TITLE: Unexpected behavior involving √2 and parity
QUESTION [24 upvotes]: This post makes a focus on a very specific part of that long post.  Consider the following map:
$$f: n \mapsto \left\{
    \begin{array}{ll}
         \left \lfloor{n/\sqrt{2}} \right \rfloor & \text{ if } n \text{ even,}  \\
         \left \lfloor{n\sqrt{2}} \right \rfloor & \text{ if } n \text{ odd.}
    \end{array}
\right.$$
Let $f^{\circ (r+1)}:=f \circ f^{\circ r}$, consider the orbit of $n=73$ for iterations of $f$, i.e. the sequence $f^{\circ r}(73)$:  $$73, 103, 145, 205, 289, 408, 288, 203, 287, 405, 572, 404, 285, 403, 569, 804, 568, 401, \dots$$ 
It seems that this sequence diverges to infinity exponentially, and in particular, never reaches a cycle. Let illustrate that with the following picture of $(f^{\circ r}(73))^{1/r}$, with $200<r<20000$:

According to the above picture, it seems that $f^{\circ r}(73) \sim \delta^r$ with $\delta \sim 1.02$.
Now consider the probability of the $m$ first terms of the sequence $f^{\circ r}(73)$ to be even: $$p_{0}(m):= \frac{|\{ r<m \ | \  f^{\circ r}(73) \text{ is even}\}|}{m}.$$
Then $p_1(m):=1-p_0(m)$ is the probability of the $m$ first terms of $f^{\circ r}(73)$ to be odd.  
If we compute the values of $p_i(m)$ for $m=10^{\ell}$, $\ell=1,\dots, 5$, we get something unexpected: 
$$\scriptsize{ \begin{array}{c|c}
\ell  & p_0(10^{\ell}) &p_1(10^{\ell}) \newline \hline
1 &0.2&0.8 \newline \hline
2 &0.45&0.55 \newline \hline
3 &0.467&0.533 \newline \hline
4 &0.4700&0.5300 \newline \hline
5 &0.46410&0.53590 \newline \hline
6 & 0.465476& 0.534524
\end{array} }$$ (the line for $\ell = 6$ was computed by Gottfried Helms, see the comments)
It is unexpected because it seems that $p_0(m)$ does not converge to $1/2$, but to $\alpha \sim 0.465$.
It matches with the above observation because $$ \delta \sim 1.02 \sim \sqrt{2}^{(0.535-0.465)} = \sqrt{2}^{(1-2 \times 0.465)} \sim \sqrt{2}^{(1-2\alpha)}.$$

Question: Is it true that $f^{\circ r}(73)$ never reach a cycle, that $(f^{\circ r}(73))^{1/r}$ converges to $\delta \sim 1.02$, that $p_0(m)$ converges to $\alpha \sim 0.465$, and that $\delta^24^{\alpha} = 2$?  What are the exact values of $\delta$ and $\alpha$?  (or better approximations?)

The following picture provides the values of $p_0(m)$ for $100 < m < 20000$:
 
Note that this phenomenon is not specific to $n=73$, but seems to happen as frequently as $n$ is big, and then, the analogous probability seems to converge to the same $\alpha$. If $n <100$, then it happens for $n=73$ only, but for $n<200$, it happens for $n=73, 103, 104, 105, 107, 141, 145, 146, 147, $ $ 148,  149, 151, 152, 153, 155, 161, 175, 199$; and for $10000 \le n < 11000$, to exactly $954$ ones.
Below is the picture as above but for $n=123456789$:


Alternative question: Is it true that the set of $n$ for which the above phenomenon happens has natural density one? Is it cofinite? When it happens, does it involves the same constant $\alpha$?

