For every positive integer $n$, let $\bmod_{5}(n)$ be the remainder obtained when $n$ is divided by 5. Define a function $f:\{0,1,2,3,\dots\}\times\{0,1,2,3,4\}\to\{0,1,2,3,4\}$ recursively as follows:
$$f(i,j)=\begin{cases}\bmod_{5}(j+1) & \text{if $i=0$ and $0\leq j\leq4$,}\\f(i-1,1) & \text{if $i\geq1$ and $j=0$,}\\f(i-1,f(i,j-1)) & \text{if $i\geq1$ and $1\leq j\leq4$.}\end{cases}$$
What is $f(2015,2)$?
对于每个正整数 $n$,令 $\bmod_{5}(n)$ 为 $n$ 除以5的余数。递归定义函数 $f:\{0,1,2,3,\dots\}\times\{0,1,2,3,4\}\to\{0,1,2,3,4\}$ 如下:
$$f(i,j)=\begin{cases}\bmod_{5}(j+1) & \text{if $i=0$ and $0\leq j\leq4$,}\\f(i-1,1) & \text{if $i\geq1$ and $j=0$,}\\f(i-1,f(i,j-1)) & \text{if $i\geq1$ and $1\leq j\leq4$.}\end{cases}$$
$f(2015,2)$ 是多少?