A shuffling theorem for lozenge tilings of doubly-dented hexagons
aa r X i v : . [ m a t h . C O ] J u l A SHUFFLING THEOREM FOR LOZENGE TILINGS OF DOUBLY-DENTEDHEXAGONS
TRI LAI AND RANJAN ROHATGI
Abstract.
MacMahon’s theorem on plane partitions yields a simple product formula for tiling numberof a hexagon, and Cohn, Larsen and Propp’s theorem provides an explicit enumeration for tilings ofa dented semihexagon via semi-strict Gelfand–Tsetlin patterns. In this paper, we prove a naturalhybrid of the two theorems for hexagons with an arbitrary set of unit triangles removed along the ahorizontal axis. In particular, we show that the ‘shuffling’ of removed unit triangles only changes thetiling number of the region by a simple multiplicative factor. Our main result generalizes a number ofknown enumerations and asymptotic enumerations of tilings. We also reveal connections of the mainresult to the study of symmetric functions and q -series. Introduction
MacMahon’s classical theorem [16] on plane partition fitting in a given box is equivalent to thefact that the number of lozenge tilings of a centrally symmetric hexagon H ( a, b, c ) of side-lengths a, b, c, a, b, c (in this cyclic order) is given by the simple product:(1.1) PP( a, b, c ) := a Y i =1 b Y j =1 c Y k =1 i + j + k − i + j + k − . This formula was generalized by Cohn, Larsen and Propp [6, Proposition 2.1] when they presented acorrespondence between lozenge tilings of a semihexagon with unit triangles removed on the base andsemi-strict Gelfand-Tsetlin patterns. In particular, the dented semihexagon S a,b ( s , s , . . . , s a ) is theregion obtained from the upper half of the symmetric hexagon of side-lengths b, a, a, b, a, a (in clockwiseorder, starting from the north side) by removing a up-pointing unit triangles along the base at thepositions s , s , . . . , s a from left to right. The number of lozenge tilings of the dented semihexagon isgiven by(1.2) M( S a,b ( s , s , . . . , s a )) = Y ≤ i Mathematics Subject Classification. Key words and phrases. perfect matchings, plane partitions, lozenge tilings, dual graph, graphical condensation.This research was was supported in part by Simons Foundation Collaboration Grant ( B l l A y + d y + u y + u y + d x+n-dx+n-u y + u y + d y + u y + d x+n-d x+n-u Figure 2.1. (A) The region H , (2 , , , , 11; 4 , , , 12) and (B) a lozenge tiling ofit. The black and shaded triangles indicate the unit triangles removed.arrays of triangles removed in [11, Theorems 2.11 and 2.12] and Ciucu’s main results about ‘ F -coredhexagons ’ in [2, Theorems 1.1 and 2.1] (see Corollary 3.1).2. Shuffling theorems Let x, y, n, u, d be nonnegative integers, such that u, d ≤ n . Consider a symmetric hexagon of side-lengths x + n − u, y + u, y + d, x + n − d, y + d, y + u in clockwise order, starting from the north side. Weremove u + d arbitrary unit triangles along the lattice line l that contains the west and the east verticesof the hexagon. Assume further that, among these u + d removed triangles, there are u up-pointingones and d down-pointing ones. Let U = { s , s , . . . , s u } and D = { t , t , . . . , t d } be, respectively, thesets of positions of the up-pointing and down-pointing removed unit triangles (ordered from left toright), such that | U ∪ D | = n (i.e., U, D ⊆ [ x + y + n ], U and D are not necessarily disjoint). Denoteby H x,y ( U ; D ) the resulting region . See Fig. 2.1 A for an example of such a region and Fig. 2.1 Bfor a sample tiling; we ignore the two horizontal unit “barriers” at the positions 6 and 13 on l at themoment.We now consider ‘shuffling’ the up- and down-pointing unit triangles in the symmetric difference U ∆ D to obtain new position sets U ′ and D ′ for the up-pointing and down-pointing removed triangles.