There are exactly $1535$ numbers $n<10000$ for which the above phenomenon does not happen. The next picture displays for such $n$ the minimal $m$ (in blue) such that $f^{\circ m}(n) = f^{\circ (m+r)}(n)$ for some $r>0$, together with the miniman such $r$ (in red):
 
In fact all these numbers (as first terms) reach the following cycle of length $33$:  
$$(15,21,29,41,57,80,56,39,55,77,108,76,53,74,52,36,25,35,49,69,97,137,193,272,192,135,190,134,94,66,46,32,22)$$ 
except the following ones: $$7, 8, 9, 10, 12, 13, 14, 18, 19, 20, 26, 27, 28, 38, 40, 54,$$ which reach $(5,7,9,12,8)$, and that ones $1, 2, 3, 4, 6$ which reach $(1)$, and $f(0)=0$.
If the pattern continues like above up to infinity, they must have infinity many such $n$.    

Bonus question: Are there infinitely many $n$ reaching a cycle? Do they all reach the above cycle of length $33$ (except the
  few ones mentioned above)? What is the formula of these numbers $n$?

Below is their counting function (it looks logarithmic):

REPLY [4 votes]: Here is a heuristic answer inspired by this comment of Lucia.
First, let assume that the probabilty for an integer $n$ to be odd is $\frac{1}{2}$, and that the probabilty for $f(n)$ to be odd when $n$ is even (resp. odd) is also $\frac{1}{2}$. We will observe that (surprisingly) it is no more $\frac{1}{2}$ for $f^{\circ r}(n)$ when $r \ge 2$ (in some sense, the probability does not commute with the composition of $f$ with itself).

if $n$ and $m=f(n)$ are even: note that $\frac{n}{\sqrt{2}} = m+\theta$ (with $0 < \theta < 1$) so that $m=\frac{n}{\sqrt{2}}- \theta$, then $$f^{\circ 2}(n) = f(m) = \left \lfloor{\frac{m}{\sqrt{2}}} \right \rfloor = \left \lfloor{\frac{\frac{n}{\sqrt{2}}- \theta}{\sqrt{2}}} \right \rfloor = \left \lfloor \frac{n}{2} - \frac{\theta}{\sqrt{2}}\right \rfloor$$ but $\frac{n}{2}$ is even with probability $\frac{1}{2}$, so in this case, $f^{\circ 2}(n)$ is odd with probability $\frac{1}{2}$.

if $n$ is even and $m=f(n)$ is odd: $$f^{\circ 2}(n) = f(m) = \left \lfloor\sqrt{2}m \right \rfloor = \left \lfloor \sqrt{2}(\frac{n}{\sqrt{2}} - \theta) \right \rfloor = \left \lfloor n - \sqrt{2} \theta) \right \rfloor$$ but $n$ is even and the probability for $0<\sqrt{2} \theta<1$ is $\frac{\sqrt{2}}{2}$ (because $\theta$ is assumed statistically equidistributed on the open interval $(0,1)$), so $f^{\circ 2}(n)$ is odd with probability
$\frac{\sqrt{2}}{2}$.

if $n$ is odd and $m=f(n)$ is even:
$$f^{\circ 2}(n) = f(m) = \left \lfloor{\frac{m}{\sqrt{2}}} \right \rfloor = \left \lfloor{\frac{\sqrt{2}n-\theta}{\sqrt{2}}} \right \rfloor = \left \lfloor n - \frac{\theta}{\sqrt{2}} \right \rfloor  $$
but $n$ is odd and $0 < \frac{\theta}{\sqrt{2}}<1$, so $f^{\circ 2}(n)$ is even.

if $n$ is odd and $m=f(n)$ is odd:
$$f^{\circ 2}(n) = f(m) = \left \lfloor \sqrt{2} m \right \rfloor = \left \lfloor \sqrt{2} (\sqrt{2}n-\theta) \right \rfloor = \left \lfloor 2n - \sqrt{2} \theta \right \rfloor $$
but $2n$ is even and the probability for $0<\sqrt{2} \theta<1$ is $\frac{\sqrt{2}}{2}$, so $f^{\circ 2}(n)$ is odd with probability $\frac{\sqrt{2}}{2}$.


