A Collatz generalization and approximation of a bounded-unbounded point of Collatz-like Functions

Question

I have been working on Collatz-like functions to test the probabilistic heuristic argument in favor of all trajectories being bounded (ending in a cycle) and have come up with a generalization to test this argument empirically. This leads me to two questions.

Let $f:\mathbb{N} \to\mathbb{N}$. The Collatz function states that the following iterated map will eventually equal to 1:

$$f(n) = \begin{cases} n/2, & \text{if}\ 2\mid n\\ 3n+1, & \text{otherwise} \\ \end{cases}$$

My generalization is as follows: Let $G:\mathbb{R} \to\mathbb{Z}$ defined as

$$G(x) = \begin{cases} x/2, & \text{if}\ 2\mid x\\ 2\lceil ax \rceil, & \text{otherwise} \\ \end{cases}$$

where $a, x \in \mathbb{R}$. The Collatz function is the case when $a = 1.5$. Likewise, the $5n+1$ problem is the case when $a=2.5$.

I conjecture in line with the heuristic argument that if $-2 \leq a \leq 2$ then for any $x_0$, repeated iteration of $G$ on $x_0$ will yield a bounded trajectory. However, if $a>2$ or $a<-2$, for some $x_0$, repeated iteration of $G$ on $x_0$ will yield an unbounded trajectory. Note, $G(x)$ is bounded if for all $x_0$ we end in a cycle whereas unbounded means that G has at least one trajectory that diverges to infinity.

1. Is this a known generalization of the Collatz function? I have check as much as I could and found only discontinuous generalizations such as 1, 2, and 3. On the other hand, $G(x)$ is continuous in the sense that it allows for both $x$ and $a$ to be any real numbers. Having this flexibility provides a way to test the probabilistic heuristic argument in favor of bounded trajectories by varying $a$ as close to 2 as possible, i.e. if $log(|a|) + log(1/2) < 0$, then for all $x_0$, iterations of $G(x)$ on $x_0$ will yield a cycle.

2. The function $H$ below is more global than $G$. Empirically, a bounded-unbounded point seems to exist for $H$ somewhere in the range $c \leq 2b^2$. Is there a better way to approximate the bounded-unbounded point for the generalization $H$ than testing different values of $c$ for a given value of $b$.

$$H(x) = \begin{cases} \frac{x}{b}, & \text{if}\ b\mid x\\ b\lceil \frac{cx}{b^2} \rceil, & \text{otherwise} \\ \end{cases}$$ where $b,c \in \mathbb{N}$. The Collatz function is when $b = 2$ and $c = 6$.

Using Excel, when $b=3$, for $c \leq 16$, H appears bounded for all $x$ but for $c \geq 17$ it appears unbounded. Similarly, when $b = 5$, for $c \leq 35$, H appears bounded for all $x$ but for $c \geq 36$ it appears unbounded. For $b=7$ the demarcation point appears to be $c=65$. Taking the ratio $\frac{c_{max}}{b^2}$ where $c_{max}$ is the bounded-unbounded demarcation point, gives

$\frac{c_{max}}{b^2}=2.0$ when $b = 2$
$\frac{c_{max}}{b^2}=1.8$ when $b = 3$
$\frac{c_{max}}{b^2}=1.5$ when $b = 4$
$\frac{c_{max}}{b^2}=1.4$ when $b = 5$
$\frac{c_{max}}{b^2}=1.4$ when $b = 6$
$\frac{c_{max}}{b^2}=1.3$ when $b = 7$

Seems that as $b \to \infty$ then, $\frac{c_{max}}{b^2}\to1$.

Any help will be greatly appreciated.

Answer 1

According to request of the asker, here is some Pari/GP -code, partly function-definitions, partly experimenting to get list of results.

\\ test collatz with *real* bases for existence of cycles and/or divergence
\\ https://math.stackexchange.com/questions/4340764/a-collatz-generalization-and-approximating-the-bounded-unbounded-point-of-collat
\\ example set global variable "base" to some example value:

  base=9/4

g(x)=if(! (x % 2),x/2,2*ceil(base*x))

\\ alternative. g1(x) works on and produces only odd numbers. May be more
\\ efficient
g1(x)=my(A);A=valuation(x,2);x>>=A;x=2*ceil(base*x);A=valuation(x,2);x>>=A

  \\ print 800 steps of iterations, distributed over 5 columns ("msep(...,5)"), 
  \\ beginning at a0=19 ("msep()" is user defined function)
printp( msep(Mat(vectorv(800,r,if(r==1,a0=19,a0=g(a0)))),5))

