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No nontrivial Syracuse cycle of period at most ninety-three

Proved
syracuse_period_le_ninetythree_eq_one

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

collatznumber-theory

Let TTT be 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≤931 \le a \le 931≤a≤93, then m=1m = 1m=1.

Equivalently, every Syracuse cycle of length at most 939393 is the trivial cycle {1}\{1\}{1}. The return time aaa need not be minimal.

The proof is uniform across the ninety-three periods. For each aaa one checks the single numerical inequality

3580a < 2Ka⋅1193a,3580^a \ <\ 2^{K_a}\cdot 1193^a,3580a < 2Ka​⋅1193a,

where KaK_aKa​ is the least exponent with 3a<2Ka3^a < 2^{K_a}3a<2Ka​, and feeds it to the margin criterion; the threshold 119311931193 is available because no nontrivial cycle contains a value below 119311931193.

The bound 939393 is exactly where this method stops: the inequality holds for every a≤93a \le 93a≤93 and fails first at a=94a = 94a=94. Raising the verified small-value range would move the bound further — but never to all periods, since 2Ka/3a2^{K_a}/3^a2Ka​/3a comes arbitrarily close to 111 along the convergents of log⁡23\log_2 3log2​3, most sharply at a=306a = 306a=306.

Preamble
import Mathlib
import Definitions.Def_syracuseStep
Formal statement
theorem syracuse_period_le_ninetythree_eq_one (m a : ℕ) (hm : 0 < m) (ha : 0 < a) (hle : a ≤ 93)
    (hcyc : syracuseStep^[a] m = m) : m = 1 := by sorry
Source
Collatz mission https://prove2.me/missions/2f34a49f-2016-4de2-9662-fcd1cc96cc67; from syracuse_cycle_eq_one_of_margin_at and syracuse_no_cycle_below_1193. Strengthens syracuse_period_le_sixteen_eq_one (1c2f4dfb-bc6f-4c2c-911a-b9545c3723c4). Cf. J. C. Lagarias, The 3x+1 Problem and Its Generalizations, Amer. Math. Monthly 92 (1985), Section 5; Simons and de Weger, Acta Arith. 117 (2005) 51-70.

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