## Enumeration of Lozenge Tilings of Hexagons with a Central Triangular Hole (0)

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Citations: | 23 - 9 self |

### BibTeX

@MISC{Ciucu_enumerationof,

author = {M. Ciucu and T. Eisenkölbl and T. Eisenk Olbl and D. Zare and C. Krattenthaler},

title = {Enumeration of Lozenge Tilings of Hexagons with a Central Triangular Hole},

year = {}

}

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### Abstract

. We deal with the 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 (\Gamma1)-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 latter formulas specialize to statements about weighted enumerations of cyclically symmetric plane partitions. One such specialization gives a proof of a conjecture of Stembridge on a certain weighted count of cyclically symmetric plane partitions. The tools employed in our proofs are nonstandard applications of the theory of nonintersecting lattice paths and determinant evaluations. In particular, we evaluate the determinants det 0i;jn\Gamma1 \Gamma !ffi ij + \Gamma m+i+j j \Delta\Delta , w...

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Citation Context ...A2, . . ., An, E1, E2, . . .,En be points of the planar integer lattice. Then the following identity holds: det 1≤i,j≤n (P(Ai → Ej)) = � (sgn σ) · P(A → Eσ, nonint.). (5.3) σ∈Sn Remark. The result in =-=[23]-=-, respectively [13], is in fact more general, as it is formulated for paths in an arbitrary oriented graph. But then the graph must satisfy an acyclicity ◦s20 M. CIUCU, T. EISENKÖLBL, C. KRATTENTHALER... |

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Citation Context ... there exists no core, we obtain enumeration results for cyclically symmetric plane partitions. Before we state these, let us briefly recall the relevant notions from plane partition theory (cf. e.g. =-=[36]-=- or [38, Sec. 1]). There are (at least) three possible equivalent ways to define plane partitions. Out of the three possibilities, in this paper, we choose to define a plane partition π as a subset of... |

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Citation Context ...ith side lengths 1 and angles of 60 ◦ and 120 ◦ . 1s2 M. CIUCU, T. EISENKÖLBL, C. KRATTENTHALER AND D. ZARE where H(n) stands for the “hyperfactorial” � n−1 k=0 k!. This follows from a bijection (cf. =-=[7]-=-) between such lozenge tilings and plane partitions contained in an a ×b×c box, and from MacMahon’s enumeration [25, Sec. 429, q → 1; proof in Sec. 494] of the latter. In [32] (see also [33]), Propp p... |

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Citation Context ...tensive theory of summation and transformation formulas (such a series is called a hypergeometric series in U(a) or an Aa hypergeometric series), mainly thanks to Milne and Gustafson (see for example =-=[14, 28, 29, 30, 34]-=-, and the references contained therein), it is only occasionally that series containing the square � 1≤i<j≤a (ki − kj) 2sLOZENGE TILINGS OF HEXAGONS WITH A CENTRAL TRIANGULAR HOLE 53 appear. Most of t... |

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Citation Context ...tensive theory of summation and transformation formulas (such a series is called a hypergeometric series in U(a) or an Aa hypergeometric series), mainly thanks to Milne and Gustafson (see for example =-=[14, 28, 29, 30, 34]-=-, and the references contained therein), it is only occasionally that series containing the square � 1≤i<j≤a (ki − kj) 2sLOZENGE TILINGS OF HEXAGONS WITH A CENTRAL TRIANGULAR HOLE 53 appear. Most of t... |

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Citation Context ...(−1) (a+1)/2 . It is then a routine computation to verify that this gives the multiplicative constant as claimed in (8.15). 9. Proof of Theorem 11 For the proof of Theorem 11, we proceed similarly to =-=[27]-=-. We define determinants Zn(x, µ) by � �n−1 � �� �� � i + µ k j − k + µ − 1 Zn(x, µ) = det −δij + x 0≤i,j≤n−1 t t j − k k−t � . (9.1) t,k=0 The only difference to the definition of Zn(x, µ) in [27] is... |

