---
abstract: |
    Let $n,k\in\mathbb{N}$ and let $p_{n}$ denote the $n$th prime number. We define $p_{n}^{(k)}$ recursively as $p_{n}^{(1)}:=p_{n}$ and $p_{n}^{(k)}=p_{p_{n}^{(k-1)}}$, that is, $p_{n}^{(k)}$ is the $p_{n}^{(k-1)}$th prime.

    In this note we give answers to some questions and prove a conjecture posed by Miska and Tóth in their recent paper concerning subsequences of the sequence of prime numbers. In particular, we establish explicit upper and lower bounds for $p_{n}^{(k)}$. We also study the behaviour of the counting functions of the sequences $(p_{n}^{(k)})_{k=1}^{\infty}$ and $(p_{k}^{(k)})_{k=1}^{\infty}$.
author:
- 'B[ł]{}ażej Żmija'
title: A note on primes with prime indices
---

[^1]

Introduction
============

Let $(p_{n})_{n=1}^{\infty}$ be the sequence of consecutive prime numbers. In a recent paper [@MT] Miska and Tóth introduced the following subsequences of the sequence of prime numbers: $p_{n}^{(1)}:=p_{n}$ and for $k\geq 2$ $$\begin{aligned}
    p_{n}^{(k)}:=p_{p_{n}^{(k-1)}}.\end{aligned}$$ In other words, $p_{n}^{(k)}$ is the $p_{n}^{(k-1)}$th prime. They also defined $$\begin{aligned}
    {{\rm Diag}\mathbb{P}}:= & \{\ p_{k}^{(k)}\ |\ k\in\mathbb{N}\ \}, \\
    \mathbb{P}_{n}^{T}:= & \{\ p_{n}^{(k)}\ |\ k\in\mathbb{N}\ \}\end{aligned}$$ for each positive integer $n$.

The main motivation in [@MT] was the known result that the set of prime numbers is ($R$)-dense, that is, the set $\{\ \frac{p}{q}\ |\ p,q\in\mathbb{P}\ \}$ is dense in $\mathbb{R}_{+}$ (with respect to the natural topology on $\mathbb{R}_{+}$). It was proved in [@MT] that for each $k\in\mathbb{N}$ the sequence $\mathbb{P}_{k}:=(p_{n}^{(k)})_{n=1}^{\infty}$ is ($R$)-dense. This result might be surprising, because the sequences $\mathbb{P}_{k}$ are very sparse. In fact, for each $k$ set $\mathbb{P}_{k+1}$ is a zero asymptotic density subset of $\mathbb{P}_{k}$. On the other hand, it was showed, that the sequences $(p_{n}^{(k)})_{k=1}^{\infty}$ for each fixed $n\in\mathbb{N}$, and $(p_{k}^{(k)})_{k=1}^{\infty}$ are not ($R$)-dense.

Results of another type that were proved in [@MT] concern the asymptotic behaviour of $p_{n}^{(k)}$ as $n\rightarrow\infty$, or as $k\rightarrow\infty$. In particular, as $n\rightarrow\infty$, we have for each $k\in\mathbb{N}$ $$\begin{aligned}
    p_{n}^{(k)}\sim n(\log n)^{k},\ \ \ \ p_{n+1}^{(k)}\sim p_{n}^{(k)}, \ \ \ \ \log p_{n}^{(k)}\sim \log n\end{aligned}$$ by [@MT Theorem 1]. Some results from [@MT] concerning $p_{n}^{(k)}$ as $k\rightarrow\infty$ are mentioned later.

For a set $A\subseteq \mathbb{N}$ let $A(x)$ be its counting function, that is, $$\begin{aligned}
    A(x):=\# \left(A\cap [1,x]\right).\end{aligned}$$ Miska and Tóth posed four questions concerning the numbers $p_{n}^{(k)}$:

1.  Is it true that $p_{k+1}^{(k)}\sim p_{k}^{(k)}$ as $k\rightarrow\infty$?

2.  Are there real constants $c>0$ and $\beta$ such that $$\begin{aligned}
            \exp\mathbb{P}_{n}^{T}(x)\sim cx(\log x)^{\beta}
        \end{aligned}$$ for each $n\in\mathbb{N}$?

