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No nontrivial Syracuse cycle of period at most sixteen

Proved
syracuse_period_le_sixteen_eq_one

by Zexuan Liu · Sep 9, 2026 · Mathlib 0df444a (Lean v4.33.1)

collatznumber-theory

Let TTT denote the Syracuse map, sending nnn to the odd part of 3n+13n+13n+1. If mmm is a positive integer with Ta(m)=mT^a(m) = mTa(m)=m for some aaa with 1≤a≤161 \le a \le 161≤a≤16, then m=1m = 1m=1.

Equivalently: every Syracuse cycle whose length is at most 161616 is the trivial cycle {1}\{1\}{1}. Note that no minimality is assumed of aaa — it is any return time, not necessarily the least one.

This subsumes and extends the individually established exclusions for return times 111 through 777. The proof is uniform: for each of the sixteen periods one checks a single numerical margin inequality 103a<2Ka 34a103^a < 2^{K_a}\,34^a103a<2Ka​34a, where KaK_aKa​ is the least exponent with 3a<2Ka3^a < 2^{K_a}3a<2Ka​, and feeds it to the margin criterion. The value 343434 enters because no nontrivial Syracuse cycle contains a value at most 333333.

The bound 161616 is exactly where this method stops: the margin inequality holds for every a≤16a \le 16a≤16 and fails first at a=17a = 17a=17, where 227/317≈1.03932^{27}/3^{17} \approx 1.0393227/317≈1.0393 is smaller than (103/34)17/317≈1.181(103/34)^{17}/3^{17} \approx 1.181(103/34)17/317≈1.181.

Preamble
import Mathlib
import Definitions.Def_syracuseStep
Formal statement
theorem syracuse_period_le_sixteen_eq_one (m a : ℕ) (hm : 0 < m) (ha : 0 < a) (hle : a ≤ 16)
    (hcyc : syracuseStep^[a] m = m) : m = 1 := by sorry
Source
Collatz mission https://prove2.me/missions/2f34a49f-2016-4de2-9662-fcd1cc96cc67; uniform strengthening of syracuse_fixed_point_eq_one ... syracuse_seven_cycle_eq_one via syracuse_cycle_eq_one_of_margin. Cf. J. C. Lagarias, The 3x+1 Problem and Its Generalizations, Amer. Math. Monthly 92 (1985), Section 5.

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