Báo cáo toán học: " Polynomial Generalizations of two-variable Ramanujan type identities" pot

... q), the electronic journal of combinatorics 18(2) (2011), #P15 7 Polynomial Generalizations of two-variable Ramanujan type identities James Mc Laughlin Department of Mathematics, 124 Anderson ... implications of Chu’s 10 ψ 10 extension of Bailey’s 6 ψ 6 summation formula, preprint, 2009. [12] P. Paule, Short and easy computer proofs of the Rogers -Ramanujan identities a...

Ngày tải lên: 08/08/2014, 14:23

... southeast corners of the components of the diagram. The essential set E(w)ofapermutationw is defined to be the set of white corners of the diagram of w. For every white corner (i, j)ofw,itsrank is ... set of distributions of n −2 chips. This completes the proof of Theorem 2.1. ✷ 3 Enumerative aspects of the essential set The combinatorial object that we are studying is the es...

Ngày tải lên: 07/08/2014, 06:20

... formed by reversing the order of the rows of M. As for the matrix P of 3.1.3(2), we have seen above that it is the integral encoding of the rectification S of any tableau T of which M is the integral ... the number of squares of row i that the picture f maps to row k, while Bin(f) k,j counts the number of squares of column j that f maps to row k. From the form of part b i...

Ngày tải lên: 07/08/2014, 13:21

... 1. This concludes the proof. We now show that almost all binary words of length n are β-images. Proof of the main Theorem 1. Let u ∈ {a, b} ∗ be a nonempty binary word of length n. We proceed ... k a = m ≥ 2 occurrences of a. Let w be the prefix of the square-free Thue word T of length m where the last letter is replaced by d, that is, w = vd, where v is the prefix of T of lengt...

Ngày tải lên: 07/08/2014, 15:22

... of finite mutation type because of Theorem 7. • Corollary 14 No graph of one of the 9 exceptional types contains X 6 or X 7 as a subgraph. 4 Conclusion We have exhibited two graphs which are of ... mutation class of Θ(m) is just the isomorphism class of Θ(m) itself. So Θ(m) is mutation-finite. Θ(3) : • GG GG GG • 1.2 Graphs from cluster algebras of finite type. An exchange ma...

Ngày tải lên: 07/08/2014, 21:21

... Numbers of chromatic polynomials of degree at most 10. (A) Total number of chromatic polynomials, (B) number of chromatic polynomials of clique-separable g r aphs and (C) number of chromatic polynomials ... chromatic polynomial gives the number of proper λ-colourings of a graph G. This paper con s iders factorisation of the chromatic polynomial as a first step in an algeb...

Ngày tải lên: 07/08/2014, 21:21

... copy of H] → 1 as n → ∞. Sketch of Proof We work with the set of graphs that don’t have any copies of H. By Lemma 4, we may assume that we have lots of pendant edges, so we may delete |H| of these ... decent proportion of graphs with lots of isolated vertices, but no components isomorphic to C. If such a graph contains lots of copies of C, then we may ‘transfer’ the edges...

Ngày tải lên: 08/08/2014, 11:20

... number of ordered pairs (π 1 , π 2 ) of partitions of [n] in k blocks such that min(π 1 ) = min(π 2 ). Proof. As U(n, k) = a (0) n,k , by Theorem 2, the integer U(n, k ) counts the number of signed k-partitions ... elements of [n] × [n]. Clearly, the union of all such Q ′ i s is a subdiagonal quasi-permutation of cardinality n − k. An example of the map ϕ is given in Figure 3....

Ngày tải lên: 08/08/2014, 12:22

... assume that Y is Q -polynomial with respect to an ordering E 0 , , E D of the primitive idem- potents. Bannai and Ito conjectured that the associated sequence of multiplicities m i (0 ≤ i ≤ D)ofY is unimodal. ... asso- ciation scheme. Our point of departure is the following well-known result of Taylor and Levingston. 1.1 Theorem. [7] If Y is P -polynomial with respect to an ordering...

Ngày tải lên: 07/08/2014, 06:22

... The number of 0-1-2 increasing trees as two different evaluations of the Tutte polynomial of a complete graph C. Merino ∗ Instituto de Matem´aticas, Universidad ... is the Tutte polynomial of the complete graph K n , we have the equal- ity T n+1 (1, 0) = T n (2, 0). This has an almost trivial proof with the right combinato- rial interpretation of T n (1, 0) ... present an algebra...

Ngày tải lên: 07/08/2014, 15:22