In this question I was wondering if the $3$ in the Collatz conjecture is arbitrary, and when I wrote that question I tried to change to $7n+1$ starting with the seed number $7$, the sequence appears to exhibit unbounded growth.

To analyze this behavior, I had a Python script developed to simulate the sequence. Remarkably, after $3000$ iterations, the maximum value is about $10^{266}$. This observation leads me to hypothesize that the sequence may either diverge to infinity or enter an extremely lengthy loop.

Here is the code for plotting the first $1000$.

import matplotlib.pyplot as plt
import numpy as np
a = 7
xpoints = []
ypoints = []

for i in range(1 , int(1000)):
    if a % 2 == 0:
        a //= 2
        a = a * 7 + 1

plt.plot(np.array(xpoints) , np.array(ypoints))

enter image description here

enter image description here

enter image description here

Does this sequence diverge to infinity?, and if so, how can we prove it? Alternatively, if the sequence does not diverge, what is the length of its loop?

Given the complexity of this problem, If the problem can't be solved with computational solution for the loop or the proof of divergence already exist then I think this question might be as challenging as solving the original Collatz conjecture .

The question has been asked on MSE here.

  • 8
    $\begingroup$ It would be much better if the question stated, in English, exactly what algorithm is being discussed, instead of expecting readers to guess correctly or read the code. $\endgroup$ Commented Jun 12 at 17:20
  • 2
    $\begingroup$ I made no mention of any "confusing part". You neglected to state clearly in English what your algorithm is. $\endgroup$ Commented Jun 12 at 17:25
  • 7
    $\begingroup$ I'd expect that in a problem like this the y-axis would have been logarithmically scaled... say like this: go.helms-net.de/math/collatz/start_at_a0_7.png $\endgroup$ Commented Jun 13 at 9:14

2 Answers 2


It is a standard conjecture that if one replaces the 3 in the Collatz function with some fixed $k>3$, then it will have sequences which go to infinity. And in fact, these seem to occur at very small values. For example, if one is using 5k+1, then it is very likely that even the orbit of 7 goes to infinity. However, we cannot prove this for any specific $k$ at this time, or even prove there is such a $k$. (See also my answer to that question and the subsequent discussion of how we can almost prove this for $1093n+1$). Even versions of Collatz where one is allowed more freedom by choosing a specific $k$ at each stage and have some finite list of $k$ are may be tough.

  • 4
    $\begingroup$ I wouldn't say we can "almost prove this for $1093n+1$". We only know it doesn't reach 1, we can prove nothing regarding divergence. $\endgroup$
    – Wojowu
    Commented Jun 14 at 12:02

(This is meant as a comment to @JoshuaZ 's answer.)

Here is a list of found cycles for a lot of $m \ne 3$ in the $m x+1$ - generalization. (I've not done any analysis about divergent sequences here, I could prove divergent sequences only in generalizations, where —for instance— $3x+1$ and $5x+1$ are intertwined in the definition.)

list of found cycles in Collatz-generalizations

  • 2
    $\begingroup$ Sorry, you mention proving divergence for when 3x+1 and 5x+1 are intertwined. What do you mean by that? $\endgroup$
    – JoshuaZ
    Commented Jun 12 at 18:16
  • $\begingroup$ JoshuaZ - I meant the operations are selected by some modular residue of the current $x$ Toy-example: if $x \equiv 3 \, or \, 5 \text{mod} 8$ then use $3x+1 $ if$x \equiv 1 \, or \, 7 \text{ mod } 8$ then use $5x+1 $ ... (not the true residues and multiplication-factors here). There have been example-questions in MSE from where I took the idea, I think on intertwining $5x+1$ and $7x+1$ if I recall right and I had done some small analyses. $\endgroup$ Commented Jun 12 at 18:24
  • 2
    $\begingroup$ Ah, so in this case, if $x \equv 3$ (mod 4) then using 3x+1, and if $x \equiv 1$ (mod 4), then use $5x+1$ will lead to divergence. $\endgroup$
    – JoshuaZ
    Commented Jun 12 at 18:26
  • 1
    $\begingroup$ yes - that's what I meant $\endgroup$ Commented Jun 12 at 18:27

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.