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Citation Context ...unts, (−1)-enumerations of plane partitions, i.e., enumerations where plane partitions are given a weight of 1 or −1, according to certain rules, have been found to possess remarkable properties (see =-=[38, 39]-=-). Motivated in part by a conjectured (−1)-enumeration on cyclically symmetric plane partitions due to Stembridge [40], in Section 2 we consider a (−1)-enumeration of the lozenge tilings of Theorems 1... |

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Citation Context ...E2, . . .,En be points of the planar integer lattice. Then the following identity holds: det 1≤i,j≤n (P(Ai → Ej)) = � (sgn σ) · P(A → Eσ, nonint.). (5.3) σ∈Sn Remark. The result in [23], respectively =-=[13]-=-, is in fact more general, as it is formulated for paths in an arbitrary oriented graph. But then the graph must satisfy an acyclicity ◦s20 M. CIUCU, T. EISENKÖLBL, C. KRATTENTHALER AND D. ZARE condit... |

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Citation Context ...ction (cf. [7]) between such lozenge tilings and plane partitions contained in an a ×b×c box, and from MacMahon’s enumeration [25, Sec. 429, q → 1; proof in Sec. 494] of the latter. In [32] (see also =-=[33]-=-), Propp posed several problems regarding “incomplete” hexagons. For example, Problem 2 in [32] (and [33]) asks for the number of lozenge tilings of a hexagon with side lengths n, n+1, n, n+1, n, n+1 ... |

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Citation Context ... (i + j + m/2)! × det (3j + m/2 + 1) . 0≤i,j≤a/2−1 (2i − j + m/2 + 1)! (2j − i)! = det 0≤i,j≤a/2−1 Both determinants on the right-hand side of this identity can be evaluated by means of Theorem 10 in =-=[17]-=-, which reads � � (x + y + i + j − 1)! det 0≤i,j≤n−1 (x + 2i − j)! (y + 2j − i)! = n−1 � i=0 This completes the proof of the theorem. i! (x + y + i − 1)! (2x + y + 2i)i (x + 2y + 2i)i . (9.6) (x + 2i)... |

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Citation Context ...binations are themselves linearly independent. (Equivalently, we find m linearly independent vectors in the kernel of the linear operator defined by the matrix underlying D1(−e, c).) See Section 2 of =-=[18]-=-, and in particular the Lemma in that section, for a formal justification of this procedure.s24 M. CIUCU, T. EISENKÖLBL, C. KRATTENTHALER AND D. ZARE To be precise, we claim that the following equatio... |

16 |
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Citation Context ... the black lozenge has weight −1, all other lozenges have weight 1.) Yet another way to obtain this weight is through the perfect matchings point of view of lozenge tilings, elaborated for example in =-=[21, 22]-=-. In this setup, the cyclically symmetric lozenge tilings that we consider heres12 M. CIUCU, T. EISENKÖLBL, C. KRATTENTHALER AND D. ZARE correspond bijectively to perfect matchings in a certain hexago... |

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Citation Context ...unts, (−1)-enumerations of plane partitions, i.e., enumerations where plane partitions are given a weight of 1 or −1, according to certain rules, have been found to possess remarkable properties (see =-=[38, 39]-=-). Motivated in part by a conjectured (−1)-enumeration on cyclically symmetric plane partitions due to Stembridge [40], in Section 2 we consider a (−1)-enumeration of the lozenge tilings of Theorems 1... |

15 |
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Citation Context ... the black lozenge has weight −1, all other lozenges have weight 1.) Yet another way to obtain this weight is through the perfect matchings point of view of lozenge tilings, elaborated for example in =-=[21, 22]-=-. In this setup, the cyclically symmetric lozenge tilings that we consider heres12 M. CIUCU, T. EISENKÖLBL, C. KRATTENTHALER AND D. ZARE correspond bijectively to perfect matchings in a certain hexago... |

15 |
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Citation Context ...ows from a bijection (cf. [7]) between such lozenge tilings and plane partitions contained in an a ×b×c box, and from MacMahon’s enumeration [25, Sec. 429, q → 1; proof in Sec. 494] of the latter. In =-=[32]-=- (see also [33]), Propp posed several problems regarding “incomplete” hexagons. For example, Problem 2 in [32] (and [33]) asks for the number of lozenge tilings of a hexagon with side lengths n, n+1, ... |

