There are some conjectures of the form: There always exist at least $X$ prime numbers between $A$ and $B$. Examples:

If we denote by $\pi(x)$ the prime-counting function we can rewrite the above conjectures in the following form:

  • Bertrand's postulate: $\pi(2n)-\pi(n) \ge 1$ for $n>1$
  • Legendre's conjecture: $\pi(n+1)^2)-\pi(n^2) \ge 1$
  • Brocard's conjecture: $\pi(p_{n+1}^2)-\pi(p_{n}^2) \ge 4$
  • Oppermann's conjecture: $\pi(n^2)-\pi(n(n-1)) \ge 1$

    • I computed and saw that $f(n) = \pi(n^2)+\pi(n)+2-\pi((n+1)^2)$ is increasing when $n$ increasing and $f(n)\ge 0$ for all $n=1, 2, \dots, 18700$ (equivalent to $n^2=1, 4, 25 \cdots , 3.5\times 10^8)$.

    • Graph of $(n,f(n))$ where $f(n) = \pi(n^2)+\pi(n)-\pi((n+1)^2); \; 370 \le n \le 1.1\times10^4$

enter image description here

  • So I proposed two conjecture as follows:

Conjecture 1: For every positive integer $n$, the number of primes between $n^2$ and $(n + 1)^2$ is less than the number of primes between $1$ and $n$ add $2$:

$$\pi((n+1)^2)-\pi(n^2) \le \pi(n)+2.$$

Conjecture 2: For every positive integer $n$ greater than $369$, the number of primes between $n^2$ and $(n + 1)^2$ is less than the number of primes between $1$ and $n$:

$$\pi((n+1)^2)-\pi(n^2) \le \pi(n).$$

Could you give a remark, comment, reference, or proof?

Noting that if the conjecture is true, it is stronger than a special case of the Second Hardy–Littlewood conjecture but this conjecture is not contradictory with the K-Tuple conjecture.

PS: In my computation I see that:

$$\lim_{n \to +large } \frac{\pi((n+1)^2)-\pi(n^2)}{\pi(n)}=1$$

What do You think with this equality?

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    $\begingroup$ Try A014085. Gerhard "The OEIS Is Your Friend" Paseman, 2018.06.25. $\endgroup$ – Gerhard Paseman Jun 26 '18 at 4:20
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    $\begingroup$ If your conjecture were stronger than the second Hardy-Littlewood conjecture, then it would contradict the k-tuple conjecture, simply because the second Hardy-Littlewood conjecture does contradict the k-tuple conjecture. What happens is that your conjecture is stronger than a special case of the second Hardy-Littlewood conjecture (the special case being $x=n^2$ and $y=2n+1$). BTW I believe that this conjecture (if true) is also out of reach. We do not know enough about the number of primes in such short intervals. $\endgroup$ – GH from MO Jul 5 '18 at 17:29
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    $\begingroup$ @ĐàoThanhOai: I think you did not understand what I said. The point is that your conjecture is not stronger than the second Hardy-Littlewood conjecture (contrary what you say in your post). Instead, your conjecture is stronger than a certain special case of the second Hardy-Littlewood conjecture. $\endgroup$ – GH from MO Jul 5 '18 at 17:58
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    $\begingroup$ @GHfromMO I am sorry, thank You very much, your remaks are true. $\endgroup$ – Đào Thanh Oai Jul 5 '18 at 18:00
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    $\begingroup$ The conjecture in your P.S. is very reasonable. We expect that $\pi(x+y)-\pi(x)\sim y/\log x$ holds as long as $y<x$ and $y$ is not too small (from the work of Maier we know that $y$ needs to grow faster than any power of $\log x$). We certainly expect the mentioned asymptotics for $y\asymp x^{1/2}$, in particular for $x=n^2$ and $y=2n+1$. That is, we should have $\pi((n+1)^2)-\pi(n^2)\sim (2n+1)/\log n^2\sim n/\log n\sim\pi(n)$. But all that is out of reach with present methods. You keep asking questions that are too difficult! $\endgroup$ – GH from MO Jul 6 '18 at 11:31

It is a folklore conjecture that for $y\le x$ one has $$ \pi(x+y) -\pi(x) = \int_{x}^{x+y} \frac{dt}{\log t} + O(y^{\frac 12} x^{\epsilon}). $$ This is only relevant for $y \ge x^{\epsilon}$, and is stronger than RH. In the case of primes in progressions, such a conjecture may be attributed to Montgomery. Probabilistic considerations might suggest a stronger error term like $O(y^{\frac 12} \log x)$, but this is known to be false thanks to the work of Maier. But the conjecture as stated above is widely believed. See, The distribution of prime numbers for example for a discussion of this and related work (in particular the discussion around (3.7) there). See also Montgomery and Soundararajan where refined asymptotics for moments of primes in short intervals are made and justified heuristically; these conjectures state that primes in short intervals have an appropriate Gaussian distribution, and imply the conjecture above.

Of course we are very far from the conjecture mentioned above, but if true it implies for large $n$ that $$ \pi((n+1)^2) - \pi(n^2) = \int_{n^2}^{(n+1)^2} \frac{dt}{\log t} +O(n^{\frac 12+\epsilon}) = \frac{n}{\log n} + O(n^{\frac 12+\epsilon}), $$ since $\log t = 2\log n + O(1/n)$ throughout the interval $n^2 \le t\le (n+1)^2$. Now we know that $\pi(n)$ is asymptoically $\text{li}(n)$ which has the asymptotic expansion $n/\log n + n/(\log n)^2 +\ldots$. The secondary term of $n/(\log n)^2$ dominates the error term in the conjectured asymptotic, and so one should certainly expect that for large $n$ one has $$ \pi((n+1)^2) - \pi(n^2) \le \pi(n). $$

An analogous question formulated for $\pi((n+1)^3) - \pi(n^3)$ is known, as mentioned by GH from MO in the comments, by Huxley's version of the prime number theorem in short intervals.

