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

The question is: does the set of prime numbers $$q$$ such that $$\sum\limits_{p\leq\sqrt{q}}p=\pi(q)$$, where $$p$$ are prime numbers, contain infinitely many elements? You can find the first elements here (http://oeis.org/A329403). Any insight on this would be welcomed.

• This question has already been posted by the same author before, though it has been deleted and reposted an hour ago. This is an unacceptable abuse of the site's functions. Feb 10 at 13:32
• The content of the question is almost identical, I don't see how it is "better expressed". Feb 10 at 13:42
• I don't know what you mean by "administrator", but MathOverflow, like Stackexchange sites generally, is community moderated. We have 'official' moderators with superpowers for various purposes, but they aren't going to step in to suggest improvements in their capacity as mods, here. More specific to the current issue, though, the system expects people to edit their questions into a better shape, rather than self-delete and start again. If people see deletions and re-asking of questions, then it raises concerns, because this is a strategy that various bad actors have used in the past. Feb 10 at 14:02
• Looking at the old question, which is almost identical to this one, Will Sawin gave some good advice that could have been used to improve that one, and make any other edits you needed at the same time (I notice that $\lt$ became $\leq$ in this version), rather than asking a new question. I also note that you got a reasonably substantial answer at math.stackexchange.com/questions/3999435/…, where Conjecture 2 looks awfully like the above question. Feb 10 at 14:07
• @JuanMoreno In general, it's better to revise questions than to repost them. I won't delete this one since it has an answer below. If you revise your post and the comments are no longer relevant, you can flag them for deletion as "no longer needed." Feb 10 at 16:56

Nice question! The answer is affirmative. If $$\sigma(x)$$ denotes the sum of primes up to $$\sqrt{x}$$, then it suffices to show that $$\pi(x)-\sigma(x)$$ changes sign infinitely often, because $$\pi(x)-\sigma(x)<0<\pi(x+1)-\sigma(x+1)$$ never holds. I thank Juan Moreno and Will Sawin for this simple but crucial observation. My streamlined argument below owes to reuns' ideas as well.
Let us introduce the notation $$\pi(x)=\mathrm{li}(x)-\mathrm{li}(1)+\rho(x),$$ then we get \begin{align*}\sigma(x) &=\int_1^\sqrt{x}t\,d\pi(t)\\ &=\int_1^\sqrt{x}\frac{t\,dt}{\log t}+\int_1^\sqrt{x} t\ d\rho(t)\\ &=\int_1^x\frac{du}{\log u}+\int_1^\sqrt{x} t\ d\rho(t)\\ &=\mathrm{li}(x)-\mathrm{li}(1)+\int_1^\sqrt{x} t\ d\rho(t). \end{align*} So the difference $$\pi(x)-\sigma(x)$$ can be directly estimated by the error term in the prime number theorem. We can analyze this difference further by considering the Mellin transform $$\int_1^\infty x^{-s}\ d(\pi(x)-\sigma(x))=\int_1^\infty(x^{-s}-x^{1-2s})\ d\rho(x).$$ We shall only need to look at the left-hand side. Integrating by parts, we see that it is holomorphic in a region containing the half-plane $$\Re(s)>1$$ and the half-line $$s>2/3$$. In fact, in this region, the left-hand side equals $$\log\zeta(s)-\log\zeta(2s-1)+\frac{1}{2}\log\zeta(4s-2)+f(s),\tag{\ast}$$ where $$f(s)$$ is holomorphic for $$\Re(s)>2/3$$. As in the proof of Theorem 15.2 in Montgomery-Vaughan: Multiplicative number theory I, this eventually yields that $$\pi(x)-\sigma(x)=\Omega_\pm(x^c)\quad\text{for any}\quad c<3/4.$$ Indeed, let us assume that this bound fails for some $$c<3/4$$. Without loss of generality, $$c>2/3$$. Then, by Landau's lemma (which is Lemma 15.1 in the same book), $$(\ast)$$ is holomorphic in the half-plane $$\Re(s)>c$$. However, this is easily seen to be false, and we are done.
• thanks for your answer! Then, as $\pi(x)-\sigma(x)$ changes sign infinitely often, can we affirm that $\pi(x)=\sigma(x)$ infinitely many times? Logic would invite to think so, as there can be no "leaps" from $\pi(x)-\sigma(x)=1$ to $\pi(x)-\sigma(x)=-1$, and viceversa. If this is so, then the set has infinitely many elements, and the question is answered. Feb 10 at 15:53
• the leap you mention at a prime square $x$ would imply (if any) a change of sign from $\pi(x)-\sigma(x)>0$ to $\pi(x)-\sigma(x)<0$, and you are right that this would not imply equality; however, I am referring to the other change of sign, from $\pi(x)-\sigma(x)<0$ to $\pi(x)-\sigma(x)>0$ in which can not be involved any prime square $x$, at least as I see it. Feb 10 at 16:43
• I think Juan Moreno is right here - every time $\pi(x)-\sigma(x)$ changes from negative to positive, it must pass through $0$, because it is impossible to have $\pi(x)-\sigma(x)< 0$ and $\pi(x+1) -\sigma(x+1)>0$. Feb 10 at 16:46