<s>[INST] How can I solve this recursive function $f(n) = f(f(n+1))$? I am trying to solve this:
$$
f(n)  =
\begin{cases} n - 1,&          n > 5\\
f(f(n+1)),&     n\leqslant  5
\end{cases}
$$
What is the technical name of this kind of function ? -->   $f(f(n+1))$
A function within another function?
\begin{align}
f(1) &= f(f(1+1)) = f(f(2)) \\
f(2) &= f(f(2+1)) = f(f(3))\\
f(3) &= f(f(3+1)) = f(f(4))\\
f(4) &= f(f(4+1)) = f(f(5))\\
f(5) &= 5 - 1 = 4\\
f(6) &= 6 - 1 = 5\\
f(7) &= 7 - 1 = 6
\end{align}
I can only find the answer when $n > 5$. When $n\leqslant5$, I can only see $f(f(k))$. How to find the answers of $f(1)$, $f(2)$, $f(3)$ and $f(4)$ ? Thank you very much.
 [/INST] An example will make it understand better
$ {let}$ $    n  = 5$ now $ f(n)=f(f(n=1))$
so, $f(5)=f(f(5+1))$
$=f(f(6)) = f(5)$
$f(5)=f(f(5+1))$
$=f(f(6)) = f(5)$
$f(5)=f(f(5+1))$
$=f(f(6)) = f(5)$
..............
 and it goes on so the function is not defined at 5
Now let $n=4$
so, $f(3)=f(f(3+1))$
$=f(f(4))$ now $f(4)$ will be $f(f(5))$
so, $f(3)=f(f(4))=f(f(f(5)))$
which again is not defined,
so we can say that the function is of indeterminate form for $n\leqslant5$
</s>