3.  Are there real constants $c>0$ and $\beta$ such that $$\begin{aligned}
            \exp{{\rm Diag}\mathbb{P}}(x)\sim cx(\log x)^{\beta}?
        \end{aligned}$$

4.  Is it true that $$\begin{aligned}
            {{\rm Diag}\mathbb{P}}(x)\sim\mathbb{P}_{n}^{T}(x)
        \end{aligned}$$ for each $n\in\mathbb{N}$?

The aim of this paper it to give answers to question B, C and D.

The main ingredients of our proofs are the following inequalities: $$\begin{aligned}
\label{ineqPrimes}
    n\log n<p_{n}<2n\log n.\end{aligned}$$ The first inequality holds for all $n\geq 2$, and the second one for all $n\geq 3$. For the proofs, see [@RS]. In Section \[Results\] we use (\[ineqPrimes\]) in order to show explicit bounds for $p_{n}^{(k)}$. In particular, for all $n>e^{4200}$ we have: $$\begin{aligned}
    \log p_{n}^{(k)}= & k(\log k+\log\log k+O_{n}(1)), \\
    \log p_{k}^{(k)}= & k(\log k+\log\log k+O(\log\log\log k)),\end{aligned}$$ as $k\rightarrow\infty$, where the implied constant in the first line may depend on $n$, see Theorem \[MAIN\] below. In consequence, we improve the (in)equalities $$\begin{aligned}
    \lim_{k\rightarrow\infty}\frac{p_{n}^{(k)}}{k\log k}= & 1, \\
    1\leq \liminf_{k\rightarrow\infty}\frac{p_{k}^{(k)}}{k\log k}\leq &  \limsup_{k\rightarrow\infty}\frac{p_{k}^{(k)}}{k\log k}\leq 2.\end{aligned}$$ that appeared in [@MT]. Then we show in Section \[Coro\] that the answers to questions B and C are negative (Corollary \[CoroBC\]), while the one for question D is affirmative (Theorem \[AsympEqThm\]). In fact, we find the following relation: $$\begin{aligned}
    \mathbb{P}_{n}^{T}(x)\sim{{\rm Diag}\mathbb{P}}(x)\sim\frac{\log x}{\log\log x}\end{aligned}$$ for all positive integers $n$.

In their paper, Miska and Tóth also posed a conjecture, that we state here as a proposition, since it is in fact a consequence of a result that had already appeared in [@MT].

Let $n\in\mathbb{N}$ be fixed. Then $$\begin{aligned}
    \frac{p_{n}^{(k)}}{p_{k}^{(k)}}\longrightarrow 0\end{aligned}$$ as $k\longrightarrow\infty$.

Let $k> p_{n}$. Then $$\begin{aligned}
    0\leq \frac{p_{n}^{(k)}}{p_{k}^{(k)}}<\frac{p_{n}^{(k)}}{p_{p_{n}}^{(k)}}=\frac{p_{n}^{(k)}}{p_{n}^{(k+1)}}.\end{aligned}$$ The expression on the right goes to zero as $k$ goes to infinity, as was proved in [@MT Corollary 3].

It is worth to note, that primes with prime indices have already appeared in the literature, for example in [@BB] and [@BKO]. However, according to our best knowledge, our paper is the second one (after [@MT]), where the number of iterations of indices, that is, the number $k$ in $p_{n}^{(k)}$, is not fixed.

Throughout the paper we use the following notation: $\log x$ denotes the natural logarithm of $x$, and for functions $f$ and $g$ we write $f\sim g$ if $\lim_{x\rightarrow\infty}\frac{f(x)}{g(x)}=1$, $f=O(g)$ if there exists a positive constants $c$ such that $f(x)<cg(x)$, and $f=o(g)$ if $\lim_{x\rightarrow\infty}\frac{f(x)}{g(x)}=0$.

