I have a recurrence relation for a countably infinite sequence that contains the integers divisible by 5 but not by 7. The relation I came up with is:
$5((n-1) + \lceil \frac{n}{6} \rceil)$
The relation seems to hold. However, I wanted to try to solve for n to find the inverse.
Now, if this is in fact the correct relation for the countable sequence then it should be invertible since a countable set is a bijection. However, I do not know how to handle floor or ceiling functions when solving for n.
Can this be solved for n? If so, how is the ceiling function handled?
Hint:
As $n$ is an integer, there are only 2 cases: $n=6k$ or $n=6k+c, 0<c<6$. For the first case, upper ceiling of $n/6$ is just $k$. While for another case, $k<n/6<k+1$,so it gives a value of $k+1$.