Irska

This is actually true for the full Fibonacci sequence, not just the iterated one. Since the Fibonacci numbers depend only on the previous two values. Fibonacci modulo $n$ must be preperiodic with period at most $n^2$. In fact it is periodic for all of the values that we examine.

Calculation shows that $f(n)$ mod $2$ has a period of $3$ and $f(n)$ mod $7$ has an "antiperiod" of $8$ (meaning $f(n+8)\equiv -f(n)$ mod $7$), and so a period of $16$. Examination of the repeated values reveals that $f(n)$ is a multiple of $2$ exactly when $n$ is a multiple of $3$, and a multiple of $7$ exactly when $n$ is a multiple of $8$. Thus $14$ divides $f(n)$ exactly when $24$ divides $n$.

Similarly, we see that both modulo $16$ and modulo $9$, $f(n)$ has a period of $24$ and is equivalent to $0$ if and only if $n$ is a multiple of $12$. The result follows.