<s>[INST] generalized binomial coefficient I know that the coefficient $$-\frac{1}{2} \choose k$$ can be simplified by multiplying both the nominator and the denominator by $$2^k$$ and then represented as $$ (-\frac{1}{4})^k {2k\choose k}$$
But what if the upper index is $$\frac{1}{3}$$it seems that our trick do not work anymore, is there anything we can do about it?
 [/INST] 
There is no nice generalisation from $\binom{-\frac{1}{2}}{k}$ to $\binom{\pm \frac{1}{n}}{k}, 0\leq k \leq n$ available.
We obtain for $n=2$:
\begin{align*}
  \binom{-\frac{1}{2}}{k}&=\frac{1}{k!}
    \left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right)\cdots\left(-\frac{1}{2}-(k-1)\right)\\
    &=\frac{1}{k!}\frac{(-1)^k}{2^k}(2k-1)!!\\
    &=\frac{1}{k!}\frac{(-1)^k}{2^k}\frac{(2k)!}{(2k)!!}\tag{1}\\
    &=\frac{(-1)^k}{2^{2k}}\frac{(2k)!}{k!k!}\\
    &=\frac{(-1)^k}{2^{2k}}\binom{2k}{k}\\
  \end{align*}
and for $n=3$:
\begin{align*}
  \binom{-\frac{1}{3}}{k}&=\frac{1}{k!}
    \left(-\frac{1}{3}\right)\left(-\frac{4}{3}\right)\cdots\left(-\frac{1}{3}-(k-1)\right)\\
    &=\frac{1}{k!}\frac{(-1)^k}{3^k}(3k-2)!!!\\
    &=\frac{1}{k!}\frac{(-1)^k}{3^k}\frac{(3k)!}{(3k)!!!(3k-1)!!!}\tag{2}\\
    &=\frac{(-1)^k}{3^{2k}}\frac{(3k)!}{k!k!}\frac{1}{(3k-1)!!!}\\
    &=\frac{(-1)^k}{3^{2k}}\binom{3k}{2k}\binom{2k}{k}\frac{1}{(3k-1)!!!}\\
  \end{align*}

Comment:


*

*In (1) we use the double factorials
\begin{align*}
  (2k)!&=(2k)!!(2k-1)!!\qquad \text{and} \qquad (2k)!!=2^kk!\\
  \end{align*}


*

*The formula with triple factorials in (2) is somewhat more complicated due to additional factors $(3k-1)!!!$


\begin{align*}
  (3k)!&=(3k)!!!(3k-1)!!!(3k-2)!!!\qquad \text{and} \qquad (3k)!!!=3^kk!\\
  \end{align*}
Note: You might find the generalisation to $\binom{\alpha}{k}$ with $\alpha \in\mathbb{C}, k\geq 0$ used in binomial series or the even more generalized binomial coefficients based upon Gamma functions useful.
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