Báo cáo toán học: "The Generating Function of Ternary Trees and Continued Fractions" doc

... proof of this symmetry will induce a bijective proof of N(i, j)=N(j, i), and then a bijective proof of Lemma 6.4. Proof of Theorem 6.3. Let G(m, n)bethenumberofK-paths starting at (−2m, −2m) and ... negative, and the verification of the recurrence is straightforward. (We will give another proof of (48) in Example 9.2.) This completes the proof of (7). For equation (8), we n...

Ngày tải lên: 07/08/2014, 13:21

... M. Stark and Audrey A. Terras. Zeta functions of finite graphs and coverings. Adv. Math., 121(1):124–165, 1996. [22] Harold M. Stark and Audrey A. Terras. Zeta functions of finite graphs and coverings. ... Haemers and Edward Spence. Enumeration of cospectral graphs. European Journal of Combinatorics, 25(2):199–211, 2004. [11] Ki-ichiro Hashimoto. Zeta functions of finite grap...

Ngày tải lên: 07/08/2014, 13:21

... one of the following cases: µ 2k−1 = 0 and µ 2k−2 =0, or µ 2k−1 = ν k and µ 2k−2 =0, or µ 2k−1 = 0 and µ 2k−2 = ν k . In any of these cases, b k − b k−1 = 5 ν k +3 2 . If m ≥ 1, at most one of ... a k−1 = 5 µ k +1 2 ,anda k =1+ k  i=2 5 µ i +1 2 . µ k =mif and only if k ∈{j3 m },where j is a positive integer and jmod3= 0. Proof. If µ k > 0, then k =(···d µ k +1 0 ···0) 3...

Ngày tải lên: 07/08/2014, 06:22

... as the union of B 1 , one vertex of R 2 ∩ B 2 and the vertex r 2 is blue, finally U 4 , defined as the union of R 2 , one vertex of R 1 ∩B 1 and the vertex b 1 is red. Now U 1 ∪U 3 and U 2 ∪U 4 are the ... Green [12]) and t = 2 (Alon and Frankl [1] and Gy´arf´as [13]). Next we state and prove the Ramsey-type form of our main result, it determines the exact value of the...

Ngày tải lên: 07/08/2014, 21:20

... (−1) j−1 B j = −B j for j>1, and (3) follows from (5) and (6). ✷ Remark. It is possible to avoid the use of complex analysis and give a purely formal power series proof of the theorem. We indicate ... (4) ✷ Proofofthetheorem.The coefficient of t n−1−m in P A (t) is by (4) the coefficient of z −n+m in (−1) n−m (n − 1 − m)! · 1 (1 − e a 1 z ) ···(1 − e a n z ) , which is the coeffici...

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