Báo cáo toán học: "A natural series for the natural logarithm" pot

... (2.2) is true for general shape o f µ, by (6.1 ) , the Hilbert series of M µ for any shape of µ would have the form in (6 .2 ) when q = 0 . Hence, the natural thing to ask is whether one can find ... q 2 ) Table 2: The Hilbert series for µ ′ = (2, 2, 1) The standard Yo ung tableaux in the first column o f the Table 2 are the possible standard Yo ung tableaux correspo...

Ngày tải lên: 08/08/2014, 12:22

... +32t − 1. Then f(n) ≥ n +32t − 1, for n ≥ n t . This completes the proof of the theorem. From the above theorem, we have lim inf n→∞ f(n) − n √ n ≥  2+ 2562 6911 , which is better than the previous ... ≤ 31t + 799). Now we define these B i ’s. These subgraphs all have a common vertex x, otherwise their vertex sets are pairwise disjoint. For 7t+1 8 ≤ i ≤ t − 742, let the subgraph...

Ngày tải lên: 07/08/2014, 06:22

... denote by u R the sequence obtained from u by reversing its order. The following theorem is due to Knuth [4, Theorem D]. The differences in notation and the opposite parity of l compared the electronic ... s 2 n . (1) where b(i) is the number of times the integer i occurs in the sequence T . Note that the sum of all b(i), 1 ≤ i ≤ n,isequaltol, the length of T . 3 A simple p...

Ngày tải lên: 07/08/2014, 08:22

... agree above row r and to the left of column c (and the definitions of r and c) we see that the only possibility for λ/µ is (r, c), completing the proof of this case and the theorem. Acknowledgements. ... . ). Finally, recall that the conjugate of a partition λ is the partition λ  obtained by flipping the diagram of λ across the NW-SE axis; it follows that λ  i counts the...

Ngày tải lên: 07/08/2014, 15:22

... (k) g (k) g . The root of the tree is (1) b , which is the label of the one-cell polyomino. the electronic journal of combinatorics 14 (2007), #R57 11 Then we are able to write down the equation for R(s, ... in detail in [2]) we look for the value of s for which the factor on the left multiplying R(s, t) is equal to zero, i.e. the solution of the kernel 1 −s + ts 2...

Ngày tải lên: 07/08/2014, 15:22

... P [k],+ j−1,i−1  The first equality comes from Lemma 12, the second by induction on i for ∆ [k],+ j,i−1 and by the induction hypothesis for ∆ [k−1],+ j−1,i−1 . We then recognize P [k+1] j,i on the right ... be the number of cells (including the possible circle at the end of the row) to the right of s. Let a lso ℓ Λ (s) be the number of cells (not including the possib...

Ngày tải lên: 07/08/2014, 21:21

... follows for (π, f ) ∈ S n,A : If R (π,f) = ∅, let φ((π, f)) = (π, f). Otherwise, let (i, j) ∈ R (π,f) be minimal under the lexicographic ordering of R (π,f) . Let ˜π = (i, j) ◦ π, the product of the ... by ˜ f(K ˜π (p)) = f(K π (p)) for all 1  p  n. Let φ((π, f)) = (˜π, ˜ f). Note that R (π,f) = R φ((π,f )) for all (π, f) ∈ S n,A , and that therefore φ is involutive. Note furthe...

Ngày tải lên: 08/08/2014, 01:20

... t) n−1−i . The above q-recurrence relation is “semi” in the sense that the summands on the right involve two factors one of which depends on q whereas the other does not. We shall establish in the present ... Q[q][[x]] in the ring of formal power series in x over Q[q]. It is easy to see that δ x (x n ) = [n] q x n−1 , so that as q → 1, δ x (x n ) → nx n−1 , the usual derivat...

Ngày tải lên: 08/08/2014, 11:20

... 42 Proof . We have the following cases: 1. Suppose that a comes from a fork. Since the degree of o is 4, o must be the white node in the fork. Thus, f is contained in the same fork. Then node 4 must ... contradiction. Therefore, the triangle must contain one of the outward edges, which forces the remaining two outward edges to be in the same block. This block must be a fork,...

Ngày tải lên: 08/08/2014, 14:23

... 321. Theorem 3 (Noonan) |S n (321; 1)| = 3 n  2n n − 3  . the electronic journal of combinatorics 18(2) (2011), #P21 1 Figure 1: An injection f : S n (321; 1) → S n+2 (321). The “3” and “1” in the ... , n} obtained by replacing the point (j, c) with (j, b − ǫ) and (b + ǫ, c) and the point (ℓ, a) with (b −ǫ, a) and (ℓ, b + ǫ) on the permutation diagram of π. In other words, let ˆ...

Ngày tải lên: 08/08/2014, 14:23