Báo cáo toán học: "On Rowland’s sequence Fernando Chamizo, Dulcinea Raboso and Seraf" ppsx

... 7 On Rowland’s sequence Fernando Chamizo, Dulcinea Raboso and Seraf´ın Ruiz-Cabello ∗ Department of Mathematics Universidad Aut´onoma de Madrid 28049 Madrid. Spain fernando. chamizo@uam.es dulcinea. raboso@ uam.es serafin.ruiz@uam.es Submitted: ... =  j<n (p j − 1) and S(1) = 0. The following is a characterization of Rowland’s chains. Proposition 3.3 A finite sequence...

Ngày tải lên: 08/08/2014, 14:23

... Project of the Ministry of Education, and the National Science Foundation of China. References [1] G.E. Andrews, The Theory of Partitions, Addison-Wesley, 1976. [2] G.E. Andrews, On a partition function ... Andrews and K. Eriksson, Integer Partitio ns, Cambridge Unversity Press, 2004. [4] B.C. Berndt, Number Theory in the Spirit of Ramanujan, Amer. Math. Soc., Provi- dence, 2004. [5] G. G...

Ngày tải lên: 08/08/2014, 11:20

... k 1+2+···+(n−1) . (2.1) It is clear that the right-hand side of (2.1) equals the number of sequences in S(n, k). As will be seen, the left hand-side of (2.1) equals the number of k-ary trees with ... that |M 1 | = |M 0 \M 1 |, and hence we obtain (3.4). In general, we can define the subset M j+1 of M j for j  1. Let (C 0 , C 1 , . . . , C 2n ) be a sequence in M j , and let T be the cor...

Ngày tải lên: 08/08/2014, 14:23

... resp ectively called the Dyson product and the q-Dyson product and denoted by D n (x, a) and D n (x, a, q) respectively, where x := (x 0 , . . . , x n ) and a := (a 0 , . . ., a n ). The Zeilberger-Bressoud ... Zeil- berger and Bressoud [17] in 1985. Recently, Gessel and Xin [4] gave a very different proof by using the properties of formal Laurent series and of polynomials. The c...

Ngày tải lên: 08/08/2014, 14:23

... c. References [1] G.W. Andrews, Plane Partitions V: the TSSCPP conjecture, J. Combin. Theory Ser. A 66 (1994), 28–39. [2] W. Doran, A connection between alternating sign matrices and totally symmetric ... (1993), 289–310. [3] I.M. Gessel and G. Xin, A short proof of the Zeilberger-Bressoud q-Dyson theorem, Proc. Amer. Math. Soc., 134 (2006), 2179–2187. [4] L. Lv, G. Xin, and Y. Zhou, A...

Ngày tải lên: 08/08/2014, 14:23

... branch-width 3 with |P |≥3 and |E(M)|≥6. Then there is a 3-separating subset Q ⊆ P in M such that Q is 3-branched, and that 1. 4 ≤|Q|≤6 and |X| =1for all X ∈ Q,or 2. 2 ≤|Q|≤4 and |X| > 1 for precisely ... T 1 ,T 2 is nice, and smaller than T 0 .IfT 0 has 5 or 6 leaves in (T ) and not all of them are single, then one of T 1 ,T 2 is nice. Finally, if T 0 has 3 or 4 leaves in (T...

Ngày tải lên: 07/08/2014, 06:23

... It is well known that a sequence is holonomic if and only if its generating series is holonomic. (See Stanley’s book [41] for the basic properties of these sequences and series.) By extension, ... [43, Chap. 4]: If both f n and 1/f n are holonomic, then f n is an interlacing of hypergeometric sequences.  2 The logarithmic sequence Let f n =logn (with log 0 ≡ 0) and let f(z) be it...

Ngày tải lên: 07/08/2014, 08:22

... contains F 1 and F 2 and which satisfies property (i): this set should consist of all the hexes which meet K 1 and K 2 . It follows that Ω i = Ω  i . By property (ii), π 1 ∩ π 2 = ∅ and π 1 ⊆ π ζ 2 where ... symplectic polarity ζ and let W (2n − 1, K) and ∆ = DW (2n − 1, K) denote the corresponding polar space and dual polar space. Let π be an (n −1)-dimensional subspace of PG(2...

Ngày tải lên: 07/08/2014, 21:21

... of graphs G x s and G y t , respectively obtained from G s and G t by adding a vertex x or y and connecting it t o the endvertices of C in G s or G t . By Menger’s theorem, x and y each has a ... function of the graphs G 1 and G 2 at vertices v 1 and v 2 . The zip product of G 1 and G 2 according to σ is the graph G 1 ⊙ σ G 2 obtained from the disjoint union of G 1 − v 1 and G...

Ngày tải lên: 08/08/2014, 12:22

... conditions (i), (ii), and (iii). Then {T n i } i≥0 gives rise to sequences {T n ′ i } i≥0 and {T n ′′ i } i≥0 of triangulations of the triangles ∆ 1 and ∆ 2 into n ′ i and n ′′ i triangles, respectively. ... that lim i→∞ n ′ i = ∞ and lim i→∞ n ′′ i = ∞. So, wlog, the sequence {T n i } i≥0 can be chosen so that n ′ 0 ≤ n ′ 1 ≤ n ′ 2 ≤ . . . n ′′ 0 ≤ n ′′ 1 ≤ n ′′ 2 ≤ . . . Note th...

Ngày tải lên: 08/08/2014, 14:23