Starting from $n=0$ you have
$$d=0,0,1,1,2,2,3,3,4,4,5,5,6,6,\cdots$$
$$k=0,0,1,1,1,1,2,2,2,2,3,3,3,3,\cdots$$
In the last row, you see that the length of the runs is $4$, but the first run has only two elements.
Hence,
$$k=\left\lfloor\frac{n+2}4\right\rfloor$$
Alternatively,
$$d=\left\lfloor\frac n2\right\rfloor\equiv n=2d+d'$$
where $d'=0,1$.
$$k=\left\lceil\frac d2\right\rceil\equiv d=2k-k'$$
where $k'=0,1$.
Then
$$n=4k-2k'+d'=4k+k''$$
where $k''=-2,-1,0,1$.
By shifting,
$$n+2=4k+k'''$$ where $k'''=0,1,2,3$ and
$$k=\left\lfloor\frac{n+2}4\right\rfloor.$$
Note there is no general method to deal with recurrence relations involving the ceiling function. It depends on things like how the ceiling function is being used, what the initial value is, etc. To handle them, I suggest you calculate a few values to see if there's any pattern you can determine, and then try to prove it.
For example, consider the equation you gave of
$$a_n=0.2*\lceil a_{n-1} \rceil \tag{1}\label{1}$$
You didn't specify an initial value, so I will show the behavior of various ranges. First, if $-1 \lt a_0 \le 0$, then $a_1 = 0.2 \times \lceil a_0 \rceil = 0.2 \times 0 = 0$. Thus, you get $0 = a_2 = a_3 = \ldots$, i.e., $a_n = 0$ for all $n \ge 0$. Note having $a_1 = 20$ is not consistent with using $a_0 = 0$ and the recurrence relation for the rest of the values. However, if you do start from $a_1 = 20$ instead, then you get $a_2 = 4$, $a_3 = 0.8$, then $0.2 = a_4 = a_5 = \ldots$, i.e., $a_n = 0.2$ for $n \ge 4$.
Next, if $0 \lt a_0 \le 1$, then you have $a_1 = 0.2 \times \lceil a_0 \rceil = 0.2 \times 1 = 0.2$. However, then $a_3 = 0.2$ as well and, in fact, $a_n = 0.2$ for all $n \ge 1$.
However, if $1 \lt a_0 \le 2$, then $a_1 = 0.2 \times \lceil a_0 \rceil = 0.2 \times 2 = 0.4$. Once again, you then get $0.2 = a_2 = a_3 \ldots$, so $a_n = 0.2$ for all $n \ge 2$.
In fact, for any positive $a_0$, you'll eventually get all later values of $a_n$ to be $0.2$, and for any $a_0$ being non-positive, you'll eventually get all later values to be $0$.
Best Answer
Hint:
$$f(3)=2f(2)-1=3,\\f(4)=2f(2)-1=3,\\f(5)=2f(3)-1=5,\\f(6)=2f(3)-1=5,\\f(7)=2f(4)-1=5,\\f(8)=2f(4)-1=5,\\f(9)=2f(5)-1=9,\\f(10)=2f(5)-1=9,\\f(11)=2f(6)-1=9,\\f(12)=2f(6)-1=9,\\\cdots$$
and a pattern emerges. To make it even more obvious, consider $g(n):=f(n)-1$, and
$$g(3)=2g(2)=2^1,\\g(4)=2g(2)=2^1,\\g(5)=2g(3)=2^2,\\g(6)=2g(3)=2^2,\\g(7)=2g(4)=2^2,\\g(8)=2g(4)=2^2,\\\cdots$$