PERIODIC P-PARTITIONS arXiv:1803.05594v2 [math.CO] 6 Aug … · BRIAN T. CHAN Abstract. In this...

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PERIODIC P-PARTITIONS

BRIAN T. CHAN

Abstract. In this paper, we introduce a class of (P, ω)-partitions that we call periodic(P, ω)-partitions, then prove that such (P, ω)-partitions satisfy a homogeneous first-ordermatrix difference equation. After defining an appropriate counting problem for the above(P, ω)-partitions, we show that as a consequence of this equation, periodic (P, ω)-partitionscan be enumerated with constant coefficient linear recurrence relations. By analysing theabove matrix difference equation, we also prove a result for the asymptotic growth ratefor the number of periodic (P, ω)-partitions. The results of this paper generalizes andstrengthens the constant coefficient linear recurrence results proved by Sun and by Lopez,Martınez, Perez, Perez, and Basova for enumerating standard Young tableaux on shiftedstrips with constant width.

1. Introduction

The problem of enumerating standard Young tableaux on non-classical shapes is a topicof recent interest [2] in which shapes other than a Young diagram or a skew shape areconsidered. For instance, a problem equivalent to the problem of enumerating the num-ber of standard Young tableaux of shifted strips was considered by Hardin and Heinz ([13],A181196). Hardin and Heinz ([13], A181196; cf. [13], A227578) conjectured constant coeffi-cient linear recurrence relations for enumerating the number of standard Young tableaux onshifted strips with constant width up to when the width is seven. Sun [17] proved Hardinand Heinz’s relations for when the constant width is four and five. Moreover, Lopez, Lopez,Martınez, Perez, Perez, and Basova [8], calling shifted strips parallelogramic shapes, gen-eralized Sun’s constant coefficient linear recurrence results to all shifted strips with a fixednumber of cells in each row. Tewari and van Willigenburg proved that the above enumerationresults are connected to the representation theory of the 0-Hecke algebra [18] and provedHardin and Heinz’ conjecture in the case where the constant width is three.

(P, ω)-partitions were first considered by MacMahon in [9]. Later on, the theory of (P, ω)-partitions was developed by Gessel and Stanley [6, 16]. For instance, this theory is knownto have applications to quasisymmetric functions as (P, ω)-partitions are essential for thetheory of quasisymmetric functions [6]. In this paper, we investigate a different aspect of

Date: August 7, 2020.2010 Mathematics Subject Classification. 05A05, 05A15, 05A16, 05E10, 05E15, 06A07, 06A11.Key words and phrases. P -partitions, standard Young tableaux, non-classicial shapes, matrix difference

equations, constant coefficient linear recurrence relations, Perron-Frobenius Theory.1

http://arxiv.org/abs/1803.05594v2

2 BRIAN T. CHAN

(P, ω)-partitions by generalizing the results described in the previous paragraph to certain(P, ω)-partitions, then prove an asymptotic property of the sequences we are enumeratingand establish a divisibility property relating (P, ω)-partitions to minimal polynomials of cer-tain matrices.

Specifically, we introduce a class of (P, ω)-partitions that we call periodic (P, ω)-partitions,define an appropriate counting problem in order to enumerate such (P, ω)-partitions, andgeneralize the coefficient linear recurrence results described in the above paragraph to thisclass. To accomplish this, we introduce the notion of a connected triple for a poset to provethe existence of a matrix difference equation that can enumerate the above class of (P, ω)-partitions. As consequences of our enumeration result, we give new proofs of Sun’s and Lopezet.al.’s constant coefficient linear recurrence results and explain how semi-standard variantsof Sun and Lopez’s results can also be derived. We also give a brief outline of how ourresults can also be applied to standard Young tableaux on certain truncated shifted shapesthat exhibit a fixed repeating pattern and for higher-dimensional analogues of such tableaux.

We then prove that the matrix in the matrix difference equation we construct, that wecall a tableau transfer matrix, is a primitive matrix in many cases. Using this result, weapply Perron-Frobenius theory to the tableau transfer matrices that are primitive to provea result for the asymptotic growth rate of the number of periodic (P, ω)-partitions. Lastly,we briefly remark on how our proof that these matrices are primitive can be used to givefurther descriptions for the constant coefficient linear recurrence relations we are analysing.

In Section 2, we give the preliminaries and describe the appropriate counting problem thatwe will use to enumerate the (P, ω)-partitions. In Section 3, we introduce connected triplesand periodic (P, ω)-partitions, then describe examples of such (P, ω)-partitions. In Section4, we introduce the tableau transfer matrices and prove that the number of periodoic (P, ω)-partitions satisfy a matrix difference equation, then we prove that the number of periodic(P, ω)-partitions can be enumerated by a constant coefficient linear recurrence relation and,as corollaries, give new proofs of Sun and Lopez et.al’s results. Lastly, in Section 5, we provea result for the asymptotic growth rate of the number of periodic (P, ω)-partitions.

2. Preliminaries and the Enumeration Problem

Let N denote the set of positive integers and let N0 denote the set of non-negative in-tegers. If f : X → Y is a function and if X ′ is a subset of X , then let f |X′ denote therestriction of f to X ′. Let X be a set, and let ≡ be an equivalence relation on X . Then forall r ∈ X , the equivalence class of r in X with respect to ≡ is the set {r1 ∈ X : r ≡ r1}.Lastly, let X/ ≡ denote the set of all equivalence classes in X with respect to ≡. If X is a(non-empty) set, then a set partition of X is a set F of non-empty subsets of X such thatevery element of X is contained in exactly one element of F . For all positive integers n,define [n] = {1, 2, . . . , n}. Moreover, for all positive integers n1 and n2 such that n1 ≤ n2,

PERIODIC P-PARTITIONS 3

define [n1, n2] = {k ∈ N : n1 ≤ k ≤ n2}. In particular, [n, n] = {n}, [n, n + 1] = {n, n + 1},[n, n+2] = {n, n+1, n+2}, and so on. Lastly, write [n,∞) = {m ∈ Z : m ≥ n}. We denoteany sequence of numbers a1, a2, . . . by (an)n=1,2,.... Such a sequence satisfies a constantcoefficient linear recurrence relation if there exist constants c1, c2, . . . , cd such that for alln ≥ 1, an+d = c1an+d−1 + c2an+d−2 + · · ·+ cdan. The characteristic polynomial of the afore-mentioned recurrence an+d = c1an+d−1+c2an+d−2+· · ·+cdan is xd−c1x

d−1−c2xd−2−· · ·−cd.

We will use posets extensively in this paper, so we adhere to the following conventions. Re-call the notion of order-preserving maps, order-reversing maps, order-isomorphisms, order-embeddings, and subposets [5, 12]. Moreover, call an order-automorphism on P [12] anorder-isomorphism of the form f : P → P . For convenience, we write P instead of (P,≤).Whether P represents a poset or its underlying set of elements will be clear from context.For instance, an order-preserving map will be written as f : P → Q. Subsets of the set ofelements of a poset P will be regarded as subposets of P and vice versa. In particular, givena poset P , we may also call a subposet of P a subset of P . We regard any poset as being asubposet of itself, and we allow for a poset to have no elements.

As we regard subposets as subsets and vice versa, we will also use set-theoretic notationfor subposets. Given a poset P and two subposets S1 and S2 of P , let S1 ∩ S2 denote thesubposet of P whose elements are the elements of P that are contained in S1 and in S2, letS1∪S2 denote the subposet of P whose elements are the elements of P that are contained inS1 or in S2, let S1\S2 denote the subposet of P whose elements are the elements of S1 thatare not in S2, and let S1 ⊆ S2 mean that every element of S1 is contained in S2. Moreover,if f : P → Q is a map between two posets, if S is a subset of P , and if S ′ is a subset of Q,then let f(S) be the subposet of Q with {f(p) : p ∈ P} as its set of elements and let f−1(S ′)be the subset of P with {p ∈ P : f(p) ∈ S ′} as its set of elements. Lastly, let ∅ denote theempty set and the poset with no elements.

Additionally, we define the following [5]. Write p � q if p ≤ q is false. Also, let ≥, >,and < take on their usual meanings. If p, q ∈ P , then write p ‖ q to mean that p � q andq � p. Furthermore, if we want to clarify which partial order we are using, we say p ≤ qin P , where P is a poset and p, q ∈ P . Let P be a poset. Then we call a subset S ofP an order ideal of P if, for all p ∈ S and q ∈ P , q < p in P implies q ∈ S. Moreover,we call a subset S of P an order filter of P if, for all p ∈ S and q ∈ P , q > p in P implies q ∈ S.

A Young diagram (cf. [2, 14, 15]) is the empty set or a finite subset X of N2 such that forsome i, j ∈ N, (1, j) ∈ X and (i, 1) ∈ X . We call the elements of a Young diagram X thecells of X . Lastly, given a Young diagram X , define a row of X to be a non-empty subset ofcells of the form {r ∈ X : ∃j ∈ N such that r = (i, j)} for some i ∈ N, and define a columnof X to be a non-empty subset of cells of the form {r ∈ X : ∃i ∈ N such that r = (i, j)}for some j ∈ N. Following convention, we will depict the cells of a Young diagram with

4 BRIAN T. CHAN

boxes. Moreover, when depicting a Young diagram X , we position (i, j + 1) immediately tothe right of (i, j) and we position (i+ 1, j) immediately below (i, j).

Example 2.1. If X1 = {(1, 1), (1, 2), (1, 3)}, X2 = {(1, 1), (1, 3), (1, 6), (1, 7), (2, 3)}, andX3 = {(1, 3), (2, 2), (3, 1), (3, 3), (3, 4)}, then the Young diagram X1 is depicted by

,

the Young diagram X2 is depicted by

,

and the Young diagram X3 is depicted by

.

Regard Z as the poset · · · < −2 < −1 < 0 < 1 < 2 < · · · . Moreover, for any integerd ≥ 2, regard the d-dimensional integer lattice Zd as a poset with the standard partial ordergiven by (k1, k2, . . . , kd) ≤ (k′

1, k′

2, . . . , k′

d) if ki ≤ k′

i for all 1 ≤ i ≤ d. We will also regardany subset S of Zd as the subposet of Zd with S as its set of elements. In particular, asX ⊂ Z2 for any Young diagram X , we will also regard Young diagrams as subposets of Z2.In a Hasse diagram, p < q implies that q is positioned above p, but in a Young diagram,p < q implies that q cannot be positioned to the left of p and q cannot be positioned abovep. We illustrate this with an example.

Example 2.2. If P is the poset depicted by the following Hasse diagram

◦◦

◦◦ ◦

◦◦ ,

then any of the Young diagrams shown below can also be used to depict P .

Some posets that are not Young diagrams can also be depicted with Young diagrams. Forinstance, note the following.

Example 2.3. If P is the subposet of Z2 where P = {(−1,−5), (−1,−4), (−1,−3)}, thenwe formally define the following Young diagram that depicts P to be {(1, 1), (1, 2), (1, 3)}.


If u, v ∈ Rd and c1, c2 ∈ R, then let c1u+ c2v and c1u− c2v take on their usual meaningsfrom linear algebra where u and v are vectors and where c1 and c2 are scalars. Moreover,if u ∈ Rd, if c ∈ R, and if S ⊆ Rd, then write S + cu = {v + cu : v ∈ S} and writeS − cu = {v − cu : v ∈ S}.

