Approximation for the $n$th nontrivial zero of $\zeta(s)$

Question

For all positive integers $n$, let $$t_n = \frac{1}{2\pi} \operatorname{Im} \rho_n,$$ where $\rho_n$ donates the $n$th nontrivial zero of the Riemann zeta function in the upper half plane (listed in increasing order of imaginary part, counted to multiplicity, and including potential zeros off of the critical line). Due to the Riemann-Von Mangoldt explicit formula for $N(T)$, one has the good approximation $$t_n \sim \frac{n-11/8}{W((n-11/8)/e)} \ (n \to \infty)$$ of $t_n$, where $W$ is the Lambert $W$ function. (The $11/8$ comes as the average of $7/8$ and $7/8+1$.) I discovered this approximation recently, but haven't found it in the literature anywhere. My question is, what is a good $O$ bound for the error $t_n-\frac{n-11/8}{W((n-11/8)/e)}$? I suspect that one ought to be able to prove that $$t_n-\frac{n-11/8}{W((n-11/8)/e)} = O_t((\log n)^t) \ (n \to \infty)$$ for all $t> 1$, and in fact it might even be true that $$t_n-\frac{n-11/8}{W((n-11/8)/e)} = O_t((\log n)^t) \ (n \to \infty)$$ for all $t>-1/2$. I am curious what you think is the infimum of all $t \in\mathbb{R}$ such that the $O$ estimate above holds. It should definitely be the case at least that $$t_n-\frac{n-11/8}{W((n-11/8)/e)} = O_t(n^t) \ (n \to \infty)$$ for all $t>0$, but I don't know how to prove this. Probably also the error is $O(1)$, and maybe even $o(1)$. I'm not great at series inversion, but my guess is that it should be verifiable from the Riemann-Von Mangoldt formula and some series inversion.

Note that, for any fixed $a \in \mathbb{R}$, the inverse of the function $T\log T-T+a$ is $\frac{n-a}{W((n-a)/e)}$, and I believe that the Riemann-Von Mangoldt explicit formula for $N(T)$ implies that $a = 11/8$ is optimal for approximating $t_n$. By contrast, $a = 7/8$ is optimal for approximating the average of $t_{n+1}$ and $t_n$. One can also use this method to well-approximate the $n$th gap $t_{n+1}-t_n$ as $$t_{n+1}-t_n \approx \frac{n-11/8+1}{W((n-11/8+1)/e)}-\frac{n-11/8}{W((n-11/8)/e)} \approx \frac{1}{1+{W((n-7/8)/e)}}.$$

In analogy with the prime counting function, I would say that $\operatorname{Ri}(x)$ and $\frac{x}{\log x}$ are to $\pi(x)$ as $\frac{n-11/8}{W((n-11/8)/e)}$ and $\frac{n}{\log n}$ are to $t_n$, where $\operatorname{Ri}(x)$ is Riemann's approximation to $\pi(x)$. Please correct me if you think I'm wrong, or point out where someone might have observed this before! I think the idea of Gram spacing is related, but he didn't quite say the same thing as I'm saying here.

Here is a related question I asked: Riemann-Von Mangoldt formula, revised question. The reason I asked that question is because I wanted to understand the error better in the approximation $N(T) \approx 1+\frac{1}{\pi}\theta(T)$, where $\theta$ is the Riemann-Siegel theta function, which is the analogue of Riemann's approximation $\pi(x) \approx \operatorname{Ri}(x)$.

An alternative approach is to utilize the inverse of the Riemann-Siegel theta function, but that I don't know how to compute. At least the Lambert $W$ function is easily computable using Mathematica. Also the error in the approximation $1+\frac{1}{\pi}\theta(2\pi T) \approx T\log T-T+\frac{7}{8}$ is $O(T^{-1})$, so the inverses are close enough that it doesn't really matter too much which you use.

This is current research of mine, and I am interested in getting some informed opinions about it. Many thanks!

Answer 1

I think I might now have an answer to my question, in that the approximation I gave should be within $O(1)$ of $t_n$, and it can be related to the function $N(T)$. However, I am not yet sure in what sense if any $11/8$ is optimal.

