Enumeration of Lozenge Tilings of Punctured Hexagons

We present a combinatorial solution to the problem of determining the number of lozenge tilings of a hexagon with sides a, b + 1, b, a + 1, b, b + 1, with the central unit triangle removed. For a = b, this settles an open problem posed by Propp 7]. Let a, b, c be positive integers, and denote by H the hexagon whose side-lengths are (in cyclic order) a, b, c, a, b, c and all whose angles have 120 degrees. The lozenge tilings (i.e., tilings by unit rhombi) of H can be regarded as plane partitions contained in an a b c box (cf. 2]), and therefore their number is given by the simple product formula 5] a Y i=1 b Y j=1 c Y k=1 Motivated by this, Propp 7] considered the problem of enumerating the lozenge tilings of a hexagon whose sides are alternately a and a + 1, from which the central unit triangle has been removed (removal of a suitable unit triangle is necessary for the remaining region to have lozenge tilings). Based on numerical evidence, he conjectured that there exists a simple product formula for the number of tilings of these regions. The more general question of nding the number of lozenge tilings of a hexagon with sides a, b+1, c, a+1, b, c+1, with the central unit triangle removed { denote it by N (a; b; c) { appeared in work of Kuperberg 4] concerning certain weighted enumerations of plane partitions. This general question has been recently settled by Okada and Krattenthaler 6], who proved that N (a; b; c) is equal to the product of four factors of type (1) (their proof relies on a new Schur function identity they prove using the minor summation formula of Ishikawa and Wakayama 3]). The purpose of this paper is to give a simple product formula (with a simple combi-natorial proof) for N (a; b; b) (this settles in particular Propp's original question; Figure 1 shows the region corresponding to a = 2, b = 4). Let SC (a; b; c) be the number of self-complementary pane partitions that t in an abc box (see 8] for the deenition). In 8] it is given a simple product formula for SC (a; b; c).