In enumerating the primes algorithmically, the "completely naive" algorithm loops through every integer, the difference between succeeding candidates being 1. The first improvement is to "pre-identify" 2 as prime, and then only consider odd integers as candidates, the difference between succeeding candidates being 2. But this type of "improvement" can be extended indefinitely. The next step is "pre-identifying" both 2 and 3, and then only considering non-multiples of 2 and 3 as candidates. The differences between candidates forms a period-2 sequence of repeating (2,4), i.e. 5,7,11,13,17,19,23,25,29,31,... with difference sequence 2,4,2,4,2,4,2,4,2,... . The next step is to bring in 5 and additionally eliminate multiples of 5. The "difference sequence" takes on a period of 8, i.e. 4,2,4,2,4,6,2,6, repeated.
Is there a formula for calculating such difference sequences for any initial set of primes? Or even just a formula for calculating what the period of the difference sequence would be?
It's called a wheel sieve as you found out in the comments. It actually can have a derivation but quite honestly not much point to it. Here's how it would go: $\gcd(a,b)=1\implies\gcd(a-b,b)=1$ that is, if $b$ is a remainder class mod $a$ that can have primes, so is $a-b$ . this is basically using the distributive property and factoring out. If we have to values that aren't relatively prime, then we can factor out a factor greater than 1 (and primes don't have factors greater than 1 that aren't themselves). So it turns out just pick all the relatively prime remainder classes. The number of relative primes to a number is always even, and the sequence will always be symmetric around the halfway point through the repeating part. You really want a formula for Euler's phi function: $$\varphi(n)=n\prod_{d\mid n,d\in \Bbb{P}}(1-{1 \over d})$$ this tells you how long the repeat part possibly goes on for. As to the values of differences, they'll typically be divisible by primes you eliminated.