(In particular, U and U ′ have the same size, and so do D and D ′ .) The following theorem shows thatthe shuffling of removed triangles only changes the tiling number by a simple multiplicative factor.Moreover, the factor can be written in a similar form to Cohn–Larsen–Propp’s formula (i.e. theproduct on the right-hand side of Eq. 1.2). Theorem 2.1 (Shuffling Theorem) . For nonnegative integers x, y, n, u, d ( u, d ≤ n ) and four orderedsubsets U = { s , s , . . . , s u } , D = { t , t , . . . , t d } , U ′ = { s ′ , s ′ , . . . , s ′ u } , and D ′ = { t ′ , t ′ , . . . , t ′ d } of [ x + y + n ] such that U ∪ D = U ′ ∪ D ′ , and U ∩ D = U ′ ∩ D ′ . Then (2.1) M( H x,y ( U ; D ))M( H x,y ( U ′ ; D ′ )) = Y ≤ i Remark . By Cohn–Larson–Propp’s theorem (see Eq. 1.2), Eq. 2.1in Theorem 2.1 can be written in terms of tiling numbers asM( H x,y ( U ; D ))M( H x,y ( U ′ ; D ′ )) = M( S u,x + y + n − u ( U )) M( S d,x + y + n − d ( D ))M( S u,x + y + n − u ( U ′ )) M( S d,x + y + n − d ( D ′ )) . (2.2)It would be interesting to have a combinatorial explanation for this identity. SHUFFLING THEOREM FOR LOZENGE TILINGS OF DOUBLY-DENTED HEXAGONS 3 x+n-u y + u y + d y + u y + d x+n-d -6 -123 qq qq Figure 2.2. Assigning weights to lozenges in a doubly-dented hexagon.Moreover, one readily sees that the two dented semihexagons in the numerator of the right-handside are obtained by dividing the region H x + y, ( U ; D ) along the horizontal axis l . Similarly, the twodented semihexagons in the denominator are obtained by dividing H x + y, ( U ′ ; D ′ ) along the horizontalaxis. This means that, identity (2.2) is equivalent to(2.3) M( H x,y ( U ; D )) M( H x + y, ( U ′ ; D ′ )) = M( H x,y ( U ′ ; D ′ )) M( H x + y, ( U ; D )) . The both sides of (2.3) count pairs of tilings of doubly-dented hexagons. It would be interesting tofind a bijective proof for this identity.We can generalize our Shuffling Theorem 2.1 by additionally allowing the unit triangles in thesymmetric difference U ∆ D to ‘flip’ (from up-pointing to down-pointing, and vice versa). It is possible,then, that the new position sets of removed up-pointing triangles and removed down-pointing triangles U ′ and D ′ may have sizes different than those of U and D . We also allow the appearance of “ barriers ”along the axis l . A barrier is a unit horizontal lattice interval which is not allowed to be containedwithin a lozenge in a tiling. Assume that we have a set of barriers at the positions B ⊆ [ x + y + n ] − U ∪ D so that vertical lozenges may not appear at the positions in B and that | B | ≤ x (see the red barriersin Fig. 2.1; B = { , } in this case). We now consider the tilings of H x,y ( U ; D ) which are compatiblewith the set of barriers B . Denote by H x,y ( U ; D ; B ) the doubly-dented hexagons with such setup ofremoved unit triangles and barrier. Theorem 2.3 (Generalized Shuffling Theorem) . For nonnegative integers x, y, n, u, d ( u, d ≤ n )and five ordered subsets U = { s , s , . . . , s u } , D = { t , t , . . . , t d } , U ′ = { s ′ , s ′ , . . . , s ′ u ′ } , D ′ = { t ′ , t ′ , . . . , t ′ d ′ } , and B := { k , k , . . . , k b } of [ x + y + n ] such that U ∪ D = U ′ ∪ D ′ , U ∩ D = U ′ ∩ D ′ , B ∪ ( U ∪ D ) = ∅ , and | B | ≤ x , we always have (2.4) M( H x,y ( U ; D ; B ))M( H x,y ( U ′ ; D ′ ; B )) = Y ≤ i We can assign to each right-tilting lozenge above l (resp., below l ) a weight q z (resp, a weight q − z ),where ( z − ) √ is the distance from its center to the axis l (see Figure 2.2). The ‘ tiling-generatingfunction ’ of a region H x,y ( U ; D ; B ) is the sum of the weights of all tilings of the region, where the weight of a tiling is the product of weights of all its constituent lozenges. We denote this tiling-generatingfunction by M q ( H x,y ( U ; D ; B )). We have a q -analog of Theorem 2.3: Theorem 2.4 ( q -Shuffling Theorem) . With the same assumptions as in Theorem 2.3, we have M q ( H x,y ( U ; D ; B ))M q ( H x,y ( U ′ ; D ′ ; B )) = q C Y ≤ i Ciucu and Krattenthaler in [4] proved a counterpart of MacMahon’s theorem (Eq. 1.1) by obtainingthe asymptotic tiling number of the exterior of a concave hexagonal contour in which we turn 120 ◦ after drawing each edge (see Fig. 3.1 B; the tiling number in MacMahon’s theorem is for the interiorof a hexagonal contour in which each turn 60 ◦ after drawing each edge as in Fig. 3.1 A). Ciucu [2]later obtained a similar counterpart of Cohn–Larsen–Propp’s theorem corresponding to the exterior ofa concave polygon with an arbitrary number sides (see the contour in Fig. 3.1 C). Recently, the firstauthor generalized the asymptotic result of Ciucu to the union of three polygons [11]. In Corollary3.1, we will show a multi-parameter generalization of the latter two asymptotic results.We now assume that the set of removed unit triangles is partitioned into k separated clusters (i.e.chains of contiguous unit triangles). Denote these clusters by C , C , . . . , C k and the distances between SHUFFLING THEOREM FOR LOZENGE TILINGS OF DOUBLY-DENTED HEXAGONS 5 A DBC Figure 2.3. Obtaining general doubly-dented hexagons from symmetric doubly-dented hexagons by removing forced lozenges. CA B Figure 3.1. Three contours: (A) the contour in MacMahon’s theorem, (B) the contourin [4], and (C) the contour in [2].them by d , d , . . . , d k − ( d i > C is attached to the west vertex of the hexagon, that C k is attached to the east vertex ofthe hexagon, and that C and C k may be empty. We use the notation H x,y ( C , . . . , C k ; d , . . . , d k − )for these regions (see Fig. 3.3 for an example; the black unit triangles indicate the ones removed). Foreach cluster C i , we use the notations U i and D i for the index sets of its up-pointing and down-pointingtriangles. For each cluster C i , we can shuffle unit triangles at the positions in U i ∆ D i to obtain a newcluster C ′ i , for i = 1 , , . . . , k . The index sets of triangles in C ′ i are denoted by U ′ i and D ′ i . Assumethat | U i | = u i , D i = d i , | U ′ i | = u ′ i , | D i | = d ′ i and | U i ∪ D i | = | U ′ i ∪ D ′ i | = f i . We call f i the length ofthe cluster C i (and also the length of C ′ i ).We now consider the behavior of the tiling number of the region when the side-lengths of the outerhexagon and the distances between two consecutive clusters get large. TRI LAI AND RANJAN ROHATGI = Figure 3.2. Illustrating Corollary 3.1. Corollary 3.1. For nonnegative integers x and y , lim N →∞ M( H Nx,Ny ( C , . . . , C k ; N d , . . . , N d k − ))M( H Nx,Ny ( C ′ , . . . , C ′ k ; N d , . . . , N d k − ))= k Y i =1 s + ( C i ) s − ( C i ) s + ( C ′ i ) s − ( C ′ i )(3.1) where s + ( C i ) = M( S u i ,f i − u i ( U i )) and s − ( C i ) = M( S d i ,f