As I discuss a similar question regarding the best upper bound for maximum gap after $n^{th}$ iteration of sieve of Eratosthenes here, I'm interested to know whether such a thing is possible for a modified sieve of Eratosthenes where every multiple of a prime number and numbers which have remainder two for that prime number will be sieved.
So any number $k$ will be sieved if $k\bmod q = 0$ or $2$, where $q$ is the prime number.
In such a modified sieve, what is the best upper bound formula for the maximum gap after the $n^{th}$ iteration?
Is it possible to prove that
$$a(n) \ll q^2 - q$$
Or
$$a(n) \ll p^2 - q$$
where
$a(n)$ is the upper bound,
$p$ is $(n+1)^{th}$ prime number and
$q$ is $n^{th}$ prime number.
After $n$ sieving steps, there is one residue class mod 2 remaining and $p-2$ residue classes mod each of $p_2=3,p_3=5,\ldots,p_n$. Hence there are $k_nx+O_n(1)$ unsieved numbers up to $x$, where $$ k_n = \frac12\prod_{3 \le q \le p_n}\frac{p-2}{p} $$
If you sieve up to $\sqrt x$ rather than to some fixed $n$, then you get $\ll x/\log^2 x$ by the Brun sieve.