Báo cáo toán học: "Enumeration of Pin-Permutations" pdf

... conversely). This gives 2 n+2 strict pin words of size n and concludes the proof. The proof of Theorem 4.1 2 follows the structure of the proof of Theorem 3.4 in [16]. Lemma 4.11. [16, Lemma ... oscillation ξ of size n − 1 by the addition of either a minim al element at the beginning of ξ or a maximal element at the e nd of ξ, followed by a flip of an element of ξ according...

Ngày tải lên: 08/08/2014, 14:22

... number of tilings of half of a hexagon or half of a diamond, with dents at given places. This is not new: see [7] and [8]. 2. Express the number of tilings of the figure as a whole as a sum of squares ... transpose, then the determinant of MM t is equal to the sum of the squares of the k by k minors of M. The number of tilings of a (k, q, k) hexagon in which the indic...

Ngày tải lên: 07/08/2014, 06:20

... Proof of Theorem 7.1. Ignoring the case with only o ne block of odd length, fix a choice of s elements from [a + 1,n] to be placed in the first two blocks. We have shown that more of these ... x k ) . The proof uses scalar products of symmetric functions. We give a more direct proof, with a combinatorial flavour. The proof uses the following definition, and the bijective result of Lem...

Ngày tải lên: 07/08/2014, 21:21

... Enumeration of perfect matchings of Q m,n,r 3.1 The sign of pfB(x 1 , x 2 ) In order to decide the sign of pfaffians of B(x 1 , x 2 )(x 1 , x 2 = ±1), we distinguish the perfect matchings of Q m,n,r into ... Proc. 1997 Symposium on the Theory of Computing (STO C). [16] H. Narumi and H. Hosoya, Proof of the generalized expressions of the number of perfect matchings of polyc...

Ngày tải lên: 08/08/2014, 12:22

... number of −1 entries (to the number of orbits of −1 entries in the case of symmetric ASM’s). For a generic value of the g lobal parameter a, when we put all variables to 1, the weight of zero ... is no QTASM of size 4N + 2. Proof. Let us suppose that M is a QTASM of even size 2L. Now we use the fact that the size of an ASM is given by the sum of its entries, and the symmet...

Ngày tải lên: 08/08/2014, 12:22

... (11) 3 Enumeration of m-tuples of Permutations More generally, we may consider the m-tuples of permutations of S n with exactly k occurrences of “ R m ”. Denote by D(n, k) the number of such multiple ... expression of Ω in terms of g, which turns to be the generating function displayed in the theorem. Furthermore, letting x = 0 in (10), we deduce that the number of permutati...

Ngày tải lên: 08/08/2014, 12:22

... hyperoperations that associate with two elements of H not an element, as in a classical structure, but a subset of H. One of the interests of the researchers in the field of hyperstructur es is to construct ... Let V n be the number of non isomorphic Varlet hyper groups of order n. Then, V n+m ≥ V n V m , for all n, m ∈ N. Using the results of [22] we can enumerat e the number...

Ngày tải lên: 08/08/2014, 14:23

... the end of the first row, κ 2 cells are deleted from the end of the second row, etc. Letting N be the size of (n m ) \ κ, a SYT of truncated shape (n m ) \ κ is a labeling of the cells of this shape ... partitions as sets, [m] is the disjoint union of the shape of T 1 and the the shape of T 2 . Proof. (i) is clear. In order to prove (ii), denote the shifted shapes of T 1 an...

Ngày tải lên: 08/08/2014, 14:23

... Classification of solutions for an integral equation. Comm. Pure Appl. Math. 59, 330–343 (2006) Nonexistence of positive solutions of an integral system with weights Zhengce Zhang School of Mathematics ... some assumptions of p, q, α, and β. Furthermore, using Doubling Lemma indicated in [14], which is an extension of an idea of Hu [15], Chen and Li [16, Theorem 4.3] obtained the...

Ngày tải lên: 20/06/2014, 21:20

Ngày tải lên: 05/08/2014, 09:46