Call a matrix an integer matrix if all entries of that matrix are integers. If M is a matrix,then let M(i, j) denote the entry in the ith row and jth column of M . A column vector isa matrix with one column, and a row rector is a matrix with one row. If v is a columnvector with N rows, then for all 1 ≤ i ≤ N , let v(i) denote the entry in the ith row ofv. If p(x) =

∑n

k=0 akxk is a polynomial for some scalars a0, a1, . . . , ak and if M is an

N by N square matrix, then write p(M) = a0IN +∑n

k=1 akMk, where IN is the N by N

identity matrix. Given functions f, g : N → R>0, R>0 denoting the set of positive reals,write f(n) ∼ g(n) as n → ∞ if limn→∞ f(n)/g(n) = 1. Lastly, given sequences (vn)n=1,2,...

and (v′n)n=1,2,... of column-vectors with positive entries such that for all n, vn and v′n have Nentries, write vn ∼ v′n as n → ∞ if for all integers 1 ≤ i ≤ N , vn(i) ∼ v′n(i) as n → ∞.

If X is a finite set, then an indexing of X is a bijection f : X → [k] where k = |X|. Next,we introduce terminology for (P, ω)-partitions from [14], but extend labellings to countablyinfinite posets and define A(Q, ω) in a non-standard way. If P is a finite poset, then alabelling of P is a bijection ω : P → [k] where k = |P |. Moreover, if P is a countably infiniteposet, then a labelling of P is a bijection ω : P → Z. Recall [14, 15] that a map f : P → Q,where P and Q are posets, is order-reversing if p1 ≤ p2 in P implies f(p2) ≤ f(p1) in Q. IfP is a finite poset and if ω is a labelling of P , then a (P, ω)-partition is an order-reversingmap f : P → N0 such that f(x) > f(y) if x < y and ω(x) > ω(y). We sometimes call a(P, ω)-partition a P -partition. Next, if P is a countable poset, if Q is a finite poset suchthat Q ⊆ P , and if ω is a labelling of P , then let A(Q, ω) denote the set of order-reversingmaps U : Q → N0 such that U(x) > U(y) if x < y and ω(x) > ω(y). If Q1 ⊆ Q, then welet U |Q1

denote the restriction of the function U to Q1. We also denote labellings with thesymbol ω′ or with the symbol ω′′. Lastly, if P is a countable poset and if ω is a labellingof P , then ω is a dual-natural labelling if ω is order-reversing and a natural labelling if ω isorder-preserving (cf. [14, 15]).

If P is a poset, then we will depict a labelling or an order-reversing map on P with a Hassediagram (or a Young diagram) whose nodes (or cells) are filled with integers. Specifically, ifP is a poset, if p ∈ P , if X ⊆ Z, and if f : P → X satisfies f(p) = k, then, when depictingf with a diagram, replace the node (or fill in the cell) corresponding to p with k.

Example 2.4. If (P,≤) is the poset depicted by the left-most diagram shown below, whereP = {p1, p2, p3, p4}, and if f : P → Z is a function such that f(p1) = 4, f(p2) = 2,f(p3) = −1, and f(p4) = 3, then we will depict f with the right-most diagram shown below.

6 BRIAN T. CHAN

p4

p1 p2 p3

3

4 2 −1

Example 2.5. If (P,≤) is the poset depicted by the left-most diagram shown below, whereP = {p1, p2, p3}, and if f : P → {1, 2, 3} is a function such that f(p1) = 3, f(p2) = 2, andf(p3) = 2, then we will depict f with the right-most diagram shown below.

p1p2 p3

3

2 2

If P is a finite poset and if ω is a labelling of P , then we would like to say that twoelements U1, U2 ∈ A(P, ω) are the same if the relative orderings of their entries are the same.Hence, we define the following equivalence relation.

Definition 2.6. Let P be a poset, let S1 and S2 be subsets of Z, and let f1 : P → S1 andf2 : P → S2 be maps. Then f1 is order equivalent to f2 if there exists an order-isomorphismg : f1(P ) → f2(P ) such that f2 = g ◦ f1. Lastly, we write f1 ≡ f2 if f1 is order equivalent tof2, and write f1 6≡ f2 otherwise.

Example 2.7. Let f1 : P → Z and f2 : P → Z be depicted by

5 2 2

6 7and 8 −1−1

9 15

respectively. Then f1(P ) = {2, 5, 6, 7}, f2(P ) = {−1, 8, 9, 15}, and g : f1(P ) → f2(P ) isdefined by g(2) = −1, g(5) = 8, g(6) = 9, and g(7) = 15. Since f2 = g ◦ f1, and since g isan order-isomorphism, f1 is order equivalent to f2.

We now formally define the enumeration problem for (P, ω)-partitions that we are in-terested in analysing. Specifically, we are interested in enumerating the number |Q, ω| ofequivalence classes as described below. For any set S, let |S| denote the cardinality of S.

Definition 2.8. Let P and Q be posets such that Q is finite and Q ⊆ P . Moreover, let ωbe a labelling P and let ≡Q,ω denote the equivalence relation on A(Q, ω) defined by f ≡Q,ω gif f is order equivalent to g. Then define

Tb(Q, ω) = A(Q, ω)/ ≡Q,ω

and define|Q, ω| = |Tb(Q, ω)|.

Moreover, if Q1 ⊆ Q and if T ∈ Tb(Q, ω), then let T |Q1be the element T ′ of Tb(Q1, ω)

such that, for all U ∈ A(Q, ω) satisfying U ∈ T , U |Q1∈ T ′.

To depict an element T of Tb(Q, ω), where Q and ω are as specified in Definition 2.8,pick the element U ∈ A(Q, ω) such that U ∈ T and the range of U : Q → N0 is [m] forsome integer m ≥ 1, display the diagram that represents the element U , then say that theresulting diagram depicts T .


Example 2.9. Let P be the six-element poset depicted by the left-most diagram shown below,let Q be the five-element subposet of P that is depicted by the cells in the left-most diagramthat are filled with bullets, and let ω : P → {1, 2, 3, 4, 5, 6} be depicted by the right-mostdiagram shown below.

• • •• •

2 1 36 4 5

For all 1 ≤ k ≤ 6, let pk ∈ P satisfy ω(pk) = k. Then A(Q, ω) consists of the order-preserving maps f : Q → N0 such that f(p2) > f(p1), f(p1) ≥ f(p3), f(p1) ≥ f(p6),f(p3) ≥ f(p4), and f(p6) > f(p4). Three of the elements in A(Q, ω) are depicted as follows.

U1 = 3 2 22 1

U2 = 5 4 32 1

U3 = 10 9 99 7

We have that U1 ≡ U3. However, U2 6≡ U1 and U2 6≡ U3.

Moreover, it can be checked that the |Q, ω| = 8 elements of Tb(Q, ω) are depicted by thebelow diagrams.

3 2 22 1

3 2 12 1

4 3 23 1

4 3 32 1

4 3 22 1

4 3 12 1

5 4 23 1

5 4 32 1

Now consider P . Note that Tb(P, ω) is well-defined because P is finite. So if T is the elementof Tb(P, ω) depicted below

4 3 3

3 2 1,

then T |Q is the element of Tb(Q, ω) depicted below

3 2 22 1

.

3. Periodic P-Partitions

In this section, we introduce a class of (P, ω)-partitions that we call periodic (P, ω)-partitions. Afterwards, we indicate how the problem of enumerating periodic (P, ω)-partitionsgeneralizes the problem of enumerating standard Young tableaux, and semi-standard Youngtableaux, of parallelogramic shapes/shifted strips when the number of cells in each row isfixed. In addition to this, we describe examples that illustrate the level of generality thatperiodic (P, ω)-partitions encompass.

We first introduce the notion of a connected triple.

8 BRIAN T. CHAN

Definition 3.1. Let P be a poset. Then a connected triple (A,B,C) of P is an orderedtriple (A,B,C) that satisfies the following two properties.

(1) The set {A,B,C} is a set partition of the set of elements of P .(2) For all p ∈ A and for all q ∈ C, there exists an element p′ ∈ B such that p < p′ < q

in P .

Moreover, if B is a non-empty subset of P , then B connects P if there exist non-emptysubsets A and C of P such that (A,B,C) is a connected triple of P .

Example 3.2. A parallelogramic shape [8] is a Young diagram X with n rows and k cellsin each row such that for any two adjacent rows, the lower row is shifted one cell to the rightof the upper row. For instance, if n = 3 and k = 5, then the corresponding parallelogramicshape X is the following.

Such shapes are also called shifted strips where the parameter k is called the width of sucha strip [17], and they were investigated in [8, 17, 18].

We make the following observation which can be generalized to any parallelogramic shape.If P is the poset corresponding to the parallelogramic shape with k = 4 and n = 4 depictedbelow, then if A is the subset of P that is depicted by the blank cells, if B is the the subset ofP that is depicted by the cells filled with bullets, and if C is the subset of P that is depictedby the cells filled with asterisks, then (A,B,C) is a connected triple of P .

•• • •∗ ∗ ∗ ∗

We extend the usual definition of the successor function from the natural numbers to theintegers. Namely, let s : Z → Z be the successor function defined by s(n) = n + 1 for allintegers n. With this function, we define periodic quadruple systems.

Definition 3.3. Let Z be a countably infinite poset, and let ω be a labelling of Z. Moreover,let π : Z → Z be a surjective order preserving map, and let θ : Z → Z be an order-automorphism on Z. Then the ordered quadruple (Z, ω, π, θ) is a periodic quadruple systemif the following three properties hold.

(1) There exists an order-automorphism α : Z → Z such that the following diagramcommutes.


◦◦

◦◦

◦◦

◦◦

◦◦

◦◦

◦◦

◦◦

.

.

.

.

.

.

0

−1

−2

−3

4

3

2

1

8

7

6

5

12

11

10

9

.

.

.

.

.

.

−6

−5

−4

−3

−2

−1

0

1

2

3

4

5

6

7

8

9

.

.

.

.

.

.

Figure 1.

ZZZ

ZZZ

ωπ

ωπ

αθs

(2) For all integers n, π−1({n}) is finite.(3) There exists a finite subset S of Z such that S connects Z.

Definition 3.3 can be applied to (collections of) parallelogramic shapes.

Example 3.4. Consider the poset Z depicted by the left-most diagram in Figure 1. More-over, let ω be the labelling of Z depicted by the middle diagram in Figure 1. For all integersn, let pn denote the element of Z such that ω(pn) = n. Next, define π : Z → Z so that forall integers n, π−1({n}) = {p1−4n, p2−4n, p3−4n, p4−4n}. Furthermore, let θ : Z → Z be theorder-automorphism on Z such that for all integers n, θ(pn) = pn−4.

To see that (Z, ω, π, θ) satisfies Property 1 of Definition 3.3, let α : Z → Z be defined byα(n) = n−4 for all integers n. Then for all p ∈ Z, s(π(p)) = π(θ(p)) and α(ω(p)) = ω(θ(p)).For instance, s(π(p7)) = s(−1) = 0, π(θ(p7)) = 0, α(ω(p7)) = α(7) = 3, and ω(θ(p7)) =ω(p3) = 3, implying that s(π(p7)) = 0 = π(θ(p7)) and α(ω(p7)) = 3 = ω(θ(p7)). Moreover,by how π is defined in this example, |π−1({n})| = 4 for all integers n. So (Z, ω, π, θ) satisfies

10 BRIAN T. CHAN

◦

◦

◦

◦

◦

◦

◦q1

◦p1

◦q0

◦p0

◦q−1

◦p−1

◦

◦

◦

◦

◦

◦

...

