On the zeros of Bessel functions of parameter $\nu=-1/3$ Working in a problem I found with the task of finding the zeros of a sum of the Airy functions ( See for example) $\sqrt{3}\text{Ai}(x)+\text{Bi}(x)$, it result that these are exactly the zeros for the modified Bessel function of the First Kind of parameter $\nu=-1/3$ this is $$I_{-1/3}(x)=\sum_{n=0}^{\infty} \frac{(-1)^{n}}{n !} \frac{(x / 2)^{2 n-1/3}}{\Gamma(n-\frac{1}{3}+1)}$$.
Do you know if there is an expression for the zeros of this function with this specific value of parameter?
Added: With computational calculation it looks like that if $\lambda_n$ are the zeros then $|\lambda_n-\lambda_{n+1}|\to \pi.$ Looking for I already found that the zeros have the following expresion $z_{\nu, k}=\left\{k+\frac{1}{2}\left(\nu-\frac{1}{2}\right)\right\} \pi-\frac{4 \nu^{2}-1}{8\left\{k+\frac{1}{2}\left(\nu-\frac{1}{2}\right)\right\} \pi}+O\left(\frac{1}{k^{3}}\right)$.

If you consider Wolfram language functions a closed form, then there is  Bessel Y Zero $\text y_{v,x}$:
$$\def\Ai{\text{Ai}} \def\Bi{\text{Bi}} \sqrt3\Ai(x)+\Bi(x)=0\implies -\sqrt3=\frac{\Bi(x)}{\Ai(x)}$$
Now use Bessel J and Bessel Y:
$$\frac{\text J_\frac13(x)}{\text Y_\frac13(x)}=\frac4{\frac{\Ai\left(-\left(\frac32\right)^\frac23x^\frac23\right)}{\Bi\left(-\left(\frac32\right)^\frac23x^\frac23\right) }+\sqrt3}-\sqrt3\implies \frac{\Bi(x)}{\Ai(x)}=\frac{\frac4{1-\sqrt3 \frac{\text J_\frac13\left(\frac23(-x)^\frac32\right)}{\text Y_\frac13\left(\frac23(-x)^\frac32\right)}}-1}{\sqrt3}=\sqrt3$$
Here is the solution
$$\begin{align}\frac{\text J_\frac13\left(\frac23(-x)^\frac32\right)}{\text Y_\frac13\left(\frac23(-x)^\frac32\right)} =\sqrt3\iff \sqrt3\Ai(x)+\Bi(x)=0\implies x=-\left(\frac32\text y_{\frac13,\Bbb N+\frac16}\right)^\frac23\end{align}$$
which is true with the $n$th natural number giving the $n$th solution.