Now the question is only asking for an upper bound on $\pi((n+1)^2) -\pi(n^2)$, and one does have unconditional upper bounds by sieves. The Brun-Titchmarsh theorem would give a bound like $4\pi (n)$ for this quantity, and one can do somewhat better than this. The best result that I know is due to Iwaniec from whose work (see Theorem 14 there) it follows that $$ \pi((n+1)^2) - \pi(n^2) \le \Big( \frac{36}{11}+ o(1)\Big) \frac{n}{\log n}. $$

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  • $\begingroup$ With my highest respect, I always look at your answers in MathOverflow to learn. Technically the last equality in your second display is wrong, because the integral has an asymptotic $=\frac{n}{\log n}-\frac{45n}{4\log^7n}+O(\log^{-1}n)$, and you retain the lesser error term $O(n^{\frac12+\varepsilon})$ Nevertheless your reasoning is still correct. I have some doubts about this conjecture, by the reason I give in my note below. There have to be some collaboration between the terms $\pi((n+1)^2)-\textrm{Li}((n+1)^2)$ and $\pi(n^2)-\textrm{Li}(n^2)$ for this to be true. $\endgroup$ – juan Jul 7 '18 at 7:29
  • $\begingroup$ So is the cojecture right or false? Or disprove? Thank You very much. $\endgroup$ – Đào Thanh Oai Jul 7 '18 at 15:08
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    $\begingroup$ @juan: I added one more line after that display to clarify why it holds. I think it is correct as I stated, and I think you must have a calculation error in the lower order term that you wrote. I also added another related reference regarding this conjecture. The error terms at $n^2$ and $(n+1)^2$ are indeed closely related, because of the explicit formula. One can write the difference between the remainder terms as a sum over zeros, and this sum should have a Gaussian distribution, which is a stronger version of pair correlation. $\endgroup$ – Lucia Jul 7 '18 at 15:13
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    $\begingroup$ @ĐàoThanhOai: I thought I had explained in my answer why the conjecture is likely true (and implied by other conjectures), but very far from known. I suggest that you look at some books in prime number theory to get an idea of the subject -- many problems have been extensively studied, and it is good to get a systematic view. An easy read is the book by Tenenbaum and Mendes--France on the Distribution of prime numbers; more advanced treatments may be found in Davenport's book on Multiplicative number theory, and the book by Montgomery and Vaughan. $\endgroup$ – Lucia Jul 7 '18 at 15:20
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    $\begingroup$ @Lucia Yes my computation was wrong. $\endgroup$ – juan Jul 7 '18 at 16:46

We know is that the difference between $\pi(x)$ y $\textrm{Li}(x)$ is less than $xe^{-c\sqrt{\log x}}$, and we have the asymptotic expansion $$\textrm{Li}(x)-\textrm{Li}((x+1)^2)+\textrm{Li}(x^2)=$$ $$x\Bigl(\frac{1}{\log^2x}+\frac{2}{\log ^3x}+\frac{6}{\log ^4x}+\frac{24}{\log ^5x}+\frac{120}{\log ^6x}+\frac{45}{\log ^7x}\Bigr)+ O\Bigl(\frac{1}{\log x}\Bigr).$$ The main term of this expansion $\frac{x}{\log^2 x}$ is less than the differences between $\pi(x)$ and $\textrm{Li}(x)$. For example $$|\textrm{Li}(x^2)-\pi(x^2)|\le x^2e^{-c\sqrt{2\log x}}.$$ We have some better bounds $xe^{-c\log^\alpha x}$, but this are equally insufficient.

Assuming the Riemann hypothesis Schoenfeld proved that $$|\pi(x)-\textrm{Li}(x)|<\frac{1}{8\pi}\sqrt{x}\log x,\qquad x>2657.$$ Therefore assuming RH we have $$|\pi(x^2)-\textrm{Li}(x^2)|<\frac{1}{4\pi}x\log x>\frac{x}{\log^2x}.$$ Therefore our knowledge, even assuming RH, is insufficient to resolve this question. I think that possibly the conjecture is not true, although the first counterexample can be very large. Because they can only happens where the Schoenfeld inequality is almost equality.

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  • $\begingroup$ Do You think $\lim_{n \to +\infty } \frac{\pi((n+1)^2)-\pi(n^2)}{\pi(n)}=1$ ?? $\endgroup$ – Đào Thanh Oai Jul 6 '18 at 9:16
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    $\begingroup$ If you substitute each $\pi(\cdot)$ by the corresponding $\textrm{Li}(\cdot)$, the resulting function tends to $1$. But the difference $\pi(x)-\textrm{Li}(x)$ has been shown to be $\Omega_{\pm}(\frac{x^{1/2}\log\log\log x}{\log x})$. Therefore one of the error $\pi(x^2)-\textrm{Li}(x^2)$ can be of order $\frac{x}{\log x}\log\log\log x$. Therefore the sup limit of this divided by $\pi(x)$ is $+\infty$. So only if the error in $\pi((x+1)^2)$ and $\pi(x^2)$ collaborate can the limit be $1$. $\endgroup$ – juan Jul 6 '18 at 9:54
  • $\begingroup$ I am sorry, could you answer that the limmit is true or not true? $\endgroup$ – Đào Thanh Oai Jul 6 '18 at 10:44
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    $\begingroup$ I don't know. Our knowledge make the assertion unlikely. $\endgroup$ – juan Jul 6 '18 at 11:08

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