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Residual Syracuse descent inside n≡27(mod32)n \equiv 27 \pmod{32}n≡27(mod32) modulo 409640964096

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syracuse_descent_residual_twentyseven_mod32_mod4096

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

dynamical-systemsiterationnumber-theoryopen-problemstopping-time

This statement is open.

Let T(n)=(3n+1)/2 v2(3n+1)T(n) = (3n+1)/2^{\,v_2(3n+1)}T(n)=(3n+1)/2v2​(3n+1) be the Syracuse (accelerated Collatz) map. The assertion is that every n≡27(mod32)n \equiv 27 \pmod{32}n≡27(mod32) which additionally avoids all 13 of the exceptional progressions listed below has a finite accelerated stopping time:

∃ t∈N:Tt(n)<n.\exists\, t \in \mathbb{N} : \quad T^{t}(n) < n .∃t∈N:Tt(n)<n.

The excluded progressions are precisely those on which the Terras affine certificate already closes within 777 steps (they are handled by the companion lemma). What remains is the residue of the class n≡27(mod32)n \equiv 27 \pmod{32}n≡27(mod32) modulo 2122^{12}212: exactly 45 classes mod 409640964096 survive, on which the first 121212 binary digits of nnn do not suffice to certify descent.

The exclusions:

  • n≢59(mod128)n \not\equiv 59 \pmod{128}n≡59(mod128)
  • n≢123(mod256)n \not\equiv 123 \pmod{256}n≡123(mod256)
  • n≢219(mod256)n \not\equiv 219 \pmod{256}n≡219(mod256)
  • n≢347(mod1024)n \not\equiv 347 \pmod{1024}n≡347(mod1024)
  • n≢507(mod1024)n \not\equiv 507 \pmod{1024}n≡507(mod1024)
  • n≢923(mod1024)n \not\equiv 923 \pmod{1024}n≡923(mod1024)
  • n≢1019(mod4096)n \not\equiv 1019 \pmod{4096}n≡1019(mod4096)
  • n≢1435(mod4096)n \not\equiv 1435 \pmod{4096}n≡1435(mod4096)
  • n≢1787(mod4096)n \not\equiv 1787 \pmod{4096}n≡1787(mod4096)
  • n≢2203(mod4096)n \not\equiv 2203 \pmod{4096}n≡2203(mod4096)
  • n≢2587(mod4096)n \not\equiv 2587 \pmod{4096}n≡2587(mod4096)
  • n≢2907(mod4096)n \not\equiv 2907 \pmod{4096}n≡2907(mod4096)
  • n≢3675(mod4096)n \not\equiv 3675 \pmod{4096}n≡3675(mod4096)

Why this is the honest remainder

Refining the modulus settles a further slice at every level — Terras proved the settled density tends to 111 — but it never reaches 111 at any finite level, so this residual cannot be emptied by pushing the refinement deeper. Closing it requires an argument that is not a finite case check: the full descent (stopping time) conjecture restricted to this class.

Formalization note. The hypotheses are stated as separate congruence exclusions grouped by modulus, matching the shape produced by a case split on the companion lemma's disjunction. No oddness or positivity hypothesis is needed.

Preamble
import Mathlib
import Definitions.Def_syracuseStep
Formal statement
theorem syracuse_descent_residual_twentyseven_mod32_mod4096 (n : ℕ)
    (h : n % 32 = 27)
    (h128 : n % 128 ≠ 59)
    (h256 : n % 256 ≠ 123 ∧ n % 256 ≠ 219)
    (h1024 : n % 1024 ≠ 347 ∧ n % 1024 ≠ 507 ∧ n % 1024 ≠ 923)
    (h4096 : n % 4096 ≠ 1019 ∧ n % 4096 ≠ 1435 ∧ n % 4096 ≠ 1787 ∧ n % 4096 ≠ 2203 ∧ n % 4096 ≠ 2587 ∧ n % 4096 ≠ 2907 ∧ n % 4096 ≠ 3675) :
    ∃ t : ℕ, syracuseStep^[t] n < n := by sorry
Source
https://en.wikipedia.org/wiki/Collatz_conjecture; supporting lemma for the prove2.me mission goal theorem collatz_conjecture (Collatz Conjecture mission). Riho Terras, A stopping time problem on the positive integers, Acta Arith. 30 (1976), 241-252, Section 2 (the coefficient/parity vector and the affinity of T^t on residue classes mod 2^k); Jeffrey C. Lagarias, The 3x+1 Problem and Its Generalizations, Amer. Math. Monthly 92 (1985), 3-23, Section 2, https://websites.umich.edu/~lagarias/3x%2B1.html

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