A recurrence is given by 
$f[0]=2x$, $f[1]=3x^3-x^2+x+1$, 
$$ 
f[n]=(x^{2^n}+1)f[n-1]+(x^{2^n}+1)(x^{2^n-1}+1)
$$


How does the PRODUCT of the nonzero coefficients of $f[n]$ scale with $n$?



Product of polynomial coefficients of a recurrence
It follows by induction from the recursion that for all $n\ge2$ the polynomial $f\_n$ writes as $f\_n=(x^{2^{n}}+1)(x^{2^{n-1}}+1)g\_n $ where $g\_n$ satisfies
$$\begin{cases} g\_2:=3x+2 \\ g\_n:=(x^{2^{n-2}}+1)g\_{n-1}+1,\quad & \mathrm{if }\; n>2\end{cases}$$
and has degree $2^{n-1}-1$. Since $f\_n=(x^{2^{n}}+1)(x^{2^{n-1}}+1)g\_n $, the list of coefficients of $f\_n$ is that of $g\_n$ repeated $4$ times, and their product is the fourth power of the product $P\_n$ of the coefficients of $g\_n$. Also, the list of coefficients of $g\_n$ is the one of $g\_{n-1}$ repeated twice, with the constant term incremented by one. Thus the constant term is $g\_n(0)=n$ and
$$\begin{cases} P\_2:=3\cdot 2, \\ P\_n:= {n\over n-1}\ P\_{n-1}^2\quad & \mathrm{if }\; n>2\end{cases}$$
whence it follows by induction, for $n\ge2$
$$P\_n=3^{2^{n-2}}n\prod\_{k=1}^{n-1}k^{2^{n-1-k}}.$$


So the product of coefficients of $f\_n$ is $P\_n^4=3^{2^{n}}n^4\prod\_{k=1}^{n-1}k^{2^{n+3-k}}$, and as to how it scales with $n$ I'd say it is quite a lot larger. For a more precise asymptotics, $$\log P^4\_n=2^{n}\bigg({\log3} +8\sum\_{k=1}^{n-1}{\log k\over 2^k}+{4\log n\over 2^{n}}\bigg)=2^{n}\bigg({\log3} +8\sum\_{k=1}^{\infty}{\log k\over 2^k} +o(1)\bigg).$$


**[edit]** sorry for mis-reading the correct definition of the iteration. The correct result is kindly given by მამუკა ჯიბლაძე in a comment below!


