# Conjecture about the density of primes

Conjecture

For any sufficiently large integer $$kn$$ , the sequence representing the number of primes in each block obtained by splitting $$kn$$ into $$k$$ equal blocks, is a strictly decreasing sequence, i.e:

$$\pi\left(n\right)>\pi\left(2n\right)-\pi\left(n\right)>\pi\left(3n\right)-\pi\left(2n\right)\ldots>\pi\left(kn\right)-\pi\left(\left(k-1\right)n\right)$$

My approach

For any given $$k$$, we need to prove that for all sufficiently large $$n$$, the last block contains less primes than the block before:

$$\underbrace{\pi\left(\left(k-1\right)n\right)-\pi\left(\left(k-2\right)n\right)}_{\text{before last}}>\underbrace{\pi\left(kn\right)-\pi\left(\left(k-1\right)n\right)}_{\text{last}}$$

Let $$f( k,n )$$ denote the difference between the prime count of before last and last block of $$kn$$:

$$f\left(k,n\right) = \left(\pi\left(\left(k-1\right)n\right)-\pi\left(\left(k-2\right)n\right)\right)-\left(\pi\left(kn\right)-\pi\left(\left(k-1\right)n\right)\right)$$ $$= 2\pi\left(\left(k-1\right)n\right)-\pi\left(\left(k-2\right)n\right)-\pi\left(kn\right)$$

Therefore, an equivalent formulation of the conjecture is that $$f(k,n)>0$$ for any integers $$k,n$$ with sufficiently large $$n$$.

Does anybody have ideas or references on how to prove this?  Computation of $$f( k,n )$$

I made some computation that strongly suggests the conjecture is true, at least for small $$k$$:

Failing case for small $$n$$

Let's take the case $$k=2$$. We see that the conjecture fails for $$n=1,2,4$$ & $$10$$:

I computed $$f( k,n )$$ for $$2 ≤ k ≤ 30$$ with $$1 ≤ n ≤ 10^{8}$$, in order to find what appears to be the last failing case of each $$k$$ , after which $$f( k,n ) >0$$ for all $$n$$:

• Prime gaps increases unboundedly(one can see that considering all integers n!+2,....n!+n are composites) is this a consequence of that? Commented Aug 9, 2021 at 9:54
• I think this should follow from the Prime Number Theorem (with error terms). For example, the explicit version of the PNT mentioned here: mathoverflow.net/questions/373737/… Commented Aug 9, 2021 at 11:26
• Note that the second Hardy Littlewood conjecture is likely false, and it is a statement of a similar flavor en.wikipedia.org/wiki/… . Commented Aug 9, 2021 at 11:51
• To make sure I understand the order of your quantifiers? One is given an $n$ and then is allowed to choose $k$ based on $n$? Commented Aug 9, 2021 at 11:56
• @JoshuaZ: I agree that this is similar, and yet while I would bet $1000 against$1 that the 2nd Hardy Littlewood conjecture is false, this seems reasonable. Being tied to the beginning rather than floating gives a LOT less flexibility. Commented Aug 9, 2021 at 16:44

The conjecture is true, and it follows from the following form of the Prime Number Theorem: $$\pi(x)=\frac{x}{\log x}+(1+o(1))\frac{x}{\log^2 x},\qquad x\to\infty.$$ Indeed, this implies for any fixed $$m\geq 1$$ and sufficiently large $$n$$ that $$2\pi(mn)>\pi(mn+n)+\pi(mn-n).$$ Why? We have \begin{align*}\pi(mn)&=\frac{mn}{\log(mn)}+(1+o(1))\frac{mn}{\log^2(mn)}\\ &=m\frac{n}{\log n}\frac{1}{1+\frac{\log m}{\log n}}+(m+o(1))\frac{n}{\log^2 n}\\ &=m\frac{n}{\log n}+(m-m\log m+o(1))\frac{n}{\log^2 n}. \end{align*} Similarly (replacing $$m$$ by $$m-1$$ and $$m+1$$, respectively), \begin{align*} \pi(mn+n)&=(m+1)\frac{n}{\log n}+(m+1-(m+1)\log(m+1)+o(1))\frac{n}{\log^2 n},\\ \pi(mn-n)&=(m-1)\frac{n}{\log n}+(m-1-(m-1)\log(m-1)+o(1))\frac{n}{\log^2 n}. \end{align*} Here, in case of $$m=1$$, we understand $$(m-1)\log(m-1)$$ as zero. It follows that $$2\pi(mn)-\pi(mn+n)-\pi(mn-n)=(f(m)+o(1))\frac{n}{\log^2 n},$$ where $$f(m):=(m+1)\log(m+1)+(m-1)\log(m-1)-2m\log m.$$ So we only need to verify that $$f(m)$$ is positive, that is, $$(m+1)\log(m+1)+(m-1)\log(m-1)>2m\log m.$$ However, this one is clear, because the function $$x\mapsto x\log x$$ is strictly convex (its derivative is strictly increasing).
• The way I read the question, OP does not ask that $n$ be sufficiently large, only that $kn$ be sufficiently large. If $n=1$ and $k$ is sufficiently large, then $kn$ is sufficiently large, but what OP wants to conclude is quite false. Commented Aug 14, 2021 at 13:05
• @GerryMyerson: The OP wrote: "For any given $k$, we need to prove that for all sufficiently large $n$, the last block contains less primes than the block before". This is what I proved. Commented Aug 14, 2021 at 17:39
• @FrançoisHuppé: The statement should begin with: for fixed $k$ and sufficiently large $n$ etc. More formally: for any integer $k\geq 1$ there exists an integer $N\geq 1$ such that for any integer $n\geq N$ etc. Commented Aug 15, 2021 at 4:47