IA idea · Pure maths, number & proof

The Collatz conjecture: what can we say about stopping times?

AA SLAA HLAI SLAI HL Solid Also in: Statistics, Probability

Research question

How are the stopping times of the Collatz sequence distributed for starting values up to 10⁶, and does a simple probabilistic model explain why sequences tend to fall?

Adapt it: change the place, the data or the comparison until the question is yours.

Why it makes a good exploration

The Collatz conjecture is unsolved, so you cannot prove it — but you can investigate it. A heuristic argument that on average each step multiplies by about 3/4 explains why sequences usually come down, and is a lovely use of expected value and logarithms.

The mathematics you'll need

  • Iteration and sequences
  • Statistics of stopping times
  • Expected value and geometric means
  • Logarithms to model average decrease
  • Honest discussion of heuristic vs proof

Course labels show where a technique sits; using maths from outside your course is fine if you explain it clearly and say it is new to you.

Where the data comes from

Generate data with a spreadsheet or short program; compare with OEIS records.

Cite every source in a footnote where you use it and in your bibliography. Check the licence of any dataset you download.

A possible outline

  1. Explain the rule and the conjecture.
  2. Generate stopping times and analyse their distribution.
  3. Build the probabilistic model.
  4. Test the model's prediction of average stopping time against log n.
  5. Reflect on why heuristics are not proofs.

Pitfalls that cost marks

  • Claiming to prove the conjecture.
  • Huge tables with no analysis.
  • Not explaining the program used.

Showing personal engagement

  • Find a starting number with a long sequence yourself.
  • Explore a variant (5n + 1) and see it fail.
  • Discuss why famous mathematicians consider it hard.

See Criterion C: personal engagement for what examiners look for.

Taking it further

Fit stopping time against ln n and compare the gradient with your model's prediction.

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