How can I find the Coefficient of $x^n$ in $\frac{1}{(1-x).(1-x^2).(1-x^4).(1-x^8).......)}$ ?
Using $\frac{1}{1-r}=1+r+r^2+....$, we can write we need
$$[x^n]\ \ in\ \ (1+x+x^2+..)(1+x^2+x^4+..)(1+x^4+x^8+..)(1+x^y+x^{2y}+..)$$ where $y=floor(log_2(n))$.
What should I do next? This is sort of generating function for number of ways of expressing a given number as a sum of powers of 2.
This is the generating function of the sequence OEIS A018819, which gives the number of partitions of $n$ into powers of $2$. The OEIS entry gives the recurrence
$$\left\{\begin{align*} a_{2m+1}&=a_{2m}\\ a_{2m}&=a_{2m-1}+a_m \end{align*}\right.$$
with the easy proof. (If $n$ is odd, there is a part of size $1$; removing it gives a partition of $n-1$ into powers of $2$. If $n$ is even, either there is a part of size $1$, whose removal gives a partition of $n-1$, or else all parts have even sizes, and dividing each part by $2$ gives a partition of $\frac{n}2$.)
There does not seem to be a nice closed form. There is a recurrence
$$a_n=\sum_{k=0}^{\lfloor n/2\rfloor}a_k$$
with initial value $a_0=1$.