The following sequence appears to be always an integer, experimentally.



> 
> **QUESTION.** Let $n\in\mathbb{Z}^{+}$. Are these indeed integers?
> $$\sum\_{k=1}^n\frac{(4k - 1)4^{2k - 1}\binom{2n}n^2}{k^2\binom{2k}k^2}.$$
> 
> 
> 


**POSTSCRIPT.** After Carlo's cute response and several useful comments, I like to ask this: is there a combinatorial proof?



Integrality of a binomial sum
$$\sum\_{k=1}^n\frac{(4k - 1)4^{2k - 1}\binom{2n}n^2}{k^2\binom{2k}k^2}=16^n \left(1-\frac{\Gamma \left(n+\frac{1}{2}\right)^2}{\pi \Gamma (n+1)^2}\right)$$
$$\qquad=2^{4n}-c\_n^2,\;\;\text{with}\;\;c\_n=2^n\frac{(2n-1)!!}{n!}={{2n}\choose n}. \qquad\qquad\text{[thanks, Pietro Majer]}$$


