I'm interested in binomial coefficients of the form $$\binom{-1/m}{k} ,$$ where $m$ is a positive integer.
For $m=2$, it holds that \begin{align} \binom{-1/2}{k} &= (-4)^{-k} \binom{2k}{k} \qquad \qquad \quad (*) \\ &= \Big{(}-\frac{1}{4}\Big{)}^{k} \binom{2k}{k} \qquad \quad \space \space \space \space(**) \end{align}
Sources for the $(*)$ and $(**)$ marked expressions can be found here and here, respectively.
I've tried obtaining a similar expression for the general case. Upon doing so, I first tried to obtain an identity for the case $m=3$ by following source of the $(**)$ approach. However, I got stuck.
Here is what I tried so far:
\begin{align} \binom{-1/3}{k} &= \frac{(-1/3)\cdot(-4/3)\dots \big{(} \frac{3k-2}{3} \big{)}}{k!} \\ &= (-1)^{k} \frac{1\cdot4\cdot7\dots(3k-2)}{3^{k}k!} \\ &= (-1)^{k} \frac{1\cdot4\cdot7\dots(3k-2)}{3\cdot6\cdot9\dots(3k)} \\ &= \dots \end{align}
I couldn't continue, because multiplying by the fraction of the denominator over itself did not yield a "clean" expression in the numerator, which contrasts with what happens when $m=2$. In that case, on obtains $2k!$ in the numerator. For $m=3$, the product terms $2, 5, 8, \dots, (3k-1) $ are missing in the numerator, so one does not obtain $3k!$.
I am therefore curious whether another approach can yield nice, closed form expressions for $\binom{-1/m}{k}$ binomial coefficients when $m \in \mathbb{Z}_{\geq 3}$. Also, I wonder whether such nice expressions can be found for all positive integers, or only a subset of them.