Exact residue-20 tail to 38 modulo 81

Proved

CollatzWork.residueAncestor_tail38

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 $T^0(n)=n$ .

For $a\in\mathbb N$ , put $m=216a+101$ and $z=81a+38$ . Then

m>0,\quad m\equiv20\pmod{27},\quad T^{3}(m)=z,\quad 3m+1=8z,\quad m\le64z.

This is one row of the finite tail selector. The bound by 64z is not a claim that the ancestor is smaller than z.

Preamble

import Std
import Init.Grind.Ordered.Module
import Definitions.Def_CollatzWork_ConvergenceStatement
import Definitions.Def_CollatzWork_InverseWordBoundaryStatement
import Definitions.Def_CollatzWork_RefinedMersenneChild

Formal statement

theorem CollatzWork.residueAncestor_tail38 (a : Nat) :
    0 < 216 * a + 101 ∧
    (216 * a + 101) % 27 = 20 ∧
    shortcutIter 3 (216 * a + 101) = 81 * a + 38 ∧
    3 * (216 * a + 101) + 1 = 8 * (81 * a + 38) ∧
    216 * a + 101 ≤ 64 * (81 * a + 38) := by sorry

Source

https://github.com/Sodelin/Collatz-Conjecture-Work/blob/026aa4ad4be6453a005ab950b160a9f2204c5271/lean/CollatzWork/ResidueAncestorTails.lean#L15-L33

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