<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] Let's replace $10$ with a lower number, $5$, and illustrate the role of partitions in the result.
Only $f+f^2+f^4$ contributes to the coefficient of $x^5$. The $f$ term has none, of course.
The $f^2=f\cdot f$ term contributes a coefficient of $2$ from $x^1\cdot x^4$ and $x^4\cdot x^1$. Note that we get two terms because one contribution has $x^4$ from the first factor and the other has the $x^4$ factor coming from the second $f$ factor. Ordering of the partitions is important.
The $f^4$ term contributes a coefficient of $4$ which comes from $x^2\cdot x^3$ and $x^3\cdot x^2$ in the square of $f^2$. We get a contribution of $4$ here because the $x^3$ factor from $f^2$ had a coefficient of $2$ in $f^2$, which in turn came from $x^1\cdot x^2$ and $x^2\cdot x^1$ in the square of $f$.
So, the total of six terms in $x^5$, therefore a coefficient of $6$, came from the following partitions:
$5 = 1+4$ (from $f^2$)
$5 = 4+1$ (from $f^2$)
$5 = (1+1)+(1+2)$ (from $f^4$, with the parentheses showing the separate inputs to $f^4$ from $f^2$)
$5 = (1+1)+(2+1)$ (from $f^4$)
$5 = (1+2)+(1+1)$ (from $f^4$)
$5 = (2+1)+(1+1)$ (from $f^4$)
In other words:

*

*We picked partitions into powers of $2$, corresponding to the powers of $x$ in $f$.


*We picked those partitions where the number of terms is also a power of $2$, from the powers of $f$ you use in the outer function evaluation.


*For each partitioning satisfying the criteria above, we count all the distinguishable orderings.
Now try this partition scheme using $10$ as the input. There could be $1,2,4,$ or $8$ powers of $2$ in the partition, and for each partition that has different powers of $2$ there will be multiple orderings to count as we saw with the $x^5$ coefficient above.

 We have $8+2$ with $2$ orderings, $4+4+1+1$ with six orderings, $4+2+2+2$ with four orderings, and $2+2+1+1+1+1+1+1$ with $28$ orderings. Thus the coeficient of $x^{10}$ will be $40$.

The entire series begins as
$x+2x^2+2x^3+3x^4+6x^5+8x^6+8x^7+16x^8+22x^9+\text{[spoiler]}x^{10}+...$
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