2011
DOI: 10.2298/aadm110717017g
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On Jensen's and related combinatorial identities

Abstract: Motivated by the recent work of Chu [Electron. J. Combin. 17 (2010), #N24], we give simple proofs of Jensen's identity 2010 Mathematics Subject Classification. Primary 05A10. Secondly 05A19.

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Cited by 6 publications
(4 citation statements)
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“…R((a + 1) 2 a 2 ) = a(a + 1)(a + 1)a 1 > s (a + 1)a(a + 1)a 2 > s (a + 1)(a + 1)aa (14) 3 > s (a + 1)aa(a + 1) R((a + 1) 2 a 3 ) = a(a + 1)a(a + 1)a 1 > s a(a + 1)(a + 1)aa 2 > s (a + 1)a(a + 1)aa (15) 3 > s (a + 1)aa(a + 1)a 4 > s (a + 1)(a + 1)aaa 5 > s (a + 1)aaa(a + 1) Remark 36. According to (2) through (5) of Corollary 16, Theorem 35 confirms that for each of the four posets above, the box diagonal diagram is indeed the unique maximal element, confirming Conjecture 12 in these cases, which turn out to be all chains.…”
Section: Totally Ordered Equitable Ribbonsmentioning
confidence: 99%
See 1 more Smart Citation
“…R((a + 1) 2 a 2 ) = a(a + 1)(a + 1)a 1 > s (a + 1)a(a + 1)a 2 > s (a + 1)(a + 1)aa (14) 3 > s (a + 1)aa(a + 1) R((a + 1) 2 a 3 ) = a(a + 1)a(a + 1)a 1 > s a(a + 1)(a + 1)aa 2 > s (a + 1)a(a + 1)aa (15) 3 > s (a + 1)aa(a + 1)a 4 > s (a + 1)(a + 1)aaa 5 > s (a + 1)aaa(a + 1) Remark 36. According to (2) through (5) of Corollary 16, Theorem 35 confirms that for each of the four posets above, the box diagonal diagram is indeed the unique maximal element, confirming Conjecture 12 in these cases, which turn out to be all chains.…”
Section: Totally Ordered Equitable Ribbonsmentioning
confidence: 99%
“…(1) Relation 3 in Chain (14) (2) Relation 4 in Chain (15) (3) Relation 5 in Chain (15) Proof. We first prove Parts 1 and 3 before proving Part 2, which is the most intricate.…”
Section: Totally Ordered Equitable Ribbonsmentioning
confidence: 99%
“…and C is a positive constant depending only on the norms of u in the spaces mentioned. [41,45] We know that, the J N (Λ)×J N (0, T ) is a finite subspace of H r (Λ)×L 2 (0, T ), according to the assumption u ∈ H r (Λ) × L 2 (0, T ), the ∥u∥ = ∫ T 0 ∥u∥ 2 r,ω dt < ∞, therefore, there is an N 0 ∈ N that for any N > N 0 , N r be bigger than ∥u∥ or equivalently C. So, if N be large enough, we can say: if N −→ ∞ then ∥u − u N ∥ H r (Λ)×L 2 (0,T ) −→ 0, hence, the error of this approximation by increasing N will be decreased. In numerical examples this principle will be shown.…”
Section: Remarkmentioning
confidence: 99%
“…The identity (4) may be obtained from Hagen and Rothe identity Guo, 2011) by setting x = −1, z = −2 and y = M + N.…”
Section: Remarkmentioning
confidence: 99%