The fractional part of $n\log(n)$

When I was thinking about my other question on the sequence

$$p(n)=\min_a\left\{a+b,\ \left\lfloor\frac {2^a}{3^b}\right\rfloor=n\right\}$$

I found an interesting link with the sequence

$$q(n)=\{n\log(n)\}=n\log(n)-[n\log(n)]$$

the fractional part of $n\log(n)$.

If we draw the sequence $q$, we get this (for $n$ up to $520$, $5\,000$ and $30\,000$ respectively):

We can see some gaps looking like parabolas.

What is causing those gaps? Why do they look like this?

In my other question, we can observe similar structures:

An answer is telling me that it is due to the continued fraction of $\log2/\log 3$.

Could the two questions be linked?

Edit.

Thanks to Gottfried Helms idea, here is some more drawings with the moduli varying.

The moduli will take those values respectively:

$$1,2,3,4,5,6,7,8,9,10,20,30,40,50,60,70,80,90,100.$$

A modulus of $1$ corresponding to the fractional part.

• Have you tried to take other moduli instead of $\pmod 1$, for instance $\pmod 2$ or $\pmod 10$? Perhaps the parabolas become even more prominent? – Gottfried Helms Feb 16 '17 at 1:14

While this is not a complete answer, in order to get some insight to this phenomenon, consider the expression $$\{q(n+1)-q(n)\}=\{(n+1)\log(n+1)-n\log(n)\}=\{n\log(1+\frac{1}{n})+\log(n+1)\}$$ The first summand $n\log(1+\frac{1}{n})\sim 1-\frac{1}{2n}$ which is almost an integer and doesn't affect too much on that expression. It follows that $\{q(n+1)-q(n)\}\sim\{\log(n+1)\}$. So if $n+1\sim e^k$ or equivalently $\log(n+1)\sim k$ then the difference above is almost zero. You can see this in your diagram at $e^4\sim 54, e^5 \sim 148, e^5 \sim 403$ which is the bottom of the "parabolas". A little bit before it $\{q(n+1)-q(n)\}\sim\{\log(n+1)\}$ is a little bit smaller than 1 so that your graph is monotone decreasing, and similarly after it the graph is monotone increasing.
If for example $n+1\sim e^{5.5}\sim 255$ then the difference above is more or less 1/2 and then you will see two "parabolas" one above the other. The same argument will work for $n\sim e^{k +\frac{p}{q}}$ when you fix the rational $\frac{p}{q}$.