. . .

Figure 2. A more exotic example of a periodic quadruple system.

Property 2 of Definition 3.3. To see that (Z, ω, π, θ) satisfies Property 3 of Definition 3.3,let S ⊂ Z be defined by S = {p1, p2, p3, p5}. The set S is finite, and S connects Z. Hence,(Z, ω, π, θ) is a periodic quadruple system.

Lastly, let ω′′ be the labelling of Z depicted by the right-most diagram in Figure 1. WithZ, π, and θ being as described in this example, one can similarly check that (Z, ω′′, π, θ) isalso a periodic quadruple system.

Remark 3.5. The labelling ω′′ of Z in Example 3.4 can be regarded as being a Schur labelling.Roughly speaking, Schur labellings are labellings for skew shapes that relate (P, ω)-partitionsto semi-standard Young tableaux of skew shapes [15].

Periodic quadruple systems can be very different from Example 3.4. The following is anexample of such a system.


Example 3.6. Let Z be the poset depicted in Figure 2. Moreover, let p1, q1, p0, q0, p−1,and q−1 be the six elements of Z that are as specified in Figure 2. Let ω : Z → Z be de-fined by ω(p0) = 0, ω(q0) = 1, and, for all p ∈ Z, ω(θ(p)) = ω(p) + 2, let π : Z → Zsatisfy π(p1) = π(q1) = 1, π(p0) = π(q0) = 0, π(p−1) = π(q−1) = −1, and so on, and letθ : Z → Z satisfy θ(p−1) = p0, θ(q−1) = q0, θ(p0) = p1, θ(q0) = q1, and so on. To check thatthe quadruple (Z, ω, π, θ) satisfies Property 1 of Definition 3.3, let α : Z → Z be defined byα(n) = n+2 for all integers n. Then for all p ∈ Z, s(π(p)) = π(θ(p)) and α(ω(p)) = ω(θ(p)).

To see that (Z, ω, π, θ) satisfies Property 3 of Definition 3.3, let Pn = π−1([n]) for allintegers n ≥ 1. We show that P15 connects Z. Consider the element q0 ∈ Z. We have theinequalities,

q0 < θ4(p0) < θ7(p0) < θ10(p0) < θ13(p0),

q0 < θ4(p0) < θ4(q0) < θ8(p0) < θ11(p0) < θ14(p0),

and

q0 < θ4(p0) < θ4(q0) < θ8(p0) < θ8(q0) < θ12(p0).

Moreover, {θ14(p0), θ13(p0), θ12(p0)} is the set of minimal elements of π−1([12, ∞)). Soas {θ14(p0), θ13(p0), θ12(p0)} ⊂ P15, it follows that for all p ∈ Z satisfying π(p) ≥ 16, thereis an element q ∈ P15 such that q0 < q < p. Since θ is an order-automorphism on Z,the same conclusions also hold for θ−1(q0) and θ−1(P15) = π−1([0, 14]), and for θ−2(q0) andθ−2(P15) = π−1([−1, 13]). So as {q0, θ−1(q0), θ

−2(q0)} is the set of maximal elements ofπ−1(Z\N), it follows that P15 connects Z. Hence, (Z, ω, π, θ) is a periodic quadruple system.

We now introduce the P -partitions that we are interested in enumerating in this paper.

Definition 3.7. Let (Z, ω′, π, θ) be a periodic quadruple system and let n ∈ N. Then a lengthn periodic (P, ω)-partition derived from (Z, ω′, π, θ) is a (P, ω)-partition (P, ω) such thatP = π−1([n]) and ω ≡ ω′|P . Moreover, a periodic (P, ω)-partition derived from (Z, ω′, π, θ)is a length n periodic (P, ω)-partition derived from (Z, ω′, π, θ) for some integer n ≥ 1. Lastly,a periodic (P, ω)-partition is a periodic (P, ω)-partition derived from (Z, ω′, π, θ) for someperiodic quadruple system (Z, ω′, π, θ). For brevity, we also call periodic (P, ω)-partitionsperiodic P -partitions.

Remark 3.8. Since P = π−1([n]), P can be informally thought of as the poset that resultsfrom pasting together n copies π−1({1}), π−1({2}), . . . , and π−1({n}) of the poset π−1({0})where the pasting is determined by the periodic quadruple system (Z, ω, π, θ).

Enumeration formulas for counting standard Young tableaux on certain non-classicalshapes was established in [8] with special cases being established in [17, 18]. We describethese objects below. Let X be a parallelogramic shape with n rows and k cells in each rowas described in Example 3.2. Then a standard Young tableau of shape X is a bijective fillingof the cells of X with integers from [nk] such that the entries increase along every row of Xfrom left to right and the entries increase along every column of X from top to bottom, and

12 BRIAN T. CHAN

we call such a standard Young tableau a standard Young tableau of parallelogramic shape.For instance, two standard Young tableaux of parallelogramic shape are displayed below.

1 2 4 6

3 5 7 108 9 11 12

1 2 4 5

3 6 8 97 10 11 12

The enumeration problem we are considering in this paper generalizes the problem of enu-merating standard Young tableaux of parallelogramic shape when k is fixed.

Example 3.9. Let Z, ω, π, and θ be as in Example 3.4. Then the length n periodic (P, ω)-partitions derived from (Z, ω, π, θ) correspond to the standard Young tableaux of parallelo-gramic shape with n rows and four cells in each row. Specifically, if n ≥ 1 is an integer andif Pn = π−1([n]), then the elements of Tb(Pn, ω) correspond to the aforementioned standardYoung tableaux. For instance, when n = 3, three elements of Tb(P3, ω) are depicted below.

12 11 9 8

10 7 6 54 3 2 1

12 11 9 8

10 7 6 54 3 2 1

12 11 10 8

9 7 5 46 3 2 1

In particular, replacing every entry m in each of the three diagrams depicted above with12 − m + 1 gives three examples of standard Young tableaux of paralellogramic shape withthree rows and four cells in each row.

Next, we explain how we can also apply our results to semi-standard Young tableaux.If X is a parallelogramic shape, then define a semi-standard Young tableau of shape X tobe a function f : X → Z where f(i1, j) < f(i2, j) if (i1, j), (i2, j) ∈ X and i1 < i2, andwhere f(i, j1) ≤ f(i, j2) if (i, j1), (i, j2) ∈ X and j1 < j2. That is, fill the cells of X so thatthe entries weakly increase along every row of X from left to right, and the entries strictlyincrease along every column of X from top to bottom. Moreover, if f is a semi-standardYoung tableau of shape X , then define a semi-standard tableau class on X to be the set F ofsemi-standard Young tableau of shape X that are order equivalent to f . For example, if Xis a parallelogramic shape with n = 3 and k = 4, then three semi-standard tableau classeson X are depicted below.

1 1 2 2

2 3 4 44 5 5 5

1 1 1 1

2 2 2 23 3 3 3

1 2 4 6

3 5 7 108 9 11 12

The periodic P -partitions that we introduced in Definition 3.7 can be regarded as a gener-alization of the semi-standard Young tableau of parallelogramic shape X if the number ofcells in each row of X is fixed. This is explained in the following example when the numberof cells in each row is four.

Example 3.10. Let Z, ω′′, π, and θ be as in Example 3.4. Then the length n periodic(P, ω)-partitions derived from (Z, ω, π, θ) correspond to the semi-standard tableau classes on


X when X is a parallelogramic shape with n rows and four cells in each row. For instance,when n = 3, three elements of Tb(P3, ω

′′) are depicted below.

6 6 6 6

5 4 3 23 2 1 1

5 5 5 5

4 3 2 22 1 1 1

5 5 4 4

3 2 2 21 1 1 1

In particular, replacing every entry k in each of the three diagrams depicted above with12 − k + 1 gives three examples of semi-standard tableaux classes of paralellogramic shapewith three rows and four cells in each row.

In the last part of this section, we describe how more examples of periodic quadruplesystems can be generated if they are subposets of Zd.

Definition 3.11. Fix a positive integer d ≥ 2 and consider Zd. Next, let X be a finite subsetof Zd. Moreover, let ∆ = (k1, k2, . . . , kd) ∈ Nd be such that (X +∆) ∩X = ∅. Then define

Z(X,∆) =⋃

n∈Z

(X + n∆).

Informally, Z(X,∆) is a pairwise-disjoint union of translates of a finite subset X of Zd

where ∆ is the translate. An important property of this translate is that all coordinates of∆ are positive.

Remark 3.12. Note that by our convention, Z(X,∆) is a subposet of Zd, where the partialorder on Zd is as described in Section 2, because Z(X,∆) is a subset of Zd.

Example 3.13. If d = 2, X = {(1, 0), (2, 0), (3, 0), (4, 0)}, and ∆ = (1, 1), then the subposetZ(X,∆) of Z2 is the poset depicted in the left-most diagram referred in Example 3.4.

The posets Z(X,∆) satisfy the following property.

Lemma 3.14. Fix a positive integer d ≥ 2 and let Z(X,∆) be as described in Definition3.11. Then there exists a finite set S that connects Z(X,∆).

Example 3.15. The poset Z in Example 3.4 is an example of a poset Z(X,∆) where X ={(1, 0), (2, 0), (3, 0), (4, 0)} and ∆ = (1, 1).

Proof. Define π : Z(X,∆) → Z by π(p + n∆) = n for all p ∈ X and for all n ∈ Z. ByDefinition 3.11, (X + ∆) ∩ X = ∅. So π is well-defined. Hence, we proceed as follows.Because all coordinates of ∆ are positive, there is a sufficiently large integer k ≥ 1 such thatfor all p, q ∈ X , the inequality

(1) p ≤ q + k∆

holds in Z(X,∆). In particular, for all p, q ∈ Z(X,∆) satisfying π(q) ≥ π(p) + k, Inequality1 implies that p < q in Z(X,∆). Let

S =

2k+1⋃

n=1

(X + n∆).

14 BRIAN T. CHAN

Since X and k are finite, S is finite. For all p, q ∈ Z(X,∆) satisfying π(p) < 1 and π(q) >2k + 1, the following holds. Let p′ ∈ Z(X,∆) satisfy π(p′) = k + 1. Since π(p′) ≥ π(p) + kand π(q) ≥ π(p′) + k, p < p′ < q in Z. Hence, by Definition 3.1, S connects Z(X,∆). �

There is at least one periodic quadruple system that can be constructed from the posetZ(X,∆) in Lemma 3.14.

Proposition 3.16. Fix a positive integer d ≥ 2, and let Z(X,∆) be as described in Definition3.11. Moreover, let ωX be a dual-natural labelling of X, let π : Z(X,∆) → Z be defined byπ(p+n∆) = n for all p ∈ X and for all n ∈ Z, and let θ : Z(X,∆) → Z(X,∆) be defined byθ(p) = p+∆ for all p ∈ Z(X,∆). Then there exists a dual-natural labelling ω : Z(X,∆) → Zof Z(X,∆) such that ωX = ω|X and (Z(X,∆), ω, π, θ) is a periodic quadruple system.

Example 3.17. The periodic quadruple system (Z, ω, π, θ) in Example 3.4 is an example ofa periodic quadruple system constructed in Proposition 3.16.

