Lozenge Tilings of Hexagons with Intrusions II: Shuffling Phenomenon

Abstract

The enumeration of lozenge tilings of hexagons with holes has been studied intensively in recent years. Researchers tried to find shapes and positions of holes in hexagonal regions so that the number of lozenge tilings of the resulting regions is given by a simple product formula. In the present work, we consider new regions that are hybrids of regions studied by the first author (hexagons with intrusions) and Ciucu (F-cored hexagons). Then, we show that the tiling generating functions of these new regions under a certain weight are given by simple product formulas. To give a proof, we present shuffling theorems for lozenge tilings of hexagons with intrusions, which give simple relations between the tiling generating functions of two related hexagonal regions with intrusions.

Figures (18)

• Figure 1.1: Examples of a hexagon with an intrusion (left) and a hexagon with a fern removed from its center (right). The numbers of lozenge tilings of these regions are given by simple product formulas.
• Figure 1.2: Examples of a hexagon with an intrusion and an asymmetric fern removed (left) and a hexagon with an intrusion and a symmetric fern removed (right).
• Figure 2.1: Intrusions of length $10$ and $9$ (top left and bottom left) and symmetric ferns $F_{sym}(6,2,3)$ (center) and $F_{sym}(5,2,3)$ (right).
• Figure 2.2: Five regions $A(x,y,z,w;a_1,a_2,a_3)$ (top), $B(x,y,z,w;a_1,a_2,a_3)$ (middle left), $C(x,y,z,w;a_1,a_2,a_3)$ (middle right), $D(x,y,z,w;a_1,a_2,a_3)$ (bottom left), and $E(x,y,z,w;a_1,a_2,a_3)$ (bottom right), where $x=3$, $y=4$, $z=7$, $w=2$, $a_1=3$, $a_2=2$, and $a_3=3$ (thus $a_o=a_1+a_3=6$ and $a_e=a_2=2$).
• Figure 2.3: A lozenge tiling of a hexagonal region on the $(i,j)$-coordinate system (left) and the symmetric fern $F_{sym}(6,2,3)$ on the $(i,j)$-coordinate system (right). In both pictures, the $i$-coordinates of the centers of horizontal lozenges are marked. If a horizontal lozenge is marked by $n$, then it is weighted by $\frac{q^{n}+q^{-n}}{2}$ and every lozenge with no mark is weighted by $1$.

Theorems & Definitions (15)

• Theorem 3.1
• Theorem 3.2
• Remark 3.3
• Theorem 3.4
• proof : Proof of Theorem \ref{['tcd']}
• Lemma 4.1
• Lemma 4.2
• Lemma 4.3
• proof : Proof of Theorems \ref{['tca']} and \ref{['tcb']}
• Theorem A.1