Upper and lower bounds for $p_{n}^{(k)}$ {#Results}
========================================

In this Section, we find explicit upper and lower bounds for $p_{n}^{(k)}$. We start with the upper bound.

\[lemUP\] Let $n\geq 9$. Then for each $k\in\mathbb{N}$ we have: $$\begin{aligned}
    p_{n}^{(k)}<2^{2k-1}\cdot n\cdot (k-1)!\cdot\big(\log(\max\{k,n\})\big)^{k}. \end{aligned}$$ In particular, $$\begin{aligned}
    p_{n}^{(k)}< \big(4\cdot k\log k\big)^{k}\end{aligned}$$ for $k\geq n$.

We proceed by induction on $k$. For $k=1$ it is a simple consequence of (\[ineqPrimes\]). Then the second induction step goes as follows: let us denote $m:=\max\{n,k\}$. Observe that $(k-1)!<(k-1)^{k-1}<m^{k-1}$ and $4\log m<m$ for $m\geq 9$. Hence, $$\begin{aligned}
    p_{n}^{(k+1)}\leq & 2p_{n}^{k}\log p_{n}^{(k)}<2\cdot 2^{2k-1}\cdot n\cdot (k-1)!\cdot(\log m)^{k}\cdot \log\left[2^{2k-1}\cdot n\cdot (k-1)!\cdot (\log m)^{k}\right] \\
    < & 2^{2k}\cdot n\cdot (k-1)!\cdot(\log m)^{k} \log\left[4^{k}\cdot m\cdot m^{k-1}\cdot (\log m)^{k}\right]=2^{2k}\cdot n\cdot (k-1)!\cdot(\log m)^{k} \log\left[4\cdot m\cdot \log m\right]^{k} \\
    \leq & 2^{2k}\cdot n\cdot k!\cdot (\log m)^{k}\cdot \log[m]^{2}\leq 2^{2k+1}\cdot n\cdot k!\cdot (\log m)^{k+1}.\end{aligned}$$

The second part of the statement is an easy consequence of the first part and the inequalities $(k-1)!<k^{k-1}$ and $n\leq k$.

In order to prove a lower bound for $p_{n}^{(k)}$ we will need the following fact.

\[lemL\] Let $$\begin{aligned}
    L(x):=\left(\frac{x}{x+1}\right)^{x+1}\left(\frac{\log x}{\log (x+1)}\right)^{x+1}.\end{aligned}$$ Then we have $$\begin{aligned}
    L(x)>0.32627\end{aligned}$$ for all $x\geq 4200$.

Observe, that the function $\left(\frac{x}{x+1}\right)^{x+1}$ is increasing. Indeed, if $$\begin{aligned}
    f(x):=\log\left(\frac{x}{x+1}\right)^{x+1}=(x+1)\left(\log x-\log (x+1)\right),\end{aligned}$$ then $$\begin{aligned}
    f'(x)=\log x-\log (x+1)+(x+1)\left(\frac{1}{x}-\frac{1}{x+1}\right)=\frac{1}{x}-\log \left(1+\frac{1}{x}\right)>0,\end{aligned}$$ where the last inequality follows from the well-known inequality $y>\log (1+y)$ used with $y=\frac{1}{x}$. Hence, we can bound $$\begin{aligned}
\label{L1}
    \left(\frac{x}{x+1}\right)^{x+1}\geq \left(\frac{4200}{4201}\right)^{4201}\end{aligned}$$ for all $x\geq 4200$.