Proof. By Definition 3.11, (X + ∆) ∩ X = ∅. So π is well-defined. Moreover, note that πis order-preserving. So we proceed as follows. For all p ∈ Z(X,∆), π(θ(p)) = π(p + ∆) =π(p) + 1. So π ◦ θ = s ◦π where s is as described in Definition 3.3. Note that θ is an order-automorphism on Z(X,∆). Hence, we define the following. Let ω : Z(X,∆) → Z be thebijection such that ω|X = ωX , ω(p) > ω(q) for all p, q ∈ Z(X,∆) such that π(p) < π(q), and(ω ◦ θn)|X ≡ ω|X for all n ∈ Z. As (ω ◦ θn)|X ≡ ωP for all n ∈ Z, ω|θn(X) is order-reversingfor all n ∈ Z. So if π(p) = π(q), then ω(p) > ω(q) because p < q in Z(X,∆). If π(p) 6= π(q),then π(p) < π(q) since π is order-preserving and since p < q. It follows from the definitionof ω that ω(p) > ω(q). So ω is dual-natural.

Because (ω ◦ θn)|X ≡ ω|X for all n ∈ Z and because ω is a bijection, it follows that forall p ∈ Z(X,∆), ω(p + ∆) = ω(p) − |X|. So for all p ∈ Z(X,∆), ω(θ(p)) = ω(p + ∆) =ω(p)− |X|. Hence, the map α : Z → Z defined by α(n) = n− |X| satisfies ω ◦ θ = α ◦ ω. Itfollows that the quadruple (Z, ω, π, θ) satisfies Property 1 of Definition 3.3. For all n ∈ Z,|π−1({n})| = |X + n∆| = |X|, implying that |π−1(n)| is finite. So (Z, ω, π, θ) satisfiesProperty 2 of Definition 3.3. Lastly, Lemma 3.14 implies that (Z, ω, π, θ) satisfies Property3 of Definition 3.3. From this, the lemma follows. �

Lastly, we make the following remarks.

Remark 3.18. Given any periodic quadruple system (Z, ω, π, θ), the quadruple (Z, ω′, π, θ),where ω′(p) = −ω(p) for all p ∈ Z, is also a periodic quadruple system. In particular, dualversions of the periodic quadruple systems in Example 3.4, Example 3.6, and Proposition3.16 also exist.

Remark 3.19. Standard Young tableaux on truncated shifted shapes have recently been atopic of interest [1, 2, 3, 11]. Using Proposition 3.16, it can be checked that by defining Xand ∆ appropriately, the results in this paper can be applied to Standard Young tableauxon truncated shifted shapes in which the numbers of cells in each row, when read from topto bottom, form (part of) a periodic sequence.


4. Matrix Difference Equations and Recurrence Relations

In this section, we construct a matrix with non-negative integer entries that we call atableau transfer matrix, then prove that the periodic (P, ω)-partitions can be enumeratedwith these matrices. We then explore some of the implications of our result by giving anew proof of Lopez et.al. and Sun’s constant coefficient linear recurrence results and byexplaining how to derive semi-standard variants of their results.

4.1. The Construction. Building from the previous section, we define the following.

Definition 4.1. Let (Z, ω, π, θ) be a periodic quadruple system. Then an index shape of(Z, ω, π, θ) is a finite subset S of π−1(Z\N) that satisfies the following two properties.

(1) (π−1(Z\N) \S , S , π−1(N)) is a connected triple of Z.(2) θ(S) ⊆ S ∪ π−1({1}).

When (Z, ω, π, θ) is not specified, we will call S an index shape.

Remark 4.2. Index shapes always exist. Let (Z, ω, π, θ) be a periodic quadruple system. Thenby Definition 3.3, there exists a finite set S that connects Z. So as θ is an order-automorphismon Z, it follows that for some integer n, θn(S) connects Z and θn(S) ⊆ π−1([n1, n2]) for someintegers n1 ≤ n2 ≤ 0. It can be checked that π−1([n1, n2]) is an index shape of (Z, ω, π, θ).

Remark 4.3. If S is an index shape of a periodic quadruple system (Z, ω, π, θ), then Sconnects Z. This fact, and the fact that S is finite, will become important later on in thissection.

Example 4.4. Let (Z, ω, π, θ) and S be as defined in Example 3.4. Moreover, for all in-tegers n, let pn ∈ Z satisfy ω(pn) = n. In Example 3.4, we saw that θ(pn) = pn−4, thatS = {p1, p2, p3, p5}, and that (π−1(Z\N) \S, S, π−1(N)) is a connected triple of Z. Hence,S satisfies Property 1 of Definition 4.1. Since θ(S) = {p−3, p−2, p−1, p1} and π−1({1}) ={p−3, p−2, p−1, p0},

θ(S) = {p−3, p−2, p−1, p1} ⊆ {p−3, p−2, p−1, p0, p1, p2, p3, p5} = S ∪ π−1({1}).Hence, θ(S) ⊆ S∪π−1({1}), implying that S satisfies Property 2 of Definition 4.1. It followsthat S is an index shape of (Z, ω, π, θ).

The following definition is an analogue of the notion of order equivalence that depends onthe order-automorphism θ on Z in a periodic quadruple system.

Definition 4.5. Let (Z, ω, π, θ) be a periodic quadruple system, and let S1 and S2 be subsetsof Z. Moreover, assume that S2 = θk(S1) for some integer k. Then for all T1 ∈ Tb(S1, ω) andT2 ∈ Tb(S2, ω), write T1 ≡θ T2 if for all U1 ∈ T1 and U2 ∈ T2, there is an order-isomorphismg : U1(S1) → U2(S2) such that g ◦ U1 = U2 ◦ θk.

Informally, the above definition says that T1 ≡θ T2 if the relative ordering of the entriesin T1 is the same as the relative ordering of the entries in T2.

16 BRIAN T. CHAN

Example 4.6. Let (Z, ω, π, θ) and S be as described in Example 4.4. The eight-elementsubposet S ∪ θ−1(S) ∪ π−1({0}) of Z is depicted by the Young diagram below. In the belowdiagram, S is depicted by the cells that are filled with circles or asterisks, and θ−1(S) isdepicted by the cells that are filled with asterisks and bullets.

◦◦ ◦ ∗

• • •Next, let T be the element of Tb(S ∪ θ−1(S) ∪ π−1({0}), ω) that is depicted by the diagramshown below. Lastly, for all integers 1 ≤ i ≤ 8, let ci denote the element of S ∪ θ−1(S) ∪π−1({0}) that is depicted by the cell in the diagram below that contains the entry i.

78 6 3

5 4 2 1

The function θ satisfies θ(c7) = c3, θ(c8) = c4, θ(c6) = c2, and θ(c3) = c1. From this,it follows that T |θ−1(S) ≡θ T |S. For instance, let U1 ∈ T |θ−1(S) be defined by U1(c7) = 7,U1(c8) = 8, U1(c6) = 6, and U1(c3) = 3, and let U2 ∈ T |S be defined by U2(c3) = 3, U2(c4) =4, U2(c2) = 2, and U2(c1) = 1. Then the order-isomorphism g : U1(θ

−1(S)) → U2(S) definedby g(3) = 1, g(6) = 2, g(7) = 3, and g(8) = 4 satisfies U2 ◦ θk = g ◦ U1 with k = 1.

We now use index shapes to define the following family of square matrices. Informally, weare defining matrices that are built from index shapes and that allow us to enumerate manydifferent P -partitions at once. Recall that we write M(i, j) to denote the entry in row i andcolumn j of a matrix M . Lastly, in the following definition, note that S ⊆ θ−1(S)∪π−1({0})due to Property 2 of Definition 4.1 and the fact that θ is an order-automorphism on Z.

Definition 4.7. Let (Z, ω, π, θ) be a periodic quadruple system, and let S be an index shapeof (Z, ω, π, θ). Moreover, let L be an indexing of Tb(S, ω), let N = |S, ω|, and let R : [N ] →Tb(S, ω) be the inverse of L. Then the tableau transfer matrix M derived from (Z, ω, π, θ),S, and L is the N by N matrix M such that for all 1 ≤ i ≤ N and 1 ≤ j ≤ N , M(i, j) isthe number of elements T of

Tb(θ−1(S) ∪ π−1({0}), ω)such that

T |S ≡θ R(i) and T |θ−1(S) ≡θ R(j).

When L is not specified, we will call M a tableau transfer matrix derived from (Z, ω, π, θ)and S, when S and L are not specified, we will call M a tableau transfer matrix derivedfrom (Z, ω, π, θ), and when (Z, ω, π, θ), S, and L are not specified, we will call M a tableautransfer matrix.

Example 4.8. Consider the periodic quadruple system (Z, ω, π, θ) and the index shape Sfrom Example 4.6. There are two elements of Tb(S, ω), and they are depicted by the belowdiagrams.


4

3 2 1

3

4 2 1

In particular, N = |S, ω| = 2. Next, define L : Tb(S, ω) → {1, 2} so that L sendsthe element of Tb(S, ω) depicted by the left-most diagram shown above to 1 and L sendsthe element of Tb(S, ω) depicted by the right-most diagram shown above to 2. Lastly, letR = L−1 be the inverse of L so that

R(1) = 4

3 2 1and R(2) = 3

4 2 1.

The tableau transfer matrix M derived from (Z, ω, π, θ), S, and L is equal to[

3 42 3

]

.

To see how to construct M , we calculate M(2, 1). The integer M(2, 1) is the number ofelements T ∈ Tb(S ∪ θ−1(S) ∪ π−1({0}), ω) such that T |S = R(2) and T |θ−1(S) ≡θ R(1).By using the depictions of S ∪ θ−1(S) ∪ π−1({0}), S, and θ−1(S) from Example 4.6, it canbe checked by an exhaustive search that the elements T of Tb(S ∪ θ−1(S) ∪ π−1({0}), ω)that satisfy T |S = R(2) and T |θ−1(S) ≡θ R(1) are the two elements depicted by the belowdiagrams.

87 6 3

5 4 2 1

87 5 3

6 4 2 1

We first prove a lemma that will be useful for the rest of this paper.

Lemma 4.9. Let P be a finite poset and let ωP be a labelling of P . Moreover, let S0 bean order ideal of P and let S1 be an order filter of P such that {S0, S1} is a set partitionof the set of elements of P . Lastly, let ω0 be a labelling of S0 such that ω0 ≡ ωP |S0

andlet ω1 be a labelling of S1 such that ω1 ≡ ωP |S1

. Then for all T0 ∈ Tb(S0, ω0) and for allT1 ∈ Tb(S1, ω1), there exists an element T ∈ Tb(S, ω) such that T |S0

= T0, T |S1= T1, and

for all U ∈ T , p ∈ S0, and q ∈ S1, U(p) > U(q).

Example 4.10. Let P be the eight-element poset depicted by the below Young diagram, letS0 be depicted by the blank cells, and let S1 be depicted by the cells that are filled with bullets.

• • • •Moreover, let ω be the following labelling of P

1 2 3 45 6 7 8

.

Next, let ω0 be a labelling of S0 such that ω0 ≡ ω|S0and let ω1 be a labelling of S1 such that

ω1 ≡ ω|S1. Lastly, let T0 ∈ Tb(S0, ω0) be depicted by the left-most diagram below and let

T1 ∈ Tb(S1, ω1) be depicted by the right-most diagram below.

18 BRIAN T. CHAN

2 2 2 1 2 1 1 1

Then an element T ∈ Tb(S, ω) as described in Lemma 4.9 is depicted below.

