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 
https://mathoverflow.net/questions/24517/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).