By combining these four cases together, we deduce that the probability for $f^{\circ 2}(n)$ to be odd is $$\frac{1}{2} \times \frac{1}{2} \times (\frac{1}{2} + \frac{\sqrt{2}}{2} + 0 + \frac{\sqrt{2}}{2}) = \frac{2\sqrt{2}+1}{8}$$
By continuing in the same way, we get that the probability for $f^{\circ 3}(n)$ to be odd is:
$$ \frac{1}{4} (\frac{1}{2}\frac{1}{2} + \frac{1}{2}\frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2}\frac{\sqrt{2}}{2} + 1\frac{1}{2} + \frac{\sqrt{2}}{2}\frac{\sqrt{2}}{2})  = \frac{\sqrt{2}+7}{16}$$
For $2 \le r \le 24$, we computed the probability $p_r$ for $f^{\circ r}(n)$ to be odd (see Appendix). It seems (experimentally) that $p_r$ converges to a number $\simeq 0.532288725 \simeq \frac{8+3\sqrt{2}}{23}$ by Inverse Symbolic Calculator. This leads to the following question/conjecture:
$$\lim_{r \to \infty}p_r = \frac{8+3\sqrt{2}}{23} \ \  ?$$
If so, consider the number $\alpha$ mentioned in the main post, then $$\alpha = 1-\frac{8+3\sqrt{2}}{23} = \frac{15-3\sqrt{2}}{23} \simeq 0.467711,$$ which matches with the computation in the main post. And next, we would have:
$$ \delta = \frac{\sqrt{2}}{2^{\alpha}}= 2^{\frac{1}{2}-\alpha} = 2^{\frac{6\sqrt{2}-7}{46}} \simeq 1.022633$$

Appendix
Computation
sage: for i in range(3,26):
....:     print(sq2(i))
....:
[1/4*sqrt(2) + 1/8, 0.478553390593274]
[1/16*sqrt(2) + 7/16, 0.525888347648318]
[3/32*sqrt(2) + 13/32, 0.538832521472478]
[15/64*sqrt(2) + 13/64, 0.534581303681194]
[5/128*sqrt(2) + 61/128, 0.531805217280199]
[39/256*sqrt(2) + 81/256, 0.531852847392776]
[93/512*sqrt(2) + 141/512, 0.532269260352925]
[51/1024*sqrt(2) + 473/1024, 0.532348527032254]
[377/2048*sqrt(2) + 557/2048, 0.532303961432938]
[551/4096*sqrt(2) + 1401/4096, 0.532283123258685]
[653/8192*sqrt(2) + 3437/8192, 0.532285334012406]
[3083/16384*sqrt(2) + 4361/16384, 0.532288843554459]
[3409/32768*sqrt(2) + 12621/32768, 0.532289246647030]
[7407/65536*sqrt(2) + 24409/65536, 0.532288816169701]
[22805/131072*sqrt(2) + 37517/131072, 0.532288667983386]
[24307/262144*sqrt(2) + 105161/262144, 0.532288700334941]
[72761/524288*sqrt(2) + 176173/524288, 0.532288728736551]
[159959/1048576*sqrt(2) + 331929/1048576, 0.532288729880941]
[202621/2097152*sqrt(2) + 829741/2097152, 0.532288725958633]
[639131/4194304*sqrt(2) + 1328713/4194304, 0.532288724978704]
[1114081/8388608*sqrt(2) + 2889613/8388608, 0.532288725350163]
[1825983/16777216*sqrt(2) + 6347993/16777216, 0.532288725570602]
[5183461/33554432*sqrt(2) + 10530125/33554432, 0.532288725561857]

Code
def sq2(n):
    c=0
    for i in range(2^n):
        l=list(Integer(i).digits(base=2,padto=n))
        if l[-1]==1:
            cc=1/4
            for j in range(n-2):
                ll=[l[j],l[j+1],l[j+2]]
                if ll==[0,0,0]:
                    cc*=1/2
                if ll==[0,0,1]:
                    cc*=1/2
                if ll==[0,1,0]:
                    cc*=(1-sqrt(2)/2)
                if ll==[0,1,1]:
                    cc*=sqrt(2)/2
                if ll==[1,0,0]:
                    cc*=1
                if ll==[1,0,1]:
                    cc=0
                    break
                if ll==[1,1,0]:
                    cc*=(1-sqrt(2)/2)
                if ll==[1,1,1]:
                    cc*=sqrt(2)/2
            c+=cc
    return [c.expand(),c.n()]