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 $(n+1)^2$?
$\begingroup$
$\endgroup$
11
asked Mar 14, 2017 at 2:22
-
3$\begingroup$ Empirically, $\sum_{k=0}^n \sum_{j=0}^k {k\ \choose j}^2 {2j \choose j} (2j+1) = (n+1)^2 \sum_{i = 0}^n {n \choose i}^2 C_i$, where $C_i$ is the $i$-th Catalan number. $\endgroup$– D. Ror.Commented Mar 14, 2017 at 3:25
-
2$\begingroup$ Also empirically, $\sum_{k=0}^n\sum_{j=0}^k\binom{k}j^2\binom{2j}j(2j+1)^{2p}$ also seems to be divisible by $\left(n+1\right)^2$ for any positive integer $p$. $\endgroup$– darij grinbergCommented Mar 14, 2017 at 19:29
-
1$\begingroup$ @darijgrinberg ...and divisible by $(n+1)$ but not $(n+1)^2$ for odd exponents greater than $1$. $\endgroup$– D. Ror.Commented Mar 15, 2017 at 4:28
-
3$\begingroup$ This appears as (part of) Conjecture 5.6 in Zhi-Wei Sun "Two new kinds of numbers and related divisibility results", arXiv:1408.5381 v8. In Remark 5.3 Zhi-Wei Sun asserts it is divisible by n+1. $\endgroup$– juanCommented Mar 15, 2017 at 20:08
-
3$\begingroup$ @juan In a recent preprint, Sun's conjecture was proven by Mao: arxiv.org/pdf/1511.06221.pdf (see Theorem 1.1). $\endgroup$– Ofir GorodetskyCommented Mar 16, 2017 at 11:20
|
Show 6 more comments
1 Answer
$\begingroup$
$\endgroup$
2
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.
answered Mar 18, 2017 at 19:26
-
1$\begingroup$ Note: My conjecture seems to be the first claim of Conjecture 5.1 of that paper. $\endgroup$ Commented Mar 18, 2017 at 20:00
-
$\begingroup$ @darijgrinberg: Yes, and this was proved by G.-S. Mao (arxiv.org/abs/1511.06221). $\endgroup$ Commented Mar 18, 2017 at 20:06