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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 $a$?

Since I am not sure did I explain this procedurally in a good way by 1), 2) and 3), I will describe here again the procedure in other words. It is very simple, choose natural numbers $a$ amd $b$ with $a<b$. We are trying to arrive at $b$ by adding to $a$ some divisor of $a$ different than $1$. Then we arrive at $a_1=a+\alpha_1$ , where $\alpha_1 | a$. Then if we did not arrive at $b$ with that single step we add to $a_1$ some divisor of $a_1$ different than $1$ to arrive at $a_2=a_1 + \alpha_2$, where $\alpha_2 | a_1$. We repeat this until we arrive at $b$, if possible. The question is of when is this possible.

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

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 to guarantee that $b$ is reachable by $a$?

Since I am not sure did I explain this procedurally in a good way by 1), 2) and 3), I will describe here again the procedure in other words. It is very simple: choose natural numbers $a$ and $b$ with $a<b$. We are trying to arrive at $b$ by adding to $a$ some divisor of $a$ different than $1$. Then we arrive at $a_1=a+\alpha_1$ , where $\alpha_1 | a$. Then if we did not arrive at $b$ with that single step, we add to $a_1$ some divisor of $a_1$ different than $1$ to arrive at $a_2=a_1 + \alpha_2$, where $\alpha_2 | a_1$. We repeat this until we arrive at $b$, if possible. The question is: when is this possible?

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 $a$?

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 $a$?

Since I am not sure did I explain this procedurally in a good way by 1), 2) and 3), I will describe here again the procedure in other words. It is very simple, choose natural numbers $a$ amd $b$ with $a<b$. We are trying to arrive at $b$ by adding to $a$ some divisor of $a$ different than $1$. Then we arrive at $a_1=a+\alpha_1$ , where $\alpha_1 | a$. Then if we did not arrive at $b$ with that single step we add to $a_1$ some divisor of $a_1$ different than $1$ to arrive at $a_2=a_1 + \alpha_2$, where $\alpha_2 | a_1$. We repeat this until we arrive at $b$, if possible. The question is of when is this possible.

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 $a<a_1<b$ it is only allowed that we sum $a$ with one of its divisors different from $1$.

2. To arrive at $a_2$ which is such that we have $a<a_1<a_2<b$ 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 $a$?

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 $a$?