On primes of the form $2\pi(n)p_n+1$, for some $n\geq 1$ being $\pi(x)$ the prime-counting function and $p_k$ the $k$th prime number

Question

I was inspired in the sequence A088348 from the OEIS to explore if there are primes of the form $$2\pi(n)p_n+1,\tag{1}$$ when $1\leq n$ runs over integers, where $\pi(x)$ denotes the prime-counting function and $p_n$ the $n$th prime number.

Previous sequence starts as $$7, 29, 67, 79, 137, 233, 311,\ldots\tag{2}$$

I think that this sequence isn't in the literature. I would like to define it and explore some properties if it is feasible.

Claim. The prime number theorem and the comparison test implies that $$\sum_{\substack{p\text{ primes of the form }\\2\pi(n)p_n+1}\\\quad\text{ for some }n\geq 1}\frac{1}{p}\tag{3}$$ is convergent.

Question. Is it known or in case that this sequence isn't in the literature, if our sequence has infinitely many terms (if there are infinitely many primes of the form $(1)$)? Can you propose as a conjecture how many terms less than $x$ should have our sequence $(1)$, for $x$ sufficiently large (I am asking if you can propose what is/a statement about the asymptotic behaviour of the counting function $\pi(x)$ of our sequence $(1)$ you can work on assuption of the conditions that you need)? Many thanks.

If our sequence and previous questions is well-knowns please answer this question as a reference request.

Have you tried running code and seeing if there is any logarithmic progression to the number of primes of this form less than x? — Sean Nemetz, Feb 09 '18 at 20:47
No, in fact I never did such test to know if a sequence seems a logarithmic progression @SeanNemetz Many thanks for your attention. — , Feb 09 '18 at 20:52
It's an interesting question to say the least. Mathematicians don't even know if there are infinitely many primes of the form $2p+1$, where $p$ is prime. — Sean Nemetz, Feb 09 '18 at 20:55

score 1 · Answer 1 · answered Feb 09 '18 at 22:03

Just an observation to the claim (3). Long time ago I asked this question leading to $\pi(x)>\sqrt{x}$ from some $x$ onwards. And $p_n>n$, so $2\pi(n)p_n+1 > 2n\sqrt{n}$ from some $n_0$ onwars. As a result $$\sum_{\substack{p\text{ primes of the form }\\2\pi(n)p_n+1}\\\quad\text{ for some }n\geq 1}\frac{1}{p}<\sum\limits_{n=1}\frac{1}{2\pi(n)p_n+1}< \sum\limits_{n=1}^{n_0-1}\frac{1}{2\pi(n)p_n+1} + \sum\limits_{n=n_0}^{\infty}\frac{1}{2n^{1+\frac{1}{2}}}$$ which indeed converges.

Many thanks for your nice proof and feedback. I am going to read your reasoning. — , Feb 09 '18 at 22:29

On primes of the form $2\pi(n)p_n+1$, for some $n\geq 1$ being $\pi(x)$ the prime-counting function and $p_k$ the $k$th prime number

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