The 3n+1 conjecture

Definitions and mathematical statement

Define a map $T:\mathbb Z_{>0}\to\mathbb Z_{>0}$ by \[ T(n)=\begin{cases} 1,&n=1,\\ n/2,&n>1\text{ even},\\ 3n+1,&n>1\text{ odd}. \end{cases} \] For an initial positive integer $n_0$, set $n_{j+1}=T(n_j)$ for $j\ge0$, or equivalently $n_j=T^j(n_0)$. The conjecture asserts \[ \forall n_0\in\mathbb Z_{>0}\ \exists j\in\mathbb Z_{\ge0} \quad T^j(n_0)=1. \] The convention $T(1)=1$ makes one absorbing. If instead one applies the odd rule also at one, the equivalent question is eventual entry into the cycle $1,4,2,1$. No bound on the stopping time $j$ is part of this statement; it may depend on the starting integer.

Because the iteration is deterministic on positive integers, a bounded orbit eventually repeats and hence becomes periodic. Thus a counterexample would either enter a different periodic cycle or have an unbounded orbit. Establishing absence of additional cycles alone would leave the unbounded possibility. Conversely, verifying that every start below a finite cutoff terminates establishes only that finite range. The problem covers every positive initial integer with exactly the map above.

Short English statement

Does repeatedly halving an even number and replacing an odd number by three times it plus one always eventually reach one?

Sources

• S1 ULAM.AI and the UnsolvedMath Contributors, supplied problems.json, record 3380 (OPG-432), OpenGarden collection. Catalogue identity and source transcription; CC BY 4.0. https://huggingface.co/datasets/ulamai/UnsolvedMath.
• S2 Open Problem Garden, The 3n+1 conjecture. Original problem and discussion; the mathematical formulation below is expanded from the supplied source record. http://www.openproblemgarden.org/op/strange_series.