Iterations of modulus operation

Question

While working on a completely unrelated task, I thought up the following problem:

Consider the following process. Let $a_0$ and $n$ be given, and determine $a_1,\ldots, a_k$ as follows: $$a_{j+1} = n \text{ mod } a_j$$ ...where the process terminates when $a_j \mid n$. That is, $a_k = 0$.

Define $f(a_0, n) = k$.

Aside from direct iteration, how might one compute or express $f(a_0, n)$?

Example: $f(124323, 938342) = 12$

68081 = 938342 mod 124323
53289 = 938342 mod 68081
32429 = 938342 mod 53289
30330 = 938342 mod 32429
28442 = 938342 mod 30330
28198 = 938342 mod 28442
 7808 = 938342 mod 28198
 1382 = 938342 mod 7808
 1346 = 938342 mod 1382
  180 = 938342 mod 1346
    2 = 938342 mod 180
    0 = 938342 mod 2

This process always terminates (as can be trivially proven).

Aside from the obvious $f(n\pm1, n)$ case, I don't really know where to begin approaching this--problems I just think up tend to be harder than the math I know. ;)

So, my questions are:
1) Obviously, if someone sees an obvious route to computing this, that would be great.
2) If this is something that is known to be impossible, that would also be good to know.
3) If someone has a hint, I'd appreciate that--I like trying to figure things out, but I know that good hints can be difficult to write.

To get an exact expression I don't think you can do much better than just doing the iterations. To get a feeling for $f(a_0, n)$, you can approximate your iteration with a random process, where $a_{j+1}$ is a random number between $0$ and $a_j - 1$. (That's also roughly what happens when $n$ is a random large number.) In that case I think you should be able to prove that $f(a_0, n) \approx \log a_0$, which seems to be a pretty good estimate for your iteration ($\log 124323\approx 11.7$). — TMM, Jan 18 '14 at 21:56

score 4 · Accepted Answer · edited Apr 13 '17 at 12:20

This is what I call Gauss's algorithm for computing inverses, when $n = p$ prime. By replacing $a$ by $a' := n\ {\rm mod}\ a = n - qa\,$ it yields $a'\equiv -qa\,$ which is a smaller multiple of $a$. Iterated, it yields a descent on the elements of the ideal $(a)$ of all multiples of $a$ mod $n$. It terminates with a factor of $n$. This factor can only be $1$ when $n = p$ is prime, so it shows that $1$ is a multiple of $a$, by the product of the sequence of quotients $q_i,\,$ so yielding $\,a^{-1}$ mod $n$.

Iterations of modulus operation

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