4 4 4 3

2 1 1 1

Proof. Let T0 ∈ Tb(S0, ω0) and let T1 ∈ Tb(S1, ω1). Because S0 and S1 are finite andbecause S0 ∩ S1 = ∅, there exist elements U0 ∈ T0 and U1 ∈ T1 such that for all p ∈ S0

and q ∈ S1, U0(p) > U0(q) ≥ 0. So define the map U : S → N0 by letting U(p) = U0(p)if p ∈ S0 and U(p) = U1(p) if p ∈ S1. This function U satisfies U |S0

≡ U0 and U |S1≡ U1.

Because S0 is an order ideal of P and because S1 is an order filter of P , the fact thatU(p) = U0(p) > U0(q) = U(q) for all p ∈ S0 and q ∈ S1 implies that U ∈ A(P, ωP ). Let Tbe the element of Tb(P, ωP ) that contains U . Then T satisfies the conditions described inthe lemma. �

To complement Definition 4.7, we define the following. Informally, we are defining thecolumn vectors that correspond to the tableau transfer matrices. Recall that we write v(i)to denote the entry in row i of a column vector v.

Definition 4.11. Let (Z, ω, π, θ) be a periodic quadruple system, let S be an index shapeof (Z, ω, π, θ), and let L be an indexing of Tb(S, ω). Moreover, let R = L−1 be the in-verse of L. Then an admissible number for (Z, ω, π, θ) and S is an integer n′ such thatn′ ≤ 0 and S ⊆ π−1([n′, 0]). Next, let n′ be an admissible number for (Z, ω, π, θ) and S, letP ′ = π−1([n′, 0]), let T0 ∈ Tb(P ′, ω), let n ∈ N, let Qn = π−1([n′, n]), and let Pn = π−1([n]).

Define the nth set derived from (Z, ω, π, θ), S, and T0 to be the set Xn(T0) of elementsT ∈ Tb(Qn, ω) such that T |P ′ = T0 and the following condition holds. If U ∈ T , p ∈ P ′, andq ∈ Pn, then U(p) > U(q).

Moreover, for all 1 ≤ i ≤ |S, ω|, let the ith part of the nth set derived from (Z, ω, π, θ),S, L, and T0 be the set Xn,i(T0) of elements T ∈ Xn(T0) such that T |θn(S) ≡θ R(i). Lastly,define the nth vector derived from (Z, ω, π, θ), S, L, and T0 to be the column vector vn with|S, ω| entries such that, for all 1 ≤ i ≤ |S, ω|, vn(i) = |Xn,i(T0)|.

Informally, the above definition describes the following. The nth sets as given in the abovedefinition are a collection of modified periodic P -partitions that will allow us to enumeratethe periodic P -partitions themselves. Moreover, the ith parts of such sets, as given in theabove definition, provides us with a set partition of such collections of modified periodic P -partitions that we will use later on in this section, and the nth vectors give the cardinalitiesof these ith parts. Lastly, admissible numbers enable us to effectively use such modifiedperiodic P -partitions.

Example 4.12. Let (Z, ω, π, θ), S, and L be as in Example 4.8, and let R = L−1 be theinverse of L. Moreover, for all integers n, let pn ∈ Z satisfy ω(pn) = n. As described


in Example 3.4, S = {p1, p2, p3, p5}. The integer n′ = −1 is an admissible number for(Z, ω, π, θ) and S because S ⊆ π−1([−1, 0]), and, as described in Example 3.4, π−1({0}) ={p1, p2, p3, p4} and π−1({−1}) = {p−3, p−2, p−1, p0}. Let P ′ = π−1([−1, 0]) and assume thatT0 is the following element of Tb(P ′, ω).

8 7 6 4

5 3 2 1

We illustrate Xn(T0), Xn,i(T0), and vn when n = 1. As described in Example 4.8, Tb(S, ω)has two elements. Hence, by Definition 2.8, |S, ω| = 2 and it follows that the range forthe index i is 1 ≤ i ≤ 2. The 1st set X1(T0) derived from (Z, ω, π, θ), S, and T0 can bedetermined as follows.

Let Q1 = π−1([−1, 1]) and let P1 = π−1({1}). Consider the element T ∈ Tb(Q1, ω)depicted by the left-most diagram below. Using Lemma 4.9 and setting P = Q1, S0 = P ′,S1 = P\P ′, and ωP = ω|P , it can be checked that the 1st set X1(T0) derived from (Z, ω, π, θ),S, and T0 is X1(T0) = {T}.

12 11 10 8

9 7 6 5

4 3 2 1

•• • •

Moreover, θ(S) is depicted by the cells in the right-most diagram above that are filled withbullets. Hence, with T ∈ X1(T0) as above, T |θ(S) ≡θ R(1) where R(1) is as described inExample 4.8. So the 1st part, X1,1(T0), of the 1st set derived from (Z, ω, π, θ), S, L, and T0,is X1(T0). Moreover, the 2nd part, X1,2(T0), of the 1st set derived from (Z, ω, π, θ), S, L,and T0, is the empty set because the only element of X1(T0) is T .

Therefore, as |X1,1(T0)| = 1 and |X1,2(T0)| = 0, the 1st vector derived from (Z, ω, π, θ), S,L, and T0 is the following column vector.

v1 =

[

10

]

The sum of the entries in the column vectors defined in Definition 4.11 gives the numberof periodic P -partitions.

Lemma 4.13. Let (Z, ω, π, θ) be a periodic quadruple system. Moreover, let S be an indexshape of (Z, ω, π, θ), let L be an indexing of Tb(S, ω), let R = L−1 be the inverse of L, letPn = π−1([n]) for all integers n ≥ 1, let n′ be an admissible number for (Z, ω, π, θ) and S,let T0 ∈ Tb(π−1([n′, 0]), ω), and let N = |S, ω|. Lastly, let vn be the nth vector derived from(Z, ω, π, θ), S, L, and T0. Then for all n ≥ 1,

|Pn, ω| =N∑

i=1

vn(i).

20 BRIAN T. CHAN

Example 4.14. Let (Z, ω, π, θ) and v1 be as described in Example 4.12. The sum of theentries of v1 is 1. Moreover, as there is exactly one standard Young tableau on a single row,|P1, ω| = 1, where P1 = π−1({1}). Hence, |P1, ω| is also the sum of the entries of v1.

Proof. Let n ∈ N, let Xn,i(T0) be the ith part of the nth set derived from (Z, ω, π, θ), S, L,and T0 for all 1 ≤ i ≤ N , and let Xn(T0) be the nth set derived from (Z, ω, π, θ), S, and T0.Recall that, by Definition 4.11, Xn,i(T0) is the set of elements T ∈ Xn(T0) such that

T |θn(S) ≡θ R(i)

for all 1 ≤ i ≤ N . Moreover, for all T ∈ Xn(T0), there is exactly one index 1 ≤ i ≤ N suchthat T |θn(S) ≡θ R(i) because R(i1) 6= R(i2) for all 1 ≤ i1 ≤ N and 1 ≤ i2 ≤ N satisfyingi1 6= i2. Hence, we have

(2) Xn(T0) =

N⋃

i=1

Xn,i(T0).

Define the map f : Xn(T0) → Tb(Pn, ω) by f(T ) = T |Pn. Let P ′ = π−1([n′, 0]). For

all T ∈ Xn(T0), for all U ∈ T , for all p ∈ P ′, and for all q ∈ Pn, U(p) > U(q). So ifT1, T2 ∈ Xn(T0), then T1 6= T2 if and only if T1|Pn

6= T2|Pn. It follows that f is injective. Let

Qn = P ′ ∪ Pn. Because π : Z → Z is order-preserving, it follows that P ′ is an order idealof Qn and that Pn is an order filter of Qn. So by Lemma 4.9 applied to P = Qn, S0 = P ′,S1 = Pn, and ωP = ω|P , f is surjective. Hence, f is a bijection, and the lemma follows fromEquation 2. �

4.2. The Matrix Difference Equation. In this subsection, we prove that periodic (P, ω)-partitions satisfy first order homogeneous matrix difference equations in which the matricesare the tableau transfer matrices. Then we prove that the periodic (P, ω)-partitions can beenumerated with constant coefficient linear recurrence relations. As a consequence, we givea new proof of Lopez et.al’s and Sun’s constant coefficient linear recurrence relation resultsby defining an appropriate periodic quadruple system (Z, ω, π, θ), then explain how to provea semi-standard variant of their results by modifying the labelling ω of Z in our new proof.

We first prove a structural property of connected triples. It is a crucial lemma that isessential for what follows. Informally, the following lemma states that for connected triplesof finite posets, we can define a notion of union for two P -partitions in a well-defined manner.

Lemma 4.15. Let Q be a finite poset, let (A,B,C) be a connected triple of Q, and letωQ be a labelling of Q. Then for all T ′ ∈ Tb(A ∪ B, ωQ) and T ′′ ∈ Tb(B ∪ C, ωQ) suchthat T ′|B = T ′′|B, there is a unique element T ∈ Tb(Q, ωQ) such that T |A∪B = T ′ andT |B∪C = T ′′.

Example 4.16. Let Q be the twelve element poset depicted by the below Young diagram.Moreover, let A be the subposet of Q depicted by the cells that are filled with asterisks, let B


be the subposet of Q depicted by the cells that are filled with bullets, and let C be the subposetof Q depicted by the blank cells. Moreover, let ωQ be a dual-natural labelling of Q, where interms of tableaux the entries in the rows decreases from left to right, and the entries in thecolumns decrease from top to bottom.

∗ ∗ ∗ •∗ • • •

From how A, B, and C are defined, (A,B,C) is a connected triple of Q by Definition 3.1.Let T ′ be the element of Tb(A ∪B, ωQ) that is depicted by the left-most diagram below, andlet T ′′ be the element of Tb(B ∪ C, ωQ) that is depicted by the right-most diagram below.

8 7 6 3

5 4 2 1

7

8 5 36 4 2 1

From the above, we see that T ′|B = T ′′

B. Moreover, the unique element T ∈ Tb(Q, ωQ) suchthat T |A∪B = T ′ and T |B∪C = T ′′ is depicted by the following diagram.

12 11 10 79 8 5 3

6 4 2 1

Proof. It is enough to prove the following. Assume that there are elements U1,1, U1,2 ∈A(A ∪ B, ωQ) and elements U2,1, U2,2 ∈ A(B ∪ C, ωQ) such that U1,1 ≡ U1,2, U2,1 ≡ U2,2,U1,1|B ≡ U2,1|B, and U1,2|B ≡ U2,2|B. Then the following two statements hold.

(1) There exist elements U1 ∈ A(Q, ωQ) and U2 ∈ A(Q, ωQ) such that U1|A∪B ≡ U1,1,U1|B∪C ≡ U2,1, U2|A∪B ≡ U1,2, and U2|B∪C ≡ U2,2.

(2) If U ′

1 ∈ A(Q, ωQ) and U ′

2 ∈ A(Q, ωQ) satisfy U ′

1|A∪B ≡ U1,1, U′

1|B∪C ≡ U2,1, U′

2|A∪B ≡U1,2, and U ′

2|B∪C ≡ U2,2, then U ′

1 ≡ U ′

2.

We first prove Statement 1. Since U1,j|B ≡ U2,j |B for all 1 ≤ j ≤ 2, there are order-embeddings g1,j : U1,j(A ∪ B) → N0 and g2,j : U2,j(B ∪ C) → N0 such that, for all p ∈ B,

g1,j(U1,j(p)) = g2,j(U2,j(p)).

