Sign of 3j symbol (in view of interpolation)

Question

Question

Is there a closed formula for the sign of a 3j symbol?

Context

I need to compute Wigner 3J symbols/Clebsch–Gordan coefficients,

$$ \begin{pmatrix} \ell_1 &\ell_2 &\ell_3\\ 0&0&0 \end{pmatrix} , $$

for all configurations of ($\ell_1,\ell_2,\ell_3$) up to $\ell_\text{max}\sim 2000$. Speed is crucial so rather than computing the 3j a billion times, I would resort to an interpolation scheme.

The raw 3j symbol is impervious to interpolation as it continuously alternates from negative to positive values; for example, consider this plot of the 3j coefficients as a function of $\ell_3$ with $\ell_1=120$ and $\ell_2=90$:

_{(source: guidowalterpettinari.eu)}

The absolute value of the 3j, on the other hand, is much smoother:

_{(source: guidowalterpettinari.eu)}

My plan is to interpolate the absolute value of the 3j symbol, and assign the sign only after the interpolation. Hence the question: is there a closed formula for the sign of a 3j symbol?

EDIT:

The question has been answered thanks to Gjergji Zaimi. I am now looking into the same problem but with a more general 3J symbol,

$$ \begin{pmatrix} \ell_1 &\ell_2 &\ell_3\\ 0&m&-m \end{pmatrix} . $$

Please feel free to have a look at the corresponding Math Overflow question.

Answer 1

Wolfram contains the following formula which should make your calculations easy:

If $\ell_1+\ell_2+\ell_3=2g$ then $$\begin{pmatrix} \ell_1 &\ell_2 &\ell_3\\ 0&0&0 \end{pmatrix}=$$ $$(-1)^g\sqrt{\frac{(2g-2\ell_1)!(2g-2\ell_2)!(2g-2\ell_3)!}{(2g+1)!}}\frac{g!}{(g-\ell_1)!(g-\ell_2)!(g-\ell_3)!}$$ and if $\ell_1+\ell_2+\ell_3=2g+1$ then $\begin{pmatrix} \ell_1 &\ell_2 &\ell_3\\ 0&0&0 \end{pmatrix}=0$.