**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 $n$:

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