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!
Best Answer
Despite the comments to the question (including mine), this is a bit easier than it seems at first sight. We can show that $L_{\max}=2$ or $3$. Almost certainly we have $L_\max=3$. However, determining which of these is actually the case seems to be beyond current technology, according to this MO answer "Are all primes in a PAP-3?".
Showing that, $L_{\max} < 4$ is easy. That is, not every odd prime is contained in an arithmetic progression of primes of length 4. More specifically, 3 is not contained in an arithmetic progression of length 4. Suppose that $\lbrace x, x+d, x+2d, x+3d \rbrace$ was such a progression for $d > 0$. Then $x\not=2$, otherwise we would have $x=2,d=1$, but $x+2d=4$ is not prime. So, $x=3$. But, then, $x+3d=3(1+d)$ is not prime.
Edit: Looking at $\tilde L_\max \equiv \max_A\liminf_i \vert A_i\vert$ might be more interesting. I expect that this is infinite but, again, showing that $\tilde L_\max > 2$ appears to be beyond current means.