I'm trying to work out a new way of visualizing the Collatz conjecture (or 3n+1 problem) using the Manipulate feature of Mathematica to show paths that numbers take in the 3n+1 problem in the form of a circle.

for those of you who don't know, the Collatz conjecture works as following, if n is even, then n is mapped to n/2, if n is odd, then n is mapped to 3n+1. The question is whether or not every number will eventually cycle into the loop [1-4-2-1]

So, when the code is working correctly, I hope that it will produce m points in a circle, and that using manipulate I will be able to choose a starting point and see that point's path around the circle. (For example, if the starting point is 3, the points path should be [3-10-5-16-8-4-2-1] Here is the current code I have

u = {};
n = 0;
t = tstart;
While[1 < t < 100,
   If[Mod[t, 2] == 0,
  t = (t/2); n = n + 1; AppendTo[u, n],
  t = 3 t + 1; n = n + 1; AppendTo[u, n]]];
p =
    With[{nmax = Length[u]}, 
  Table[{-Cos[N[Mod[u[[nos]], m]]*(2 Pi/m)], 
    Sin[N[Mod[u[[nos]], m]]*(2 Pi/m)]}, {nos, 0, nmax}]];
q = Table[{-Cos[nos*(2 Pi/m)], Sin[nos*(2 Pi/m)]}, {nos, 0, m - 1, 
r = Table[
  Text[Style[ToString@nos, Medium], q[[nos + 1]] 1.1], {nos, 0, 
   m - 1, 1}];
Graphics[{{Blue, Circle[{0, 0}, 1.3]}, {AbsoluteThickness[2], 
  s]},*), {Red, AbsolutePointSize[5], Point[q]}, {Brown, r}}, 
 ImageSize -> {500, 500}], {{m, 10}, 10, 20, 2, 
 Appearance -> "Labeled"}, {{tstart, 2}, 2, 10, 1, 
 Appearance -> "Labeled"}]

where the Print[u] and Print [p] lines are only there to try and see what the problem is with the code. (right now when the code is executed a circle with m points shows up on the screen, but no lines appear and there is the message from mathematica

{{-Cos[1/5 π Mod[List,10.]],Sin[1/5 π Mod[List,10.]]},{-0.809017,0.587785}} 

is not a point that can be plotted.)

Any help would be awesome since this is my first project ever using Mathematica. Thank you so much!

  • $\begingroup$ Your tables should go from 1,m not 0,m-1 mathematica indexes lists starting with 1. You're getting List because u[[0]] is the head of the expression, which in this case is List. $\endgroup$
    – N.J.Evans
    Aug 25, 2017 at 17:52

1 Answer 1


Here's a more idiomatic way to get your result:

Define a collatz function that takes an integer n checks it for even, or oddness and applies the appropriate transformation:


Then use NestWhileList to calculate the sequence u conditionally. (This is one idiomatic replacement for loops that you should get familiar with.)

Then use map(i.e. /@ operator) to apply your transformation to geometric coordinates using the fact that Mod is Listable. Map is another idiomatic loop replacement that lets you dispense with indices.

 u = NestWhileList[collatz, tstart, # > 1 &];
 ln = {Cos[-#], Sin[#]} & /@(Mod[u, m]*2 π/m);
   Blue, Circle[{0, 0}, 1.3]
  , PlotRange -> All
 , {{tstart, 2}, 2, 10, 1, Appearance -> "Labeled"}
 , {{m, 10}, 2, 20, 2, Appearance -> "Labeled"}

enter image description here

You can add the other embellishments to the image.

I'd suggest reading this post what-are-the-most-common-pitfalls-awaiting-new-users to get an idea of where to start with more idiomatic MMA constructions. And of course, welcome to MMA!

  • $\begingroup$ Thank you so much!! This worked perfectly, thank you for the idea of making a specific function for the Collatz conjecture. This really helped. $\endgroup$
    – JonHales
    Aug 25, 2017 at 22:36

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