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Footnote 1 — consecutive distances less than double, ratio tending to 222

Proved
TitiusBode.ratio_to_previous

by Lucas · Oct 9, 2026 · Mathlib 0df444a (Lean v4.33.1)

limitstitius-bode

Let a(n)=0.4+0.3⋅2na(n) = 0.4 + 0.3\cdot 2^na(n)=0.4+0.3⋅2n for n∈{−∞,0,1,… }n \in \{-\infty, 0, 1, \dots\}n∈{−∞,0,1,…} and ana_nan​ the same expression for n∈Nn \in \mathbb Nn∈N. Then

  1. a(0)<2 a(−∞)a(0) < 2\,a(-\infty)a(0)<2a(−∞), i.e. 0.7<0.80.7 < 0.80.7<0.8;
  2. an+1<2ana_{n+1} < 2a_nan+1​<2an​ for every n∈Nn \in \mathbb Nn∈N;
  3. lim⁡n→∞an+1an=2\displaystyle \lim_{n\to\infty} \frac{a_{n+1}}{a_n} = 2n→∞lim​an​an+1​​=2.

This is the first sentence of footnote 1 to the data table of the source: for large kkk each Titius–Bode distance is approximately twice the preceding value, while for small kkk the predicted distances do not fully double.

Preamble
import Definitions.Def_TitiusBode_Defs
import Mathlib
open Filter Topology
Formal statement
namespace TitiusBode
theorem ratio_to_previous :
    tbAU ((0 : ℕ) : WithBot ℕ) < 2 * tbAU ⊥ ∧
    (∀ n : ℕ, tbCanonical (n + 1) < 2 * tbCanonical n) ∧
    Tendsto (fun n : ℕ => tbCanonical (n + 1) / tbCanonical n) atTop (𝓝 2) := by sorry
end TitiusBode
Source
Wikipedia, "Titius–Bode law", revision oldid=1372822920, https://en.wikipedia.org/w/index.php?title=Titius%E2%80%93Bode_law&oldid=1372822920, section "Data", footnote 1 to the table
Read-back

What the Lean code literally says, in plain math · Aristotle (Harmonic) — non-blind, same agent as the drafter

Non-blind read-back — not independent testimony. This read-back was written by the same agent that drafted the Lean statements, at the explicit request of the proposal owner. Its author knew the intended meaning and the source material, so it is not a blind audit and must not be mistaken for independent testimony; reviewers should compare the Lean code against the source themselves.

No hypotheses. With AAA the AU form (A(⊥)=0.4A(\bot) = 0.4A(⊥)=0.4, A(n)=0.4+0.3⋅2nA(n) = 0.4 + 0.3\cdot 2^nA(n)=0.4+0.3⋅2n) and an=0.4+0.3⋅2na_n = 0.4 + 0.3\cdot 2^nan​=0.4+0.3⋅2n for n∈Nn \in \mathbb Nn∈N, the statement asserts the conjunction of:

  1. A(0)<2 A(⊥)A(0) < 2\,A(\bot)A(0)<2A(⊥) (strict);
  2. for every n∈Nn \in \mathbb Nn∈N: an+1<2ana_{n+1} < 2a_nan+1​<2an​ (strict);
  3. the real sequence n↦an+1/ann \mapsto a_{n+1}/a_nn↦an+1​/an​ converges to 222 as n→∞n \to \inftyn→∞ (convergence in the usual topology of R\mathbb RR along n→∞n \to \inftyn→∞ in N\mathbb NN).
Human review
  • Endorsed by Shuze Chen · Oct 10, 2026

    Confirmed by the moderator at approval.

  • Endorsed by Lucas · Oct 10, 2026

    Confirmed by the mission captain (proposal self-audit).

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