Number Theory and Its Applications 2020
DOI: 10.5772/intechopen.92046
|View full text |Cite
|
Sign up to set email alerts
|

Moments of Catalan Triangle Numbers

Abstract: In this chapter, we consider the Catalan numbers, C n ¼ 1 nþ1 2n n , and two of their generalizations, Catalan triangle numbers, B n,k and A n,k , for n, k ∈ . They are combinatorial numbers and present interesting properties as recursive formulae, generating functions and combinatorial interpretations. We treat the moments of these Catalan triangle numbers, i.e., with the following sums: P n k¼1 k m B j n,k , P nþ1 k¼1 2k À 1 ðÞ m A j n,k , for j, n ∈  and m ∈  ∪ 0 fg. We present their closed expressions f… Show more

Help me understand this report

Search citation statements

Order By: Relevance

Paper Sections

Select...
1

Citation Types

0
1
0

Year Published

2023
2023
2023
2023

Publication Types

Select...
1

Relationship

1
0

Authors

Journals

citations
Cited by 1 publication
(1 citation statement)
references
References 22 publications
0
1
0
Order By: Relevance
“…An alternative proof appears in Shapiro [4, Proposition 3.1]. Similarly for z=1$$ z=-1 $$, we get that k=1nfalse(1false)kBn,k=Cn1,1emk=1n+1false(1false)kAn,k=0,1emn1,$$ \sum \limits_{k=1}^n{\left(-1\right)}^k{B}_{n,k}=-{C}_{n-1},\kern1em \sum \limits_{k=1}^{n+1}{\left(-1\right)}^k{A}_{n,k}=0,\kern1em n\ge 1, $$ see, for example, Miana and Romero [11, Theorems 2.1 and 2.2] and references therein. For z=14$$ z=\frac{1}{4} $$, we follow similar ideas by induction method to get that k=1nBn,k()14k=afalse(nfalse)4n,2emk=1n+1An,k()14k=bfalse(nfalse)4n+1,$$ \sum \limits_{k=1}^n{B}_{n,k}{\left(\frac{1}{4}\right)}^k=\frac{a(n)}{4^n},\kern2em \sum \limits_{k=1}^{n+1}{A}_{n,k}{\left(\frac{1}{4}\right)}^k=\frac{b(n)}{4^{n+1}}, ...…”
Section: Some New Results About Catalan Triangle Numbersmentioning
confidence: 99%
“…An alternative proof appears in Shapiro [4, Proposition 3.1]. Similarly for z=1$$ z=-1 $$, we get that k=1nfalse(1false)kBn,k=Cn1,1emk=1n+1false(1false)kAn,k=0,1emn1,$$ \sum \limits_{k=1}^n{\left(-1\right)}^k{B}_{n,k}=-{C}_{n-1},\kern1em \sum \limits_{k=1}^{n+1}{\left(-1\right)}^k{A}_{n,k}=0,\kern1em n\ge 1, $$ see, for example, Miana and Romero [11, Theorems 2.1 and 2.2] and references therein. For z=14$$ z=\frac{1}{4} $$, we follow similar ideas by induction method to get that k=1nBn,k()14k=afalse(nfalse)4n,2emk=1n+1An,k()14k=bfalse(nfalse)4n+1,$$ \sum \limits_{k=1}^n{B}_{n,k}{\left(\frac{1}{4}\right)}^k=\frac{a(n)}{4^n},\kern2em \sum \limits_{k=1}^{n+1}{A}_{n,k}{\left(\frac{1}{4}\right)}^k=\frac{b(n)}{4^{n+1}}, ...…”
Section: Some New Results About Catalan Triangle Numbersmentioning
confidence: 99%