A sequence of primes i.e. $2,3,5,7,11,\dots$ could be described with other offsets than zero, especially with a prime number as offset:
$2:1,3,5,9,\dots$
$3:2,4,8,10,\dots$
$5:2,6,8,12,\dots$ etc
Intuitively those sequences are unique for each prime number and that for each prime number there exist an unique minimal finite sequence for that prime number. Example: for $2$ that unique sequence is just $1$ and for $3$ the sequence $2,4$ is unique. No other primes has offset sequences starting with those finite sequences.
But what about $5$? There is a lot of primes with sequences starting at $2,6,8,12$, but there is no prime less than $100,000,000$ having the same five first number in it's sequence:
$5:2,6,8,12,14$
Can it be proved that the offset sequence $2,6,8,12,14$ is unique for $5$ among primes? What discipline in number theory deal with questions like this?
Examples:
5:2,6,8,12,14,18,24,26,32,36,38,42,48,54,56,62,66,68,74,78,...
7:4,6,10,12,16,22,24,30,34,36,40,46,52,54,60,64,66,72,76,82,...
11:2,6,8,12,18,20,26,30,32,36,42,48,50,56,60,62,68,72,78,86,...
13:4,6,10,16,18,24,28,30,34,40,46,48,54,58,60,66,70,76,84,88,...
17:2,6,12,14,20,24,26,30,36,42,44,50,54,56,62,66,72,80,84,86,...
19:4,10,12,18,22,24,28,34,40,42,48,52,54,60,64,70,78,82,84,88,...
23:6,8,14,18,20,24,30,36,38,44,48,50,56,60,66,74,78,80,84,86,...
29:2,8,12,14,18,24,30,32,38,42,44,50,54,60,68,72,74,78,80,84,...
Note that $$\{2,6,8,12, 14\}\equiv \{2,1,3,2,4\}\pmod 5$$
It is obvious that for any prime $p\neq 5$ one of $p+1,p+2,p+3,p+4$ is divisible by $5$. The desired claim follows immediately.
Note: In general, it is clear that the infinite difference sequence determines the prime uniquely. After all, the sequence for $p$ can not contain any multiples of $p$ while for any other prime $q$ there are infinitely many primes of the form $q+pk$. As with $5$, to be sure your finite starting list determines the prime, you need one difference of each of the non-zero residues.