TITLE: Is the maximal temperature of the curlicue fractal acheived by $e\times\gamma$?
QUESTION [6 upvotes]: The Curlicue Fractal is defined as follows:
Choose an irrational number $s$ and a horizontal unit segment with angle $\phi_0 = 0$. Define $\theta_{n+1} = \theta_{n} + 2 \pi s \pmod{2 \pi}$, with $\theta_0=0$. To the previous segment, add a new unit segment with angle $\phi_{n+1} = \theta_{n} + \phi_{n} \pmod{2 \pi}$. The resulting series of line segments is the curlicue fractal. The "temperature" of these fractals measures the boundedness of these curves.
$s = e\times\gamma$ creates a very high temperature for this fractal, far greater than all the millions of other irrational numbers I've tried.  Does any other irrational number beat it?

REPLY [2 votes]: Refined Solution -- A Red Hot Curlicue
A more elegant variation on my previous solution is 
$s=55\sqrt{2}+92\sqrt{5}\>$ 
which reaches a temperature of approx $T=226\,000$ near $l=1\,060\,000$.
Note that numerically $s=(567/2)-\!1.39499...\!\times\!10^{-7}$.

And yes, that is an actual plot -- I didn't just make a huge Point[].
At $l=2\,000\,000$ I get the following. (Which for some odd reason differs from what Ed Pegg gets for what should be the same thing. I'm wondering if there are round-off issues somewhere.)


Previous Solution
Sort of a cheat, perhaps, but, assuming I've made no mistakes, $s=1/2+\pi/10^6$ will give a temperature of approx $T=10\,000$ at $l=47\,000$

Note that the slight deviation from $1/2$ will eventually come further into play, dropping the temperature. Eg, at $l=100\,000$ we have (not to the same scale)

The temperature for the curve for the first $100\,000$ steps is

Similar curves w/ increasing powers dividing the $\pi$ should allow arbitrarily high temperatures to be reached.