There is no simple equation that given $n$ gives us $\gamma_n$, with sufficient approximation. Given a constant $c>0$, the error in both equations are at times greater than $c/\log n$ for any given constant $c$.
The main equation of Fran\c ca-LeClair
$$ n=\vartheta(\gamma_n)+\lim_{\delta\to0^+}\arg\zeta(\tfrac12+\delta+i\gamma_n)+\tfrac{3\pi}{2}$$
where $\rho=\frac12+i\gamma_n$ is the $n$-th zero of zeta is wrong.
First, the limit always exist, for example if $\frac12+i\gamma_n$ is a simple zero we have $\zeta'(\frac12+i\gamma_n)\ne0$ and (assuming, for simplicity, that $\zeta'(\frac12+it)$ is not a negative real number)
$$\lim_{\delta\to0^+}\arg\zeta(\tfrac12+\delta+i\gamma_n)=\lim_{\delta\to0^+}\arg\frac{\zeta(\tfrac12+\delta+i\gamma_n)-\zeta(\frac12+i\gamma_n)}{\delta}=\arg\zeta'(\tfrac12+i\gamma_n).$$
Let $z\mapsto\arg(z)$ be the harmonic function with $|\arg(z)|<\pi$ defined on the plane with a cut along the negative real axis $(-\infty,0]$, then the function $t\mapsto\arg(\zeta(\frac12+it))$ extend uniquely to right continuos function. Call it $\arg\zeta'(\frac12+it)$. Then for any zero $\rho_n=\frac12+i\gamma_n$ that is on the critical line, there is a integer $\mathop{\rm depth}(\rho_n)$ such that
$$ n=\vartheta(\gamma_n)+\arg\zeta'(\tfrac12+i\gamma_n)+\tfrac{3\pi}{2}+\mathop{\rm depth}(\rho_n),\qquad \mathop{\rm depth}(\rho_n)\in\mathbf{Z}$$
this is the true equation.
It is true that $\mathop{\rm depth}(\rho_n)=0$ has only 17399 exceptions for $1\le n\le 10^7$ the first one being for $n=28813$. But this is somewhat misleading. First for any $T>0$ there is a zero with $\gamma_n>T$ and such that $\mathop{\rm depth}(\rho_n)>c\sqrt{\log T/\log\log T}$. This implies that to get a value near $\gamma_n$ we must put $n-\mathop{\rm depth}(\rho_n)$ instead of $n$ in the approximations of Fran\c ca-LeClair. This gives us the error I indicated at first.
Besides, under the Riemann hypothesis, we have always $|\mathop{\rm depth}(\gamma_n)-S(\gamma_n)|\le 3/2$. And since $S(t)/\sqrt{\log\log t}$ is distributed for $t$ large as a gaussian, we may expect that for $n$ really large we will have $\mathop{\rm depth}(\rho_n)$ usually large. Therefore, it is likely that the probability to have $\mathop{\rm depth}(\rho_n)=0$ tends to $0$.