14 |
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Citation Context ...tensive theory of summation and transformation formulas (such a series is called a hypergeometric series in U(a) or an Aa hypergeometric series), mainly thanks to Milne and Gustafson (see for example =-=[14, 28, 29, 30, 34]-=-, and the references contained therein), it is only occasionally that series containing the square � 1≤i<j≤a (ki − kj) 2sLOZENGE TILINGS OF HEXAGONS WITH A CENTRAL TRIANGULAR HOLE 53 appear. Most of t... |

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Citation Context ...d. This problem was solved in [4, Theorem 1], [15, Theorem 20] and [31, Theorem 1] (the most general result, for a hexagon with side lengths a, b + 1, c, a + 1, b, c + 1, being contained in [31]). In =-=[5]-=-, the first author considers the case when a larger triangle (in fact, possibly several) is removed. However, in contrast to [31], the results in [5] assume that the hexagon has a reflective symmetry,... |

9 | Enumeration of lozenge tilings of punctured hexagons - Ciucu - 1998 |

9 | Enumeration of rhombus tilings of a hexagon which contain a fixed rhombus in the centre
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Citation Context ...= 1, 2, . . ., n. (The only exceptions that we are aware of, i.e., applications of the above formula in the case where the sum on the right-hand side does not reduce to a single term, can be found in =-=[8]-=-, [23], and [41].) This is, however, not exactly the situation that we encounter in our problem. Therefore, it seems that Lemma 14 is not suited for our problem. However, our choice of starting and en... |

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8 |
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Citation Context |

8 |
Ueber die Funktionen, welche durch Reihen von der Form dargestellt werden 1+ p p q ′ p ′′ q ′′ + p p+1 p q
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Citation Context ...a+m (1 + a + c − e + 2 m − s)−1+e−m+s (1 + a − e + 2 m − s)−1+e−m+s � � 1 − c − i − m, 1 − e + m − s, 1 − a − m × 3F2 ; 1 . 1 − a − c − m, 2 − e − i Next we use a transformation formula due to Thomae =-=[42]-=- (see also [10, (3.1.1)]), � � A, B, −n 3F2 ; 1 = D, E (E − B)n � � −n, B, D − A 3F2 ; 1 , (7.7) (E)n D, 1 + B − E − n where n is a nonnegative integer. This gives (1 + a + c − e + 2 m − s)e−m+s−1 (1 ... |

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Citation Context ...gle removed. This problem was solved in [4, Theorem 1], [15, Theorem 20] and [31, Theorem 1] (the most general result, for a hexagon with side lengths a, b + 1, c, a + 1, b, c + 1, being contained in =-=[31]-=-). In [5], the first author considers the case when a larger triangle (in fact, possibly several) is removed. However, in contrast to [31], the results in [5] assume that the hexagon has a reflective ... |

7 | Multidimensional matrix inversion and A r and D r basic hypergeometric series - Schlosser - 1997 |

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5 |
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Citation Context ...n. (The only exceptions that we are aware of, i.e., applications of the above formula in the case where the sum on the right-hand side does not reduce to a single term, can be found in [8], [23], and =-=[41]-=-.) This is, however, not exactly the situation that we encounter in our problem. Therefore, it seems that Lemma 14 is not suited for our problem. However, our choice of starting and end points (see Fi... |

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3 | Plane partitions II: 5 1 symmetry classes, in: Combinatorial Methods in Representation Theory - Ciucu, Krattenthaler |

3 | U(n) very-well-poised 10 - Milne, Newcomb - 1996 |

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1 |
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(Show Context)
Citation Context ...rding to certain rules, have been found to possess remarkable properties (see [38, 39]). Motivated in part by a conjectured (−1)-enumeration on cyclically symmetric plane partitions due to Stembridge =-=[40]-=-, in Section 2 we consider a (−1)-enumeration of the lozenge tilings of Theorems 1 and 2. The corresponding results are given in Theorems 4 and 5. In Section 3, we restrict our attention to cyclically... |

1 | Plane partitions II: 5 2 symmetry classes, in: Combinatorial Methods in Representation Theory - Ciucu, Krattenthaler |