Enumeration of lozenge tilings of hexagons with cut off corners

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 generalize it by making some holes in the region and counting its tilings. In this paper, we provide new regions whose numbers of lozenges tilings are given by simple product formulas. The regions we consider can be obtained from hexagons by removin...

We deal with unweighted and weighted enumerations of lozenge tilings of a hexagon with side lengths $a,b+m,c,a+m,b,c+m$, where an equilateral triangle of side length $m$ has been removed from the center. We give closed formulas for the plain enumeration and for a certain $(-1)$-enumeration of these lozenge tilings. In the case that $a=b=c$, we also provide closed formulas for certain weighted enumerations of those lozenge tilings that are cyclically symmetric. For $m=0$, 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 indeed is not --- given by a simple product formula. However, when considering a certain natural normalized counterpart of any such region, we prove that the ratio between the number of tilings of the original and the number of tilings of the no...

We $q$-enumerate lozenge tilings of a hexagon with three bowtie-shaped regions have been removed from three non-consecutive sides. The unweighted version of the result generalizes a problem posed by James Propp on enumeration of lozenge tilings of a hexagon of side-lengths $2n,2n+3,2n,2n+3,2n,2n+3$ (in cyclic order) with the central unit triangles on the $(2n+3)$-sides removed.

We investigate a new family of regions that is the universal generalization of three well-known region families in the field of enumeration of tilings: the quasi-regular hexagons, the semi-hexagons, and the halved hexagons. We prove a simple product formula for the number of tilings of these new regions. Our main result also yields the enumerations of two special classes of plane partitions with restricted parts.

In their 2002 paper, Ciucu and Krattenthaler proved several product formulas for the number of lozenge tilings of various regions obtained from a centrally symmetric hexagon on the triangular lattice by removing maximal staircase regions from two non-adjacent corners. For the case when the staircases are removed from adjacent corners of the hexagon, they presented two conjectural formulas, whose proofs, as they remarked, seemed at the time "a formidable task". In this paper w...

In a recent paper, Byun presented nice formulas for the enumeration of lozenge tilings of certain hexagonal regions with intrusions. This paper attempts to generalise some of Byun's investigations.

In this paper we enumerate the centrally symmetric lozenge tilings of a hexagon with a fern removed from its center. The proof is based on a variant of Kuo's graphical condensation method. An unexpected connection with the total number of tilings is established~---~when suitably normalized, the number of centrally symmetric tilings is equal to the square root of the total number of tilings. The results we present can be regarded as a new extension of the enumeration of self-c...

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 this paper, we generalize this to hexagons with arbitrary collinear holes. It turns out that, by using same approach, we can also generalize Ciucu's work on the number and the number of centrally symmetric tilings of a hexagon with a fern re...

A rhombus tiling of a hexagon is said to be centered if it contains the central lozenge. We compute the number of vertically symmetric rhombus tilings of a hexagon with side lengths $a, b, a, a, b, a$ which are centered. When $a$ is odd and $b$ is even, this shows that the probability that a random vertically symmetric rhombus tiling of a $a, b, a, a, b, a$ hexagon is centered is exactly the same as the probability that a random rhombus tiling of a $a, b, a, a, b, a$ hexagon ...