Now we need to find a lower bound for $\left(\frac{\log x}{\log (x+1)}\right)^{x+1}$. Let us write $$\begin{aligned}
    \left(\frac{\log x}{\log (x+1)}\right)^{x+1}=\left[\left(1-\frac{1}{\frac{\log (x+1)}{\log (x+1)-\log x}}\right)^{\frac{\log (x+1)}{\log (x+1)-\log x}}\right]^{\log\left(1+\frac{1}{x}\right)^{x}\cdot\frac{1}{\log (x+1)}\cdot\left(1+\frac{1}{x}\right)}.\end{aligned}$$ At first, we prove that functions $g(t):=\left(1-\frac{1}{t}\right)^{t}$ and $h(x):=\frac{\log (x+1)}{\log (x+1)-\log x}$ are increasing. For the function $g(t)$ it is enough to observe, that $\log g(t)=f(t-1)$ and the function $f(x)$ is increasing. For the function $h(x)$ we have: $$\begin{aligned}
    h'(x)= & \frac{1}{(\log (x+1)-\log x)^{2}}\left[\frac{\log (x+1)-\log x}{x+1}-\log (x+1)\left(\frac{1}{x+1}-\frac{1}{x}\right)\right] \\
    = & \frac{1}{(\log (x+1)-\log x)^{2}}\left[\frac{\log (x+1)}{x}-\frac{\log x}{x+1}\right] >0.\end{aligned}$$

The fact that the functions $g(t)$ and $h(x)$ are increasing, together with the properties $g(h(4200))\in (0,1)$ and $\log\left(1+\frac{1}{x}\right)^{x}<1$, give us $$\label{L2}
    \begin{aligned}
        \left(\frac{\log x}{\log (x+1)}\right)^{x+1}\geq & \big[g(h(4200))\big]^{\log\left(1+\frac{1}{x}\right)^{x}\cdot\frac{1}{\log (x+1)}\cdot\left(1+\frac{1}{x}\right)} 
        > \big[g(h(4200))\big]^{\frac{1}{\log (x+1)}\cdot\left(1+\frac{1}{x}\right)} \\
        \geq & \big[g(h(4200))\big]^{\frac{1}{\log (4200+1)}\cdot\left(1+\frac{1}{4200}\right)}=\left(\frac{\log 4200}{\log 4201}\right)^{\frac{4201}{4200}\cdot\frac{1}{\log 4201 -\log 4200}}
    \end{aligned}$$ for all $x\geq 4200$.

Combining (\[L1\]) and (\[L2\]) we get the inequality $$\begin{aligned}
    L(x)\geq \left(\frac{4200}{4201}\right)^{4201}\left(\frac{\log 4200}{\log 4201}\right)^{\frac{4201}{4200}\cdot\frac{1}{\log 4201 -\log 4200}}\approx 0.3262768>0.32627.\end{aligned}$$ The proof is finished.

In the next lemma we provide a lower bound for $p_{n}^{(k)}$.

\[lemDOWN\] If $n>e^{4200}$, then for all $k\geq\lfloor\log n\rfloor$ we have $$\begin{aligned}
    p_{n}^{(k)}>\left(\frac{e\cdot k\log k}{\log\log n}\right)^{k}.\end{aligned}$$

First, let us observe that a simple induction argument on $k$ implies the inequality $$\begin{aligned}
\label{ineqDOWN}
    p_{n}^{(k)}>n(\log n)^{k}.\end{aligned}$$ Indeed, for $k=1$ this follows from left inequality in (\[ineqPrimes\]). Using the same inequality we get also $$\begin{aligned}
p_{n}^{(k+1)}>p_{n}^{(k)}\log p_{n}^{(k)}>n (\log n)^{k}\log(n (\log n)^{k})>n(\log n)^{k+1},\end{aligned}$$ and hence (\[ineqDOWN\]).

