Set of prime numbers $q$ such that $\sum\limits_{p\leq\sqrt{q}}p=\pi(q)$

Question

If we define the sum of consecutive prime numbers $p$ up to the square root of a given positive integer $n$ as $S(n)=\sum\limits_{p\leq\sqrt{n}}p$, and the prime counting function up to a given positive integer $n$ as $\pi(n)=\#\left\{p\leq n\right\}$, it can be showed that $S(n)\sim\pi(n)$ (for a proof, you can check http://vixra.org/abs/1911.0316). This led me to study the positive integers for which $S(n)=\pi(n)$.

If we define set $Q$ as the set of values of $n$ such that $\pi\left(n\right)=S\left(n\right)$, the first values are $Q=\{11,12,29,30,59,60,179,180,...\}$, for $\pi(n)=5,5,10,10,17,17,41,41,...$

It can be conjectured that the first value of $n$ with a concrete $\pi\left(n\right)=S\left(n\right)$ will be always a prime number. This conjecture is a corollary of the following

Conjecture 1.

It does not exist any squared prime number $p^{2}$ such that $\pi\left(p^{2}\right)=S\left(p^{2}\right)$ except of $p_{1}=2$. That is, $$\pi\left(p_{n}^{2}\right)\neq\sum_{k=1}^{n}p_{k}$$

If the Conjecture were false, we could have that, being $p_m$ and $p_{m+1}$ prime numbers such that $p_m<p_n^2<p_{m+1}$, then $S\left(p_{m}\right)=S\left(p_{m}+1\right)=S\left(p_{m}+2\right)=...=S\left(p_n^{2}-1\right)=S\left(p_n^{2}\right)-p_n$; and as $\pi\left(p_n^{2}\right)=S\left(p_n^{2}\right)$, then $p_n^{2}\in Q$,and $p_n^{2}$ would be the first of a series of consecutive elements of $Q$ until $p_{m+1}$.

The conjecture has been tested and found to be true for the first thousands of primes.

If we define set $M$ as the set of values of $n$ such that $\pi\left(n\right)=S\left(n\right)$ and $n$ is some prime number, we find that $M=\{11,29,59,179,389,541,...\}$. You can see more terms in https://oeis.org/A329403. And if we define $p_{k}$ as the last prime number which is a term of $S\left(n\right)$, we can see that the first values of $k$ are $2,3,4,6,8,9,...$, which led me to formulate the following

Conjecture 2

Set $M$ has infinitely many elements

If the conjecture were true, as $M\subset Q$, the Conjecture implies that $\pi\left(n\right)$ intersects $S\left(n\right)$ infinitely many times, so $S\left(n\right)$ is not only asymptotically equivalent to $\pi\left(n\right)$: it is infinitely many times equal to $\pi\left(n\right)$. And it would show that the number of primes between $p_{n}^{2}$ and $p_{n+1}^{2}$ , on average, do not differ much from $p_{n+1}$.

Questions

(1) Any idea of how could be proved Conjecture 1, if it is already proved? Or maybe, with more computational power than mine, if anyone could verify if the conjecture supports itself also for big numbers, it would be great.

(2) With my computational power, I was able only to identify the first 19 prime numbers and $k$ values of the set (last one was $1756397$, for $k=216$). If anyone could calculate more of them to see if the conjecture supports itself also for big numbers, it would be great.

(3) Any idea of how to show that Conjecture 2 is true? I guess that it could be done with some theorem of the kind of Littlewood's oscillation theorem, but as an amateur mathematician obtaining such a result is beyond my actual knowledge.

Thanks in advance!

Answer 1

My answer was wrong.

If $\zeta(s)$ has a zero of real part $> 3/4$ then $\sum_{p\le x}1-\sum_{p\le x^{1/2}} p$ changes of sign infinitely often. If $\zeta(s)$ has no zero of real part $> 3/4-\epsilon$ then $\sum_{p\le x}1-\sum_{p\le x^{1/2}} p$ still changes of sign infinitely often but due to more complicated reasons (similar to the prime number races theorem).

See https://mathoverflow.net/questions/383610/set-of-prime-numbers-q-such-that-sum-limits-p-leq-sqrtqp-piq-where?noredirect=1#383620

Answer 2

Conjecture 2 has finally been proved as correct here , thanks to GH from MO. Conjecture 1 still remains open.