Suppose that we take some two natural numbers $a$ and $b$ and that $a<b$.

Then, we may try to arrive at $b$ by starting at $a$ by this set of rules:

1. To arrive at $a_1$ which is such that we have $a< a_1 \leq b$ it is only allowed that we sum $a$ with one of its divisors different from $1$. If $a_1=b$ we are done.

2. If $a_1<b$ then to arrive at $a_2$ which is such that we have $a<a_1<a_2\leq b$ we perform step 1) with $a_1$.

3. Repeat 2) until we arrive at $a_{k(a,b)}=b$

As an example, take $a=12$ and $b=100$, we have $12 \to 16 \to 20 \to 40 \to 80 \to 100$.

As an example where $\gcd (a,b)=1$ take $a=15$ and $b=22$. Then we have $15 \to 18 \to 20 \to 22$.

As an example where $\gcd (a,b)=1$ and where we cannot arrive at $b$ take $a=18$ and $b=23$.

Of course, $k(a,b)$ can be multi-valued since for the example $a=12$ and $b=100$ we also have $12\to 24\to 48\to 96\to 100$. But this question is not about behaviour of $k(a,b)$ (although it could be that its behaviour has some awesome features).

If this procedure exists for some $a$ and $b$ and $a<b$ we may say that $b$ is reachable by $a$.

Can we find some condition(s) on $a$ and $b$ that is(are) both necessary and sufficient which guarantee that $b$ is reachable by $a$?