Báo cáo toán học: "Proof of the (n/2 − n/2 − n/2) Conjecture for large n" pps

... +1)=0=⇒AT(2n) =0.Together, these imply the truth of the Alon-Tarsi conjecture for n =2r(p+1) for any r ≥ 0 and any odd prime p (and, by the table of known values, for p =2also).Our goal ... to N,so there is only one isotopy class in the sum (8), and its size is not divisible by p.Therefore,fdels(p) − fdols(p) ≡|NG|sgn(N)(modp). (9)Proof of the Alon-Tarsi Conjecture for n =2rpArthur ... [1] conjectured: Conjecture 1 (Alon-Tarsi) If n is even then els(n) =ols(n).Equivalently, the sum of the signs of all L ∈ LS(n) is nonzero. This conjecture isrelated to several other conjectures...

... − αi)a+c+1−mj=c+ 1− (z − βj)(3.7)To connect with Eq. (2.9), setP(z) = (z − α1) · · ·(z − α −1 ) (z − β1) · · ·(z − βc−)Qm(z) = (z − αb+m+1) · · ·(z − αb+c) (z − ... Fa,b;1satisfies the following simple recurrence formula:Fa,b;1=αb+1 − γa+bαb+1 − βaFa−1,b;1+γa+b − βa+1αb − βa+1Fa,b−1;1(3.6)and by using z − γa+b=αb+1 − a+bαb+1 − a+1(z − βa+1) ... Fig. 6. The weight of a lozenge is equal to q u − q −1 v where u and v are the spectral parameters crossing at the center of the lozenge in such a way that the line of v forms an angle of +π/3...

... substantially), and it is the purpose of the presentnote to try to lay bare the basic mechanism of Kouba’s proof.2 Proof of Theorem 2 The proof of the Theorem 2 will be based on the following simple ... questionIs there any interesting use for the fact that even in the case of integral domains the coefficients of Kouba’s map can be explicitly expressed in terms of cofactors of the matrices(si−1j)?Acknowledgements The ... (3)where{s,s′}∈(Sk2)(s − s′) = rkis not a well-defined element of R but only defined up to asign (since (s−s′) = −( s′−s) = (−1 )·(s′−s) and (−1 )· (−1 ) = −( −1 ) = 1, in every ring),depending on how the elements...

... |Y′| ≥ n/2 − n, then |Y | ≥ (1 − √3θ ) (n/2 − n). Since δ(X, V (G1)) ≥ n − n, we haveδ(X, Y ) ≥ |Y |− n≥ (1 − √3θ ) (n/2 − n) − θn>12 − √3θ2 − 2θn≥n2 − √θn,where the last ... graphs withdegree constraints. the electronic journal of combinatorics 18 (2011), #P27 5Proof of the (n/2 n/2 − n/2) Conjecture for large nYi Zhao∗Department of Mathematics and StatisticsGeorgia ... components C of H −S. It suffices to prove the following claim.Claim. Either |V (M)| ≥ 2 (1 − c)k − 1, or there is a component C in H − S thatcontains two adjacent vertices of L. The former case of the...