Collatz variant $7 x + 1$?

Question

Let $n$ be a positive integer

Now define the collatz variant

1. if $n=2m$ divide by $2$ as often as possible.
2. if $n=3m$ divide by $3$ as often as possible.
3. if $n=5m$ divide by $5$ as often as possible.
4. else take $7n + 1$

Notice the order of steps 1,2,3 does not matter.

Now statistically we divide on average by $4\cdot 3^{\frac{3}{4}} \cdot 5^{\frac{5}{16}} = 15.077...$ wich is larger than $7$ , so we should not go to infinity ?

( Notice that the analogue case with 11x + 1 does not work out since there are sequences that are never divisible by $5$ ! Also 11 is slightly larger than $4\cdot 3^{\frac{3}{4}} = 9.118...$ so disaster strikes hard. Also I want to keep it simple )

Correct me if I said wrong things.

What is known about this ? How many cycles ?

By the way this is not the 7x+1 collatz I found online at arxiv or here, this one is different.

edit

I changed the statistical estimate because it was wrong.

Allow me to explain.

Take division by $3$ and powers of $3$. On average we can divide by $3$ with probability $1/3$.

And on average we can divide by $9$ with probability $1/9$. And by $27$ with probability $1/27$. This means that on average we divide by

$$ 3^{1/3} \cdot 9^{1/9} \cdot 27^{1/27}...$$

Or equivalent

$$ 3^{1/3} \cdot 3^{2/9} \cdot 3^{3/27}...$$

The same applies for every prime $p$.

So for a prime $p$ we divide on average by

$$ p^{1/p} \cdot p^{2/p^2} \cdot p^{3/p^3}...$$

What equals ( by the taylor series $x + 2x^2 + 3x^3 + ... = \frac{x}{(x-1)^2}$ )

$$p^{\frac{p}{(p-1)^2}}$$

Therefore the acurate estimate is

$4\cdot 3^{\frac{3}{4}} \cdot 5^{\frac{5}{16}}$

Sorry for the mistake.

---

Answer 1

history:

• 1st version
• update : frequency tables (on $N=800\,000$)
• update2: reliability of estimate of $\hat q_\text{go}$ for $N \to \infty$
• update3: estimates for $\hat q_{\text{go},p}$ for general $p$ instead of $7$

---

A heuristic frequencies list using only $a_{r}$ from the first $800\,000$ numbers $a_{r} \in \mathbb N^+ \mathbb{/2/3/5}$ beginning with $[1, 7, 11, 13, 17, 19, 23, 29, 31, 37]$.

Using $p=7$ I compute $w_{r} = p \cdot a_{r}+1$ then $b_{r}= \{w_{r}\}_{2,3,5} $ and $q_{r}=w_{r}/b_{r}$ . Here the notation $ \{n\}_s$ means complete extraction of all primefactors given in the list $s$ from natural number $n$.
In the following the absolute and the relative frequencies of $q_{r}$ (partially reordered hoping this gives a better sense for the patterns):

       q_r  frq(q_r)    relfrq(q_r) %    N=800 000 
  -------------------------------------------------
         2  150000      18.7500000000
         4   75000      9.37500000000
         8   37500      4.68750000000
        16   18750      2.34375000000
        32    9375      1.17187500000
        64    4687     0.585875000000
       128    2343     0.292875000000
       ...
         6  100000      12.5000000000
        12   50001      6.25012500000
        24   25000      3.12500000000
        48   12500      1.56250000000
        96    6249     0.781125000000
       ...
        18   33332      4.16650000000
        36   16666      2.08325000000
        72    8334      1.04175000000
       144    4166     0.520750000000
       ...
        10   40000      5.00000000000
        20   20000      2.50000000000
        40   10000      1.25000000000
        80    5000     0.625000000000
       ...
        50    8000      1.00000000000
       100    4000     0.500000000000
       200    2000     0.250000000000
       400    1000     0.125000000000
       ...
        54   11112      1.38900000000
       108    5556     0.694500000000
       ...
        30   26667      3.33337500000
        60   13333      1.66662500000
       120    6667     0.833375000000
       ...
        90    8889      1.11112500000
       150    5334     0.666750000000
       160    2500     0.312500000000
       162    3704     0.463000000000
       180    4445     0.555625000000
       192    3125     0.390625000000
       ...
       216    2777     0.347125000000
       ...
       ...

