Báo cáo toán học: "Subsequence containment by involutions" ppt

... 5, 6, 7} 8 1 2 5 14 {0, 1, 2, 3, 4, 5, 6, 7} 64 1 2 5 14 Table 5: Classifying S 7 by subsequence containment by involutions. the electronic journal of combinatorics 12 (2005), #R14 9 Before proving ... 5} 2 1 2 4 11 31 102 {0, 1, 2, 3, 4, 5} 16 1 2 5 14 43 142 Table 3: Classifying S 5 by subsequence containment by involutions. J (τ) |{τ}| I 6 (τ) I 7 (τ) I 8 (τ) I 9 (τ) I 10 (τ) {...

Ngày tải lên: 07/08/2014, 08:22

... f −1 (z)=σ z ( )byf 0 (z)=σ z ( ) means finding the unique coefficients α, β such that  σ z ( ) − (1 + αz) σ z ( )  1 β z −2 (1) is a unitary series f 1 (z)=σ z ( ). Dividing in turn f 0 (z)byf 1 (z), ... (−1) k−1 . These terms sum up to (−z) k S k k ( +1/z), and the required series f k (z) is obtained by dividing by (−1) k−1 S k k−1 ( ). QED The first defining equations for the f i ’s are,...

Ngày tải lên: 07/08/2014, 06:22

... maintained by 2.5% sevofluran. Normoglycemia was confirmed by determining blood glucose content (plasma glucose concentration 4.4 to 7.6 mM). The animal experiments were approved by the local ... constant determined by the ratio between B max and K d . Therefore, BP can be altered without any change in receptor affinity (K d ). By contrast, B max can be affected by several f...

Ngày tải lên: 20/06/2014, 21:20

... were obtained approximately 2- to 5-folds with yields of 40–70% within 2 d by the simple purification procedure as judged by SDS-acrylamide gel electrophoresis (see Additional file 2). Assay ... in a final volume of 1.0 ml. After the reactions were terminated by boiling for 10 min, the products formed were quantified by high-performance liquid chromatography (HPLC) using a pack...

Ngày tải lên: 20/06/2014, 21:20

Ngày tải lên: 05/08/2014, 15:20

... classes of polyhedra tile R d if translated by an appro- priate lattice. The notched cube (see Figure 1) has already been shown by Stein [St] to tile R d by a lattice (Conlan [Con] has done this ... notched cube, which is a very explicit function (see (11)). We find all the tilings discovered by Stein, which, by a deeper theorem of Schmerl [Sch], is the complete list of possible tran...

Ngày tải lên: 07/08/2014, 06:22

... number of states of an incompletely specified Mealy type automaton A by finding a closed irreducible covering of the set of states of A by ”maximal compatible sets of states”, which are cliques in the graph ... from an irreducible covering by cliques of the complementary graph G. 2 Main result We will evaluate the clique number ω(G)whenG of order n has an irreducible covering by n − k c...

Ngày tải lên: 07/08/2014, 06:23

... alternative, bijective proof of this by generalizing our earlier bijection ρ : [k] n (121) → [k] n (112). Let α ∈ [k] n (v m,l ). Recall that α j is a word obtained by deleting all letters 1 through ... Alternatively, multiplying the recursive formula (3) by x n−1 /(n − 1)! and summing over n ≥ 1 yields d dx F 112 (x; k) = F 112 (x; k) + (1 + x) d dx F 112 (x; k − 1). Multiplying this...

Ngày tải lên: 07/08/2014, 07:21

... avoids B then by Lemma 8 it can be generated by Algorithm 4, so it can be generated by the stacks. If σ contains a permutation from B then it cannot be generated by the stacks by Lemma 2, so ... B then it cannot be generated by a stack of depth 2 followed by an infinite stack. Proof: It suffices to prove that none of the permutations in B can be generated by the two stacks. It th...

Ngày tải lên: 07/08/2014, 13:21

... d 1 d r be the maximal 2-path containing d i . Either d 1 is pointed to by ≥ 2 diedges, and ¯ d 1 can be determined by lemma 15, o r d r points to ≥ 2 diedges, meaning that ¯ d r can be determined. In ... zeta function. Lemma 5. If G is md2, then the number of edges e of G is given by e = 1 2 deg(Z G (u) −1 ) Proof. Because G is md2, det(Q) = 0. By equation (2), the degree of Z G (u...

Ngày tải lên: 08/08/2014, 01:20