Consider two positive integers $2 < a < x$. The following two statements would seem to follow intuitively from the PNT, but I'm wondering if either has been proven or discussed. Here, $\pi(n)$ is the prime-counting function.

Conjecture 1: For any $x$, there exists some $a_0$ such that if $a \geq a_0$, the following inequality is always true:

$$\tag{1} \pi(x+a) - \pi(x) \leq \pi(x) - \pi(x-a)$$

In plainer English, consider two intervals of length $a$ on either side of $x$. This conjecture suggests that the earlier interval (which contains smaller numbers) contains more prime numbers than the later interval (which contains larger numbers).

Shown in comments to be trivially true but left for context: For every $a$ there is some large enough $x$ such that the following inequality is always true:

$$\tag{2} \frac{\pi(x) - \pi(a)}{x-a} \leq \frac{\pi(a)}{a}$$

Here, the question is about the density of primes on the two intervals. Are the primes more dense below $a$ than above $a$?

(The inequality can be written as $\frac{\pi(x)}{x} < \frac{\pi(a)}{a}$, which is true for any sufficiently different $a$ and $x$ with $a < x$ via the PNT).

We can consider an extreme example that applies to both statements. If we set $x = a$ in conjecture (1), or $x = 2a$ in conjecture (2), both conjectures simply ask whether there are more (or equal) primes between $0$ and $n$ than between $n$ and $2n$. We know this is true for $n > 2$. Outside of the extreme case, however, the PNT would suggest that equal-length intervals should have more primes if they begin at a lower value, and that the density of primes decreases as $n$ increases. I also suspect that if $a$ is very small, or in conjecture (2) if $a$ is close to $x$, that these would fail.

It seems these conjectures must have been floated at some point, but I'm wondering if they've been proven. I don't think they're just special cases of the second Hardy-Littlewood conjecture, but I'm also very much an amateur. Does anyone know of proofs/disproofs/discussions of these?

Edit: Restatement of conjecture 1, Note on conjecture 2 as it is even more trivial than I thought it even could be.

  • $\begingroup$ Be careful with your quantifiers. For every $x$ if $a$ is large enough then there are more primes between $x$ and $x-a$ than between $x$ and $x+a$. $\endgroup$ – Gerry Myerson Feb 7 at 11:27
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    $\begingroup$ The second inequality can be rewritten as ${\pi(x)\over x}\le{\pi(a)\over a}$. Since $\pi(x)/x\to0$ as $x\to\infty$, this is true for any $a$, for all sufficiently large $x$. $\endgroup$ – Gerry Myerson Feb 7 at 11:31
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    $\begingroup$ Well, $y$ should approach 2, but it converges with glacial slowness. Using $p_n=n\log n$ (which underestimates $p_n$), we get $y=2(1+\log 2/\log n)=2(1+1/\log_2 n)$. OTOH, using $\pi(x)=x/\log x$, we get $y=2(1+\log y/\log x)$, and we can use the previous expression as the $y$ value on RHS. For large $x$, the arithmetic mean of those two expressions is a good estimate of $y$. I'll post a Sage / Python script that does these calculations, as well as exact $y$ calculations for $x\le10^{10}$, in the next comment. $\endgroup$ – PM 2Ring Feb 8 at 11:40
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    $\begingroup$ y estimator script $\endgroup$ – PM 2Ring Feb 8 at 11:41
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    $\begingroup$ Ah, I see. And as $x \to \infty$, in both expressions everything but the $2$ cancels out. $\endgroup$ – Eric Snyder Feb 8 at 21:31

For any $a \geq 2$ Let $ x = (a+2)! +a +2$ and since there is no prime in the interval $ [(a+2)!+2,(a+2)!+a+2]$ we have that $ \pi(x) = \pi(x-a)$ and so $ \pi(x)-\pi(x-a)=0$

Now since $\pi(y)$ is non-decreasing function we have that $\pi(x+a)-\pi(x) \geq 0$, if we have $ \pi(x+a)-\pi(x)>0$ we are done otherwise $ \pi(x+a)-\pi(x) = 0$, now let $x_{new} = x_{old}+a$ and so $ \pi(x_{new})-\pi(x_{new}-a) = \pi(x_{old}+a)-\pi(x_{old}+a-a) = \pi(x+a)-\pi(x) = 0$ (Our Assumption) and so $ \pi(x_{new})-\pi(x_{new}-a)=0$ and we repeat until $\pi(x_{new}+a)-\pi(x_{new})>0$ (Guaranteed because the infinitude of the primes).

Which implies that $ \pi(x+a)-\pi(x) > \pi(x)-\pi(x-a)$ holds infinitely many times.

With a little tweaking of the above argument we get that $ \pi(x+a)-\pi(x) < \pi(x)-\pi(x-a)$ holds infinitely many times.

So the first conjecture for any $a\geq 2$ holds infinitely many times and violated for infinitely many times for different sequences of $x$ as $ x \to \infty$.

  • $\begingroup$ That is brilliant and elegant. I had to write it out on paper on my own to fully follow it, and which point it was obvious and simple. That is a really interesting result, though! I had expected it would switch signs with $a<<x$ but having it switch signs infinitely many times just seems crazy. And cool. $\endgroup$ – Eric Snyder Feb 26 at 3:33
  • $\begingroup$ You are very welcome. $\endgroup$ – Ahmad Feb 26 at 13:53

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