In the "relfrq%" columns in relative fractional expression:

       q_r  frq(q_r) relfrq(q_r)     N=800 000 
  -------------------------------------------------
         2  150000   3/16
         4   75000   3/32
         8   37500   3/64
        16   18750  3/128
       ...
         6  100000    1/8
        12   50001   1/16
        24   25000   1/32
        48   12500   1/64
        96    6249  1/128
       ...
        10   40000   1/20
        20   20000   1/40
        40   10000   1/80
        80    5000  1/160
       ...
        18   33332   1/24
        36   16666   1/48
        72    8334   1/96
       144    4166  1/192
       ...
        30   26667   1/30
        60   13333   1/60
       120    6667  1/120
       ...
        50    8000  1/100
       100    4000  1/200
       ...
        54   11112   1/72
       108    5556  1/144
       ...
        90    8889   1/90
       150    5334  1/150
       180    4445  1/180
       ...

Update

---

Here are my frequencies-tables. Each table is understood as increasing exponents at $3$ along rows(downwards) and increasing exponents at $5$ along columns (rightwards), each beginning with exponents $0$.
The separate tables are according to increasing exponents of $2$. The first table is empty because they are no cases, where the exponents at $2$ are zero. The last row,column,table are accumulated frequencies for exponents larger or equal than $6$:

       0      0     0     0    0   0   0  |
       0      0     0     0    0   0   0  |
       0      0     0     0    0   0   0  |
       0      0     0     0    0   0   0  |       for exponent 2^0
       0      0     0     0    0   0   0  |
       0      0     0     0    0   0   0  |
       0      0     0     0    0   0   0  |
       -      -     -     -    -   -   -  +
  150000  40000  8000  1600  320  64  16  | Example 320 cases with 2^1*3^0*5^4
  100000  26667  5334  1067  213  42  11  |         213 cases with 2^1*3^1*5^4
   33332   8889  1777   356   71  15   3  |
   11112   2962   593   118   24   5   1  |       for exponent 2^1
    3704    988   197    39    8   1   1  |
    1234    330    66    13    2   1   0  |
     618    164    33     7    2   0   0  |
       -      -     -     -    -   -   -  +
   75000  20000  4000   800  160  32   8  |
   50001  13333  2666   533  107  22   5  |
   16666   4445   889   178   35   7   2  |
    5556   1482   296    59   12   2   1  |       for exponent 2^2
    1851    494   100    20    3   1   0  |
     617    164    33     7    2   0   0  |
     309     82    16     3    1   0   0  |
       -      -     -     -    -   -   -  +
   37500  10000  2000   400   80  16   4  |
   25000   6667  1333   266   53  11   3  |
    8334   2222   445    89   18   3   1  |
    2777    740   149    30    6   1   0  |       for exponent 2^3
     925    247    49    10    3   0   0  |
     309     82    16     3    0   1   0  |
     155     42     8     2    0   0   0  |
       -      -     -     -    -   -   -  +
   18750   5000  1000   200   40   8   2  |
   12500   3333   667   134   27   5   1  |
    4166   1111   222    45    9   2   0  |
    1390    370    74    14    3   1   0  |       for exponent 2^4
     463    124    24     4    1   0   1  |
     155     42     8     2    0   0   0  |
      76     20     5     1    0   0   0  |
       -      -     -     -    -   -   -  +
    9375   2500   500   100   20   4   1  |
    6249   1666   333    67   14   3   0  |
    2085    556   111    21    4   1   1  |
     694    186    37     8    1   0   0  |       for exponent 2^5
     231     62    13     3    0   0   0  |
      77     20     5     1    0   0   0  |
      39     10     1     0    1   0   0  |
       -      -     -     -    -   -   -  +
    9375   2500   500   100   20   4   1  |
    6250   1668   333    67   12   3   1  |
    2083    554   112    22    6   0   0  |       accumulated
     694    186    36     8    1   1   0  |       for exponents 2^(6..oo)
     233     60    13     2    1   0   0  |
      76     22     3     1    0   0   0  |
      39     10     3     0    0   0   0  |
       -      -     -     -    -   -   -  +

The counting the occurences of primefactors 2,3,5 (respecting their multiplicities as well) I get the average quotient as $\hat q_\text{go} = 2^2 \cdot 3^{3/4} \cdot 5^{5/16} \approx 15.077$ This is little but systematically different from yours where you have $\hat q_\text{mick} = 2^2 \cdot 3^{2/4} \cdot 5^{4/16} \approx 10.36$.