Now we show that the inequality from the statement is true for $k=\lfloor \log n\rfloor$. Because of (\[ineqDOWN\]) it is enough to show: $$\begin{aligned}
    n(\log n)^{k}>\left(\frac{e\cdot k\log k}{\log\log n}\right)^{k},\end{aligned}$$ or equivalently, after taking logarithms we get $$\begin{aligned}
    \log n+\lfloor\log n\rfloor\log\log n>\lfloor\log n\rfloor+\lfloor\log n\rfloor\log\lfloor\log n\rfloor+\lfloor\log n\rfloor\log\log\lfloor\log n\rfloor -\lfloor\log n\rfloor\log\log\log n.\end{aligned}$$ This is equivalent to the inequality $$\begin{aligned}
    \big(\log n-\lfloor\log n\rfloor\big)+\lfloor\log n\rfloor\big(\log\log n-\log\lfloor\log n\rfloor\big)+\lfloor\log n\rfloor\big(\log\log\log n-\log\log\lfloor\log n\rfloor\big)>0,\end{aligned}$$ which is obviously true.

In order to finish the proof, we again use the induction argument. The inequality from the statement of our lemma is true for $k=\lfloor \log n\rfloor$. Assume it holds for some $k\geq \lfloor\log n\rfloor$. Then by (\[ineqPrimes\]) and the induction hypothesis we get $$\begin{aligned}
    p_{n}^{(k+1)}> & p_{n}^{(k)}\log p_{n}^{(k)}>\left(\frac{e\cdot k\log k}{\log\log n}\right)^{k}\log \left(\frac{e\cdot k\log k}{\log\log n}\right)^{k}.\end{aligned}$$ It is enough to show that for all $n>e^{4200}$ and all $k\geq\lfloor\log n\rfloor$ we have $$\begin{aligned}
    \left(\frac{e\cdot k\log k}{\log\log n}\right)^{k}\log \left(\frac{e\cdot k\log k}{\log\log n}\right)^{k}>\left(\frac{e\cdot (k+1)\log (k+1)}{\log\log n}\right)^{k+1}.\end{aligned}$$ This is equivalent to $$\begin{aligned}
    k^{k+1}(\log k)^{k}\left[\log k+\log\left(\frac{e\log k}{\log\log n}\right)\right]>\frac{e}{\log\log n}(k+1)^{k+1}(\log(k+1))^{k+1}.\end{aligned}$$ Recall, that we assume that $k\geq \lfloor\log n\rfloor$. Thus $k^{e}>\log n$, that is, $e\log k>\log\log n$. Therefore, it is enough to show the following inequalities: $$\begin{aligned}
    k^{k+1}(\log k)^{k+1}>\frac{e}{\log\log n}(k+1)^{k+1}(\log(k+1))^{k+1},\end{aligned}$$ or equivalently $$\begin{aligned}
    \left(\frac{k}{k+1}\right)^{k+1}\left(\frac{\log k}{\log (k+1)}\right)^{k+1}>\frac{e}{\log\log n}.\end{aligned}$$ Notice that the left-hand side expression of the last inequality is equal to $L(k)$, where the function $L(x)$ is defined in the statement of Lemma \[lemL\]. If $n>e^{4200}$, then $k\geq \lfloor\log n\rfloor \geq 4200$, and Lemma \[lemL\] implies $L(k)>0.32627$. Therefore, if $N:=\max\left\{\lfloor e^{4200}\rfloor,\lceil e^{e^{e/0.32627}}\rceil\right\}=\lfloor e^{4200}\rfloor$, then for all $n>N$ and $k\geq \lfloor\log n\rfloor$ we have: $$\begin{aligned}
    L(k)>0.32627\geq \frac{e}{\log\log N}>\frac{e}{\log\log n}.\end{aligned}$$ This finishes the proof.

Main results {#Coro}
============

We begin this section by a theorem that provides good information about asymptotic growth of $\log p_{n}^{(k)}$ for large fixed $n$, and for $\log p_{k}^{(k)}$ as $k\rightarrow\infty$.

\[MAIN\]

1.  Let $n>e^{4200}$. Then $$\begin{aligned}
            \log p_{n}^{(k)}=k(\log k+\log\log k+O_{n}(1))
        \end{aligned}$$ as $k\rightarrow\infty$, where the implied constant may depend on $n$.

