Start with any positive integer n. If even, divide by 2; if odd, compute 3n+1. Repeat. Erdős said "mathematics is not yet ready for this problem." Lothar Collatz proposed it in 1937. Every starting value tested (over 2⁶⁸ of them) eventually reaches 1 — but no one has proved this for all n.
Single trajectory
Stopping times for all n ≤ N
starting n
N max (stopping plot) 5000
log scale on trajectory
why this is hard
The Collatz map n → 3n+1 if odd, n/2 if even is so simple to state that schoolchildren can compute it. But its dynamics seem to lie at the statistical sweet spot: each odd step multiplies by ~3/2 (then halves), so trajectories grow on average like (3/2)^(half-steps) — but with enough variance to always eventually hit a power of 2, fall fast, and reach 1.
Tao 2019 made the largest recent progress: for almost every starting n, the trajectory eventually descends below log(log n). But "almost every" isn't "every" — and infinitely many integers could in principle escape to infinity or cycle. The conjecture is verified up to about 2⁶⁸ ≈ 3·10²⁰, but each new digit takes exponentially more computation.
The right panel plots stopping times (number of steps to reach 1) for every n ≤ N. Specific starting values shoot up — 27 reaches a peak of 9232 before falling — and the upper envelope grows roughly logarithmically. The fine structure shows correlations between n and its stopping time that nobody fully understands.