Unexpected behavior involving √2 and parity
A list of predecessors as mentioned in my comment.
I document pairs of $(m,n)$ for consecutive $m$ and their 1-step predecessors $n$ such that $f(n)=m$. The value $n=0$ indicates, that $m$ has no predecessor. I didn't reflect, that one $m$ can have two predecessors, but if $n/2$ is odd, then $n/2$ is a second predecessor.(This makes the table more interesting, because all odd predecessors $n$ are overwritten by the even predecessors $2n$...
Moreover, a nearly periodic structure occurs. I tried to resemble this by the arrangement of three or four columns of $(m,n)$ such that the first column contains all $m$ which have no predecessor. The basic pattern is not really periodic, but has super-patterns which again seem to be periodic but actually aren't. This pattern-superpattern-structure is also recursive. It reminds me of a similar structure when I looked at $\beta=\log_2(3)$ and found a similar style of pattern-superpattern-supersuperpattern-... and is there related to the continued fraction of $\beta$.
So I think we'll get no nice description for the cases $m$ which have no predecessor...
m n m n m n m n
-------------------------------------------------------------
1 2 2 4
3 0 4 6 5 8
6 0 7 10 8 12 9 14
10 0 11 16 12 18
13 0 14 20 15 22 16 24
17 0 18 26 19 28
20 0 21 30 22 32
23 0 24 34 25 36 26 38
27 0 28 40 29 42
30 0 31 44 32 46 33 48
34 0 35 50 36 52
37 0 38 54 39 56
40 0 41 58 42 60 43 62
44 0 45 64 46 66
47 0 48 68 49 70 50 72
51 0 52 74 53 76
54 0 55 78 56 80 57 82
58 0 59 84 60 86
61 0 62 88 63 90
64 0 65 92 66 94 67 96
Update Some more explanation on the idea of "recursive aperiodic pattern". If we list the values $m$ which have no predecessor, we get
m_k: 3, 6,10,13, 17,20,23,27,30,...
Writing the differences (I have prepended a zero-value to the above list of $m_k$)
,3,3,4 ,3,4 ,3,3,4 ,3,4 ,3,3,4 ,3,4 ,3,4 ,3,3,4 , ...
We note, that we have a pattern of two different words: 3,3,4 and 3,4 repeating, but aperiodical. Let's denote the longer one with the capital A and the shorter one with the small a (and A means a difference of 10 and a of 7).
We get
Aa Aa Aaa
Aa Aaa
Aa Aa Aaa
Aa Aaa
Aa Aa Aaa
Aa Aaa
Aa ...
Again we find only two kind of "words". Let's them shorten by Aaa=B and Aa=b. B means now a difference of 24, b of 17.
Then we get
bbB bB
bbB bB
bbB bB bB
bbB bB
bbB bB bB
bbB bB
bbB bB
bbB bB bB
...
Next obvious step gives
Cc Cc Ccc
Cc Ccc
Cc Cc Ccc
Cc Ccc
Cc Cc Ccc
Cc Ccc
...
with c representing a difference of 17+24=41 and C of 17+17+24=58.
And so on.
If I recall correctly, then with the mentioned case of working with $\beta = \log_2(3)$ the same style of recursive pattern reflected the convergents of the continued fractions of $\beta$.
The first few differences here match the convergents of the continued fraction of $\sqrt2$ so far:
a b c short patterns
-------------------------------------
[1 1 3 7 17 41 99 239 577 ... ] convergents of contfrac(sqrt(2))
[0 1 2 5 12 29 70 169 408 ... ]
-------------------------------------...
A/2 B/2 C/2 long patterns
Update 2 The above can be explained by the following:
- a number of the form $\lfloor2k\sqrt2\rfloor$ has exactly one predecessor $4k$;
- a number of the form $\lfloor(2k-1)\sqrt2\rfloor$ has exactly two predecessors $2k-1$ and $4k-2$;
- a number has no predecessors iff it has form $\lfloor n(2+\sqrt2)\rfloor$.
The first two statements are easily checked, while the third follows from the Beatty theorem, as explained in another answer by @Dattier
Update 3 Using a back-step algorithm (recursive) it seems I've got the predecessing tree of $m=73$. If no bugs, then this tree would also be complete. (But my routine may still be buggy, please check the results!)
The back-steps go from top-right south-west (antidiagonal) downwards. When there are two possible predecessors, they occur in the same column, but on separate rows.
If there is a predecessor without further predecessor, a short line (---) is printed.
73 <--- start
104
148
105 ---
210
149
212
300 ---
298
211 ---
422
299
424
600 ---
598
423 ---
846 ---
---------------------------- tree seems to be complete (please check for errors!)
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()]