2.  We have $$\begin{aligned}
            \log p_{k}^{(k)}=k(\log k+\log\log k+O(\log\log\log k))
        \end{aligned}$$ as $k\rightarrow\infty$.

If $n>e^{4200}$ and $k\geq n$, Lemmas \[lemUP\] and \[lemDOWN\] give us: $$\begin{aligned}
    \left(\frac{e\cdot k\log k}{\log\log n}\right)^{k}<p_{n}^{(k)}<\left(4\cdot k\log k\right)^{k}.\end{aligned}$$ After taking logarithms, we simply get the first part of our theorem. In order to get the second part, we need to put $n=k$ and repeat the reasoning.

Now we give the answer to Question D.

\[AsympEqThm\] For each $n\in\mathbb{N}$ we have $$\begin{aligned}
    {{\rm Diag}\mathbb{P}}(x)\sim \mathbb{P}_{n}^{T} (x)\end{aligned}$$ as $x\rightarrow\infty$.

From [@MT Theorem 6] we know that $$\begin{aligned}
    \mathbb{P}_{m}^{T}(x)\sim \mathbb{P}_{n}^{T}(x)\end{aligned}$$ for each $m,n\in\mathbb{N}$. Therefore, it is enough to prove ${{\rm Diag}\mathbb{P}}(x)\sim \mathbb{P}_{n}^{T} (x)$ for some sufficiently large $n$.

Let $n=\lfloor e^{4200}\rfloor +100$. We use the idea from the proof of [@MT Theorem 17]. Let $x$ be a large real number. Let $k$ be such that $p_{k}^{(k)}\leq x<p_{k+1}^{(k+1)}$. Then ${{\rm Diag}\mathbb{P}}(x)=k$. By [@MT Theorem 8] and Theorem \[MAIN\] above we have $$\begin{aligned}
    \frac{\mathbb{P}_{n}^{T}(x)}{{{\rm Diag}\mathbb{P}}(x)}\leq & 1+\frac{\log p_{k+1}^{(k+1)}}{k\log\log p_{n}^{(k)}}-\frac{\log p_{n}^{(k)}}{k\log\log p_{n}^{(k)}} \\
    = & 1+\frac{(1+o(1))(k+1)\log (k+1)}{k\log \big[(1+o(1))k\log k\big]}-\frac{(1+o(1))k\log k}{k\log\big[(1+o(1))k\log k\big]} \\
    = & 1+(1+o(1))\left(1+\frac{1}{k}\right)\frac{\log (k+1)}{\log k+\log\left[(1+o(1))\log k\right]}-(1+o(1))\frac{\log k}{\log k+\log\left[(1+o(1))\log k\right]}.\end{aligned}$$ The whole last expression goes to $1$ as $k$ goes to infinity. On the other hand, ${{\rm Diag}\mathbb{P}}(x)\leq \mathbb{P}_{n}^{T}(x)$ for $x\geq p_{n}^{(n)}$ and we get the result.

The answers to Questions B and C will follow from our next result, which is of independent interest.

\[AsympDiagP\]

1.  Let $n\in\mathbb{N}$. Then $$\begin{aligned}
            \mathbb{P}_{n}^{T}(x)\sim\frac{\log x}{\log\log x}.
        \end{aligned}$$

2.  We have $$\begin{aligned}
            {{\rm Diag}\mathbb{P}}(x)\sim \frac{\log x}{\log\log x}.
        \end{aligned}$$

