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Related papers: Enumeration of lozenge tilings of hexagons with cu…

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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…

Combinatorics · Mathematics 2015-10-16 Ranjan Rohatgi

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…

Combinatorics · Mathematics 2017-09-08 Tri Lai

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…

Combinatorics · Mathematics 2015-04-28 Tri Lai

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…

Combinatorics · Mathematics 2017-09-08 Tri Lai

We generalize a special case of a theorem of Proctor on the enumeration of lozenge tilings of a hexagon with a maximal staircase removed, using Kuo's graphical condensation method. Additionally, we prove a formula for a weighted version of…

Combinatorics · Mathematics 2015-10-16 Ranjan Rohatgi

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…

Combinatorics · Mathematics 2007-05-23 Mihai Ciucu , Theresia Eisenkölbl , C. Krattenthaler , D. Zare

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…

Combinatorics · Mathematics 2015-10-06 Tri Lai

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…

Combinatorics · Mathematics 2018-08-02 Tri Lai , Ranjan Rohatgi

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.…

Combinatorics · Mathematics 2019-05-21 Tri Lai

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…

Combinatorics · Mathematics 2013-09-19 Mihai Ciucu , Ilse Fischer

Inspired by Propp's intruded Aztec diamond regions, we consider halved hexagons in which two aligned arrays of triangular holes have been removed from their boundaries. Unlike the intruded Aztec diamonds (whose numbers of domino tilings…

Combinatorics · Mathematics 2019-02-12 Tri Lai

Eisenk"olbl gave a formula for the number of lozenge tilings of a hexagon on the triangular lattice with three unit triangles removed from along alternating sides. In earlier work, the first author extended this to the situation when an…

Combinatorics · Mathematics 2014-12-15 Mihai Ciucu , Ilse Fischer

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…

Combinatorics · Mathematics 2020-02-13 Daniel Condon

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…

Combinatorics · Mathematics 2022-07-01 Seok Hyun Byun

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…

Combinatorics · Mathematics 2019-06-12 Seok Hyun Byun

The number of plane partitions contained in a given box was shown by MacMahon to be given by a simple product formula. By a simple bijection, this formula also enumerates lozenge tilings of hexagons of side-lengths $a,b,c,a,b,c$ (in cyclic…

Combinatorics · Mathematics 2007-05-23 Mihai Ciucu

MacMahon proved a simple product formula for the generating function of plane partitions fitting in a given box. The theorem implies a $q$-enumeration of lozenge tilings of a semi-regular hexagon on the triangular lattice. In this paper we…

Combinatorics · Mathematics 2017-04-12 Tri Lai

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…

Combinatorics · Mathematics 2019-05-20 Mihai Ciucu , Tri Lai

In this paper we present a combinatorial generalization of the fact that the number of plane partitions that fit in a $2a\times b\times b$ box is equal to the number of such plane partitions that are symmetric, times the number of such…

Combinatorics · Mathematics 2017-02-13 Mihai Ciucu , Christian Krattenthaler

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…

Combinatorics · Mathematics 2020-01-08 Mihai Ciucu , Tri Lai , Ranjan Rohatgi
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