TITLE: Divisibility of a binomial sequence
QUESTION [8 upvotes]: Convincing numerical evidence prompts me to ask:

Question. Is $\sum_{k=0}^n\sum_{j=0}^k\binom{k}j^2\binom{2j}j(2j+1)^2$ divisible by $(n+1)^2$?

REPLY [3 votes]: The answer is yes, and the proof can be found in V. J. W. Guo, J.-C. Liu, Proof of some conjectures of Z.-W. Sun on the divisibility of certain double sums, Int. J. Number Theory 12 (2016), 615-623. An arXiv version is also available.