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Congruences for the second-order Catalan numbers


Authors: Li-Lu Zhao, Hao Pan and Zhi-Wei Sun
Journal: Proc. Amer. Math. Soc. 138 (2010), 37-46
MSC (2000): Primary 11B65; Secondary 05A10, 11A07
DOI: https://doi.org/10.1090/S0002-9939-09-10067-9
Published electronically: September 4, 2009
MathSciNet review: 2550168
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Abstract | References | Similar Articles | Additional Information

Abstract: Let $ p$ be any odd prime. We mainly show that

$\displaystyle \sum_{k=1}^{p-1}\frac{2^{k}}k\binom{3k}k\equiv 0 (\operatorname{mod} p) $

and

$\displaystyle \sum _{k=1}^{p-1}2^{k-1} C_{k}^{(2)}\equiv (-1)^{(p-1)/2}-1 (\operatorname{mod} p),$

where $ C_{k}^{(2)}=\binom{3k}k/(2k+1)$ is the $ k$th Catalan number of order 2.


References [Enhancements On Off] (What's this?)

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Additional Information

Li-Lu Zhao
Affiliation: Department of Mathematics, Nanjing University, Nanjing 210093, People’s Republic of China
Email: zhaolilu@gmail.com

Hao Pan
Affiliation: Department of Mathematics, Nanjing University, Nanjing 210093, People’s Republic of China
Email: haopan79@yahoo.com.cn

Zhi-Wei Sun
Affiliation: Department of Mathematics, Nanjing University, Nanjing 210093, People’s Republic of China
Email: zwsun@nju.edu.cn

DOI: https://doi.org/10.1090/S0002-9939-09-10067-9
Received by editor(s): April 7, 2009
Published electronically: September 4, 2009
Additional Notes: The third author is the corresponding author. He was supported by the National Natural Science Foundation (grant 10871087) and the Overseas Cooperation Fund of China.
Communicated by: Ken Ono
Article copyright: © Copyright 2009 American Mathematical Society
The copyright for this article reverts to public domain 28 years after publication.

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