i − d i ( D i )) are respectively the tiling numbersof the dented semihexagons whose dents are defined by the up-pointing triangles and down-pointingtriangles in the cluster C i , and where s + ( C ′ i ) and s − ( C ′ i ) are defined similarly with respect to C ′ i . Corollary 3.1 can be visualized as in Fig. 3.2, for k = 3. The dented semihexagons correspondingto s + ( C i ) and s − ( C i ) are the upper and lower halves of the ‘numerator hexagon’ in the i th fractionon the right-hand side; the dented semihexagons corresponding to s + ( C ′ i ) and s − ( C ′ i ) are the upperand lower halves of the ‘denominator hexagon’ in the i th fraction, for i = 1 , , Sketch of the proof. Assume that U i = { a ( i )1 , . . . , a ( i ) u i } and U ′ i = { e ( i )1 , . . . , e ( i ) u i } . Applying Theorem2.1 to the regions H Nx,Ny ( U ; D ) and H Nx,Ny ( U ′ ; D ′ ), for U := S i U i , D := S j D j , U ′ := S i U ′ i , and D ′ := S j D ′ j , we get(3.2) M( H Nx,Ny ( C , . . . , C k ; N d , . . . , N d k − ))M( H Nx,Ny ( C ′ , . . . , C ′ k ; N d , . . . , N d k − )) = ∆( U )∆( D )∆( U ′ )∆( D ′ ) , where the operation ∆ is defined as ∆( S ) := Q ≤ i Obtained a hexagon with ferns removed from the region H x,y ( C , . . . , C k ; d , . . . , d k − ).obtain a new region with the same tiling number in which each cluster is replaced by a chain ofremoved equilateral triangles of alternating orientations (see Fig. 3.3; the forced lozenges are coloredwhite). Each such chain of triangles is called a ‘ fern ’ (see e.g. [2, 11]); the side-lengths of trianglesin a fern are equal to the lengths of the intervals of unit triangles of the same orientation in thecorresponding cluster. Denote by E x,y ( F , . . . , F k ; d , . . . , d k − ) the corresponding hexagon with fernsremoved (the fern F i corresponds to the cluster C i ; and the fern F ′ i corresponds to the cluster C ′ i ).By setting k = 3, d = d or d = d − 1, and specifying that the cluster C ′ i has all its up-pointingtriangles on the left and all its down-pointing triangles on the right, for i = 1 , , F ′ i consists of two triangles, an up-pointing one followed by a down-pointing one), our Corollary3.1 implies the first author’s work in [11, Theorem 2.11]. Similarly, we can recover the work of Ciucuin [2, Theorem 1.1] by further specifying that C = C = ∅ (i.e. we actually have only a non-emptyfern F in the center of the region).4. Proof of the main theorem In general, the lozenges in a region can carry weights. In the weighted case, M( R ) denotes the sumof weights of the tilings in R , where the weight of a tiling is the product of weights of its constituentlozenges. The removal of one or more forced lozenges changes the number of tilings of the region bya factor equal to the reciprocal of the weighted product of the forced lozenges. In particular, if weremove the lozenges l , l , . . . , l k from a region R and get a new region R ′ , then(4.1) M( R ′ ) = k Y i =1 wt ( l i ) ! − M( R ) , where wt ( l i ) is the weight of the forced lozenge l i .A (perfect) matching of a graph is a collection of disjoint edges that covers all the vertices of thegraph. When edges of a graph carry weights, we denote by M( G ) the weighted sum of matchings in G ,where the weight of a matching is the product of weights of its edges. (In the unweighted case, M( G )is exactly the number of matchings of G .) The (planar) dual graph of the region R on the triangularlattice is the graph whose vertices are the unit triangles in