Equal shortcut parity prefixes force equal dyadic residues

Proved

CollatzWork.prefixCollision

by Sodelin · Sep 8, 2026 · Mathlib 0df444a (Lean v4.33.1)

collatz-work-import

Let $T:\mathbb N\to\mathbb N$ be the shortcut Collatz map: $T(n)=n/2$ for even $n$ and $T(n)=(3n+1)/2$ for odd $n$ . Write $T^k$ for its $k$ -fold iterate, with $T^0(n)=n$ . The starts $n,m\in\mathbb N$ have the same parity prefix of length $k\in\mathbb N$ when $T^i(n)\equiv T^i(m)\pmod2$ for every $i<k$ .

If the parity prefixes of $n$ and $m$ agree for $k$ steps, then

n\equiv m\pmod{2^k}.

This constrains finite parity-prefix collisions without a convergence assumption.

Preamble

import Std
import Init.Grind.Ordered.Module
import Definitions.Def_CollatzWork_ConvergenceStatement
import Definitions.Def_CollatzWork_PrefixCollisionStatement
import Theorems.Thm_CollatzWork_parityPrefix_dvd_sub_of_le

Formal statement

theorem CollatzWork.prefixCollision : PrefixCollisionStatement := by sorry

Source

https://github.com/Sodelin/Collatz-Conjecture-Work/blob/026aa4ad4be6453a005ab950b160a9f2204c5271/lean/CollatzWork/PrefixCollision.lean#L95-L95

View graph