Let $\tau$ be $(-1+\sqrt{5})/2$, let $f(x)$ be $\lfloor (x+1)\tau \rfloor$, let $s_n$ be $\tau n (n+1) (n+2) / 6$, and let $S_n$ be $$\sum_{k=0}^{n} (n−2k)f(n−k) = n f(n) + (n-2) f(n-1) + (n-4) f(n-2) + \dots - (n-2) f(1) - n f(0).$$

Is $(S_n-s_n)/(n \log n)$ bounded for $n > 1$?
(It stays between -0.35 and +0.30 for all $n$ between 2

and $10^6$.)

This is a specific instance of the question Dedekind-esque sums that I posted a few weeks ago. It may be an atypical instance in some ways (since $\tau$ is a pretty atypical real number for Diophantine approximation problems) but it's the one that interests me most right now. An affirmative answer to my question would have implications concerning the "Goldbug machine" described in http://front.math.ucdavis.edu/0501.5497 .

The plot at the bottom of http://www.cs.uml.edu/~jpropp/Phi-short.pdf is a histogram of $(S_n - s_n)/n$ for $n$ going from 1 to a million. As you can see, it doesn't stray very far away from 0. So perhaps that $\log n$ in the denominator could be replaced by something smaller, like $\sqrt{\log n}$ or even $\log \log n$ (or maybe even 1, though I doubt it).