak{A}_1 = 1$ and

\begin{equation*} \mathfrak {A}_{n} = \sum _{j=1}^{n-1} \mathfrak {A}_{n-j} \mathfrak {A}_{j} + \mathfrak {A}_{n-j} \mathrm{Cat}(j-1) \end{equation*}

where $\mathrm{Cat}(j)$ is the $j$ -th Catalan number.

Computation 6.18. Table 6 provides a calculation of $\mathfrak{A}_n$ for small values of $n$ .

Table 6. Values of $\mathfrak{A}_n$ computed via Proposition6.17.

Numerical experiments lead us to conjecture that maximally extendable transfer systems in $T(n)$ are counted by $\mathfrak A_{n+2}$ . By Corollary6.14, we conjecture the same count for maximally stationary transfer systems in $T(n)$ and maximally stationary transfer systems in $L(n)$ .

The authors’ attempts to prove this conjecture by standard enumerative methods (both bijective and inductive) have been spoiled by the subtle effects of restriction on relations joining the bottom and top rows. Following the argument of [Reference Klazar11], we note that it would suffice to decompose the top row $r$ of each maximally extendable transfer system in $T(n)$ into “primary components” $r_1,\ldots, r_k$ for which

\begin{equation*} w(r) = \prod _{i=1}^k (1+w(r_i)) \end{equation*}

where $w(s)$ measures the number of maximally extendable transfer systems with top row $s$ .

6.1. Figures

We conclude by plotting of the behavior of our enumerations related to transfer systems for $D_{p^n}$ and $C_{qp^n}$ . All of the data for these figures was produced using code available at https://github.com/bifibrant/recursion/.

The graph in Figure 1 suggests that $|L(n)|$ and $|T(n)|$ might have subexponential growth rates of a shape similar to the asymptotics of $|L^{\mathrm{max}}(n)|$ from Proposition6.11.

Finally, in Figure 2, we plot $|L(n)|/|T(n)|$ and observe that the fraction of liftable transfer systems appears to approach $0$ .

Figure 2. A semilog plot of the values of $|L(n)|$ (blue dots), $|T(n)|$ (red dots), $|L^{\mathrm{max}}(n)|$ (blue squares), and $|T^{\mathrm{max}}(n)|$ (red squares) for $0\leqslant n\leqslant 80$ . Here, the superscript $\mathrm{max}$ indicates maximally extendable transfer systems.