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David Roberts
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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).

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).

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

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).

replaced http://mathoverflow.net/ with https://mathoverflow.net/
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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 sumsDedekind-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).

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).

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).

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Gerry Myerson
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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 $\sum_{k=0}^{n} (n−2k)f(n−k) = n f(n) + (n-2) f(n-1) + (n-4) f(n-2) + \dots - (n-2) f(1) - n f(0)$.$$\sum_{k=0}^{n} (n−2k)f(n−k) = n f(n) + (n-2) f(n-1) + (n-4) f(n-2) + \dots - (n-2) f(1) - n f(0).$$

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
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).

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).

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).

Bounty Ended with Charles Matthews's answer chosen by James Propp
Bounty Started worth 50 reputation by James Propp
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James Propp
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