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Numerical evidence suggests that \begin{align} &\operatorname{li}^{-1}(n)=x_n+\mathcal{O}\left(\log^2 n\right)\\ \end{align} where \begin{align} &x_n=x_{n-1}+\log \left(x_{n-1}\right)\\ \end{align} and $\operatorname{li}^{-1}$ is the inverse logarithmic integral function.

Transpose[{(InverseFunction[LogIntegral][N[#]]), 
(Nest[# + Log[#] &, N[Pi/2], #] + Divide[(Log@#)^2, 10.5] + 7.5)} 
& /@ Range[3, 1000]] // ListLinePlot

It this the case? If so, can the relationship be described more precisely?

martin
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1 Answers1

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$\DeclareMathOperator{\li}{li}$The claim is correct and can be improved.
Theorem: Consider any primitive $\li:\,]1,\infty[\mathbb{R}$ of $1/\log(x)$ and any sequence $x_n$, $n=0,1,\dots$, of real numbers satisfying $x_0>1$ and $x_n=x_{n-1}+\log(x_{n-1})$ for $n\geq1$. Then there is a constant $C$ such that $$ \li(x_n)+\tfrac12\log(\log(x_n))=n+C+O\left(\frac1{n\log(n)}\right)\mbox{ as }n\to\infty.$$ As a consequence, with the above constant, $$x_n=\li^{-1}(n)+\log(n\log(n))\cdot(C-\tfrac12\log(\log(n)))+o(1)$$ and, in particular, $x_n=\li^{-1}(n)+O(\log(n)\log(\log(n)))$ for large $n$.

Proof: Clearly, the sequence $x_n$ is stricly increasing and tending to $\infty$, even $x_n/n$ tends to infinity.

As for other recurrence relations, it is useful to have a function $F$ with $$F(x+\log(x))\approx F(x)+1\mbox{ for large }x$$ because then $F(x_{n+m})\approx F(x_m)+n$. Solving approximately $F'(x)\log(x)+\frac12F''(x)\log(x)^2\approx 1$, we find $F'(x)\approx \frac 1{\log(x)}+\frac1{2x\log(x)}$.

We first use $F(x)=\li(x)$ and find that $F(x+\log(x))-F(x)=\frac{\log(x)}{\log(x+d(x))}$ with some $d(x)$ between $0$ and $\log(x)$. Hence $F(x+\log(x))-F(x)\to1$ as $x\to+\infty$ and $F(x_{n+1})-F(x_n)\to 1$ as $n\to\infty$. By Cesaro's theorem we obtain that also $$\frac1n(F(x_n)-F(x_0))=\frac1n((F(x_n)-F(x_{n-1})+\cdots+(F(x_1)-F(x_0)))\to1.$$ This means that $\li(x_n)\sim n$ as $n\to\infty$. Since $\li(x)\sim \frac x{\log(x)}$ as $x\to+\infty$ by L'Hôpital's rule this implies that $\frac{x_n}{\log(x_n)}\sim n$. Estimating the unique zero of $x-\alpha\log(x)$ in $]e,\infty[$ for large $\alpha$, this implies that $x_n\sim n\log(n)$ as $n\to\infty$.

Now define $F(x)=\li(x)+\frac12\log(\log(x))$ for $x>e$. Then a small calculation using Taylor's formula yields $F(x+\log(x))-F(x)=1+O\left(\frac{\log (x)}{x^2}\right)$ for $x>1$. Hence $F(x_{n+1})=F(x_n)+1+O\left(\frac{\log (x_n)}{x_n^2}\right)$ for all $n$. Since $x_n\sim n\log(n)$ we have $F(x_{n+1})=F(x_n)+1+O\left(\frac1{n^2\log(n)}\right).$ Hence $F(x_n)-n$ is a Cauchy sequence and tends to some constant we call $C$. We find $$F(x_n)-n-C=O\left(\sum_{k=n}^\infty\frac1{k^2\log(k)}\right)=O\left(\frac1{n\log(n)}\right).$$ This proves the first statement of the Theorem.

For the second statement, we first write $x_n=\li^{-1}\left(n+C-\frac12\log(\log(x_n))+O\left(\frac1{n\log(n)}\right)\right)$. Then we use a Taylor expansion around $n$ and that $x_n\sim n\log(n)$, $\li^{-1}(n)= n(\log(n)+O(\log(\log(n)))$ $\left(\li^{-1}\right)'(n)=\log\left(\li^{-1}(n)\right)= \log(n\log(n))+O\left(\frac{\log(\log(n))}{\log(n)}\right)$ and $\left(\li^{-1}\right)''(n)=\frac{1}{\li^{-1}(n)}\log\left(\li^{-1}(n)\right)\sim \frac1n$ and find the wanted approximation.

Edit: In the same way, a third term in $F$ can be found and we obtain an improvement of the first statement of the theorem. With $F(x)=\li(x)+\frac12\log(\log(x))-\frac1{12x}- \frac1{12}\li(\frac1x)$, we find $F(x+\log(x))-F(x)=1+O\left(\frac{\log (x)}{x^3}\right)$. As above, this yields $$\li(x_n)+\frac12\log(\log(x_n))-\frac1{12x_n}- \frac1{12}\li\left(\frac1{x_n}\right)=n+C+O\left(\frac1{n^2\log(n)^2}\right).$$

Helmut
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