A conjecture that I believe is possible to prove:
For any $N\in\mathbb N$ there is a $d\in\mathbb N$ such that $p,p+d,p+2d,\dots,p+N\cdot d$ all are primes (for some prime $p$).
My context: it seems that greater $\omega(d)$ (number of different prime factors) gives greater $N$.
This has already been proven, and is called the Green-Tao theorem.
And yes, the longer chain you want, the more prime factors you require of $d$. Basically, if you want a chain to be $n$ long, $d$ must be divisible by all primes less than or equal to $n$.
The only exception to the above rule is if $p=n$, in which case $d$ doesn't have to be divisible by $n$ itself. For instance, $$5,11,17,23,29$$ is an arithmetic sequence of five primes, with $d=6$, and since the first prime in the sequence is equal to the number of terms in the sequence, it works out without $d$ being divisible by $5$. But for any other sequence of five primes, $d$ must be divisible by $5$ as well as $2$ and $3$.