Here are some pictures for your/our intuition. I graphed the trajectories for initial values $x=5,15,25,...$ for the first $256$ steps of $x_{k+1}=(5x_k+1)/2^A$.
To get the curves to a meaningfully visual interval I show logarithmic scales. The pictures show how most trajectories begin to diverge (not really a safe indication of what characteristic the infinite curves really have) but some show cycling already at early iteration indexes $k$ .
I find $2$ cycles besides the "trivial" one.
$x=5,15,25,35,...,95$ detail of the first few iterations . At the bottom we see the "trivial" cyle (brown curve):
$x=5,15,25,35,...,95$ first $2^8 = 256$ iterations. At later iteration-indexes $k$ a first "non-trivial" cycle occurs (red line):
$x=105,115,125,135,...,195$ first $2^8 = 256$ iterations .
$x=205,215,225,235,...,295$ first $2^8 = 256$ iterations . Here a second "non-trivial" cycle becomes visible:
$x=205,215,225,235,...,295$ first $2^{11} = 2048$ iterations
It seems really that all trajectories which are divergent up to iteration $k=256$ are also divergent up to iteration $k=2048$ . In general: I doubt that there are "later" cycles:
This is intended as a comment, but too long for the comment-field
A simple table gives infinite modular classes of odd numbers $a_1$, which always decrease by one step of the $(3x+1)/\nu_2(3x+1)$ iteration to $a_2$. This is, with $k \in \{0,1,2,3,...\}$:
Table 1:
$$ \begin{array}{r|r|r||r|r|r}
A &a_1 &a_2&& A & a_1 & a_2\\ \hline
\color{red} {1} & \color{red} { 2^1\cdot 2k+3\phantom {1}} & \color{red} {3\cdot 2k+5} && \color{green} {2}& \color{green} {2^2\cdot 2k+1 \phantom {1}} &\color{green} { 3\cdot 2k+1} \\
3 & 2^3\cdot 2k+13 & 3\cdot 2k+5 && 4& 2^4\cdot 2k+5 \phantom {1} & 3\cdot 2k+1 \\
5 & 2^5\cdot 2k+53 & 3\cdot 2k+5 && 6& 2^6\cdot 2k+21 & 3\cdot 2k+1 \\
\vdots & \vdots &\vdots && \vdots &\vdots & \vdots & \\
A & 2^A\cdot 2k+r_A & 3\cdot 2k+5 && A& 2^A\cdot 2k+r_A & 3\cdot 2k+1 \\
\vdots & \vdots &\vdots && \vdots &\vdots & \vdots & \\
\end{array}$$
where $r_A = { 2^A-1\over 3}$ if $A$ is even, and $r_A = { 5 \cdot 2^A-1\over 3}$ if $A$ is odd.
From this table it is easily visible, that
- only for numbers $\color{red} {a_1}$ with $\color{red} {A=1}$ the iteration to $a_2$ is increasing.
- For $k=0$ we have with $A=2$ the trivial cycle with $a_1=a_2=1$ and
- for all other cases the iteration decreases.
This shows that the least element $a_1$ of a divergent sequence must be of the form $4k+3$ .
Out of couriosity I've tried to tabular the subforms of $a_1=4k+3$ in a comparable way($ b_1=16k+3$ and $c_k=16k+11$ and so on) to pinpoint the distinction between increasing and decreasing iterations - but this leads soon to another representation scheme, which becomes more complicated. (And I've not yet put that together in a nicely-readable essay.)
update: I see that the list (or better:tree) that I've created for this aspect is just that list, that user @Vepier mentions in his/her comment.
The beginning of this list, ordered for the powers-of-2 in the $a_1$ variable, looks like this. It is a systematic list of modular classes for $a_1$ which are decreasing after a small number $n$ of iterations (here $n$ is also the length of the exponents-vector $E$). The exponents-vector $E$ (here printed as compacted string) of the exponents $A_1,A_2,A_3,...A_n$ for "dropping time" $n$ is generated by a recursive tree structure.
Table 2:
E a_1 --> a_j
----------------------------------------------
2 1 +k* 4 --> 1 +k* 3 a_1 mod 4 = 1
- - - - - - - - - - - - - - - - - - - - -
13 3 +k* 16 --> 2 +k* 9 a_1 mod 4 = 3, but decreasing
122 11 +k* 32 --> 10 +k* 27
113 23 +k* 32 --> 20 +k* 27
1123 7 +k* 128 --> 5 +k* 81
1114 15 +k* 128 --> 10 +k* 81
1213 59 +k* 128 --> 38 +k* 81
11213 39 +k* 256 --> 38 +k* 243
11132 79 +k* 256 --> 76 +k* 243
11114 95 +k* 256 --> 91 +k* 243
12122 123 +k* 256 --> 118 +k* 243
11123 175 +k* 256 --> 167 +k* 243
11222 199 +k* 256 --> 190 +k* 243
12113 219 +k* 256 --> 209 +k* 243
111124 287 +k* 1024 --> 205 +k* 729
121123 347 +k* 1024 --> 248 +k* 729
111214 367 +k* 1024 --> 262 +k* 729
112123 423 +k* 1024 --> 302 +k* 729
121213 507 +k* 1024 --> 362 +k* 729
111115 575 +k* 1024 --> 410 +k* 729
... ... ...
The still increasing (and so hopefully diverging) cases are that where the last entry in $E$ is smaller than the shown one.
Here is a table, where in the above list I measure the covering of the odd numbers in residue-classes sequentially up to a given residue class. From this I calculate the relative number of residue classes not covered so far, which means, that part of a residue-class whose members $a_1$ increase after that $n$ steps. Over the residueclasses we find, that the number of those potential divergent $a_1$ decreases with $n$ (P.s.: this is just barely observation and no final interpretation so far):
- $cov =$ all displayed (weighted) occurences of residues in class $2^S$
- $miss= 1 - cov/(2^S/2)$ - the relative number of missing residue classes (relative number of potential divergent $a_1$)
- $crit= S \cdot miss $ - just something which might be a significant coefficient
Table 3:
S 2^S cov: miss: crit:
--------------------------------------------------------------
2, 4, 1, 0.500000000000, 1.00000000000
4, 16, 5, 0.375000000000, 1.50000000000
5, 32, 12, 0.250000000000, 1.25000000000
7, 128, 51, 0.203125000000, 1.42187500000
8, 256, 109, 0.148437500000, 1.18750000000
10, 1024, 448, 0.125000000000, 1.25000000000
12, 4096, 1822, 0.110351562500, 1.32421875000
13, 8192, 3729, 0.0895996093750, 1.16479492188
15, 32768, 15089, 0.0790405273438, 1.18560791016
16, 65536, 30654, 0.0645141601563, 1.03222656250
18, 262144, 123577, 0.0571823120117, 1.02928161621
...
Best Answer
$$f(n)=\frac74n+\frac12+(-1)^{n+1} \left(\frac54n+\frac12\right)$$