The value of $\lim\frac{a_{k+1}}{a_k}$ if $a_k=\min\{n\in \mathbb{N}\,|\,n=k\pi(n)\}$ Solomon W. Golomb proved that the function $\frac{n}{\pi(n)}$ takes on every integer value
greater than 1 [The American Mathematical Monthly, Vol. 69, No. 1 (Jan., 1962), pp. 36-37].
Let's consider the sequence (OEIS A038625)
$$a_k=\min\{n\in \mathbb{N}\space|\space n=k\pi(n)\}, \space k\gt1$$
The first terms are
$$2, 27, 96, 330, 1008, 3059, 8408, 23526, 64540, 175197, 480852, 1304498, 3523884, 9557955,...$$
Is it possible to prove that
$$\lim_{k\rightarrow\infty}\frac{a_{k+1}}{a_k}=e\;?$$
 A: OEIS suggests

It appears that $a_k$ is asymptotic to $e^2 \cdot e^k$

However, I can't find a proof of the "it appears" part. Let's try using Vallée Poussin implication of the PNT proof instead.

For $\forall\varepsilon >0$ and for large enough $k$'s we should have
$$\frac{a_k}{\log{a_k}-(1-\varepsilon)}<\pi(a_k)<\frac{a_k}{\log{a_k}-(1+\varepsilon)}\Rightarrow \\
\frac{k a_k}{\log{a_k}-(1-\varepsilon)}<k\pi(a_k)<\frac{ka_k}{\log{a_k}-(1+\varepsilon)}\Rightarrow \\
\frac{k a_k}{\log{a_k}-(1-\varepsilon)}<a_k<\frac{ka_k}{\log{a_k}-(1+\varepsilon)}\Rightarrow \\
\frac{k}{\log{a_k}-(1-\varepsilon)}<1<\frac{k}{\log{a_k}-(1+\varepsilon)}\Rightarrow\\
\log{a_k}-(1+\varepsilon)<k<\log{a_k}-(1-\varepsilon)$$
or
$$\log{a_k}-\varepsilon<k+1<\log{a_k}+\varepsilon\tag{1}$$
Now
$$\left\{\begin{matrix}
\log{a_{k+1}}-\varepsilon<k+2<\log{a_{k+1}}+\varepsilon\\ 
-\log{a_k}-\varepsilon<-(k+1)<-\log{a_k}+\varepsilon
\end{matrix}\right.$$
Altogether
$$\log{\frac{a_{k+1}}{a_k}}-2\varepsilon<1<\log{\frac{a_{k+1}}{a_k}}+2\varepsilon\Rightarrow\\
\left|\log{\frac{a_{k+1}}{a_k}}-1\right| <2\varepsilon\tag{2}$$

Of course all of the above works if $(a_k)_{k\in\mathbb{N}}$ is a subsequence of $(k)_{k\in\mathbb{N}}$. (now I am using it) It appears that $a_k < a_{k+1}$.
It is clear though that

Pr1. For $\forall i\ne j \Rightarrow a_i\ne a_j$

Proof by contradiction. Let's assume $\exists i_0<j_0$ s.t. $\color{blue}{a_{i_0}=a_{j_0}}$, then $\pi(a_{i_0})=\pi(a_{j_0})$ and
$$\color{red}{a_{i_0}}=i_0 \pi(a_{i_0})=i_0 \pi(a_{j_0})\color{red}{<}
j_0 \pi(a_{j_0})=\color{red}{a_{j_0}}$$
contradiction.

Update (2022-04-22). To the question of $a_k < a_{k+1}$ ...

Pr2. For $\forall x,y$ s.t. $60184\le x< y$ we have
$$\frac{x}{\pi(x)} < \frac{y}{\pi(y)} + 1$$

The same Vallée Poussin article has references to
$$\frac{x}{\log{x}-1} <\pi(x) < \frac{x}{\log{x} - 1.1}, \forall x\ge 60184 \tag{3}$$
Now, let's assume the contrary, $\exists (60184\le x_0< y_0)$ s.t.
$$\frac{x_0}{\pi(x_0)} \ge \frac{y_0}{\pi(y_0)} + 1 \overset{(3)}{\Rightarrow}\\
\log{x_0}-1>\frac{x_0}{\pi(x_0)} \ge \frac{y_0}{\pi(y_0)} + 1 > \log{y_0}-1.1 + 1$$
or
$$\log{x_0}-1>\log{y_0}-1.1 + 1 \Rightarrow \\
0=\log{1} > \log{\frac{x_0}{y_0}}>0.9$$
which is a contradiction.

Remark. Generally $x<y \Rightarrow \frac{x}{\pi(x)} <\frac{y}{\pi(y)}$is not true! A counterexample is $x=p_{n}-1$ and $y=p_n$ ($p_n$ is the $n$-th prime). Then, obviously $x<y$ and
$$\frac{p_n-1}{\pi(p_n-1)}>\frac{p_n}{\pi(p_n)} \iff
\frac{p_n-1}{n-1}>\frac{p_n}{n} \iff
(p_n-1)n>p_n (n-1) \iff\\
p_n>n$$
which, the latter, is always true. These are the "falls" (discontinuities) of the function $f(x)=\frac{x}{\pi(x)}$ at the prime values, after continuous ascending in between prime values.

Back to $a_k < a_{k+1}$ part. From the definition of the sequence we have
$$\frac{a_{k+1}}{\pi(a_{k+1})}=k+1=\frac{a_k}{\pi(a_k)}+1$$
i.e. $\frac{a_{k+1}}{\pi(a_{k+1})}$ "jumps" over $\frac{a_{k}}{\pi(a_{k})}$ by $1$. According to Pr2, for values $\ge 60184$ this can only happen for $x>a_k$, i.e. on the RHS of $a_k$. Thus, $a_{k+1}> a_k\ge 60184$ and we are done with this part. The values $< 60184$ were already listed in the OP's question and are ascending too.

Summary. From $(2)$, Pr2 and previous paragraph concluding the ascendance of $(a_k)_{k\in\mathbb{N}}$, we conclude that
$$\lim\limits_{k\to\infty}\frac{a_{k+1}}{a_k}=e$$
