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 $n$:
We know that $ \frac{n}{\ln n} ( 1+ \frac{1}{\ln n}) < \pi(n) < \frac{n}{\ln n} ( 1+ \frac{1}{\ln n}+\frac{3}{\ln^2 n})$ for all $n \geq 90,000$ so for $k \geq a\geq 1$ we need to prove that $ 2\pi((a+1) n) \geq \pi(a n)+\pi((a+2)n)$ (for the case $a=0$ you need to prove that $2\pi(n) > \pi(2n)$)
If we prove that $2 \frac{(a+1)n} {\ln{(a+1)n}}( 1+ \frac{1}{\ln{(a+1)n}}) > \frac{a n}{\ln an} (1+\frac{1}{\ln a n}+\frac{3}{\ln^2 an}) +\frac{(a+2) n}{\ln (a+2)n} (1+\frac{1}{\ln (a+2) n}+\frac{3}{\ln^2 (a+2)n}) $
With algebraic manipulation of the above inequality we get that your conjecture is true given $k$ there is $N$ such that for all $n>N$ we have that $[1,n]$ contains more primes than $[n+1,2n]$ which is containing more primes than $[2n+1,3n]$ and so on...