Here I'll give a helpful scheme of notation for the analysis of existence of cycles, which is a simple generalization of that same scheme which I used for the discussion of the cycles in the Collatz-problem.
Notation
Let us introduce the notation for the extraction of primefactors from a number $n$:
$$ \{n\}_{2,3,5,7} : \text{ extracts all primefactors $2,3,5,7$ from $n$} \tag {1.a} $$
We would also write the relevant primefactors of $n$:
$$ n=2^A 3^B 5^C 7^D m \implies m=\{n\}_{2,3,5,7} \tag {1.2}$$
For a 1-step-transformation we may write:
$$ b= \{11a+1\}_{2,3,5,7} \\ \text{ or } \\
b\cdot 2^A3^B5^C7^D = 11 a+1 \tag {2.1}
$$
Note, that values of $b$ cannot have the primefactors $(2,3,5,7)$, but even more, cannot have the primefactor $11$.
If we discuss cycles, and $a=b$ then this is also valid for $a$, and thus:
$$ \text{no member $a,b,c,...$ of a cycle can have a primefactor $(2,3,5,7,11)$} \tag {2.2}$$
Existence of cycles
Existence of 1-step-cycles
For a 1-step-cycle $b=a$ from (2.1) thus follows
$$ a\cdot 2^A3^B5^C7^D = 11 a+1 \\
a ( 2^A3^B5^C7^D - 11) = 1 $$
$$ a = {1 \over 2^A3^B5^C7^D - 11} \tag {3.1}$$
Theorem 1: Solutions for 1-step cycle are:
$$
a = 1 \qquad \text{for} (A,B,C,D)=(2,1,0,0)\\
a =-1 \qquad \text{for} (A,B,C,D)=(1,0,1,0)\\
$$
Existence of 2-step-cycles
For a 2-step cycle we'll have $2$ members $a,b$.
$$ b= \{11a+1\}_{2,3,5,7} \qquad a= \{11b+1\}_{2,3,5,7}$$
For the following we assume $a \lt b$ (and in generalizations for more steps: $a$ is the minimal element)
It is not so simple as before, but a "product-equation" can be stated:
$$ a \cdot b = \{11a+1\}_{2,3,5,7} \cdot \{11b+1\}_{2,3,5,7} \tag{4.1}$$
from which the following can be formulated:
$$ a b \cdot 2^{A_1+A_2}3^{B_1+B_2}5^{C_1+C_2}7^{D_1+D_2} = (11 a+1)(11 b+1) \tag {4.2} $$
$$ 2^{A_1+A_2}3^{B_1+B_2}5^{C_1+C_2}7^{D_1+D_2} = (11 +1/a)(11 +1/b) \tag {4.3}$$
where the rhs must remain integer and equal to a valid expression in the lhs, which has the additional restriction that at least $A_1+A_2 \ge 2$ because by each transfromation-step (at least) one primefactor $2$ occurs.
We can reformulate the rhs a bit more:
$$ \qquad\qquad\qquad =121+11(\frac1a+\frac1b)+\frac1a\frac1b $$
$$ \qquad\qquad\qquad =121+{11(a+b)+1 \over ab} \tag {4.4} $$
We see by the fractional term in (4.4), that $a,b$ have upper bounds, because with sufficiently increase of $a$ and $b$, the last term in the rhs decreases below $1$ and shall be unconditionally fractional, preventing any further solution for the 2-step cycles in integers.
For instance, heuristically,
- assuming $a=13$ we find, that $13 \lt b \le 71$,
- assuming $a=17$ we find, that $17 \lt b \le 31$,
- assuming $a=19$ we find, that $19 \lt b \le 23$,
- and find that no further test is needed.
Thus we have only to test few cases for $(a,b)$.
Heuristically, all that cases lead to fractional values in the rhs, so ...
Theorem 2: no $2$-step-cycle is possible.
Existence of 3- and more steps cycles
This can be generalized in the obvious way, but which I did not expand here.
For $3,4,5$ or other small number step-cycles, I think this can be done in Pari/GP by direct examination of not too many cases. My intention with this all is just to give a tool for a more algebraic analysis of this problem, which immediately allows generalization to other sets of primes-to-be-extracted.