I'm not completely sure of my result because of finitety of the frequencies tables (and maybe overlooking something). The extrapolation of the heuristical tables using the rules of geometric series (to fill up the missing frequencies for higher exponents) give a change only less than $0.01 \text{%} $ and so I'm now somehow confident of my estimate.

---

Update 2 Improving reliability of estimate for $\hat q_\text{go}$ .

---

To see, how increase of $N$ (number of used initial numbers $a_r$) increases accuracy of estimate of $\hat q_\text{go}$ I've now made a small but efficient program which shows, that my given estimates for the quotient $q$ are very likely correct.

Pari/GP-program:

p=7;
N=30000000;
{s2=s3=s5=0;n7=0;
  forstep(a1=1,N ,[6,4,2,4,2,4,6,2],n7++; \\ a1<--[1,7,11,13,17,...]
     w1=p*a1+1;
     e2=valuation(w1,2);s2+=e2;
     e3=valuation(w1,3);s3+=e3;
     e5=valuation(w1,5);s5+=e5;
 );
print(n7,"   ",[s2,s3,s5]/n7*1.0);}

This gives for increasing $N$:

 n7        expon at 2       expon at 3         expon at 5       
-------------------------------------------------------------
8000    [2.00025000000, 0.750625000000, 0.312500000000]
80000   [2.00007500000, 0.750037500000, 0.312525000000]
800000  [2.00000500000, 0.750001250000, 0.312501250000]
8000000 [2.00000087500, 0.750000000000, 0.312500000000]

This is obvious convergence to $\hat q_{go}=2^2 \cdot 3^{3/4} \cdot 5^{5/16}$

---

Update 3: Generalizing to $px+1$ with $p<>7$ .

---

If we insert any $p>2$ into the basic formula $px+1$ and define the divisor by the sequence of all smaller primes then we find heuristically $$ \begin{array} {} \hat q_{\text{go},3} & = 2^ 2 \\ \hat q_{\text{go},5} & = 2^ 2 \cdot 3^{3/4} \\ \hat q_{\text{go},7} & = 2^ 2 \cdot 3^{3/4} \cdot 5^{5/16}\\ \hat q_{\text{go},11} & = 2^ 2 \cdot 3^{3/4} \cdot 5^{5/16}\cdot 7^{7/36}\\ \end{array} \tag {3.1} $$ from where I guess, we can continue this in the obvious way.

This gives as decreasion-ratio per step for $p=3$ : $ r_p = p/ \hat q_{\text{go},3} = 0.75 $, for $p=5$ : $ r_p = p/ \hat q_{\text{go},5} = 0.548364172064 $ and so on, which for higher $p$ seems to converge to some value near $0.5221...$ (but convergence is extremely slow and possibly no convergence occurs at all)
Here are some few first decrease-ratios:

p    r_p
--------------------------
   3 0.750000000000
   5 0.548364172064
   7 0.464268246566
  11 0.499734125829
  13 0.453666258060
  17 0.470627952355
  19 0.435785623530
  23 0.443872763259
  29 0.482190280028
  31 0.455080623128
  37 0.482568810773
  41 0.482358783004
  43 0.459967559885
  47 0.458710305317
  53 0.474871755038
  59 0.489052691745
  61 0.470733529185
...
9941 0.514336091023
9949 0.514273517069
9967 0.514727391276
9973 0.514561607904
...
99961 0.520162302125
99971 0.520154427675
99989 0.520188173244
99991 0.520138684311
...
999931  0.521839942933
999953  0.521844214134
999959  0.521840135480
999961  0.521833969462
999979  0.521836153082
999983  0.521831030904
...
9999929  0.522079299112
9999931  0.522078562031
9999937  0.522078033784
9999943  0.522077505537
9999971  0.522078125867
9999973  0.522077388790
9999991  0.522077487040
...
99999773  0.522130942161
99999787  0.522130919079
99999821  0.522131000424
99999827  0.522130935571
99999839  0.522130902047
99999847  0.522130847637
99999931  0.522131190048
99999941  0.522131146080
99999959  0.522131143884
99999971  0.522131110360
99999989  0.522131108163
...
905041847  0.522168528916
905055527  0.522168532733
905055559  0.522168539297
905055623  0.522168540525
905055691  0.522168544062
...

I took this up to $N =100\,000\,000$ , $p_N=2\,038\,074\,743 $ giving the following picture for $r_p-0.5$ depending on $p$ (for the higher primes I only show the values near $p \approx 10^k$ and linear interpolation by Excel (dotted lines)):

And the upper & lower hullcurve (upper hullcurve up to $p_N=