Let's find all sequences of integers $a_{1}<a_{2}<\cdots<a_{n}<\ldots$ such that:

- for every positive integer $n$, $a_{2 n}=a_{n}+n$

- if $a_{n}$ is a prime number, then $n$ is also a prime number.
From the condition $a_{2 n}-a_{n}=n$, it follows that $n-1$ integers fall between $a_{2 n}$ and $a_{n}$. At the same time, exactly this many elements of the sequence can be found between them, and since the sequence consists of integers and is strictly increasing, this can only be true if the sequence increases by one between $a_{n}$ and $a_{2 n}$. This is true for all $n$, hence

$$
a_{n}=a_{1}+n-1
$$

By introducing the notation $a_{0}=a_{1}-1$, we get $a_{n}=a_{0}+n$.

We will show that $a_{0}$ is divisible by every number, i.e., $a_{0}=0$. Assume, for some prime $q$, that $q \mid a_{0}$, and thus $a_{0} \neq 0$. Let's first consider $a_{0}>0$. Choose a prime $p>(a_{0}+1)q$. Then, from the equality $p=a_{p-a_{0}}$, it follows that $p-a_{0}$ is also prime, and continuing this, $p-2a_{0}, \ldots, p-(q-1)a_{0}>a_{0}q+q-qa_{0}+a_{0}>q$ are also primes. This is a sequence of $q$ primes, none of which can be divisible by $q$ since they are all greater than $q$; among them, there exist two whose difference is divisible by $q$:

$$
q \mid (p-i a_{0})-(p-j a_{0})=(j-i) a_{0}
$$

Since $(q, a_{0})=1$, it follows that $q \mid (j-i)$, but $0<j-i<q-1$, which is a contradiction.

If, on the other hand, $a_{0}<0$, the same reasoning can be applied to the sequence $p, p+a_{0}, \ldots, p+(q-1)a_{0}$. There is no other case, so we conclude that the only possibility is $a_{n}=n$, which is clearly a valid solution.