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For $x>0$ let $\pi(x)$ denote the number of primes not exceeding $x$. A well-known conjecture of Legendre states that $\pi(n^2)<\pi((n+1)^2)$ for any positive integer $n$. Here I ask the following natural question.

QUESTION: Are all the numbers $\pi(n^2)/n^2\ (n=1,2,3,\ldots)$ pairwise distinct?

Based on my computation, in 2015 I conjectured that the answer is yes. Moreover, my computation suggests that the sequence $$\frac{\pi(n^2)}{n^2}\ \ (n=15647,15648,\ldots)$$ is strictly decreasing. For each $k=3,4,5,\ldots$, I also conjectured that the sequence $$\frac{\pi(n^k)}{n^k}\ \ (n=2,3,\ldots)$$ is strictly decreasing.

Any comments on the question are welcome!

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  • $\begingroup$ By the Prime Number Theorem, $\lim_{n\to\infty}\pi(n^k)/n^k=0$. But this does not provide an answer to the question. $\endgroup$
    – Zhi-Wei Sun
    Commented May 31, 2018 at 16:56
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    $\begingroup$ But unconditional bounds such as those of Dusart should show many of your sequences are indeed decreasing. Gerhard "Have You Tried These Bounds?" Paseman, 2018.05.31. $\endgroup$
    – Gerhard Paseman
    Commented May 31, 2018 at 17:32
  • $\begingroup$ @Gerhard Paseman I know P. Dusart's results in Math. Comp. [68(1999), 411-415]. But it seems to me that Dusart's bounds don't imply that $\pi(n^2)/n^2>\pi((n+1)^2)/(n+1)^2$ for sufficiently large $n$. Even RH (Riemann Hypothesis) might not be helpful to my question. $\endgroup$
    – Zhi-Wei Sun
    Commented May 31, 2018 at 22:04
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    $\begingroup$ Under RH it is easy to see that $\pi(n^2)(n+k)^2=\pi((n+k)^2)n^2$ can only hold for $k\ll\log^2 n$. If on the other hand $k$ is small, then the greatest common divisor of $n^2$ and $(n+k)^2$ is small, thus $n^2$ essentially divides $\pi((n+k)^2)$. This gives a contradiction, if $k^2<(1-\epsilon)\log n$. Both bounds can probably be improved. For the first note that we do not really need PNT for short intervals, but only an upper bound which may be of by a factor of 2. For the second some elementary fiddling should lead to something. Note $\endgroup$
    – Jan-Christoph Schlage-Puchta
    Commented Jun 1, 2018 at 9:21
  • $\begingroup$ I think eventual monotonicity of $\pi(n^2)/n^2$ may also be difficult to establish unconditionally, it fails frequently early on. $\endgroup$
    – kodlu
    Commented Jun 3, 2018 at 23:46

1 Answer 1

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Edit: This is not quite an answer, leaving it up as a long comment for now after fixing some arithmetic errors.

As suggested in the comments, using Dusart's results from his paper (see here) where he proves unconditionally

$$ \pi(x)\geq \frac{x}{\log x}\left(1+ \frac{1}{\log x}+\frac{2}{\log^2 x}\right)\stackrel{\triangle}{=}xL(x), \quad x\geq 88~783, $$ and

$$ \pi(x)\leq \frac{x}{\log x}\left( 1+\frac{1}{\log x}+\frac{2.334}{\log^2 x}\right)\stackrel{\triangle}{=}xU(x), \quad x\geq 2~953~632~287, $$ one can obtain, for $n$ larger than the second inequality cutoffs (if I haven't made a mistake) and for $k=2$ (higher $k$ will be similar):

$$ \frac{\pi(n^2)}{n^2} -\frac{\pi((n+1)^2)}{(n+1)^2} \geq L(n^2) - U((n+1)^2),\quad (0) $$ where the right hand side equals $$ \left(\frac{1}{2 \log n}-\frac{1}{2 \log(n+1)}\right)+ \left(\frac{1}{4 \log^2 n}-\frac{1}{4\log^2(n+1)}\right)+ \left(\frac{2.334}{8 \log^3 n}-\frac{2}{8 \log^3(n+1)}\right).\quad\quad (1) $$ The first term in (1) can be rewritten as $$ \frac{\log^2(n+1)-\log^2(n)}{ 2\log^2(n+1)\log^2(n)}= \frac{(\log(n)+\log(1+\frac{1}{n}))^2-\log^2(n)}{2 \log^2(n+1)\log^2(n)}= $$

$$ =\frac{2\log(n)\log(1+\frac{1}{n})+\log^2(1+\frac{1}{n})}{2\log^2(n+1)\log^2(n)} \sim \frac{1}{ n\log^2(n+1)\log(n)} $$ since the first term in the last fraction dominates, and it is $O(1/(n\log^3 n))$.

A similar argument for the second term in (2) shall give a positive quantity with a constant on top and an $n$ and a higher power of $\log n$ in the denominator, which will be smaller than the first term.

The third term in (2) is unfortunately negative due to the mismatched coefficients and
dominates the right hand side of (0), so this effort fails.

If that worked, it would have proved that that the numbers $\pi(n)^2/n^2$ are pairwise distinct beyond $n\geq 2~953~632~287.$ The differences up to then can be checked computationally. There are some extensive tables of the prime counting function, e.g., see this link here.

Actually, the Magma online calculator here can return lists of primes from $2$ to $10^9$, if they are split into sublists to avoid memory overflow as I just confirmed.

Remark: Using the simpler Dusart formulas in the paper above these don't seem to work.

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  • $\begingroup$ There are some typos, and possibly arithmetic errors. In (0), you should have L(n^2) ,not L(n)^2, and similarly for U. I expect to see more 2's and 4's in the denominators. Hopefully the conclusion holds. Gerhard "Glad Someone Is Doing Calculations" Paseman, 2018.06.01. $\endgroup$
    – Gerhard Paseman
    Commented Jun 2, 2018 at 2:29
  • $\begingroup$ @GerhardPaseman, thanks, I'll have a look. $\endgroup$
    – kodlu
    Commented Jun 2, 2018 at 2:33
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    $\begingroup$ Unfortunately the 2.334 is the numerator of the other fraction, and the third term is negative in general. If you take a factor of (1/log n - 1/log(n+1)) out of the terms, and have a -1.334/8log^3(n+1) as a negative term, you might get the other terms to dominate it, but now I don't think so. Thanks for the attempt. Gerhard "Doesn't Know Enough Prime Gaps" Paseman, 2018.06.01. $\endgroup$
    – Gerhard Paseman
    Commented Jun 2, 2018 at 2:58
  • $\begingroup$ @GerhardPaseman, it might well be that an unconditional proof is quite difficult, I guess. $\endgroup$
    – kodlu
    Commented Jun 2, 2018 at 3:48

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