Hello,


let's define the notion of order of primality in such a way:
$n$ is a prime of order $k$ if and only if $k$ is the smallest non negative integer such that $\pi^{(k)}(n)$ is not a prime number, where $\pi^{(0)}(n)=n$ and for all $m$, $\pi^{(m+1)}(n)=\pi(\pi^{(m)}(n))$.
Now let's write $\pi\_{k}(x)$ for the number of primes of order $k$ not exceeding $x$.
It is quite easy to see that every non negative integer $n$ verifies the equality $n=\displaystyle{\sum\_{k\geq 0}\pi\_{k}(n)}$ and that $\pi(n)=n-\pi\_{0}(n)$. Moreover, for all $(k,n)\in\mathbb{N}^{2}$, $\pi\_{k+1}(n)\leq\pi\_{k}(n)$.


My question is: is there a function $f$ such that for all $(k,n)\in\mathbb{N}^{2}$, $\pi\_{k+1}(n)=f(\pi\_{k}(n))+O(1)$?


Thank you in advance.



order of primality of an integer
I think your argument may be generalized to any order. If $m \in \mathbb{N}$, then $\pi^{(m)}(x)$ is the number of primes of order $\ge k$ not exceeding $x$. That is


$$\pi^{(m)}(x) = \sum\_{k \ge m} \pi\_k(x)$$


And we get


$$\pi\_m(x) = \pi^{(m)}(x) - \pi^{(m+1)}(x)$$


Thus for $n \in \mathbb{N}$, we get


$$\pi\_m(\pi^{(n)}(x)) = \pi^{(m)}(\pi^{(n)}(x)) - \pi^{(m+1)}(\pi^{(n)}(x)) = \pi^{(m+n)}(x) - \pi^{(m+n+1)}(x) = \pi\_{m+n}(x)$$