So for all 1 ≤ j ≤ 2, define Uj : Q → N0 by

Uj(p) =

{

g1,j(U1,j(p)) if p ∈ A ∪B

g2,j(U2,j(p)) if p ∈ B ∪ C.

For all 1 ≤ j ≤ 2, the above map Uj is well-defined because of the definition of g1,j andg2,j. Moreover, for all 1 ≤ j ≤ 2, the map Uj satisfies

Uj|A∪B = g1,j ◦ U1,j ≡ U1,j and Uj |B∪C = g2,j ◦ U2,j ≡ U2,j.

22 BRIAN T. CHAN

Hence, to prove Statement 1, it is enough to prove that Uj ∈ A(Q, ωQ) for all 1 ≤ j ≤ 2. Solet j ∈ {1, 2}. To see that Uj is order reversing as required by the definition of A(Q, ωQ) inSection 2, suppose otherwise.

Because Uj |A∪B ∈ A(A∪B, ωQ) and Uj |B∪C ∈ A(B∪C, ωQ), Uj|A∪B and Uj |B∪C are orderreversing.

So, as we are supposing that Uj is not order reversing, there are elements p ∈ A andq ∈ C such that p < q but Uj(p) < Uj(q). By Property 2 of Definition 3.1, there existsan element p′ ∈ B such that p < p′ < q in Q. As p, p′ ∈ A ∪ B, as p < p′ in A ∪ B, andas Uj |A∪B is order reversing, it follows that Uj(p) ≥ Uj(p

′). Moreover, as p′, q ∈ B ∪ C, asp′ < q in B ∪ C, and as Uj |B∪C is order reversing, it follows that Uj(p

′) ≥ Uj(q). But then,Uj(p) ≥ Uj(p

′) ≥ Uj(q), which is contrary to the assumption that Uj(p) < Uj(q).

So Uj is order reversing. Suppose that Uj /∈ A(Q, ωQ). Then there are elements p, q ∈ Qsuch that p < q in Q, ωQ(p) > ωQ(q), and Uj(p) = Uj(q). Because Uj |A∪B ∈ A(A ∪ B, ωQ)and Uj|B∪C ∈ A(B ∪ C, ωQ), it follows that p ∈ A and q ∈ C. Since p ∈ A and q ∈ C,Property 2 of Definition 3.1 implies that there exists an element p′ ∈ B such that p < p′ < qin Q.

If ωQ(p) ≤ ωQ(p′) and ωQ(p

′) ≤ ωQ(q), then ωQ(p) ≤ ωQ(p′) ≤ ωQ(q), implying that

ωQ(p) ≤ ωQ(q). But that is contrary to the assumption that ωQ(p) > ωQ(q). Hence,ωQ(p) > ωQ(p

′) or ωQ(p′) > ωQ(q). So as p, p′ ∈ A∪B, p < p′ in A∪B, p′, q ∈ B ∪C, p′ < q

in B ∪ C, Uj |A∪B ∈ A(A ∪B, ωQ), and Uj |B∪C ∈ A(B ∪ C, ωQ), it follows that

Uj(p) > Uj(p′) ≥ Uj(q) or Uj(p) ≥ Uj(p

′) > Uj(q).

But then, Uj(p) > Uj(q), which is contrary to the assumption that Uj(p) = Uj(q). Hence,Uj ∈ A(Q, ωQ) and Statement 1 follows.

To prove Statement 2, let U1,1, U1,2, U′

1, U1,2, U2,2, and U ′

2 be as described in the beginningof the proof, and suppose that U ′

1 is not order equivalent to U ′

2. Because U ′

1|A∪B ≡ U1,1 ≡U1,2 ≡ U ′

2|A∪B and because U ′

1|B∪C ≡ U2,1 ≡ U2,2 ≡ U ′

2|B∪C , we have that U ′

1|A∪B ≡ U ′

2|A∪B

and U ′

1|B∪C ≡ U ′

2|B∪C . So there are elements p, q ∈ Q such that p ∈ A, q ∈ C, and exactlyone of the following holds.

• U ′

1(p) < U ′

1(q) and U ′

2(p) > U ′

2(q)

• U ′

1(p) > U ′

1(q) and U ′

2(p) < U ′

2(q)

• U ′

1(p) = U ′

1(q) and U ′

2(p) 6= U ′

2(q)


• U ′

1(p) 6= U ′

1(q) and U ′

2(p) = U ′

2(p)

Suppose that U ′

1(p) < U ′

1(q) and U ′

2(p) > U ′

2(q). Then, as U ′

1 and U ′

2 are order reversingmaps, it follows that p ‖ q in Q. But as (A,B,C) is a connected triple of Q, that violatesProperty 2 of Definition 3.1. Similarly, if U ′

1(p) > U ′

1(q) and U ′

2(p) < U ′

2(q), then Property 2of Definition 3.1 would be violated. So without loss of generality, suppose that U ′

1(p) = U ′

1(q)and U ′

2(p) 6= U ′

2(q).

Since p ∈ A and q ∈ C, Property 2 of Definition 3.1 implies that there exists an elementp′ ∈ B such that p < p′ < q in Q. Hence, as U ′

1 is order reversing and as U ′

1(p) = U ′

1(q),U ′

1(p) = U ′

1(p′) = U ′

1(q). So as p, p′ ∈ A ∪ B and U ′

1|A∪B ≡ U ′

2|A∪B, it follows that U′

2(p) =U ′

2(p′). Similarly, as p′, q ∈ B ∪ C and U ′

1|B∪C ≡ U ′

2|B∪C , it follows that U ′

2(p′) = U ′

2(q).But then, U ′

2(p) = U ′

2(q), which is contrary to the assumption that U ′

2(p) 6= U ′

2(q). Hence,Statement 2 follows.

�

Remark 4.17. Note that the converse of Lemma 4.15 is also true. If T ∈ Tb(Q, ω), then Tuniquely determines T |A∪B ∈ Tb(A ∪B, ω) and T |B∪C ∈ Tb(B ∪ C, ω).

In preparation for the main result of this section, we introduce the following technicaldefinition. It defines a positive integer that depends on the periodic quadruple system beingconsidered.

Definition 4.18. Let (Z, ω, π, θ) be a periodic quadruple system and let S be an index shapeof (Z, ω, π, θ). Then the minimum number of (Z, ω, π, θ) and S is the smallest positive integerm such that if n ∈ N satisfies n ≥ m, then for all p ∈ Z satisfying π(p) = n+1, there existsan element q ∈ θn(S) such that q < p in Z and π(q) ≥ 1.

Example 4.19. Let (Z, ω, π, θ) and S be as in Example 3.4. We explain why the minimumnumber m for (Z, ω, π, θ) and S is 1. Let the twelve-element poset π−1([0, 2]) be depicted bythe Young diagram below. Moreover, let π−1({0}) be depicted by the four cells in the toprow of the diagram, let θ(S) be depicted by the cells filled with asterisks or circles, and letπ−1({2}) be depicted by the cells filled with bullets.

∗◦ ◦ ◦• • • •

Set m = 1 and set n = m. From the above diagram, it can be checked that if p ∈ Z satisfiesπ(p) = n+1 = 2, then p is represented by one of the cells filled with bullets. Fixing any suchelement p ∈ Z satisfying π(p) = 2, it can also be checked, from the same diagram, that thereexists an element q ∈ θn(S) = θ(S), specifically one of the cells filled with a circle, such thatq < p in Z and π(q) ≥ 1. Hence, as it can be checked that the above is also true when m = 1and n > 1, it follows that m = 1 is the minimum number of (Z, ω, π, θ).

24 BRIAN T. CHAN

Remark 4.20. The minimum number of a periodic quadruple system always exists. Let(Z, ω, π, θ) be a periodic quadruple system and let S be an index shape of (Z, ω, π, θ). SinceS is finite, there is a positive integer n such that, for all p ∈ θn(S), π(p) ≥ 1. Moreover, Sconnects Z, so as θ is an order-automorphism on Z, θn(S) connects Z. Hence, by Property2 of Definition 3.1, it follows that for all p ∈ Z satisfying π(p) = n + 1 and for all p′ ∈ Zsatisfying π(p′) ≤ 0, there exists an element q ∈ θn(S) such that p′ < q < p. But as π(q) ≥ 1for all q ∈ θn(S), it follows from Definition 4.18 that the minimum number of (Z, ω, π, θ)exists and is at most n.

We now prove that periodic P -partitions satisfy a first order homogeneous matrix differ-ence equation involving tableau transfer matrices. For the proof, recall the definitions of thesets Xn(T0) and Xn,i(T0) in Definition 4.11.

Theorem 4.21. Let (Z, ω, π, θ) be a periodic quadruple system, let S be an index shape of(Z, ω, π, θ), and let L be an indexing of Tb(S, ω). Moreover, let n′ be an admissible numberfor (Z, ω, π, θ) and S, let T0 ∈ Tb(π−1([n′, 0]), ω), and let (vn)n=1,2,... be a sequence such thatfor all n ≥ 1, vn is the nth vector derived from (Z, ω, π, θ), S, L, and T0. Lastly, let M bethe tableau transfer matrix derived from (Z, ω, π, θ), S, and L, and let m be the minimumnumber of (Z, ω, π, θ) and S. Then for all n ≥ m,

vn+1 = M vn.

Example 4.22. Let (Z, ω, π, θ), S, L, T0 be as in Example 4.12. By Example 4.19, theminimum number m for (Z, ω, π, θ) and S is 1. So consider the 1st vector v1 derived from(Z, ω, π, θ), S, and L. As shown in Example 4.12,

v1 =

[

10

]

.

Next, let M be the tableau transfer matrix from Example 4.8. What this theorem allows usto do is to determine vn from M and v1 for any n ≥ 1. For instance,

v2 = Mv1 =

[

3 42 3

] [

10

]

=

[

32

]

.

Proof. Let n ∈ N. Define Q = π−1([n′, n + 1]), C = π−1({n + 1}), B = θn(S), andA = π−1([n′, n])\B. Then (A,B,C) is a connected triple of Q. Note that A∪B = π−1([n′, n])and θ(B) = θn+1(S).

If T ∈ Xn+1(T0), then T |A∪B ∈ Tb(A ∪ B, ω) and T |B∪C ∈ Tb(B ∪ C, ω). SinceT ∈ Xn+1(T0), Definition 4.11 implies that for all U ∈ T , p ∈ π−1([n′, 0]), and q ∈ π−1([n+1]),U(p) > U(q). It follows that T |A∪B ∈ Xn(T0) and T |B∪C ∈ Tb(B ∪ C, ω). Moreover, seeRemark 4.17, T |A∪B and T |B∪C are uniquely determined by T ∈ Xn+1(T0).

Next, define the map

f : Xn+1(T0) → {(T ′′, T ′) ∈ Tb(B ∪ C, ω)×Xn(T0) : T′′|B = T ′|B}


by

f(T ) = (T |B∪C , T |A∪B)

for all T ∈ Xn+1(T0). By what we just showed, f is well-defined and injective. We will provethat f is also a bijection. To that end, it is enough to show that f is surjective.

Let T ′ ∈ Xn(T0), let T ′′ ∈ Tb(B ∪ C, ω), and assume that T ′|B = T ′′|B. Recall thatTb(Q, ω) = Tb(Q, ωQ), where ωQ is the labelling of Q such that ω|Q ≡ ωQ. Hence, byLemma 4.15, there is a unique element T ∈ Tb(Q, ω) such that T |A∪B = T ′ and T |B∪C = T ′′.So to prove that f is surjective, it is enough to prove that T ∈ Xn+1(T0).