R and whose edges connect precisely twounit triangles sharing an edge. The edges of the dual graph inherit the weights from the lozenges ofthe corresponding region. There is a natural (weight-preserving) bijection between tilings of a regionand matchings of its dual graph.Our proof is based on the following powerful graphical condensation lemma first introduced byKuo [10]: TRI LAI AND RANJAN ROHATGI Lemma 4.1. Let G = ( V , V , E ) be a (weighted) planar bipartite graph with the two vertex classes V and V such that | V | = | V | + 1 . Assume that u, v, w, s are four vertices appearing in this cyclicorder around a face of G , such that u, v, w ∈ V and s ∈ V . Then M( G − v ) M( G − { u, w, s } ) = M( G − u ) M( G − { v, w, s } ) + M( G − w ) M( G − { u, v, s } ) . (4.2)If a region admits a lozenge tiling, then it must have the same number of up-pointing and down-pointing unit triangles. We call such a region balanced . The following lemma allows us to decomposea region into smaller regions when enumerating tilings in certain situations. Lemma 4.2 (Region-splitting Lemma [12, 13]) . Let R be a balanced region on the triangular lattice.Assume that a balanced sub-region Q of R satisfies the condition that the unit triangles in Q that areadjacent to some unit triangle of R − Q have the same orientation. Then M( R ) = M( Q ) M( R − Q ) . Given k positive integers λ ≥ λ ≥ · · · ≥ λ k , a plane partition of shape ( λ , λ , . . . , λ k ) is an arrayof non-negative integers n , n , n , . . . . . . . . . n ,λ n , n , n , . . . . . . n ,λ ... ... ... ... ... n k, n k, n k, . . . n k,λ k so that n i,j ≥ n i,j +1 and n i,j ≥ n i +1 ,j (i.e. all rows and all columns are weakly decreasing from left toright and from top to bottom, respectively). The sum of all entries of a plane partition π is called the volume (or the norm ) of the plane partition, and denoted by | π | .A column-strict plane partition is a plane partition having columns strictly decreasing. A column-strict plane partition is a plane partition having columns strictly decreasing. We now assign to eachright-tilting lozenge in the dented semihexagon similarly to the upper half of the doubly dentedhexagon. In particular, a eight-tilting lozenge is weighted by q z , where the distance between itscenter and the base of the region is ( z − ) √ . Denote M q ( S a,b ( s , s , . . . , s a )) the correspondingtiling generating function of the dented semihexagon S a,b ( s , s , . . . , s a ).There is a well-known (weightpreserving) bijection between the lozenge tilings of S a,b ( s , s , . . . , s a ) and the column-strict planepartitions of shape ( s a − a, s a − − a + 1 , . . . , s − 1) with positive entries at most a (see e.g. [6] and [1]).The following weighted enumeration of the dented semihexagon follows directly from the bijection andequation (7.105) in [18, pp. 375]. Lemma 4.3. For nonnegative a, b and positive integers ≤ s < s < · · · < s a (4.3) M q ( S a,b ( s , s , . . . , s a )) = X π q | π | = q P ai =1 ( s i − i ) Y ≤ i We prove (4.5) by induction on x + y . The base cases are the situations when x = b and when y = 0.If y = 0, then we apply Region-splitting Lemma 4.2 to the region R = H x, ( U ; D ; B ) with thesubregion Q the portion above the horizontal axis l . The subregion Q is congruent with the dentedsemihexagons S u,x + n − u ( U ) weighted as in Lemma 4.3, and the R − Q , after rotated 180 ◦ , is congruentwith S d,x + n − d ( r ( D )) weighted similarly with q replaced by