Related papers: Rhombus Tilings of a Hexagon with Three Fixed Bord…
We compute the number of rhombus tilings of a hexagon with side lengths a,b,c,a,b,c which contain the central rhombus and the number of rhombus tilings of a hexagon with side lengths a,b,c,a,b,c which contain the `almost central` rhombus…
We compute the number of rhombus tilings of a hexagon with sides $N,M,N,N,M,N$, which contain a fixed rhombus on the symmetry axis. A special case solves a problem posed by Jim Propp.
We compute the number of rhombus tilings of a hexagon with sides $N,M,N,N,M,N$, which contain a fixed rhombus on the symmetry axis that cuts through the sides of length $M$.
We compute the number of all rhombus tilings of a hexagon with sides $a,b+1,c,a+1,b,c+1$, of which the central triangle is removed, provided $a,b,c$ have the same parity. The result is a product of four numbers, each of which counts the…
We compute the number of rhombus tilings of a hexagon with sides n, n, N, n, n, N, where two triangles on the symmetry axis touching in one vertex are removed. The case of the common vertex being the center of the hexagon solves a problem…
We compute the number of rhombus tilings of a hexagon with side lengths N,M,N,N,M,N, with N and M having the same parity, which contain a particular rhombus next to the center of the hexagon. The special case N=M of one of our results…
Propp conjectured that the number of lozenge tilings of a semiregular hexagon of sides $2n-1$, $2n-1$ and $2n$ which contain the central unit rhombus is precisely one third of the total number of lozenge tilings. Motivated by this, we…
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…
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…
The work of Mills, Robbins, and Rumsey on cyclically symmetric plane partitions yields a simple product formula for the number of lozenge tilings of a regular hexagon, which are invariant under roation by $120^{\circ}$. In this paper we…
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…
We state, discuss, provide evidence for, and prove in special cases the conjecture that the probability that a random tiling by rhombi of a hexagon with side lengths $2n+a,2n+b,2n+c,2n+a,2n+b,2n+c$ contains the (horizontal) rhombus with…
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''…
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…
The rhombus tilings of a simply connected domain of the Euclidean plane are known to form a flip-connected space (a flip is the elementary operation on rhombus tilings which rotates 180{\deg} a hexagon made of three rhombi). Motivated by…
It is shown that the descending plane partitions of Andrews can be geometrically realized as cyclically symmetric rhombus tilings of a certain hexagon where an equilateral triangle of side length 2 has been removed from its centre. Thus,…
Convex hexagons that can tile the plane have been classified into three types. For the generic cases (not necessarily convex) of the three types and two other special cases, we classify tilings of the plane under the assumption that all…
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…
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.…
We give a simple proof of T. Stehling's result, that in any normal tiling of the plane with convex polygons with number of sides not less than six, all tiles except the finite number are hexagons.