相关论文: Tiling enumeration of doubly-intruded halved hexag…

Motivated in part by Propp's intruded Aztec diamond regions, we consider hexagonal regions out of which two horizontal chains of triangular holes (called ferns) are removed, so that the chains are at the same height, and are attached to the…

We use the subgraph replacement method to investigate new properties of the tilings of regions on the square lattice with diagonals drawn in. In particular, we show that the centrally symmetric tilings of a generalization of the Aztec…

Proctor's work on staircase plane partitions yields an enumeration of lozenge tilings of a halved hexagon on the triangular lattice. Rohatgi recently extended this tiling enumeration to a halved hexagon with a triangle removed from the…

Ciucu showed that the number of lozenge tilings of a hexagon in which a chain of equilateral triangles of alternating orientations, called a `\emph{fern}', has been removed in the center is given by a simple product formula (Adv. Math.…

Proctor's work on staircase plane partitions yields an enumeration of lozenge tilings of a halved hexagon on the triangular lattice. Rohatgi later extended this tiling enumeration to a halved hexagon with a triangle cut off from the…

In this paper we consider arbitrary hexagons on the triangular lattice with three arbitrary bowtie-shaped holes, whose centers form an equilateral triangle. The number of lozenge tilings of such general regions is not expected --- and…

In the prequel of the paper (arXiv:1803.02792), we considered exact enumerations of the cored versions of a doubly-intruded hexagon. The result generalized Ciucu's work about $F$-cored hexagons (Adv. Math. 2017). In this paper, we provide…

In their paper about a dual of MacMahon's classical theorem on plane partitions, Ciucu and Krattenthaler proved a closed form product formula for the tiling number of a hexagon with a "shamrock", a union of four adjacent triangles, removed…

Motivated by a conjecture posed by Fulmek and Krattenthaler, we provide product formulas for the number of lozenge tilings of a semiregular hexagon containing a horizontal intrusion. As a direct corollary, we obtain a product formula for…

The enumeration of lozenge tilings of hexagons with holes has received much attention during the last three decades. One notable feature is that a lot of the recent development involved Kuo's graphical condensation. Motivated by Ciucu, Lai…

Rosengren found an explicit formula for a certain weighted enumeration of lozenge tilings of a hexagon with an arbitrary triangular hole. He pointed out that a certain ratio corresponding to two such regions has a nice product formula. In…

Motivated by the enumeration of a class of plane partitions studied by Proctor and by considerations about symmetry classes of plane partitions, we consider the problem of enumerating lozenge tilings of a hexagon with ``maximal staircases''…

It has been proven that the lozenge tilings of a quartered hexagon on the triangular lattice are enumerated by a simple product formula. In this paper we give a new proof for the tiling formula by using Kuo's graphical condensation. Our…

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…

Proctor proved a formula for the number of lozenge tilings of a hexagon with side-lengths $a,b,c,a,b,c$ after removing a "maximal staircase." Ciucu then presented a weighted version of Proctor's result. Here we present weighted and…

MacMahon's classical theorem on the number of boxed plane partitions has been generalized in several directions. One way to generalize the theorem is to view boxed plane partitions as lozenge tilings of a hexagonal region and then…

We give a formula for the number of lozenge tilings of a hexagon on the triangular lattice with unit triangles removed from arbitrary positions along two non-adjacent, non-opposite sides. Our formula implies that for certain families of…

We show how to count tilings of Aztec diamonds and hexagons with defects using determinants. In several cases these determinants can be evaluated in closed form. In particular, we obtain solutions to problems 1, 2, and 10 in James Propp's…

We investigate the connection between lozenge tilings and domino tilings by introducing a new family of regions obtained by attaching two different Aztec rectangles. We prove a simple product formula for the generating functions of the…

We get four quartered Aztec diamonds by dividing an Aztec diamond region by two zigzag cuts passing its center. W. Jockusch and J. Propp (in an unpublished work) found that the number of tilings of quartered Aztec diamonds is given by…