q − (see Fig. 4.1 A). Here we use thenotation r ( S ) for the reflection of the index set S , i.e. the set ( x + y + n + 1) − S = { ( x + y + n + 1) − s : s ∈ S } . We have M q ( H x, ( U ; D ; B )) = M q ( S u,x + n − u ( U )) M q − ( S d,x + n − d ( r ( D ))) . (4.6)Similarly, the tiling number of R ′ = H x, ( U ′ ; D ′ ; B ) is also written by a product of tiling numbers oftwo dented semihexagons asM q ( H x, ( U ′ ; D ′ ; B )) = M q ( S u,x + n − u ( U ′ )) M q − ( S d,x + n − d ( r ( D ′ ))) , (4.7)and (4.5) follows from Lemma 4.3.If x = b , consider the subregion Q of R = H b,y ( U ; D ; B ) that is obtained from the portion abovethe axis l by removing all up-pointing unit triangles in ( U ∪ D ∪ B ) c (we note that in this case | ( U ∪ D ∪ B ) c | = y ). Q is the dented semihexagon S y + u,b + n − u (( U ∪ D ∪ B ) c ∪ U ) weighted as in Lemma4.3, and its complement, after removing forced lozenges at the positions in ( U ∪ D ∪ B ) c and rotating180 ◦ , is the dented semihexagon S y + d,b + n − u (( U ∪ D ∪ B ) c ∪ D ) weighted similarly with q replaced by q − (see Fig. 4.1 B). This way, the tiling number of R is written as the product of tiling numbers oftwo dented semihexagons. Similarly, the tiling number of R ′ = H b,y ( U ′ ; D ′ ; B ) is also written by aproduct of tiling numbers of two dented semihexagons. Then (4.5) also follows from Lemma 4.3.For the induction step, we assume that x > b , y > 0, and that (4.5) holds for any H -type regionswhose sum of x - and y -parameters is strictly less than x + y . We will use Kuo condensation in Theorem4.1 to obtain a recurrence for the tiling generating function on the left-hand side of (4.5), and we showthat the expression on the right-hand side satisfies the same equation. Then (4.5) follows from theinduction principle.We apply Kuo condensation to the dual graph G of the region R obtained from H x,y ( U ; D ; B )by adding a layer of unit triangle on the top of the hexagon, with the four vertices u, v, w, s as inFig. 4.2 (the region restricted by the bold contour indicates the region H x,y ( U ; D ; B )). In particular,the vertices w and v correspond to the up-pointing triangles at the first and the last positions in( U ∪ D ∪ B ) c , the vertex v corresponds to the up-pointing triangle on the northeast corner of theregion, and the vertex s corresponds to the down-pointing triangle on the southeast corner of theregion.We consider the region corresponding to G − v (i.e., the region obtained from R by removing the v -triangle as in Fig. 4.3 A). The removal of the v -triangle yields several forced lozenges along the topof the region. After removing these forced lozenges (this lozenge removal changes the tiling numberof the region by a factor q ( y + u +1)( x + n − u − ), we get back the region H x,y ( U ; D ; B ). This means thatwe get(4.8) M( G − v ) = q ( y + u +1)( x + n − u − M q ( H x,y ( U ; D ; B )) . By considering forced lozenges in the regions corresponding to the graphs G − { u, w, s } , G − u, G −{ v, w, s } , G − w, and G − { u, v, s } (as shown in Fig. 4.3 B – F, respectively), we get(4.9) M( G − { u, w, s } ) = M q ( H x − ,y − ( αβU ; D ; B )) , (4.10) M( G − u ) = M q ( H x − ,y ( βU ; D ; B )) , (4.11) M( G − { v, w, s } ) = q ( y + u +1)( x + n − u − M q ( H x,y − ( αU ; D ; B )) , (4.12) M( G − w ) = M q ( H x − ,y ( αU ; D ; B )) , (4.13) M( G − { u, v, s } ) = q ( y + u +1)( x + n − u − M q ( H x,y − ( βU ; D ; B )) , A B d d d x+n-d u x+n-u y + u b+n-u y + uy + d