$\pi((n+1)^2)-\pi(n^2) \le \pi(n)$ for all $n \ge 370$?

Question

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

• Bertrand's postulate: for every $n>1$ there is always at least one prime $p$ such that $n < p < 2n$.

• Legendre's conjecture: there is a prime number between $n^2$ and $(n + 1)^2$ for every positive integer $n$.

• Brocard's conjecture: there are at least four prime numbers between $p_{n}^2$ and $p_{n+1}^2$.

• Oppermann's conjecture: there is at least one prime number between $n(n-1)$ and $n^2$.

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 $f(n) = \pi(n^2)+\pi(n)-\pi((n+1)^2); \; 370 \le n \le 1.1\times10^4$

• 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?

Answer 1

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 $$ \pi((n+1)^2) - \pi(n^2) \le \Big( \frac{36}{11}+ o(1)\Big) \frac{n}{\log n}. $$

Answer 2

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.