Let $ a_i$ denote terms in the said sequence ($ a_1 = 1$, $ a_2 = 24$), and let $ S_n : = \{a_1,\ldots,a_n\}$, $ S_\infty : = \bigcup_{n \in \mathbb{Z^+}} S_n$ = the set of ultimately $ \textit reachable$ numbers, and $ U : = \mathbb{Z^+} - S_\infty$ = the set of forever $ \textit unreachable$ numbers. Suppose not all positive integers occur in the sequence i.e. $ U \not = \emptyset$. $ U$ cannot contain any composite number: Suppose $ u \in U$ is composite, and by the Well-Ordering Principle (WOP), we may assume this is the minimal composite. Then among $ \Psi (u) = \{$ numbers $ < u$ having a common factor with $ u\}$, $ \exists v = a_j \in \Psi (u)$ with $ j$ maximal (last number having common factor with $ u$ to appear). But the next number $ a_{j + 1} = u \in S_\infty$. (contradiction) $ \clubsuit$ $ U$ cannot contain any prime number: Let $ p$ be the minimal (prime) element of $ U$ (using WOP again). (Note $ p \not = 2$ because we have $ a_3 = 2$). Since the composite number $ 2p \in S_\infty$, $ \exists j > 3$ such that $ 2p = a_j$. But the next number would be $ a_{j + 1} = p \in S_\infty$ (contradiction) $ \clubsuit$ (And obviously $ 1 \notin U$). We conclude that $ U = \emptyset$ and that $ S_\infty = \mathbb{Z^+}$ $ \blacksquare$

I believe the above proof is not correct. If we assume for a moment that 8 is the smallest integer in the set of unreachable numbers, then the last (and largest) number smaller than 8 which has a common factor with 8 is 6, which is the fifth number in the sequence. So, according to YeOldeMan's proof, the next number in the sequence must be 8, but it is 3. I'll post my proof later.

Let $U$ be the set of forever unreachable numbers and $n$ be the smallest member of $U$. If $(a_k, n) > 1$, then $a_{k+1} < n$ and, since $n$ is finite, this can only happen finitely often; $M$ exists such that $(M, n) > 1$ and for all $m \ge M$ with $(m, n) > 1$, we have $m \in U$. In particular, $Mp \in U$ for all primes $p$. If now $p | a_k$ for some prime $p$, then $a_{k+1} < Mp$ and this too, can happen at most finitely often. Thus, $M_p$ exists for which $p$ divides $M_p$ and if $p$ divides $m$ for any $m \ge M_p$, $m \in U$. Now, let $k$ be such that $a_k > \max_{p < M_2} M_p$. That is, all prime divisors of $a_k$ are larger than $M_2$. In particular, let $p_1$ be the smallest prime divisor of $a_k$. Then, since $2p_1 > M_2$, we must have $a_{k+1} < 2p_1$. In other words, $a_{k+1} = p_i$ for some prime $p_i$ dividing $a_k$. But then we have $a_{k+2} = 2p_i > M_2$; contradiction. Note that it's not necessary that $a_2 = 24$; it could be any positive integer larger than $1$. Furthermore, when $a_2 = 4$, I conjecture the following: 1) If $a_k = p$ ($p$ prime), then $a_{k-1} = 2p, a_{k+1} = 3p$. This only holds for $p$ large enough if $a_2 \neq 4$ 2) If $a_k = n$ and $n \neq p,3p$, then $k = n + o(n)$. For $n = p$ we have $k = 2n + o(n)$ and for $n = 3p$ we have $k = 2n/3 + o(n)$. (this might be hard to prove, although I have very little idea) Also, I can imagine the possibility that the sequence is exactly the same for all large enough $k$, whatever the value of $a_2$ is. Where the meaning 'large enough' does depend on the value of $a_2$, of course. I am hesitant, but tempted to call this a conjecture too. For example, whether $a_2 = 4$, or $a_2 = 8$, we have that the set $\{a_1, a_2, .., a_8\}$ equals the set $\{1,2,3,4,6,8,9,12\}$. And in both cases $a_9 = 10$. So, for $k \ge 9$, the sequences are indentical. I do have more conjectures/questions concerning these sequences, but the above I believe to be the most important.

See http://www.mimuw.edu.pl/~malcin/ekg.pdf for some serious stuff