Báo cáo toán học: "A two-dimensional pictorial presentation of Berele’s insertion algorithm for symplectic tableaux" doc

... electronic journal of combinatorics 12 (2005), #R4 17 A two-dimensional pictorial presentation of Berele’s insertion algorithm for symplectic tableaux Tom Roby Department of Mathematics California State ... serve as the inverse of the column insertion instead of the row insertion. It is also a semistandard version of a bijective tool used by Sundaram [Sun1, Pro...

Ngày tải lên: 07/08/2014, 08:22

... length of the longest N-row of Π the height of Π. For Π of height h, we call the vector (n 0 ,n 1 , ,n h )whoseith coordinate is the number of N rows of Π of length i the N-area vector of Π. Similarly ... equals the number of inversion pairs of τ involving 2’s and 1’s or 2’s and 0’s. Similarly, the number of inversion pairs of Π of the form (x, y)withrow x of len...

Ngày tải lên: 07/08/2014, 07:21

... They hold for subsets of any abelian group, but here we state them only for subsets of F p . Lemma 1. For any subsets X, Y, Z of F p we have |X − Z| ≤ |X − Y ||Y − Z| |Y | . Lemma 2. For any subsets ... C||AB|  min  p|A|, |A| 2 |B||C| p  . 3 Proof of Theorem 1 If G ⊂ X × Y then for a given x ∈ X we denote by G(x) the set of all elements y ∈ Y for which (x, y) ∈ G. The...

Ngày tải lên: 07/08/2014, 15:23

... the definition of J, one has Jx = Jy. Therefore, we have x = y; see [19,21] for more details. We also need the following lemmas for the proof of our main results: Lemma 2.2. [20]Let E be a uniformly ... subset of a uniformly convex and u niformly smooth Banach space E. Let {T n } ∞ n = 1 be a countable family of relatively quasi-nonexpansive mappings of C to E satis fy the (*) c...

Ngày tải lên: 20/06/2014, 21:20

... = dr dτ (≥ 0) = marginal response of cold-drink consumption to a change in τ(temperature) C 1 = opportunity cost of money tied up in inventory. C 2 = rate of change of inventory cost with respect ... − τ ∗ τ  2 p(τ) Vietnam Journal of Mathematics 33:4 (2005) 437–442 A Stochastic EOQ Policy of Cold-Drink -For a Retailer Shib Sankar Sana 1 and Kripasindhu Chaudhuri 2 1 Departme nt...

Ngày tải lên: 06/08/2014, 05:20

... prove the second statement of Theorem 1. ProofofStatement(ii) of Theorem 1. By the same argument in the proof of Stament (i) of Theorem 1, we can assume without loss of generality that A is complete. Assume ... 1  (H i m (M)), for every system of parameters x =(x 1 , ,x d ) of M, where H i m (M) stands for the i th local cohomology mo dule of M with respect to the maxima...

Ngày tải lên: 06/08/2014, 05:20

... concrete example, let S bethesetofallwordsoflength7overΣ={a, b}. Each word of S is 2-avoidable, but any binary word of length 7 or more encounters an element of S. An image of xxx or xyxyx under a non-erasing ... palindrome, we need only show that no prefix of m n is an overlap. First note that no prefix of m n is a square of length 12 or greater; otherwise the prefix of m n of...

Ngày tải lên: 07/08/2014, 06:20

... is not of the form mτ 2 +0.5 for an integer m. Then the singleton  u τu+c  is a generating set for a solution of (1). Proof. Observe that, since k =1,ρ=τ. There is nothing to check for (i). ... sets of seeds in order to find a b that decreases infinitely often. For an infinite set of seeds, ensuring that b is well-defined is problematic. On the one hand, descendants of any...

Ngày tải lên: 07/08/2014, 06:20

... must have r k ≡ g (mod 2). The choice of residue for the modulus 3 is forced. There are then six choices for the modulus 9, two for 18, three for 27, and two for the second 27. Thus, there are 6 ... congruent modulo 4, we see that there are precisely five choices for the modulus 5, four for the modulus 10, three for 15, two for 30, and two for 20. So there are 5 × 4 × 3...

Ngày tải lên: 07/08/2014, 06:20

... of the colorings of the edges of S and reverses the colors in the other 14. Figure 6 shows the effect of H S on S for a complete (up to rotations) set of colorings of S. From the perspective of ... through the same sequence of squares of parity k before and after the application of H k .All that may change is how the path moves inside each of those squares and what it d...

Ngày tải lên: 07/08/2014, 06:20