Further disproving of N-step-cycles and reducing searchspace
Equation(4.3) gives a tool to estimate a mean-value for all members $(a,b,c,...n)$ of a N-step cycle.
For instance for $N=3$ and asking for a mean value $x$ of all members, we have
$$ 2^{S_2}3^{S_3}5^{S_5}7^{S_7} = (11 +\frac1x)^N \tag {5.1}$$
(where we write $S_2$ for $A_1+A_2+A_3$, $S_3$ for $B_1+B_2+B_3$ and so on)
and there must exist a lowest number above $L>11^N$ which has only primefactors as on the left hand side. We introduce the notation
$$ L=\text{valid_lhs}(x) : \min(L>x: \{L \in \mathbb N \}_{2,3,5,7}=1 \land \nu_2(L) \ge N) \tag {5.2}$$
The last condition is required because we know by the above analysis, that the primefactor $2$ has to occur at least with exponent $N$.
So let's assume there is a
$$L=\text{valid lhs}(11^N) \gt 11^N \tag {5.2a} $$
and from the ansatz with an unknown average-value $x$
$$ L = 2^{S_2}3^{S_3}5^{S_5}7^{S_7} = (11 +\frac1x)^N$$
we calculate $x$
$$ L^{1/N} = (11 +1/x) \\
L^{1/N} -11 = 1/x $$
and
$$ x= {1 \over L^{1/N} -11} \tag {5.3} $$
As $x$ is a mean value of all involved members of a cycle (here $(a,b,c)$) and we assume always $a$ being the minimal member, we know that $a \lt x$.
Our searchspace for finding the smallest member of an assumed N-step-cycle $a$ is thus now nicely bounded from above, and sometimes that result shows immediately that a cycle cannot exist: since if for some $N$ we get that $x<13$ then $a$ must be smaller than $13$ but there is no number free of the primefactors $(2,3,5,7,11)$ below $13$ .
Of course, this is except $1$ - but for the current set of primes $(2,3,5,7)$ this meaned we'd heve the 1-step-cycle and $a=b=c=1$, which is what we exclude here.
A short heuristic:
N x upper bound for a
------------------------------------------
2 3.18767264271 1 no 2 step cycle
3 28.0137896567 23 -case test for a=13,17,19,23 required-
4 11.2502716388 1 no 4 step cycle
5 319.854255596 317 -case test up to a=317 required-
6 147.732223336 139 -case test up to a=139 required-
7 114.725982882 113 -case test up to a=113 required-
8 247.003937577 247 -case test up to a=247 required-
... ... ...
For the first few generalizations $f_2()$ to $f_{11}()$
Here the so-far found cycles for some generalizing $f_k()$ (the searchspace for initial $a$ (or better: $a_1$) is up to some 100 000 although very likely this is not needed when looking at the small table above):
"selection multi document of found cycles
primes" plicator
-----------------------------------------------------------
f_1() [2,3, ] [ 5]: no cycle found
f_2() [2,3, 5 ] [ 7]: no cycle found
f_3() [2,3, 5, 7] [11]: 3-step: 17 \to 47 \to ... \to 17
f_4() [2, ..., 11] [13]: 4-step: 19 \to 31 \to ... \to 19
f_5() [2, ..., 13] [17]: 22-step: 43 \to 61 \to ... \to 43
f_6() [2, ..., 17] [19]: 16-step: 46063 \to 437599 \to ... \to 46063
f_7() [2, ..., 19] [23]: 4-step: 179 \to 2059 \to ... \to 179
f_8() [2, ..., 23] [29]: no cycle found
f_9() [2, ..., 29] [31]: 30-step: 67 \to 1039 \to ... \to 67
f_10() [2, ..., 31] [37]: 17-step: 2173 \to 5743 \to ... \to 2173
f_11() [2, ..., 37] [41]: no cycle found
Conclusion
The algebraic tools shown here are easily extensible for your type of generalization to more primes in the primeslist $(2,3,5,7)$ without needing any modification.
Moreover, surely more of that tools of the $3x+1$-analysis can be adapted for this $3x+1$-extensions, but before going into deeper water, I'll wait whether this all here has been found helpful or you'll go further on your own...
P.S.: I don't know whether this formalism also helps for the discussion of the existence of a divergent trajectory.
