How to solve this recurrence relation $ f(x, 0)=3 x \text { and } f(x, n+1)=f(f(x, n), n)$?

Question

Given that $f(x, 0)=3 x \text { and } f(x, n+1)=f(f(x, n), n)$, find an explicit formula for $f(x, y)$. Use this formula to calculate $f(2005,3)$.

What topics do I need to learn to solve this problem?

Answer 1

Take $x=2005$ and $n=2$, then $f(2005,3)=f(f(2005, 2), 2)$

Take $x=2005$ and $n=1$, then $f(2005,2)=f(f(2005, 1), 1)$

Take $x=2005$ and $n=0$, then $f(2005,1)=f(f(2005, 0), 0)$

Now $f(2005,0)=3 \times 2005=6015$

Thus $f(2005,1)=f(f(2005, 0), 0)=f(6015, 0)=3 \times 6015=18045$

$f(2005,2)=f(f(2005, 1), 1)=f(18045, 0)=3 \times 18045=54135$

$f(2005,3)=f(f(2005, 2), 2)=f(54135, 0)=3 \times 54135=162405$

In general: $$f(x, 0)=3x, f(x, 1)=3^2x, ... , f(x, n)=3^{n+1}x$$

$$f(2005, 3)=3^{4}*2005=162405$$