## The Number of Centered Lozenge Tilings of a Symmetric Hexagon (0)

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Citations: | 14 - 7 self |

### BibTeX

@MISC{Ciucu_thenumber,

author = {M. Ciucu and C. Krattenthaler},

title = {The Number of Centered Lozenge Tilings of a Symmetric Hexagon},

year = {}

}

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### Abstract

Propp conjectured [15] that the number of lozenge tilings of a semiregular hexagon of sides 2n \Gamma 1, 2n \Gamma 1 and 2n which contain the central unit rhombus is precisely one third of the total number of lozenge tilings. Motivated by this, we consider the more general situation of a semiregular hexagon of sides a, a and b. We prove explicit formulas for the number of lozenge tilings of these hexagons containing the central unit rhombus, and obtain Propp's conjecture as a corollary of our results.

### Citations

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(Show Context)
Citation Context ...f simply-connected regions. This is described in Section 3. Then, in Section 4, lozenge tilings are translated into nonintersecting lattice paths. By the main theorem on nonintersecting lattice paths =-=[3, 4]-=-, the number(s) of nonintersecting lattice paths that we are interested in can be immediately written down in form of a determinant (see Lemmas 11, 12, 13). Finally, in Section 5, these determinants a... |

177 | The method of creative telescoping
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(Show Context)
Citation Context ...mma1 i=1 (6i \Gamma 1)(6i + 1) (2n \Gamma 1)!! 2 (2.1) (where the empty product is defined to be 1). Let us denote the sum by S(n) and its summand by F (n; i). We use the Gosper--Zeilberger algorithm =-=[12, 17, 18]-=- to obtain the relation n(2n + 1) 2 F (n + 1; i) \Gamma 3n (6n \Gamma 1) (6n + 1) F (n; i) = G(n; i + 1) \Gamma G(n; i); (2.2) with G(n; i) = (\Gamma1) n\Gammai \Gamma \Gamma3 + 9 i \Gamma 6 i 2 \Gamm... |

162 |
A fast algorithm for proving terminating hypergeometric identities
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(Show Context)
Citation Context ...mma1 i=1 (6i \Gamma 1)(6i + 1) (2n \Gamma 1)!! 2 (2.1) (where the empty product is defined to be 1). Let us denote the sum by S(n) and its summand by F (n; i). We use the Gosper--Zeilberger algorithm =-=[12, 17, 18]-=- to obtain the relation n(2n + 1) 2 F (n + 1; i) \Gamma 3n (6n \Gamma 1) (6n + 1) F (n; i) = G(n; i + 1) \Gamma G(n; i); (2.2) with G(n; i) = (\Gamma1) n\Gammai \Gamma \Gamma3 + 9 i \Gamma 6 i 2 \Gamm... |

129 | Nonintersecting paths, Pfaffians, and plane partitions - Stembridge - 1990 |

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- 1995
(Show Context)
Citation Context ...m i = 0 to i = n, little rearrangement, and division by n on both sides, leads to the recurrence (2n + 1) 2 S(n + 1) \Gamma 3 (6n \Gamma 1) (6n + 1) S(n) = 0 for the sum in (2.1). (Paule and Schorn's =-=[11]-=- Mathematica implementation of the Gosper-- Zeilberger algorithm, which is the one we used, gives this recurrence directly.) Since S(1) = 1, and since the right-hand side of (2.1) satisfies the same r... |

58 |
Combinatory Analysis vol
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(Show Context)
Citation Context ...call such a rhombus a lozenge and such tilings lozenge tilings) is equal to the number P (a; b; c) of plane partitions contained in an a \Theta b \Theta c box. In turn, by a famous result of MacMahon =-=[10]-=-, the latter is given by the product P (a; b; c) = a Y i=1 b Y j=1 c Y k=1 i + j + k \Gamma 1 i + j + k \Gamma 2 : (1.1) The starting point of this paper is a conjecture of Propp [13] stating that for... |

49 | Enumeration of perfect matchings in graphs with reflective symmetry
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- 1997
(Show Context)
Citation Context ... clearly be identified with the centered tilings we are concerned with) this is not quite the case. However, one can get around this using the Factorization Theorem for perfect matchings presented in =-=[1]-=-. Consider the tiling of the plane by unit equilateral triangles, illustrated in Figure 3.1. Define a region to be the union of finitely many such unit triangles. Suppose the region R is symmetric wit... |

