Let $n$ be a positive odd integer, let $$n_j = \Bigl\{\frac{n}{2^{j+1}}\Bigr\}\,,$$ where $\{x\}$ denotes the fractional part of $x$, and finally let $k = \lceil \log_2 n\rceil$. Consider the function $$ f(n) = \prod_{j=1}^{k} \bigl[1- 4n_j(1-n_j)\bigr]\,. $$ Each term in this product is bounded in the interval $(0,1)$, so $f(n)$ will tend to $0$ for large $n$. My question is how one can
Prove that $f(n) \ge \frac{c}{n^a}$ for some absolute positive constants $a$, $c$.
It is easy to show (using bounds on the fractional part) that there exist $a,c>0$ such that $$ f(n) \ge \frac{c}{n^{a \log n}} \,, $$ so that the product decays only subexponentially slowly, but a heuristic argument as well as numerics indicates that the true scaling is actually a much slower inverse polynomial. The heuristic argument is, roughly, that the terms in the product are highly correlated, and if one term is small, the following term must be large. I have been unable to leverage this into a proof, however.
I indicated in the title that this is a "number theoretic" function because the behavior depends heavily on the bit-wise structure of $n$, and it has some interesting fractal features as a result.
Numerical experimentation suggests that the fastest-decaying values occur at $n_k=[2^k/3]$, where $[\cdot]$ indicates rounding off to the nearest integer. These integers have binary expansion $101010\cdots$. It probably is possible to work out very good bounds for $f(n_k)$. Numerically, we seem to have $f(n_k)$ asymptotic to a constant times $1/n_k^\alpha$ where $\alpha = \log 9/\log 2$.
(Similarly, the slowest-decaying values are almost surely $m_k = 2^k-1$, for which we seem to have $f(m_k) \sim c/m_k^2$ for some constant $c$.)