Let P ′ = π−1([n′, 0]) and let P = π−1([n+1]). Because T |A∪B = T ′ ∈ Xn(T0) and becauseP ′ ⊆ A ∪ B, Definition 4.11 implies that T |P ′ = T0. Hence, by Definition 4.11, it is enoughto show that for all U ∈ T , p ∈ P ′, and q ∈ P , U(p) > U(q). To that end, let U ∈ T , letp ∈ P ′, and let q ∈ P .

If π(q) ≤ n, then p, q ∈ A∪B, implying, as T |A∪B ∈ Xn(T0), that U(p) > U(q) by Defini-tion 4.11 applied to T |A∪B. So assume without loss of generality that π(q) = n+1. Becausen ≥ m, by hypothesis where m is the minimum number of (Z, ω, π, θ) and S, Definition 4.18implies that there exists an element p′ ∈ θn(S) such that p′ < q in Z and π(p′) ≥ 1. Inparticular, as p′ < q and U is order reversing, U(p′) ≥ U(q).

Because B = θn(S), p′ ∈ B. Moreover, T |A∪B ∈ Xn(T0) and p ∈ P ′. Furthermore, byDefinition 4.11, U(p′′) > U(q′′) for all p′′ ∈ P ′ and q′′ ∈ π−1([n]). Lastly, p′ ∈ π−1([n])because p′ ∈ θn(S) and π(p′) ≥ 1. So as p ∈ P ′ and p′ ∈ π−1([n]), we have U(p) > U(p′).Hence,

U(p) > U(p′) ≥ U(q),

implying that U(p) > U(q). From this, it follows that T ∈ Xn+1(T0). Hence, f is a bijection.

Whence, for all T2 ∈ Tb(S, ω), the number of elements T ∈ Xn+1(T0) satisfying T |θn+1(S) ≡θ

T2 is

(3)∑

T1∈Tb(S,ω)

M(T2, T1) |{T ∈ Xn(T0) : T |θn(S) ≡θ T1}|

for the following reasons.

Since B = θn(S), the fact that f is a bijection implies that the following is true forall T2 ∈ Tb(S, ω). If g is the restriction of f to the set of elements T ∈ Xn+1(T0) sat-isfying T |θn+1(S) ≡θ T2, then g is injective and the range of g is the set of ordered pairs(T ′′, T ′) ∈ Tb(B ∪ C, ω)×Xn(T0) satisfying T ′′|θn+1(S) ≡θ T2 and T ′′|θn(S) = T ′|θn(S).

26 BRIAN T. CHAN

Moreover, as θ is an order-automorphism on Z, as

B ∪ C = θn(S) ∪ π−1({n+ 1}) = θn+1(θ−1(S) ∪ π−1({0})),as θn(S) = θn+1(θ−1(S)), and as M is the tableau transfer matrix derived from (Z, ω, π, θ),S and L, Definition 4.7 implies that for all T1, T2 ∈ Tb(S, ω), the number of elementsT ′′ ∈ Tb(B ∪ C, ω) satisfying T ′′|θn+1(S) ≡θ T2 and T ′′|θn(S) ≡θ T1 is M(T2, T1).

Hence, the number of elements T ∈ Xn+1(T0) satisfying T |θn+1(S) ≡θ T2 is given by Ex-pression 3.

Lastly, by Definition 4.11,

|Xn,L(T1)(T0)| = |{T ∈ Xn(T0) : T |θn(S) ≡θ T1}|for all T1 ∈ Tb(S, ω), and

|Xn+1,L(T2)(T0)| = |{T ∈ Xn+1(T0) : T |θn+1(S) ≡θ T2}|for all T2 ∈ Tb(S, ω).

Therefore, for all T2 ∈ Tb(S, ω), Definition 4.11 implies that

vn+1(L(T2)) = |Xn+1,L(T2)(T0)|= |{T ∈ Xn+1(T0) : T |θn+1(S) ≡θ T2}|=

∑

T1∈Tb(S,ω)

M(T2, T1) |{T ∈ Xn(T0) : T |θn(S) ≡θ T1}|

=∑

T1∈Tb(S,ω)

M(T2, T1) |Xn,L(T1)(T0)|

=∑

T1∈Tb(S,ω)

M(T2, T1) vn(L(T1)).

From this, the theorem follows from the definition of matrix multiplication. �

We now establish the following enumeration result for periodic P -partitions.

Theorem 4.23. Let (Z, ω, π, θ) be a periodic quadruple system, let S be an index shape of(Z, ω, π, θ), let L be an indexing of Tb(S, ω), let M be the tableau transfer matrix derivedfrom (Z, ω, π, θ), S, and L, let m be the minimum number of (Z, ω, π, θ) and S, let ν(x) be theminimal polynomial of M , and let d be the largest non-negative integer d such that xd dividesν(x). Lastly, let Pn = π−1([n]) for all integers n ≥ 1. Then the sequence (an)n=m+d,m+d+1,...

defined by an = |Pn, ω| for all n ≥ m + d, satisfies a constant coefficient linear recurrencewhose characteristic polynomial is ν(x)/xd.

Example 4.24. Let (Z, ω, π, θ), S, L, T0, and M be as in Example 4.22. Moreover, let(an)n=1,2,... be defined by an = |Pn, ω| for all n ≥ 1 where Pn = π−1([n]). The minimal


4 5 5 6 6 7 73 4 4 5 5 6 63 4 4 5 5 6 62 3 3 4 4 5 52 3 3 4 4 5 50 0 2 0 3 0 40 0 2 0 3 0 4

Figure 3.

polynomial ν(x) of M is x2 − 6x + 1; in particular, d = 0 and ν(x)/xd = x2 − 6x + 1.Moreover, by Example 4.19, the minimum number m of (Z, ω, π, θ) and S is 1. So, as d = 0and m = 1, the sequence (an)n=1,2,... satisfies the recurrence

an+2 = 6an+1 − an.

This gives a new proof of one of the recurrence relation results from Sun [17] and Lopez et.al[8].

Proof. Let N = |S, ω|, let n′ be an admissible number for (Z, ω, π, θ) and S, and let T0 ∈Tb(π−1([n′, 0]). Since ν(x) is the minimal polynomial of M , ν(M) = 0. So by Theorem 4.21,

ν(M) vn = 0

for all n ≥ m. Hence, the theorem follows from Lemma 4.13. �

We now give new proofs of the constant coefficient linear recurrence relations from [8, 17,18]. For all integers n ≥ 1 and k ≥ 1, let Xn,k denote the parallelogramic shape with nrows and k cells in each row. Moreover, let T (n, k) denote the number of standard Youngtableaux of shape Xn,k.

Corollary 4.25. [Lopez et.al. (2017, [8]), cf. Sun [17], [18]] Fix a positive integer k ≥ 3.Moreover, let an = T (n, k) for all n ≥ 1. Then the sequence (an)n=1,2,... satisfies a constantcoefficient linear recurrence.

Example 4.26. For k = 5, consider the periodic quadruple system (Z(X,∆), ω, π, θ) fromProposition 3.16 when X = {(1, 0), (2, 0), (3, 0), (4, 0), (5, 0)} and ∆ = (1, 1). Moreover, let

S = {(1, 0), (2, 0), (3, 0), (4, 0)} ∪ {(2,−1), (3,−1)} ∪ {(2,−2)}.This is depicted by the seven-cell Young diagram in Figure 3. It can be checked that Sis an index shape of (Z(X,∆), ω, π, θ). Moreover, there is a labelling L of Tb(S, ω) suchthat the tableau transfer matrix derived from (Z(X,∆), ω, π, θ), S, and L is the matrix inFigure 3. The characteristic polynomial of this matrix is x4 − 24x3 + 40x2 + 8x, so by theCayley-Hamilton Theorem, the minimal polynomial of this matrix divides the aforementionedpolynomial. Lastly, it can be checked, in a manner similar to what was done in Example

28 BRIAN T. CHAN

4.19, that the minimum number m of (Z, ω, π, θ) and S is m = 1. Hence, by Theorem4.23, the subsequence (an)n=2,3,... of the sequence (an)n=1,2,... satisfies a constant coefficientlinear recurrence. So from Hardin and Heinz’s verification that T (1, 5), T (2, 5), T (3, 5),and T (4, 5) satisfy the above recurrence, we obtain a new proof of the recurrence relationempirically observed by Hardin and Heinz ([13], A181196) for T (n, 5).

Proof. Consider the periodic quadruple system (Z(X,∆), ω, π, θ) from Proposition 3.16 whenX = {(i, 0) : 1 ≤ i ≤ k} and ∆ = (1, 1). Next, let S be defined by

S = {(i, 0) : i ∈ [n]} ∪n−2⋃

k=1

{(i,−k) : i ∈ [2, n− k]}.

It can be checked that S is an index shape of Z(X,∆), so the corollary follows from Theorem4.23. �

Lastly, we observe that by replacing the labelling ω of Z(X,∆) in the proof of Corollary4.25 with a labelling ω′′ of Z(X,∆) that is analogous to the labelling ω′′ in Example 3.4, wealso obtain the following corollary.

Corollary 4.27. Fix a positive integer k ≥ 3. Moreover, let bn denote the number of semi-standard tableau classes on Xn,k for all n ≥ 1. Then the sequence (bn)n=1,2,... satisfies aconstant coefficient linear recurrence relation.

5. Asymptotic Growth Rates

In this section, we prove that many tableau transfer matrices are primitive. Using this fact,we prove that the asymptotic growth rate of the number |P, ω| of periodic (P, ω)-partitionsis given by c rn, where c and r are constants. Afterwards, we briefly remark on how ourproof that the above matrices are primitive can give further descriptions for the constantcoefficient linear recurrence relations we are analysing.

Frobenius [10] introduced the notion of a primitive matrix for Perron-Frobenius theory andestablished that such matrices are equivalent to the following, which we give as a definition.

Definition 5.1 (Frobenius [10]). A primitive matrix is a square matrix M such that M isa non-negative matrix and such that, for some integer k ≥ 1, Mk is a positive matrix.

Motivated by the above definition, we define the following index shapes and tableau trans-fer matrices.

Definition 5.2. Let (Z, ω, π, θ) be a periodic quadruple system and let S be an index shapeof (Z, ω, π, θ). Then S is a separable index shape of (Z, ω, π, θ) if S is an order filter ofπ−1(Z\N) and if S is an order ideal of S ∪π−1(N). When (Z, ω, π, θ) is not specified, we callS a separable index shape.

Remark 5.3. The index shape constructed in Remark 4.2 is also a separable index shape.From this, it follows that separable index shapes always exist.


Example 5.4. The index shapes described in Example 3.4 and Example 4.26, and the indexshapes described in the proof of Corollary 4.25 are examples of separable index shapes.

Definition 5.5. Let (Z, ω, π, θ) be a periodic quadruple system, let S be an index shape of(Z, ω, π, θ), and let M be a tableau transfer matrix derived from (Z, ω, π, θ) and S. Then callM separable if S is separable index shape of (Z, ω, π, θ).

Example 5.6. The tableau transfer matrices in Example 4.8 and Example 4.26, and thetableau transfer matrices described in the proof of Corollary 4.25 are examples of tableautransfer matrices that are separable.

We prove the following property of separable tableau transfer matrices.

Lemma 5.7. Any tableau transfer matrix that is separable is also primitive.

