Define the *length* of a set of arithmetic progressions
of natural numbers
$A=\lbrace A_1, A_2, \ldots \rbrace$
to be $\min_i | A_i |$: the length of the shortest sequence
among all the progressions.
Say that $A$ *exactly covers* a set $S$
if $\bigcup_i A_i = S$.
Let $P'$ be the primes excluding 2.

What is the longest set of arithmetic progressions that exactly covers the primes $P'$?

In other words, I want to maximize the length of a set of such arithmetic progressions. Call this maximum $L_{\max}$.

$L_{\max} \ge 2$ because $$ P' \;=\; \lbrace 3,5 \rbrace \cup \lbrace 7,11 \rbrace \cup \cdots \cup \lbrace 521,523 \rbrace \cup \cdots $$ Perhaps it is possible that $$P' \;=\; \lbrace 3, 11, 19 \rbrace \cup \lbrace 5, 17, 29,41,53 \rbrace \cup \lbrace 7,19,31,43 \rbrace \cup \cdots \;,$$ but I cannot get far with sequences of length $\ge 3$. (I know Green-Tao establishes that there are arbitrarily long arithmetic progressions in $P$, but I don't know if that helps with my question.)

I am number-theoretically naïve, and apologize if this question is nonsensical or trivial. In any case, I appreciate the tutoring!

**Addendum**. Although my question should be revised (as Idoneal suggests)
in light of George Lowther's proof that 3
cannot be in a progression of length 4, George has shown that it is likely that $L_{\max}=3$
but certification requires resolving an open problem. So I've added the *open-problem* tag.
Thanks for everyone's interest!

5more comments