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$\binom{-1/2}{n}=\left (\frac{1}{n!}\right )\left (\frac{-1}{2}\right )\left (\frac{-3}{2}\right )\dots\frac{-(2n-1)}{2}=(-1)^n\frac{(2n-1)!!}{2n!!}$

The first three values are 1, -1/2, 3/8. 

binom[n_] := Binomial[1/2, n]
Table[binom[i], {i, 0, 2}]

{1, 1/2, -(1/8)}

According to the documentation:

In general, (n m) is defined by $\Gamma$(n+1)/($\Gamma$(m+1)$\Gamma$(n-m+1)) or suitable limits of this.

However ${-1/2 \choose 1}$= Gamma[1/2]/(Gamma[2] Gamma[-1/2]) gives the correct value: -(1/2) Is this a bug?

user5389726598465
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1 Answers1

2

By definition, $$ \binom{1/2}{0}=1 \qquad \binom{1/2}{1}=\frac{1/2}{1}=\frac{1}{2} \qquad \binom{1/2}{2}=\frac{(1/2)\cdot(1/2-1)}{2!}=-\frac{1}{8} $$ whereas $$ \binom{-1/2}{0}=1 \qquad \binom{-1/2}{1}=\frac{-1/2}{1}=-\frac{1}{2} \qquad \binom{-1/2}{2}=\frac{(-1/2)\cdot(-1/2-1)}{2!}=\frac{3}{8} $$

egreg
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