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?
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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 $n$:
Best Answer
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).