Cipolla and Césaro both gave expansions of $\operatorname{li}^{-1}$ in tems of nested $\log$ functions. I think it can be written in terms of the Lambert-W function in the form:


$\operatorname{li}^{-1}(n)=n\sum \_{i=1} a\_i(-1)^{i+1} W\_{-1}\left(-e/n\right){}^{2-i} $


where $a\_i$ begins: $\small{-1, 0, 1, 3, 11, 105/2, 613/2, 12635/6, 99677/6, 1774391/12, \dots}$


where the next one $\small{\approx 1465235 + \epsilon, \ \epsilon =-1/12?)}$.


The above should be accurate to $o(1/\log(n)^{10})$ - is this correct? Is there a nice way to obtain the coefficients other than by computation?


For convenience:



```
f[n_] := With[
{a = {-1, 0, 1, 3, 11, 105/2, 613/2, 12635/6, 99677/6, 1774391/12, 17582819/12}}, 
n Sum[(-1)^(1 + k) a[[k]] ProductLog[-1, -E/n]^(2 - k), {k, 1, 11}]];
g[n_] := Quiet[x /. FindRoot[LogIntegral@x == n, {x, N[n Log[n], 200]}, 
WorkingPrecision -> 200]];

Abs@Log@N[1/Log[#]^10] &[10^10^7]
Abs@Log@N[1 - f@#/g@#] &[10^10^7]

```


Expansion of inverse logarithmic integral in terms of lambert w
Let me elaborate on reuns' idea and show how one can find coefficients $a\_i$ by solving a certain ODE.


Let's define:
$$f(z) := \sum\_{i\geq 3} a\_i z^{i-3}$$
so that we get a functional equation:
$$\mathrm{li}\big( -x(\frac{1}{t}+tf(t)) \big) = x,$$
where $t=t(x):=-\frac{1}{W\_1(-e/x)}$.
Differentiating this equation with respect to $x$, and then substituting $x=\frac{et}{\exp(-1/t)}$, we obtain a differential equation:
$$(\star)\qquad t^3f'(t) - t(1-2t)f(t) - (1-t)\log(1-t^2f(t)) + t = 0$$
with the initial condition $f(0)=1$.


I've played with ODE $(\star)$ in Maple. In principle, Maple can solve it in the following form:
$$f(t) = \frac{1-\exp\big(r-1/t\big)}{t^2},$$
where $r$ is the root of the equation: $\mathrm{Ei}(1,-1-r)=-\frac{et}{\exp(-1/t)}$. However, I'm not sure if this implicit form is that useful.


On the other hand, Maple can solve $(\star)$ in power series of given order, thus computing the coefficients $a\_i$. For example,



```
Order:=15: dsolve( { t^3*diff(f(t),t) - t*(1-2*t)*f(t) - (1-t)*log(1-t^2*f(t)) + t = 0, f(0)=1 }, f(t), series);

```

gives


$$f \left( t \right) =1+3\,t+11\,{t}^{2}+{\frac{105}{2}}{t}^{3}+{\frac{613}{2}}{t}^{4}+{\frac{12635}{6}}{t}^{5}+{\frac{99677}{6}}{t}^{6}+{\frac{1774391}{12}}{t}^{7}+{\frac{17582819}{12}}{t}^{8}+{\frac{1919343719}{120}}{t}^{9}+{\frac{22882040099}{120}}{t}^{10}+{\frac
{295793507053}{120}}{t}^{11}+{\frac{1373607474819}{40}}{t}^{12}+{\frac{323119030735871}{630}}{t}^{13}+{\frac{20600974525589671}{2520}}{t}^{14}+O \left( {t}^{15} \right).$$


I've added the sequences of coefficients numerators / denominators to the OEIS as [A337734](https://oeis.org/A337734) and [A337735](https://oeis.org/A337735), respectively.