30 |
The problem of the calissons
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(Show Context)
Citation Context ...nd c be positive integers, and consider a semiregular hexagon of sides a, b and c (i.e., all angles have 120 degrees and the sides have, in order, lengths a, b, c, a, b, c). By a well-known bijection =-=[2]-=-, the number of tilings of this hexagon by rhombi of unit edge-length and angles of 60 and 120 degrees (we call such a rhombus a lozenge and such tilings lozenge tilings) is equal to the number P (a; ... |

29 | Generating functions for plane partitions of a given shape - Krattenthaler - 1990 |

22 | Determinant identities and a generalization of the number of totally symmetric self-complementary plane partitions, Electron
- Krattenthaler
- 1997
(Show Context)
Citation Context ...=1 (2m + i + 1) i \Theta n\Gamma1 X i=0 (\Gamma1) n\Gammai\Gamma1 (2n \Gamma 2i \Gamma 1) (m + n \Gamma i) 2i i! 2 : (5.3) Proof. The method that we use for this proof is also applied successfully in =-=[8, 5, 6, 7, 9]-=- (see in particular the tutorial description in [7, Sec. 2]). First of all, as in the proof of Lemma 15, we take appropriate factors out of the determinant. To be precise, we take (2n +m \Gamma i \Gam... |

22 | Determinants, paths, and plane partitions. Preprint, available at http://people.brandeis.edu/ gessel/homepage/papers/pp.pdf - Gessel, Viennot - 1989 |

17 | An alternative evaluation of the Andrews–Burge determinant
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- 1999
(Show Context)
Citation Context ...=1 (2m + i + 1) i \Theta n\Gamma1 X i=0 (\Gamma1) n\Gammai\Gamma1 (2n \Gamma 2i \Gamma 1) (m + n \Gamma i) 2i i! 2 : (5.3) Proof. The method that we use for this proof is also applied successfully in =-=[8, 5, 6, 7, 9]-=- (see in particular the tutorial description in [7, Sec. 2]). First of all, as in the proof of Lemma 15, we take appropriate factors out of the determinant. To be precise, we take (2n +m \Gamma i \Gam... |

15 |
Twenty open problems on enumeration of matchings, manuscript
- Propp
- 1996
(Show Context)
Citation Context ...as.edu Institut fur Mathematik der Universitat Wien, Strudlhofgasse 4, A-1090 Wien, Austria. e-mail: KRATT@Pap.Univie.Ac.At WWW: http://radon.mat.univie.ac.at/People/kratt Abstract. Propp conjectured =-=[13]-=- that the number of lozenge tilings of a semiregular hexagon of sides 2n \Gamma 1, 2n \Gamma 1 and 2n which contain the central unit rhombus is precisely one third of the total number of lozenge tilin... |

13 |
plane partitions, preprint
- Gessel, Viennot, et al.
- 1988
(Show Context)
Citation Context ... to simply-connected regions. One useful way to approach certain tiling enumeration problems is to biject them with non-intersecting lattice paths, and then use the Gessel-Viennot determinant theorem =-=[3]-=-. This approach seems to be especially appropriate if the entries of the Gessel-Viennot matrix have a simple expression. In the case of the (2n \Gamma 1) \Theta (2n \Gamma 1) \Theta 2m hexagon with th... |

11 | Some q-analogues of determinant identities which arose in plane partition enumeration
- Krattenthaler
- 1996
(Show Context)
Citation Context ...=1 (2m + i + 1) i \Theta n\Gamma1 X i=0 (\Gamma1) n\Gammai\Gamma1 (2n \Gamma 2i \Gamma 1) (m + n \Gamma i) 2i i! 2 : (5.3) Proof. The method that we use for this proof is also applied successfully in =-=[8, 5, 6, 7, 9]-=- (see in particular the tutorial description in [7, Sec. 2]). First of all, as in the proof of Lemma 15, we take appropriate factors out of the determinant. To be precise, we take (2n +m \Gamma i \Gam... |

8 | A new proof of the M–R–R conjecture — including a generalization
- Krattenthaler
(Show Context)
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8 | Proof of a determinant evaluation conjectured by Bombieri, Hunt and van der Poorten
- Krattenthaler, Zeilberger
- 1997
(Show Context)
Citation Context |

4 | Exact enumeration of certain tilings of diamonds and hexagons with defects, Elect - Helfgott, Gessel - 1999 |