Báo cáo toán học: "Ap´ry’s Double Sum is Plain Sailing Indeed e" pps

... Ap´ery’s Double Sum is Plain Sailing Indeed Carsten Schneider ∗ Research Institute for Symbolic Computation J. Kepler University Linz A-4040 Linz, Austria Carsten.Schneider@risc.uni-linz.ac.at Submitted: ... Symbolic summation assists combinatorics. S´em. Lothar. Combin., 56:1–36, 2007. Article B56b. [7] K. Wegschaider. Computer generated proofs of binomial multi -sum identities....

Ngày tải lên: 07/08/2014, 13:21

... S(n+k)−S(n) is constant for 1 ≤ n ≤ k. Then Proposition 3.3 a) implies that this value is divisible by every p n , and hence λ is a multiple of p 1 p 2 . . . p k . But this is impossible, since this product ... first case there exists a solution (mod q) if and only if S(m) does not cover all residue classes. This is c). In the second case we also need S(m) ≡ S(n) (mod p n ) and this...

Ngày tải lên: 08/08/2014, 14:23

... is incongruent (or regular ). If (1) is satisfied by exactly one value of i for each n then the system is disjoint (or exact). Two celebrated questions of Erd˝os are “Do there exist incongruent systems with ... of word called a Fibonacci word [15]. It is a privilege to know and work with Aviezri. I am grateful for his friendship, his advice, his imagination and his gentle sense of humo...

Ngày tải lên: 07/08/2014, 06:23

... rise to a monomial in at least two variables. So since each Schur function in the sum has one parameter, the terms in the sum are nonzero only if ρ is a row and (ν/ρ)  is a skew-row, that is, ... this, we return to the expression s λ [s (1 a ,b,b) ][1 + x − y]=  x |λ|(b−1) y |λ|a s λ [x + y 2 − y − xy]ifa is even x |λ|(b−1) y |λ|a (−1) |λ| s λ  [x + y 2 − y − xy]ifa is odd Whe...

Ngày tải lên: 07/08/2014, 08:20

... H is a bisubdivision of K,oraK bisubdivision 2 (F. 1). Note that a bisubdivision of a bipartite graph is also bipartite. The next concept is a generalization of alternating cycles. Suppose H is a ... (Remember, G is not an H 10 .) Hence, we will assume G is not a 4-cycle sum of three K 3,3 graphs. If we can prove G − E (C) is a brace, then G is also a brace. Hence, we can al...

Ngày tải lên: 07/08/2014, 08:22

... remark that this independence may also be proved directly. Indeed, it is not hard to show that if F is a face of A 0 that is confined to a single face of  H,then((  W.θ) S , ≺) is independent ... (Wθ) I is isomorphic to the Bruhat ordering of the double quotient I W J , and that this subposet may be generated by the application of certain reflections. Proposition 1.5. Assume θ ∈...

Ngày tải lên: 07/08/2014, 08:22

... ordinary partition function p(n). A simple combinatorial explanation of this fact is also provided. 1 Introduction This note is concerned with the partition function t(n) introduced by Stanley [8, ... integer n is a nonin- creasing sequence of nonnegative integers such that the sum of the components λ i equals n. A part is meant to be a positive component, a nd the number of parts o...

Ngày tải lên: 08/08/2014, 11:20

... + 1, 2). This completes the proof. Acknowledgments. We wish to thank the referee for helpful suggestions. This work was supported by the 973 Project, the PCSIRT Project of the Ministry of Education, ... under the assumption that u is a descendant of itself. Substituting (2.5) into (2.4), we arrive at (2.2). This completes the proof. We have the following correspondence. Theorem 2.1 There i...

Ngày tải lên: 08/08/2014, 14:23

... ression χ(S) is 1 if the stateme nt S is true, and 0 otherwise. In trying to prove Conjecture 1.4, we find that the conjectured formula is incorrect. One way to modify the conjecture is to evaluate ... for mula. Our second objective is to contribute a q-analogous formula of (1.1), which is motivated by the proof of (1.1), and is stated in Theorem 4.1. This paper is organized as...

Ngày tải lên: 08/08/2014, 14:23

... In this paper, only Identity 2 is given a direct constant term proof. In addition, a conjecture is given as a generalization of Identity 1. The paper is organized as follows. Section 1 is this ... #P11 7 Proof. Assume Q is given by (6). We claim that Q is proper in x 1 . This can be easily checked by observing that for m being free of x 1 , the degree (in x 1 ) of (1 −x 1 m) is...

Ngày tải lên: 08/08/2014, 14:23