I'm supposed to do a Mathematica function that selects the smallest number not less or equal to 4 after 125 iterations. I am not good at coding, so some help/advice would be awesome.

My code so far.

f[m_, n_] := 
   f[m, n] = If[Divisible[f[m, n - 1], 2], f[m, n - 1]/2, 3*f[m, n - 1] + 1];
Table[f[m, i], {i, 1, 125}]

So, any tips on how to write code that lists the numbers and returns anything that gives back 4 after 125 iterations?

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    – Michael E2
    Jan 19, 2016 at 12:13
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    – Michael E2
    Jan 19, 2016 at 12:15
  • $\begingroup$ Please edit your question to improve it. Provide your code in formatted form, as sugested by Michael_E2. Please also explain better what is that you need, what you have, what is the problem with what you have and an example of the desired output. People will (should?) be reluctant to just give you a straight answer for your homework without you showing due diligence first. $\endgroup$
    – rhermans
    Jan 19, 2016 at 12:46
  • $\begingroup$ If you need to apply the same function 125 times, you may want to use Nest. I still don't understand what is that function that you need to iterate. Help us to help you, edit to explain better. $\endgroup$
    – rhermans
    Jan 19, 2016 at 12:56
  • $\begingroup$ It's not clear to me what role m plays in your question above. You don't appear to use it as part of your recursion. $\endgroup$
    – user1722
    Jan 19, 2016 at 16:24

2 Answers 2


There are many questions about the Collatz sequence on this site, for example here. I interpret your question to ask for the smallest integer n that, after 125 iterations, has not settled into the 4-2-1 loop that is conjectured to terminate all Collatz iterations. You can run 125 iterations and return the result with the following definition.

collatz125[n_Integer] := Nest[If[EvenQ[#], #/2, 3*# + 1] &, n, 125]

Make a table of integer n and its result after 125 iterations, then select from the table those values of n which are not yet in the 4-2-1 loop.

Select[Table[{n, collatz125[n]}, {n, 1, 500}], #[[2]] > 4 &]

The smallest n is 313. After 125 iterations, the result is 5.

  • $\begingroup$ Very nice. Very succinct. $\endgroup$
    – user1722
    Jan 19, 2016 at 16:57

I decided to do this in the old-fashioned procedural way.

collatz[k_Integer?OddQ] := 3 k + 1
collatz[k_Integer?EvenQ] := k/2

collatzIter[k_ /; k > 2, n_: 125] :=
  Module[{i = 0, c0 = k, c1},
    While[++i < n,
      c1 = collatz[c0];
      If[c1 < 2, Break[]];
      c0 = c1];
    {i, c1}]

 Module[{m = 2, r = {0, 0}},
   While[r[[2]] <= 4, r = collatzIter[++m]];
   {m, r}]

{231, {125, 8}}

Thus 231 is smallest number the where Collatz sequence has an element greater then 4 at position 125, and that element is 8.

Let us check this answer.

<< "ExampleData/Collatz.m"


This of course means Collatz[231][[126]] is 4 and, therefore, fails by KennyColnago's interpretation of the OP's question, because he looks at the 126th element in the Collatz sequence.

Either answer can be easily modified to accommodate the other interpretation of what the OP is asking. The OP will have to tell us which interpretation is correct. In my case, I need only evaluate

Module[{m = 2, r = {0, 0}},
  While[r[[2]] <= 4, r = collatzIter[++m, 126]];
  {m, r}]

{313, {126, 5}}

to get an answer in agreement with KennyColnago's.


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