  \\ print the same, with a0=86, in one column but showing the index as well:
for(k=1,800,if(k==1,a0=86,a0=g(a0));if(a0==86,print([k,a0])))


  \\ print 800 steps of iterations, distributed over 10 columns, 
  \\ beginning at a0=19 ("msep()" is user defined function)
  \\ we see, that a0=19 is not itself cyclic! but g(19)=86 is!
printp(msep(Mat(vectorv(800,r,if(r==1,a0=19,a0=g(a0)))),10))

Here we use two example bases, and call a loop, which checks for cycles in $a_0=1..999$ (only odd $a_0$) and cycle-lengthes of at most $10\,000$ iterations steps:

fmt(2000);base=sqrt(5);fmt(200) \\ use another base. Take care, that
 \\ over many iterations numbers with many digits shall occur; choose
 \\ accordingly *big* decimal precision! (here 2000 dec digits)


fmt(2000);base=Pi-1;fmt(200)
  \\ or try this base near 2

\\ a loop to find cycles and protocol them:
{print("Base:",base);cyclist=List();
 forstep(a0=1,999,2,
   dir=Map();
   a1=a2=a0;found=0;ix1=ix2=0;mapput(dir,a1,0);
   for(k=1,10000,         \\ test for cycles with length up to 10 000
          a1=g1(a1);      \\ use g1() to document only odd steps
          if(!mapisdefined(dir,a1,&ix1),mapput(dir,a1,k);next());
          a2=a1;found=ix2=k;break();
       );
   if(!found, listput(cyclist,Str(a0," --- nothing found"));next());
     \\ find minimal element of the found cycle 
   len=ix2-ix1; mina1=a1;for(k=1,len,a1=g1(a1);mina1=min(a1,mina1));
     \\ document the parameters for this a0, be it cycle or not:
   listput(cyclist,
      strprintf("%3d : %3d-->%3d cyclen:%3d enter cyc at val:%4d at pos:%3d \n",a0,mina1,mina1,len,a2,ix1));
   next();
   );
   cyclist=Mat(Col(cyclist)); \\ convert "cyclist" from Map to Mat
   printp(cyclist[1..8,]) } \\ show the first 8 entries.

This gave the following protocol (base = $\pi-1$). Only odd numbers were documented (using function $g_1(x)$) :

    1 :   3-->  3 cyclen:  8 enter cyc at val:   3 at pos:  1 \n
    3 :   3-->  3 cyclen:  8 enter cyc at val:   3 at pos:  0 \n
    5 :   3-->  3 cyclen:  8 enter cyc at val:  11 at pos:  1 \n
    7 :   3-->  3 cyclen:  8 enter cyc at val:   7 at pos:  0 \n
    9 :   3-->  3 cyclen:  8 enter cyc at val:  11 at pos:  2 \n
   11 :   3-->  3 cyclen:  8 enter cyc at val:  11 at pos:  0 \n
   13 :   3-->  3 cyclen:  8 enter cyc at val:   7 at pos:  1 \n
   15 :   3-->  3 cyclen:  8 enter cyc at val:  15 at pos:  0 \n
   17 :   3-->  3 cyclen:  8 enter cyc at val:  11 at pos:  3 \n
   19 :   3-->  3 cyclen:  8 enter cyc at val:  41 at pos:  1 \n
   21 :   3-->  3 cyclen:  8 enter cyc at val:   7 at pos:  4 \n
   23 :  63--> 63 cyclen: 10 enter cyc at val:  63 at pos:  4 \n
   25 :  63--> 63 cyclen: 10 enter cyc at val:  63 at pos:  3 \n
   27 :  63--> 63 cyclen: 10 enter cyc at val:  63 at pos:  2 \n
   29 :  63--> 63 cyclen: 10 enter cyc at val:  63 at pos:  1 \n
   31 :   3-->  3 cyclen:  8 enter cyc at val:  11 at pos:  4 \n
   33 :   3-->  3 cyclen:  8 enter cyc at val:  33 at pos:  0 \n
   35 :  63--> 63 cyclen: 10 enter cyc at val:1639 at pos: 46 \n
   37 :   3-->  3 cyclen:  8 enter cyc at val:  11 at pos:  2 \n
   39 :   3-->  3 cyclen:  8 enter cyc at val:   7 at pos:  5 \n
   41 :   3-->  3 cyclen:  8 enter cyc at val:  41 at pos:  0 \n
   43 :  63--> 63 cyclen: 10 enter cyc at val:  63 at pos:  5 \n
   45 :   3-->  3 cyclen:  8 enter cyc at val:   7 at pos:  3 \n
   47 :   3-->  3 cyclen:  8 enter cyc at val:  33 at pos: 25 \n
   49 :   3-->  3 cyclen:  8 enter cyc at val:  11 at pos: 18 \n
   51 :   3-->  3 cyclen:  8 enter cyc at val:  11 at pos:  7 \n
   53 :   3-->  3 cyclen:  8 enter cyc at val:  33 at pos:  3 \n
   55 :   3-->  3 cyclen:  8 enter cyc at val:  11 at pos:  6 \n
   57 :   3-->  3 cyclen:  8 enter cyc at val:  33 at pos:  2 \n
   59 :   3-->  3 cyclen:  8 enter cyc at val:  11 at pos:  5 \n
                                            61 --- nothing found
   63 :  63--> 63 cyclen: 10 enter cyc at val:  63 at pos:  0 \n
   65 :  63--> 63 cyclen: 10 enter cyc at val:1639 at pos: 47 \n
   ...

