I raised this question in my answer to [On the rank of a matrix $S$ with coefficients in $\mathbb F\_{2^m}$](https://mathoverflow.net/questions/154861). Let $s\_1,s\_3,s\_5,\dots$ be indeterminates over the field $\mathbb{F}\_2$, and recursively set $s\_{2n}=s\_n^2$, with $s\_0=1$. Let 
 \begin{eqnarray\*} F(x) & = & \frac{\sum\_{n\geq 1} s\_{2n-1}x^n}{\sum\_{n\geq 0} s\_{2n}x^n}\\
 & = & \sum\_{n\geq 0} p\_n(s\_1,s\_3,\dots)x^n.
 \end{eqnarray\*}
Is it true that the number of terms of the polynomial $p\_n(s\_1,s\_3,\dots)$ is equal to the number of ways to write $n$ as a sum of powers of 2, without regard to order? For instance, the number of terms of $p\_5$ is four, corresponding to $4+1=2+2+1=2+1+1+1=1+1+1+1+1$. A bijective proof would be especially interesting. Is there a generalization to $\mathbb{F}\_p$?



Number of terms in certain polynomials over $\mathbb{F}_2$
The claim is true. Lets start by simplifying the expression as follows:
$$x\left(1+\frac{1}{1+s\_1x+s\_2x^2+s\_3x^3+\cdots}\right)=\frac{\sum\_{n\geq 1}s\_{2n-1}x^{2n}}{\sum\_{n\geq 0} s\_{2n} x^{2n}}.$$
This follows because $\sum\_{n\geq 0}s\_{2n}x^{2n}=\left(\sum\_{n\geq 0}s\_nx^n\right)^2$, and some basic manipulations over $\mathbb F\_2$.


So we are interested in the coefficient of $x^{2n-1}$ in 
$$1+\frac{1}{1+s\_1x+s\_2x^2+\cdots}=\left(\sum\_{n\geq 0}s\_nx^n\right)+\left(\sum\_{n\geq 0}s\_nx^n\right)^2+\left(\sum\_{n\geq 0}s\_nx^n\right)^3+\cdots,$$
but by taking binary expansions we see that each term is of the form
$$\left(\sum\_{n\geq 0}s\_{2^{i\_1}n}x^{2^{i\_1}n}\right)\cdots \left(\sum\_{n\geq 0}s\_{2^{i\_k}n}x^{2^{i\_k}n}\right),$$
where $i\_1,i\_2,\dots,i\_k\geq 1$. It follows that the terms in the coefficient of $x^{m}$ correspond exactly to the solutions of $m=j\_1+2j\_2+\cdots 2^rj\_{r+1}$. The ones that appear with a nonzero coefficient in $\mathbb F\_2$ are precisely the ones where the $j\_1,j\_2,\dots$ are all odd. So the number of terms in $p\_n$, with your notation, is the number of ways we can write 
$$2n-1=\sum\_{r= 0}^k (2i\_r+1)2^{r}$$
 for some k. This is the same as $$n=2^k+\sum\_{r= 0}^k i\_r 2^r$$ 
which gives exactly the number of ways to partition $n$ as powers of $2$. For example when $n=6$, we can write $11$ as $1+10, 1+6+4, 5+6, 5+2+4, 9+2,11$ each corresponding to the terms in $p\_6=s\_1s\_5^2+s\_1s\_3^2s\_1^4+s\_5s\_3^2+s\_5s\_1^2s\_1^4+s\_9s\_1^2+s\_{11}$.


