The Number of Rhombus Tilings of a Symmetric Hexagon Which Contain a Fixed Rhombus on the Symmetry Axis, I
نویسنده
چکیده
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 .
منابع مشابه
The Number of Rhombus Tilings of a Symmetric Hexagon which Contain a Fixed Rhombus on the Symmetry Axis, II
Let a, b and c be positive integers, and consider a hexagon with side lengths a, b, c, a, b, c whose angles are 120◦ (see Figure 1.a). The subject of enumerating rhombus tilings of this hexagon (cf. Figure 1.b; here, and in the sequel, by a rhombus we always mean a rhombus with side lengths 1 and angles of 60◦ and 120◦) gained a lot of interest recently. This interest comes from two facts. Firs...
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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 posed by Propp.
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Let a, b and c be positive integers and consider a hexagon with side lengths a,b,c,a,b,c whose angles are 120◦ (see Figure 1). The subject of our interest is rhombus tilings of such a hexagon using rhombi with all sides of length 1 and angles 60◦ and 120◦. Figure 2 shows an example of a rhombus tiling of a hexagon with a = 3, b = 5 and c = 4. A first natural question to be asked is how many rho...
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Abstract. Propp conjectured [15] 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 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 the...
متن کاملRhombus Tilings of a Hexagon with Three Missing Border Triangles
The interest in rhombus tilings has emerged from the enumeration of plane partitions in a given box (which was first carried out by MacMahon [5]). The connection comes from looking at the stacks of cubes of a plane partition from the right direction and projecting the picture to the plane. Then the box becomes a hexagon, where opposite sides are equal, and the cubes become a rhombus tiling of t...
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تاریخ انتشار 1998