Let $\tau_n = 2\pi t_n$ denote the imaginary part of the $n$th zero in the upper half plane, and let $$r_n = 2\pi \frac{n-11/8}{W((n-11/8)/e)}$$ for all $n$. One then has \begin{align*} n-1-\frac{1}{\pi}\theta(r_n) & = n- \frac{r_n}{2\pi} \log \frac{r_n}{2\pi}-\frac{r_n}{2\pi}-\frac{7}{8} + O\left( \frac{1}{r_n} \right)\\ & = \frac{1}{2}+ O\left(\frac{\log n}{n} \right) \ (n \to \infty), \end{align*} while also $$S(\tau_n) = -\frac{1}{2}+n-1-\frac{1}{\pi}\theta(\tau_n) = O(\log \tau_n) = O(\log n) \ (n \to \infty),$$ and therefore $$\theta(\tau_n)-\theta(r_n) = O(\log n) \ (n \to \infty).$$ Since $2\theta(t)$ is approximated by $t\log t-t+\log 2\pi -\frac{\pi}{4}$ to within $O(T^{-1})$, where the latter function has derivative $\log t$, by the mean value theorem it follows that \begin{align*} (\tau_n-r_n)\log u_n & = (\tau_n \log \tau_n - \tau_n)-(r_n \log r_n -r_n) \\ & = O(\log n) \ (n \to \infty) \end{align*} for some $u_n$ lying between $r_n$ and $\tau_n$ and therefore $$\tau_n-r_n = O(1) \ (n \to\infty).$$ More generally, the argument above shows that $$\tau_n-r_n = O\left(\frac{S(\tau_n)}{\log n}\right) \ (n \to\infty),$$ so that various conjectures bounding $S(T)$ yield bounds on the approximation of $\tau_n$. I suspect the converse is true as well.

I still think that $a = 11/8$ is optimal, but the argument above does not prove that. Note that if $$s_n = 2\pi \frac{n-11/8}{W((n-11/8)/e)},$$ then $$n-1-\frac{1}{\pi}\theta(s_n) = \frac{1}{2}+O\left(\frac{\log n}{n} \right) \ (n \to \infty),$$ so the same argument above implies that $\tau_n-s_n = O(1)$. I still think that $\tau_n-r_n = O((\log n)^t)$ for some $t < 0$, but I still don't know how to prove that. A weaker conjecture would be that $\tau_n-s_n = o(1)$.

ADDITIONAL EDIT: I now know at least one sense in which $a = \frac{11}{8}$ is optimal. I proved the following.

Proposition. Let $a \in \mathbb{R}$, and let $$r_n = r_n(a) = 2\pi \frac{n-a}{W((n-a)/e)}.$$ Then one has \begin{align*} r_n-\tau_n = 2\pi\frac{\frac{11}{8}-a +S(\tau_n)}{\log r_n} + O \left( \frac{1}{r_n \log r_n}\right) \ (n \to \infty). \end{align*} Consequently, one has $$ \frac{n-a}{W((n-a)/e)} = \frac{\tau_n}{2\pi} + O\left(\frac{S(\tau_n)}{\log \tau_n} \right) \ (n \to \infty)$$ and $$ \frac{n-a}{W((n-a)/e)} = \frac{\tau_n}{2\pi} + O(1) \ (n \to \infty).$$ Moreover, for $a = \frac{11}{8}$, one has \begin{align*} r_n-\tau_n = 2\pi\frac{S(\tau_n)}{\log r_n} + O \left( \frac{1}{r_n \log r_n}\right) \ (n \to \infty). \end{align*}

It follows from the proposition above and known $\Omega$ results on $S(T)$ that $2\pi\frac{S(\tau_n)}{\log r_n}$ is the true order of growth of $r_n-\tau_n$ when $a = \frac{11}{8}$. Another consequence of the proposition is that $r_n-\tau_n = o(1)$ if and only if $S(T) = o(\log T)$.