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](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


* Michael Kleber, *Goldbug Variations*, Mathematical Intelligencer 27 #1 (Winter 2005), pp. 55–63, <https://arxiv.org/abs/math/0501497>.


The plot at the bottom of [http://www.cs.uml.edu/~jpropp/Phi-short.pdf](http://www.cs.uml.edu/%7Ejpropp/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).



A specific Dedekind-esque sum
I can't help thinking that the answer ought to come out of a combination of standard techniques. First, indeed, reduce to fractional parts: but with a caveat, that it would be sensible to do some manipulation first. I'm thinking about "summation by parts". 


**Edit**: There is a classic reference: G. H. Hardy & J. E. Littlewood, "Some problems of Diophantine approximation: The lattice points of a right-angled triangle," (1st memoir), Proc. London Math. Soc. (2), v. 20, 1922, pp. 15-36. They consider the modified fractional part sum, with {x} set as x - [x] - 1/2, of the ${k\theta}$ up to n, where to be compatible with their notation $\theta$ would be the reciprocal of $\tau$, not that this matters at all. The bound they get is O(log n) (Hardy's *Works* vol. I p. 145), which depends only on the continued fraction having bounded partial quotients. The particular case relevant to $\tau$ is worked out in detail over the next few pages. The result is sharp. Off the top of my head this looks enough to get the error term O(nlog n) for the sum as posted, by breaking into at most n sums of this type.


