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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?

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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:

\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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