Integrality of a binomial sum 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?

There is a way to prove Zhi-Wei Sun's identity as well as Carlo Beenakker's identity. Of course, both can be treated in accord with Fedor Petrov's induction. Let's focus on Sun's identity. Divide through by $\binom{2n}n\binom{3n}n$ to write
$$A_n:=\sum_{k=1}^n\frac{(9k-2)27^{k-1}}{k^2\binom{2k}k\binom{3k}k}=\frac{27^n}{3\binom{2n}n\binom{3n}n}-\frac13. \tag1$$
so that
$$A_n-A_{n-1}=\frac{(9n-2)27^{n-1}}{n^2\binom{2n}n\binom{3n}n}.$$
Let $a_n=\binom{2n}n\binom{3n}nA_n$ (which is exactly Sun's LHS) to get the recursive equation
$$n^2a_n-3(3n-1)(3n-2)a_{n-1}=(9n-2)27^{n-1}.\tag2$$
First, we find a solution to the homogeneous equation $n^2a_n-3(3n-1)(3n-2)a_{n-1}=0$ as follows
$$a_n^{(h)}=\binom{2n}n\binom{3n}n. \tag4$$
A particular solution to the non-homogeneous equation (2) can be determined by mimicking the RHS as $a_n^{(p)}=(bn+c)27^n$. Now, plug this back in (2) to solve for $b$ and $c$:
\begin{align*} n^2(bn+c)27^n-3(3n-1)(3n-2)(bn-b+c)27^{n-1}&=(9n-2)27^{n-1} \\
\iff 27n^2(bn+c)-3(3n-1)(3n-2)(bn-b+c)&=9n-2 \\
\iff \qquad b=0 \qquad \text{and} \qquad c=\frac13.
\end{align*}
Therefore, the general solution takes the form
$$a_n=a_n^{(p)}+\beta\,a_n^{(h)}=\frac{27^n}3+\alpha\binom{2n}n\binom{3n}n.$$
Since $a_0=A_0=0$, we compute $\beta=-\frac13$ and hence
$$a_n=\frac{27^n}3-\frac13\binom{2n}n\binom{3n}n=\frac{27^n}3-\binom{2n}n\binom{3n-1}{n-1}. \qquad \square$$
