Showing $\frac{r}{2(n+r)}\binom{2n}{n}\binom{2r}{r}\in \Bbb N^{+}$

Question

For any integers $n\ge 0,$ and $r\ge 1$,show that $$f(n,r)=\dfrac{r}{2(n+r)}\binom{2n}{n}\binom{2r}{r}\in \Bbb N^{+}$$

I know that $f(0,r)=\binom{2r-1}{r}$ are integers and that $f(n,1)=\dfrac{1}{n+1}\binom{2n}{n}$ are Catalan numbers.

Answer 1

We may consider $$ \sum_{r\geq 1}\sum_{n\geq 0}\frac{r}{2}\binom{2r}{r}\binom{2n}{n}x^r y^n z^{n+r-1} = \frac{x}{(1-4xz)^{3/2}(1-4yz)^{1/2}}\tag{1}$$ which by applying $\int_{0}^{1}(\ldots)\,dz$ to both sides leads to $$ \sum_{r\geq 1}\sum_{n\geq 0}\frac{r}{2(n+r)}\binom{2r}{r}\binom{2n}{n}x^r y^n = \frac{2x}{\sqrt{1-4x}(\sqrt{1-4x}+\sqrt{1-4y})}.\tag{2}$$ It is enough to show that all the coefficients of the (bivariate) Maclaurin series of the RHS are integers.
This clearly happens for $\frac{1}{\sqrt{1-4x}}=\sum_{m\geq 0}\binom{2m}{m}x^m$, and the RHS of $(2)$ can be written as

$$ \frac{2x}{\sqrt{1-4x}}\cdot\frac{\sqrt{1-4x}-\sqrt{1-4y}}{4(y-x)} \tag{3}$$ where $\sqrt{1-4x}=\sum_{m\geq 0}\binom{2m}{m}\frac{x^m}{1-2m}$ also has integer coefficients. The claim easily follows.