Example 5.8. The matrices in Example 5.6 are tableau transfer matrices that are primitive.

Proof. Let (Z, ω, π, θ) be a periodic quadruple system, let S be a separable index shape of(Z, ω, π, θ), let L be a labelling of Tb(S, ω), let M be the tableau transfer matrix derivedfrom (Z, ω, π, θ), S, and L, let R = L−1 be the inverse of L, and let N denote the numberof rows of M . Because S is a separable index shape of (Z, ω, π, θ), M is separable. ByDefinition 4.7, any tableau transfer matrix is a non-negative square matrix. Hence, it isenough to prove that Mk is positive for some integer k ≥ 1. We will first prove by inductionon k that for all 1 ≤ i ≤ N and 1 ≤ j ≤ N , Mk(i, j) equals to the number of elementsT ∈ Tb(θ−k(S) ∪ π−1([−k + 1, 0]), ω) such that T |S ≡θ R(i) and T |θ−k(S) ≡θ R(j).

The base case k = 1 follows from Definition 4.7. So let k ≥ 1 and assume that the claimholds for all integers in [k]. Consider the matrix Mk+1. Then for all 1 ≤ i ≤ N and for all1 ≤ j ≤ N ,

(4) Mk+1(i, j) =N∑

i′=1

Mk(i, i′)M(i′, j).

For all indices i′, i, and j in the summation in Equation 4, we note the following. Definition4.7, and the fact that θ is an order-automorphism on Z, implies that M(i′, j) is the numberof elements T ∈ Tb(θ−k−1(S) ∪ π−1({−k}), ω) such that T |θ−k(S) ≡θ R(i′) and T |θ−k−1(S) ≡θ

R(j). By the induction hypothesis applied to the entries Mk(i, i′) of Mk, Mk(i, i′) equalsto the number of elements T ∈ Tb(θ−k(S) ∪ π−1([−k + 1, 0]), ω) satisfying T |S ≡θ R(i) andT |θ−k(S) ≡θ R(i′). So as

((θ−k−1(S) ∪ π−1({−k}))\θ−k(S), θ−k(S), π−1([−k + 1, 0]))

is a connected triple of θ−k−1(S) ∪ π−1([−k, 0]), it follows from Lemma 4.15, Remark 4.17,and Equation 4 that Mk+1(i, j) equals to the number of elements T ∈ Tb(θ−k−1(S) ∪π−1([−k, 0]), ω) that satisfy T |S ≡θ R(i) and T |θ−k−1(S) ≡θ R(j). From this, the claimfollows. Note that the claim just proved does not depend on the assumption that S is a

30 BRIAN T. CHAN

separable index shape of (Z, ω, π, θ).

Next, we prove the following. By Definition 3.3, π(θ(p)) = π(p) + 1 for all p ∈ Z. So as Sand π(S) are finite, there exists a sufficiently large integer k ≥ 1 such that π(S)∩π(θ−k(S)) =∅. Hence, for some integer k ≥ 1,

(5) S ∩ θ−k(S) = ∅.Let S∗ = θ−k(S)∪π−1([−k+1, 0]) and let Xk = S∗\(S∪θ−k(S)). By Property 2 of Definition4.1 and the fact that θ is an order-automorphism on Z, S ⊆ S∗ since

S ⊆ θ−1(S) ∪ π−1({0}) ⊆ θ−2(S) ∪ π−1({−1, 0}) ⊆ · · · ⊆ θ−k(S) ∪ π−1([−k + 1, 0]) = S∗.

Because S is a separable index shape of (Z, ω, π, θ), Definition 5.2 implies that S is an orderideal of S ∪ π−1(N). So as θ is an order-automorphism on Z, θ−k(S) is an order ideal of S∗.Moreover, as S is a separable index shape of (Z, ω, π, θ), it follows from Definition 5.2 thatS is an order filter of π−1(Z\N). This fact, combined with Equation 5, implies that S is anorder filter of Xk ∪S. So as Equation 5 is true, the following holds. By applying Lemma 4.9to S∗, ω|S∗, θ−k(S), and Xk ∪S, then by applying, if Xk 6= ∅, Lemma 4.9 to Xk ∪S, ω|Xk∪S,Xk, and S, we conclude that there exists an element T ∈ Tb(S∗, ω) such that T |S ≡θ R(i)and T |θ−k(S) ≡θ R(j). Hence, M(i, j) ≥ 1 and the lemma follows. �

A consequence of the above is that we can combinatorially describe the entries of anypower of a tableau transfer matrix.

Corollary 5.9. Let (Z, ω, π, θ) be a periodic quadruple system, let S be an index shape of(Z, ω, π, θ), let L be an indexing of Tb(S, ω), and let R = L−1 be the inverse of L. Lastly,let M be the N by N tableau transfer matrix derived from (Z, ω, π, θ), S, and L. Then forall integers n ≥ 1 and for all 1 ≤ i ≤ N and 1 ≤ j ≤ N , Mn(i, j) equals to the number ofelements T of

Tb(θ−n(S) ∪ π−1([−n+ 1, 0]), ω)

that satisfy

T |S ≡θ R(i) and T |θ−n(S) ≡θ R(j).

Proof. This follows from the proof of Lemma 5.7. �

A vector is non-negative if all entries of the vector are non-negative and a vector is positiveif all entries of the vector are positive [10]. Moreover, letMT denote the transpose of a matrixM and let 〈u, v〉 denote the standard inner product of vectors u and v. Recall from spectraltheory that the spectrum X ⊂ C of a matrix M is the set of distinct eigenvalues of M andthat the spectral circle of a matrix M is the set of distinct eigenvalues of M of maximummodulus [10]. We state the part of the Perron-Frobenius theorem that we will need.

Theorem 5.10 (Perron, Frobenius [10]). Let M be a non-negative square matrix that isprimitive. Then M has exactly one eigenvalue r on its spectral circle. Moreover, r is a


positive real number, r is a simple eigenvalue of M , and an eigenvector v of M is non-negative if and only if v is positive and r is the eigenvalue corresponding to v.

Now, we prove the main result of this section.

Theorem 5.11. Let (Z, ω, π, θ) be a periodic quadruple system. Moreover, let Pn = π−1([n])for all integers n ≥ 1. Then there are constants c > 0 and r > 1 such that

|Pn, ω| ∼ c rn

as n → ∞.

Proof. Let (Z, ω, π, θ) be as described in Theorem 5.11. By Remark 5.3, we can assume thatS is a separable index shape of (Z, ω, π, θ). Next, let L be an indexing of Tb(S, ω). Moreover,let n′ be an admissible number for (Z, ω, π, θ) and S, let T0 ∈ Tb(π−1([n′, 0]), ω), and let(vn)n=1,2,... be a sequence such that for all n ≥ 1, vn is the nth vector derived from (Z, ω, π, θ),S, L, and T0. Lastly, let M be the tableau transfer matrix derived from (Z, ω, π, θ), S, and L.

Because S is a separable index shape of (Z, ω, π, θ), M is separable so by Lemma 5.7, Mis primitive. Hence, the Perron-Frobenius Theorem, and in particular Theorem 5.10, appliesto M . Let r be the eigenvalue of M in the spectral circle of M and let u be a positiveeigenvector of M that is an eigenvector of r. Lastly, let N denote the number of rows of M .

Because M is primitive, and because r is as described in Theorem 5.10, it follows fromTheorem 5.10 and well-known results on spectral projectors of matrices ([10] p. 674, p.630, p. 518) that there is a positive eigenvector v of M corresponding to r and a positiveeigenvector u of MT corresponding to r such that 〈u, v〉 > 0 and

(6) limn→∞

Mn0 =

v uT

〈u, v〉 ,

where M0 = r−1M .

Consider the vector v1. We show that 〈u, v1〉 > 0. By Lemma 4.13, the sum of the entriesof v1 is |P1, ω|, where P1 = π−1({1}). If Q is a finite poset with at least one element andif ωQ is a labelling of Q, then any injective order-reversing map f : Q → N0 is an elementof A(Q, ωQ). In particular, A(P1, ω) 6= ∅. So by Definition 2.8, it follows that |P1, ω| > 0,implying that v1 has at least one positive entry. Hence, as u is positive, 〈u, v1〉 > 0.

Write M1 = limn→∞Mn0 . Because 〈u, v1〉 > 0, it follows from Equation 6 that

M1 v1 =v uT v1〈u, v〉 =

〈u, v1〉〈u, v〉 v.

So as 〈u, v1〉/〈u, v〉 > 0, the following holds. Write v∗ = M1 v1. Then by Theorem 4.21,it follows from Equation 6 that vn ∼ rn v∗ as n → ∞. Hence, the Theorem follows fromLemma 4.13. �

32 BRIAN T. CHAN

A consequence of Theorem 5.11 is the following, which adds to what is known aboutstandard Young tableaux of parallelogramic shapes. Recall that T (n, k) denotes the numberof standard Young tableaux of parallelogramic shape Xn,k, where Xn,k has n rows and k cellsin each row.

Corollary 5.12. Fix an integer k ≥ 3. Then for some constants c > 0 and r > 1,

T (n, k) ∼ c rn

as n → ∞.

Example 5.13. In [8], Lopez et.al. computed the characteristic polynomials for the constantcoefficient linear recurrence relations of minimum order for the sequences (T (n, k))n=1,2,... fork up to twelve. From these polynomials, they computed closed-form expressions [8] for T (n, k)when k = 4, 5, 6. Using their closed-form expression for k = 4, we conclude that

T (n, 4) ∼(

1

2− 1

2√2

)

(3 + 2√2)n

as n → ∞.

Proof. As described in the proof of Corollary 4.25, Proposition 3.16 implies that collectionsof parallelogramic shapes constitute examples of periodic quadruple systems if the numberof cells in each row is fixed. Hence, the corollary follows from Theorem 5.11. �

Another corollary of Theorem 5.11 is the following. Recall that Xn,k denotes a parallelo-gramic shape with n rows and k cells in each row.

Corollary 5.14. For integers n ≥ 1 and k ≥ 3, let T ′(n, k) denote the number of semi-standard tableau classes of shape Xn,k. Then for some constants c > 0 and r > 1,

T ′(n, k) ∼ c rn

as n → ∞.

Proof. The proof is similar to the proof of Corollary 5.12. �

Lastly, we remark on another application of the results in this section. Let trM denotethe trace of a square matrix M .

Remark 5.15. If we enumerate the sequences in Theorem 4.23 with characteristic polynomialsof tableau transfer matrices instead of minimal polynomials of transfer transfer matrices, thenwe can describe the coefficients of the resulting constant coefficient linear recurrence relations.Recall that Corollary 5.9 gives combinatorial descriptions for the entries of arbitrary powersof transfer matrices. In particular, using the notation in Corollary 5.9, the traces of thesematrices, for all integers n ≥ 1, can be combinatorially described as being the number ofelements T in

Tb(θ−n(S) ∪ π−1([−n+ 1, 0]), ω)

such that T |θ−n(S) ≡θ T |S. Hence, we can describe the aforementioned coefficients usingknown results for describing the coefficients of a characteristic polynomial. For instance,


we can use Lewin’s result [7] that expresses the coefficients of the characteristic polynomialof a square matrix M solely in terms of parameters of the form trMn and that involves asummation over the (integer) partitions of a fixed integer.

Acknowledgements

The author would like to thank Stephanie van Willigenburg for her guidance and adviceduring the development of this paper.

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