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import FormalConjecturesUtilErdős Problem 26
References:
Te19 G. Tenenbaum, Some of Erdős' unconventional problems in number theory, thirty-four years later, arXiv:1908.00488 [math.NT] (2019)
namespace Erdos26A sequence of naturals $(a_i)$ is thick if their sum of reciprocals diverges: $$ \sum_i \frac{1}{a_i} = \infty $$
def IsThick {ι : Type*} (A : ι → ℕ) : Prop := ¬Summable (fun i ↦ (1 : ℝ) / A i)@[category test, AMS 11]
theorem not_isThick_of_finite {ι : Type*} [Finite ι] (A : ι → ℕ) : ¬IsThick A := ι:Type u_1inst✝:Finite ιA:ι → ℕ⊢ ¬IsThick A
All goals completed! 🐙@[category test, AMS 11]
theorem not_isThick_of_geom_one_lt (r : ℕ) (hr : r > 1) : ¬IsThick fun n : ℕ ↦ r ^ n := r:ℕhr:r > 1⊢ ¬IsThick fun n ↦ r ^ n
simpa [IsThick] using summable_geometric_of_lt_one (r := 1 / r) (r:ℕhr:r > 1⊢ 0 ≤ 1 / ↑r All goals completed! 🐙)
(div_lt_self zero_lt_one (mod_cast hr))@[category test, AMS 11]
theorem isThick_const {ι : Type*} [Infinite ι] (r : ℕ) (h : r > 0) : IsThick fun _ : ι ↦ r := ι:Type u_1inst✝:Infinite ιr:ℕh:r > 0⊢ IsThick fun x ↦ r
ι:Type u_1inst✝:Infinite ιr:ℕh:r > 0⊢ ¬r = 0
All goals completed! 🐙The set of multiples of a sequence $(a_i)$ is ${na_i | n \in \mathbb{N}, i}$.
def MultiplesOf {ι : Type*} (A : ι → ℕ) : Set ℕ := Set.range fun (n, i) ↦ n * A i@[category test, AMS 11]
theorem multiplesOf_eq_univ {ι : Type*} (A : ι → ℕ) (h : 1 ∈ Set.range A) :
MultiplesOf A = Set.univ := ι:Type u_1A:ι → ℕh:1 ∈ Set.range A⊢ MultiplesOf A = Set.univ
ι:Type u_1A:ι → ℕi:ιhi:A i = 1⊢ MultiplesOf A = Set.univ
exact top_unique fun n hn ↦ ⟨(n, i), ι:Type u_1A:ι → ℕi:ιhi:A i = 1n:ℕhn:n ∈ ⊤⊢ (fun x ↦
match x with
| (n, i) => n * A i)
(n, i) =
n All goals completed! 🐙⟩A sequence of naturals $(a_i)$ is Behrend if almost all integers are a multiple of some $a_i$. In other words, if the set of multiples has natural density $1$.
def IsBehrend {ι : Type*} (A : ι → ℕ) : Prop := (MultiplesOf A).HasDensity 1A sequence of naturals $(a_i)$ is weakly Behrend with respect to $\varepsilon \in \mathbb{R}$ if at least $1 - \varepsilon$ density of all numbers are a multiple of $A$.
def IsWeaklyBehrend {ι : Type*} (A : ι → ℕ) (ε : ℝ) : Prop := 1 - ε ≤ (MultiplesOf A).lowerDensityι:Type u_1A:ι → ℕh:1 ∈ Set.range A⊢ Filter.Tendsto (fun b ↦ (MultiplesOf A).partialDensity Set.univ b) Filter.atTop (nhds 1)
exact tendsto_atTop_of_eventually_const (i₀ := 1) fun n hn ↦ ι:Type u_1A:ι → ℕh:1 ∈ Set.range An:ℕhn:n ≥ 1⊢ (MultiplesOf A).partialDensity Set.univ n = 1
ι:Type u_1A:ι → ℕh:1 ∈ Set.range An:ℕhn:n ≥ 1⊢ ¬n = 0
All goals completed! 🐙@[category test, AMS 11]
theorem isWeaklyBehrend_of_ge_one {ι : Type*} (A : ι → ℕ) {ε : ℝ} (hε : 1 ≤ ε) :
IsWeaklyBehrend A ε := ι:Type u_1A:ι → ℕε:ℝhε:1 ≤ ε⊢ IsWeaklyBehrend A ε
All goals completed! 🐙@[category test, AMS 11]
theorem not_isWeaklyBehrend_of_neg {ι : Type*} (A : ι → ℕ) {ε : ℝ} (hε : ε < 0) :
¬IsWeaklyBehrend A ε := ι:Type u_1A:ι → ℕε:ℝhε:ε < 0⊢ ¬IsWeaklyBehrend A ε
ι:Type u_1A:ι → ℕε:ℝhε:ε < 0⊢ (MultiplesOf A).lowerDensity + ε < 1
All goals completed! 🐙Let $A\subset\mathbb{N}$ be infinite such that $\sum_{a \in A} \frac{1}{a} = \infty$. Must there exist some $k\geq 1$ such that almost all integers have a divisor of the form $a+k$ for some $a\in A$?
This was formalized in Lean by Alexeev using Aristotle.
@[category research solved, AMS 11, formal_proof using lean4 at "https://github.com/plby/lean-proofs/blob/main/src/v4.24.0/ErdosProblems/Erdos26.lean"]
theorem erdos_26 : answer(False) ↔ ∀ A : ℕ → ℕ, StrictMono A → IsThick A →
∃ k, IsBehrend (A · + k) := ⊢ False ↔ ∀ (A : ℕ → ℕ), StrictMono A → IsThick A → ∃ k, IsBehrend fun x ↦ A x + k
All goals completed! 🐙If we allow for $\sum_{a\in A} \frac{1}{a} < \infty$ then Rusza has found a counter-example.
@[category research solved, AMS 11]
theorem erdos_26.variants.rusza : ∃ A : ℕ → ℕ,
StrictMono A ∧ ¬IsThick A ∧ ∀ k, ¬IsBehrend (A · + k) := ⊢ ∃ A, StrictMono A ∧ ¬IsThick A ∧ ∀ (k : ℕ), ¬IsBehrend fun x ↦ A x + k
All goals completed! 🐙Tenenbaum asked the weaker variant where for every $\epsilon>0$ there is some $k=k(\epsilon)$ such that at least $1-\epsilon$ density of all integers have a divisor of the form $a+k$ for some $a\in A$.
The DeepMind prover agent has found a formal disproof of this statement.
@[category research solved, AMS 11,
formal_proof using formal_conjectures at "https://github.com/mo271/formal-conjectures/blob/09c54540aa51cb40dff73660c94a82e2631386f8/FormalConjectures/ErdosProblems/26.lean#L625"]
theorem erdos_26.variants.tenenbaum : answer(False) ↔ ∀ᵉ (A : ℕ → ℕ), StrictMono A → IsThick A →
(∀ ε > (0 : ℝ), ∃ k, IsWeaklyBehrend (A · + k) ε) := ⊢ False ↔ ∀ (A : ℕ → ℕ), StrictMono A → IsThick A → ∀ ε > 0, ∃ k, IsWeaklyBehrend (fun x ↦ A x + k) ε
All goals completed! 🐙end Erdos26