The case of factor $3$ is more interesting because with factor $1$ it's easy to prove that the sequence hits $1$. We can prove this by induction: For $n_0=1,2$ this is clearly true. Suppose that it is true for all starting values $1,2,\dots,n_0-1$. Then if $n_0$ is even, the next number will be $n_0 / 2 < n_0$ and thus the sequence will hit $1$. If $n_0$ is odd, the next number will be $n_0 + 1$ and the number after it will be $\frac{n_0 + 1}{2} < n_0$ and thus the sequence will hit $1$ again by induction.