y + d b+n-d Figure 4.1. Applications of the Region-splitting Lemma 4.2 in the base cases: (A) y = 0, (B) x = b . l w s y + d y + d u x+n-u-1x+n-d y + u + y + u + v Figure 4.2. Applying Kuo condensation to a hexagon.where we use the notations αU , βU and αβU for the unions U ∪ { α } , U ∪ { β } and U ∪ { α, β } ,respectively. Plugging Eqs. 4.8–4.13 into the equation in Kuo’s Lemma 4.1, we get the recurrence (allpowers of q cancel out):M q ( H x,y ( U ; D ; B )) M q ( H x − ,y − ( αβU ; D ; B )) = M q ( H x − ,y ( βU ; D ; B )) M q ( H x,y − ( αU ; D ; B ))+ M q ( H x − ,y ( αU ; D ; B )) M q ( H x,y − ( βU ; D ; B )) . (4.14)Denote by h x,y ( U ; D ; U ′ ; D ′ ) := g x,y ( U ; D ; U ′ ; D ′ ) M( H x,y ( U ′ ; D ′ ; B )) the expression on the right-hand side of (4.5). To finish the proof, we shall show that the h x,y ( U ; D ; U ′ ; D ′ ) also satisfies recurrence(4.14). Equivalently, we need to verify h x − ,y ( βU ; D ; βU ′ ; D ′ ) h x,y − ( αU ; D ; αU ′ ; D ′ )) h x,y ( U ; D ; U ′ ; D ′ ) h x − ,y − ( αβU ; D ; αβU ′ ; D ′ ) + h x − ,y ( αU ; D ; αU ′ ; D ′ ) h x,y − ( βU ; D ; βU ′ ; D ′ )) h x,y ( U ; D ; U ′ ; D ′ ) h x − ,y − ( αβU ; D ; αβU ′ ; D ′ ) = 1 . (4.15) SHUFFLING THEOREM FOR LOZENGE TILINGS OF DOUBLY-DENTED HEXAGONS 11 l ll ll l y + u + y + u + y + u + x+n-d x+n-dx+n-dx+n-d y + u + y + u + y + u + y + u + y + u + y + u + y + u + A x+n-u-1 y + u + y + d y + u + y + d x+n-d B y + d y + u y + d C y + d y + d D y + d y + d E y + d y + d F y + d y + d uu u v vv sss x+n-u-1x+n-u-1x+n-u-1 x+n-dx+n-u-1 x+n-u-1 Figure 4.3. Obtaining the recurrence for the tiling numbers.We claim Claim 4.4. (4.16) g x − ,y ( βU ; D ; βU ′ ; D ′ ) g x,y − ( αU ; D ; αU ′ ; D ′ )) g x,y ( U ; D ; U ′ ; D ′ ) g x − ,y − ( αβU ; D ; αβU ′ ; D ′ ) = 1 and (4.17) g x − ,y ( αU ; D ; αU ′ ; D ′ ) g x,y − ( βU ; D ; βU ′ ; D ′ )) g x,y ( U ; D ; U ′ ; D ′ ) g x − ,y − ( αβU ; D ; αβU ′ ; D ′ ) = 1 . Proof. We have from the definitionPP q ( u, d, y )PP q ( u ′ , d ′ , y ) = q − udy + u ′ d ′ y ∆ q ([ u ])∆ q ([ d ])∆ q ([ u ′ ])∆ q ([ d ′ ]) ∆ q ([ y + u ′ ])∆ q ([ y + d ′ ])∆ q ([ y + u ])∆ q ([ y + d ]) , (4.18)where, for any ordered index set S = { s , s , . . . , s k } , we define ∆ q ( S ) := Q ≤ i A B Figure 5.1. Obtaining cored hexagons from hexagons with three ferns removed.By specifying that C = C ′ and C = C ′ , in addition to the six conditions above, we recover Ciucu’sTheorem 2.1 in [2] about F -cored hexagons (after removing forced lozenges, the region H x,y ( U ; D )becomes an F -cored hexagon and H x,y ( U ′ ; D ′ ), as in the previous case, becomes a cored hexagon).Similarly, if we require(1) k = 2,(2) all unit triangles in C ′ have the same orientation, and(3) all unit triangles in C ′ have opposite orientation to those in C ′ ,then we get Theorem 1.1 in [5] about hexagons with two ferns removed, called ‘ doubly–intrudedhexagons ’ (the region H x,y ( U ; D ) is now a doubly–intruded hexagon and H x,y ( U ′ ; D ′ ) becomes ahexagon in MacMahon’s theorem, after removing forced lozenges).(2). By equation (7.105) of [17], we have (5.1) Y ≤ i The authors would like to thank Brendon Rhoades for fruitful discussions. 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Combin. Theory Ser. A 43 (1986): 103–113. Department of Mathematics, University of Nebraska – Lincoln, Lincoln, NE 68588, U.S.A. E-mail address : [email protected] Department of Mathematics and Computer Science, Saint Mary’s College, Notre Dame, IN 46556,U.S.A. E-mail address ::