and this is the version using $g(x)$ which allows to display even values in the orbit (which also means, the documented cycle-lengthes are longer than with $g_1(x)$):

    3 :   3-->  3 cyclen: 25 enter cyc at val:   3 at pos:  0 \n
    5 :   3-->  3 cyclen: 25 enter cyc at val:  22 at pos:  1 \n
    7 :   3-->  3 cyclen: 25 enter cyc at val:   7 at pos:  0 \n
    9 :   3-->  3 cyclen: 25 enter cyc at val:  22 at pos:  5 \n
   11 :   3-->  3 cyclen: 25 enter cyc at val:  11 at pos:  0 \n
   13 :   3-->  3 cyclen: 25 enter cyc at val:  14 at pos:  3 \n
   15 :   3-->  3 cyclen: 25 enter cyc at val:  15 at pos:  0 \n
   17 :   3-->  3 cyclen: 25 enter cyc at val:  22 at pos:  9 \n
   19 :   3-->  3 cyclen: 25 enter cyc at val:  82 at pos:  1 \n
   21 :   3-->  3 cyclen: 25 enter cyc at val:  14 at pos: 13 \n
   23 :  63--> 63 cyclen: 31 enter cyc at val: 126 at pos: 10 \n
   25 :  63--> 63 cyclen: 31 enter cyc at val: 126 at pos:  7 \n
   27 :  63--> 63 cyclen: 31 enter cyc at val: 126 at pos:  4 \n
   29 :  63--> 63 cyclen: 31 enter cyc at val: 126 at pos:  1 \n
   31 :   3-->  3 cyclen: 25 enter cyc at val:  22 at pos: 13 \n
   33 :   3-->  3 cyclen: 25 enter cyc at val:  33 at pos:  0 \n
   35 :  63--> 63 cyclen: 31 enter cyc at val:3278 at pos:136 \n
   37 :   3-->  3 cyclen: 25 enter cyc at val:  22 at pos:  7 \n
   39 :   3-->  3 cyclen: 25 enter cyc at val:  14 at pos: 17 \n
   41 :   3-->  3 cyclen: 25 enter cyc at val:  41 at pos:  0 \n
   43 :  63--> 63 cyclen: 31 enter cyc at val: 126 at pos: 14 \n
   45 :   3-->  3 cyclen: 25 enter cyc at val:  14 at pos: 11 \n
   47 :   3-->  3 cyclen: 25 enter cyc at val:  66 at pos: 77 \n
   49 :   3-->  3 cyclen: 25 enter cyc at val:  22 at pos: 57 \n
   51 :   3-->  3 cyclen: 25 enter cyc at val:  22 at pos: 23 \n
   53 :   3-->  3 cyclen: 25 enter cyc at val:  66 at pos:  9 \n
   55 :   3-->  3 cyclen: 25 enter cyc at val:  22 at pos: 20 \n
   57 :   3-->  3 cyclen: 25 enter cyc at val:  66 at pos:  6 \n
   59 :   3-->  3 cyclen: 25 enter cyc at val:  22 at pos: 17 \n
                                            61 --- nothing found
   63 :  63--> 63 cyclen: 31 enter cyc at val:  63 at pos:  0 \n
   65 :  63--> 63 cyclen: 31 enter cyc at val:3278 at pos:140 \n
  ...