In view of Theorem \[AsympEqThm\], it is enough to show the statement for the function ${{\rm Diag}\mathbb{P}}(x)$. Let us fix an arbitrarily small number $\varepsilon >0$ and take a sufficiently large real number $x$ and find $k$ such that $p_{k}^{(k)}\leq x<p_{k+1}^{(k+1)}$. Then ${{\rm Diag}\mathbb{P}}(x)=k$ and by Lemmas \[lemUP\] and \[lemDOWN\] we have $$\begin{aligned}
    k^{k}<x<k^{(1+\varepsilon )k}.\end{aligned}$$ Let us write $x=e^{y}$. Then $$\begin{aligned}
    k\log k< y<(1+\varepsilon )k\log k.\end{aligned}$$ If $y$ is sufficiently large, this implies $$\begin{aligned}
\label{ineqyk}
    (1-\varepsilon )\frac{y}{\log y}< k<(1+\varepsilon )\frac{y}{\log y}.\end{aligned}$$ Indeed, if $k\leq (1-\varepsilon )\frac{y}{\log y}$, then $$\begin{aligned}
    y<(1+\varepsilon)k\log k\leq (1-\varepsilon^{2})\frac{y}{\log y}\log\left((1-\varepsilon )\frac{y}{\log y}\right)<(1-\varepsilon^{2})\left(1-\frac{\log\log y}{\log y}\right)y,\end{aligned}$$ which is impossible. Similarly, if $k\geq (1+\varepsilon )\frac{y}{\log y}$, then $$\begin{aligned}
    y>k\log k\geq (1+\varepsilon )\frac{y}{\log y}\log\left((1+\varepsilon )\frac{y}{\log y}\right)>(1+\varepsilon )\left(1-\frac{\log\log y}{\log y}\right)y.\end{aligned}$$ The above inequality cannot hold if $y$ is sufficiently large.

If we go back to $k={{\rm Diag}\mathbb{P}}(x)$ and $y=\log x$ in (\[ineqyk\]), we get $$\begin{aligned}
    (1-\varepsilon )\frac{\log x}{\log\log x}<{{\rm Diag}\mathbb{P}}(x)<(1+\varepsilon )\frac{\log x}{\log\log x}.\end{aligned}$$ The number $\varepsilon >0$ was arbitrary, so the result follows.

\[CoroBC\] There do not exist constants $c>0$ and $\beta$ such that $$\begin{aligned}
    \exp\mathbb{P}_{n}^{T}(x)\sim cx(\log x)^{\beta}\end{aligned}$$ for some $n$, or $$\begin{aligned}
    \exp{{\rm Diag}\mathbb{P}}(x)\sim cx(\log x)^{\beta}.\end{aligned}$$

We prove the result only for ${{\rm Diag}\mathbb{P}}(x)$. The case of $\mathbb{P}_{n}^{T}(x)$ is analogous.

Assume to the contrary, that $\exp{{\rm Diag}\mathbb{P}}(x)\sim cx(\log x)^{\beta}$ for some $c>0$ and $\beta$. Then $\exp{{\rm Diag}\mathbb{P}}(x)=(1+o(1))cx(\log x)^{\beta}$. This, after taking logarithms on both sides, implies ${{\rm Diag}\mathbb{P}}(x)=\log x+O(\log\log x)$, contradicting Theorem \[AsympDiagP\].

Acknowledgement {#acknowledgement .unnumbered}
===============

I would like to express my gratitude to Piotr Miska, who informed me about the problem and helped to improve the quality of the paper. I would also like to thank Carlo Sanna, whose comments allowed me to simplify a part of the paper.

[20]{} K. Broughan, R. Barnett, [*On the subsequence of primes having prime subscripts*]{}, Journal of Integer Sequences **12**, article 09.2.3, 2019. J. Bayless, D. Klyve, T. Oliveira e Silva, [*New bounds and computations on prime-indexed primes*]{}, Integers **13**: A43:1–-A43:21, 2013. P. Miska, J. Tóth, [*On interesting subsequences of the sequence of primes*]{}, preprint: arXiv:1908.10421. J. B. Rosser, L. Schoenfeld, [*Approximate formulas for some functions of prime numbers*]{}, Illinois J. Math. **6**, no. 1 , 64–94, 1962. doi:10.1215/ijm/1255631807.

[^1]: During the preparation of the work, the author was a scholarship holder of the Kartezjusz program funded by the Polish National Center for Research and Development.
