Take any whole number. If it's even, halve it; if it's odd, triple it and add one. Repeat. The Collatz conjecture says that every starting number, no matter how large, eventually tumbles down to 1 — and nobody has proved it in 90 years. Run the rule backwards instead: start at 1 and ask "what could have flowed into here?" The answer branches, and branches again, into a vast tree whose leaves are every integer in existence. Drawn outward it looks exactly like a river delta seen from orbit — countless tributaries all draining to one mouth. Grow it with the depth slider, then click any node: the path back down to 1 is that number's hailstone flight.
The rule and its shadow. The Collatz map is \(T(n)=n/2\) for even \(n\) and \(T(n)=3n+1\) for odd \(n\). The conjecture (Lothar Collatz, 1937) is that iterating \(T\) from any positive integer reaches the cycle \(4\to2\to1\). It has been checked by computer for every \(n\) up to about \(2^{68}\approx 3\times10^{20}\), yet remains open; Paul Erdős said "mathematics is not yet ready for such problems."
Why a tree grows. Reverse the arrows. A number \(m\) can be reached from \(2m\) (undoing a halving) and, when \(m\equiv 4 \pmod 6\), also from the odd number \((m-1)/3\) (undoing a \(3n+1\)). So every node has one or two predecessors, and starting from 1 the predecessors of predecessors sweep out — if the conjecture is true — every positive integer exactly once. That infinite acyclic graph is the Collatz tree; rendered with tapering, meandering edges it is indistinguishable from a satellite photo of a braided river delta. The path from any node down to the root is precisely the forward orbit — the "hailstone sequence," so named because the values bounce up and down like a hailstone in a cloud before finally falling (Lagarias, The 3x+1 Problem, 2010). Deeper reading: J. C. Lagarias' annotated bibliography, and Terence Tao's 2019 result that "almost all" orbits attain almost bounded values.