<s>[INST] Coefficient of $x^{10}$ in $f(f(x))$ Let $f\left( x \right) = x + {x^2} + {x^4} + {x^8} + {x^{16}} + {x^{32}}+ ..$, then the coefficient of $x^{10}$ in $f(f(x))$ is _____.
My approach is as follow
$f\left( {f\left( x \right)} \right) = f\left( x \right) + {\left( {f\left( x \right)} \right)^2} + {\left( {f\left( x \right)} \right)^4} + {\left( {f\left( x \right)} \right)^8} + ..$
Let $T = f\left( x \right);U = {\left( {f\left( x \right)} \right)^2};V = {\left( {f\left( x \right)} \right)^4};W = {\left( {f\left( x \right)} \right)^8}$
Let the coefficient of $x^{10}$ in $T $ is zero
Taking U
${\left( {f\left( x \right)} \right)^2} = {\left( {x + {x^2} + {x^4} + {x^8} + ..} \right)^2}$
Hence the coefficient of $x^{10}$ in  $U$ is $2x^{10}$ hence $2$
For $V$ and $W$ it is getting complicated hence any short cut or easy method to solve it
 [/INST] We have
$$f(f(x))=f(x)+f(x)^2+f(x)^4+f(x)^8+f(x)^{16}+\ldots$$
To find the coefficient of $x^{10}$ in this expansion, note that the smallest power of $f(x)^k$ for some $k$ is $x^k$.
Hence, to find the coefficient of $x^{10}$, we only need to consider
$$f(x)+f(x)^2+f(x)^4+f(x)^8$$
The coefficient of $x^{10}$ in $f(x)^k$ represents the number of solutions to
$$\sum_{i=1}^k a_i=10$$
Where $a_i$ are integral powers of $2$.
We can systematically find these solutions using the binary representation of $10_{10}=1010_2$
If we are a little hand wavy with how we use binary representation, we can say that $10_{10}$
$$=1010_2$$
$$=210_2=1002_2$$
$$=130_2=202_2$$
Note that the expressions in the $i^\text{th}$ line represent the different ways to represent $10$ as a sum of $i+1$ integral powers of $2$ e.g. in the second line we have $210_2\implies 10=4+4+2$. We created the expressions in a line by recursively taking the expressions from the previous line and splitting a single power of $2$ in half.
We then have to account for permutations. Clearly $f(x)$ has no coefficient of $x^{10}$
The coefficient of $x^{10}$ in $f(x)^2$ is the number of permutations of $\{8,2\}$, which is $2!=\boxed{2}$.
The coefficient of $x^{10}$ in $f(x)^4$ is the number of permutations of $\{4,2,2,2\}$ and $\{4,4,1,1\}$, which is $\frac{4!}{3!}+\frac{4!}{2!2!}=\boxed{10}$.
The coefficient of $x^{10}$ in $f(x)^8$ is not as easily found using the recursive method we used before. However, it is easily found that the only way to express $10$ as the sum of $8$ powers of $2$ is $1+1+1+1+1+1+2+2$. There are $\frac{8!}{6!2!}=\boxed{28}$ permutations of this.
Hence, the total coefficient of $x^{10}$ is $2+10+28=\boxed{40}$
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